Koncepts of LR - Logical Reasoning for CAT, XAT, IIFT, MAT, CMAT, NMAT & other MBA Exams 9386146134, 9789386146137

Table of contents :
Cover
Title page
Copyright page
Foreword
Dedication
Preface to the 3rd edition
Contents
1. Linear Arrangement
2. Matrix Arrangement
3. Circular Arrangement
4. Set Theory
5. Group/ Team Formation
6. Syllogism
7. Logical Connectivity
8. Input Output
9. Games and Tournaments
10. Cubes
11. Calendar
12. Path and Route
13. Clocks
14. Blood Relation and Family Tree
15. Coding and Series
16. Statement Conclusion and Binary Logic
17. Direction
18. Data Sufficiency
19. Puzzles
20. Mathematical Reasoning
21. Critical Reasoning
Miscellaneous Questions
1. Mock Test with Solutions
2. Mock Test with Solutions
3. Mock Test with Solutions
4. Mock Test with Solutions

Citation preview

Gajendra Kumar

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Malviya Nagar, New Delhi-110017



Tel. : 011-49842349 / 49842350

Typeset by Disha DTP Team

DISHA PUBLICATION All RightS Reserved © Copyright Publisher No part of this publication may be reproduced in any form without prior permission of the publisher. The author and the publisher do not take any legal responsibility for any errors or misrepresentations that might have crept in. We have tried and made our best efforts to provide accurate up-to-date information in this book.

For further information about the books from DISHA, Log on to www.dishapublication.com or email to [email protected] (ii)

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•

Foreword I was helping oversee the affairs of IMS, the management aptitude test preparation pioneers, in Kolkata for a brief period when I met this interesting person who wanted to give up a career in shipping to teach MBA aspirants. Having grown up at a time when ‘shipees’ were synonymous with wealth and glamour, I understood that Gajendra had, like me, been bitten by the teaching bug. The characteristic of this bug is that it consumes everything that comes in its way, including conventional wisdom and counsel. Having myself given up a corporate career, which could be termed ‘promising’, for the love of teaching, I instantly empathized with Gajendra and we have been co-travellers, albeit in slightly different ways, in this journey of learning and knowledge sharing. So when Gajendra asked me to write a foreword for his latest book, his third, (for me who has never dared to even conceive of writing a book, this is most impressive), I was keen. Having gone through most of the contents of ‘Koncepts of LR’, I am now sure that I can recommend this book to the large multitude of people attempting to get into a variety of professional institutions. A good book on any aspect of an ‘aptitude test’ should have the following characteristics – it should present a strong conceptual foundation, offer a large enough number of fitting questions for students to practise and carry useful explanations to the solutions of the questions. In essence, the book should guide the students to think in a manner that prepares them to address new similar or related problems with confidence. I believe that this book manages to do that in a user-friendly manner, with its format that leads the students through a graded approach in ascending order of difficulty. Given the significant part questions based on Logical Reasoning play in aptitude tests today, this book should prove to be useful to aspirants. It is easy to see that a lot of hard work has gone into the writing of ‘Koncepts of LR’ and I wish its author, Gajendra and all of you who wish to make use of this book to achieve you career aspirations, all the very best!

Charanpreet Singh, Associate Dean, Praxis Business School.

Charanpreet Singh is the CEO of AIM Education and the Associate Dean of Praxis Business School and visiting faculty of IIM Lucknow and IIM Raipur. A B. Tech from IIT Kanpur and an MBA from the University of Iowa, Charanpreet is a winner of the Chevening Scholarship of the British Government. He has worked for about two decades with corporate entities like British Oxygen, PwC and HP and presently runs Praxis Business School in Kolkata. (iii)

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Dedicated TO My father (Mr. Ramashankar Lal Karn, my only maths teacher) My Mother (Mrs. Kishori Karn, my guide) and little angels in my family Master Divyanshu (My cute Nephew) and Master Sanyukt (My son).

(iv)

Preface to the 3rd edition Tremendous response from students to the 1st and 2nd edition of the book encouraged me to come up with its 3rd edition in a very short span of time.

Feedback of the previous editions in the form of valuable suggestions, reviews from teachers and students, helped me to come out with this latest edition which I strongly believe is error free. In this edition, I have also added recent questions of XAT/IIFT/SNAP exams. Questions based on current day CAT pattern are also added in this book. Questions are divided in to 5 different difficulty levelsConcept Applicator (CA):- Everyone should attempt this section. Concept Builder (CB):- For CAT/GMAT/BANK PO/Govt Job, etc. Concept cracker (CC) :- For CAT (Who aim more than 90%) Concept Deviator(CD):- For CAT (Who aim more than 95%) Concept Eliminator (CE):- For those who are in search of excellence.

I would be really happy to receive critical review and suggestion from the students and esteemed teachers for further improvement of the book. You can reach out to me on email id given below. Gajendra Kumar [email protected]

You can contact me for any doubt(s) at

[email protected] [email protected]

For Online Audio-Visual solutions:

http://www.youtube.com/gajendra591

(v)

(vi)

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Contents

1.

Linear Arrangement

1 — 24



2.

Matrix Arrangement

25 — 66



3.

Circular Arrangement

67 — 76



4.

Set Theory



5.

Group/ Team Formation

77 — 104 105 — 116

6. Syllogism

117 — 140



7.

Logical Connectivity

141 — 150



8.

Input Output

151 — 168



9.

Games and Tournaments

169 — 190

10. Cubes

191 — 204

11. Calendar

205 — 216

12.

217 — 228

Path and Route

13. Clocks

229 — 234

14.

Blood Relation and Family Tree

235 — 248

15.

Coding and Series

249 — 260

16.

Statement Conclusion and Binary Logic

261 — 266

17. Direction

267 — 274

18.

275 — 290

Data Sufficiency

19. Puzzles

291 — 304

20.

Mathematical Reasoning

305 — 324

21.

Critical Reasoning

325 — 350 (vii)

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Miscellaneous Questions



1.

Mock Test with Solutions



2.

Mock Test with Solutions

MT-7

— MT-12



3.

Mock Test with Solutions

MT-13

— MT-18



4.

Mock Test with Solutions

MT-19

— MT-24

M-1-M-46 MT-1

(viii)

— MT-6

Linear Arrangement

1

1

Linear Arrangement Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt Exams

Very Important

Other Aptitude Test

Very Important

In Linear arrangement problems we are generally given a set of information about positioning of different elements with respect to other elements. From the given set of information we have to use the given information systematically to nd the actual arrangement of the elements. The arrangements can be in a straight line, on chair, in rooms or in a row. Another type of arrangement is arrangement in two rows parallel to each other. Left and Right: We can use Left and Right as per Information that generally is given and its interpretation is as followsÞ

Left and Right: We can use Left and Right as per our convenience. Generally (and in this book) we will use as followLeft End

Þ

Right End

A is 2 places right of B: Generally students used to get confuse that how many gaps are there between A and B. Here in this case there is only 1 gap between A and B. As it is explained in the diagram below. Left End

1st Place B

2ndPlace

3rd Place A

Right End

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2 Koncepts of Logical Reasoning Þ

A is 3 places right of B: Here in this case there is only 2 gaps between A and B. As it is explained in the diagram below. If B is at 1st place then A ia at 4th place. 1st Place

Left End Þ

3rd Place

4th Place

B

A

Right End

A stays 2 places away of B: Here in this case it is not given who is in right and who is in left so we have two different cases1st place

Left End Þ

2nd Place

2nd place

3rd Place

B/A

A/B

Right End

A stays 2 places away of B who is 3 place left of C: In this case, we can assume that B is at 3rd place then C is at 6th place, Left End

1st place A

2nd place

3rd Place B

4th Place

5th Place A

6th Place C

Right End

EXAMPLE 1 If four students are standing in a row such that A and B are always together, then draw a diagram and state all the possible cases.

Solution: It is given that 4 students are in a row. A is left of B but right of C: Here it is not mentioned that right or immediate right, as shown in the gure we have assumed that these are immediate right or left. 1st Place C

Left End

2nd Place A

3rd Place B

Right End

A is left of B but Right of C, while D is right of E but left of C We can represent this as, C > A > B, E > D > C by combining these we will get, E>D>C>A>B

EXAMPLE 2 If four persons A, B, C and D are arranged in a row such that A and B are always together while C and D are never together. How many arrangements are possible.

Solution: Here left and right of any arrangement is not mentioned hence we have following casesC

A

B

D

D

A

B

C

C

B

A

D

D

B

A

C

There are four such cases, we can represent these four cases in a line asC/d

A/B

B/A

D/C

Linear Arrangement

1

Directions for Question Nos. 1 to 3

A goldsmith has ve gold rings, each having a different weight: Statement 1: Ring D is weighing twice as much as ring E. Statement 2: Ring E is weighing four and one-half times as much as ring F. Statement 3: Ring F is weighing half as much as ring G. Statement 4: Ring G is weighing half as much as ring H. Statement 5: Ring H is weighing less than ring D but more than ring F. Based on the above statements, answer the following questions:

1.

2.

3.

Which of the following represents the descending order of weights of the rings? (a)

H, F, G, D and E

(b)

D, E, H, G and F

(c)

D, E, G, H and F

(d)

E, G, H, D and F

Wof the numbered statements above is not necessary to determine the correct order of the rings according to their weights? (a) Statement 3

(b)

statement 5

(c)

(d)

statement 4

statement 1

If these rings are sold according to their weights, which ring will fetch the highest value in rupees? (a) F

(b)

(c)

(d) H

G

4.

What is the earliest time when Ajay would complete C? (a) 11 am (b) 12 noon (c) 1 pm (d) 2 pm

5.

If Ajay takes 2 hours for B and completes other preceding tasks without delay, when can E starts? (a) 12 noon (b) 3 pm (c) 2 pm (d) 1 pm

6.

What is the earliest time when Ajay can complete all tasks? (a) 4 pm (b) 3 pm (c) 2 pm (d) 1 pm Directions for Question Nos. 7 to 9

Six products - Ariel, Vivel, Rin, Nirma, Gillette Gel and Pepsodent - are to be placed in six display windows’ of a shop numbered 1-6 from left to right of a shopper standing outside the shop. As per the company requirements, Rin and Ariel should be displayed next to each other, but Ariel should be at least three windows away from Nirma. Pepsodent is preferred to be kept between Gillette Gel and Rin but away from Vivel at least by two windows. Vivel cannot be displayed next to Rin for the reasons of mixed-product identity. Also Vivel cannot be displayed in window 1.

7.

Which of the following products is displayed left to Ariel? (a) Vivel (b) Nirma (c) Rin (d) Pepsodent

8.

If the positions of Rin and Ariel are interchanged, which item will be displayed in window 5? (a) Ariel (b) Nirma (c) Rin (d) Vivel

9.

Which of the following products except Rin will be displayed left of Ariel but right of Gillette Gel? (a) Vivel (b) Rin (c) Pepsodent (d) None of these

D

Directions for Question Nos. 4 to 6 Ajay would do ve tasks: A, B, C, D and E, starting at 9 am in the morning. A is the rst task and takes two hours. B can be done after A is completed and requires 1 hour. Work on C which would take 1 hour can start only after A and B are complete. Ajay can do task D along with B and C and would take 3 hours for that. Activity E with duration of 1 hour can start on completion of A, B, C and D.

3

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4 Koncepts of Logical Reasoning Directions for Question Nos. 10 to 12 In a cycling race, ve participants from various nations–Chinese, Nepalese, Indian, Iraqi and English–take part. Track 1 is extreme left and Track 5 is extreme right. The following conditions exist: (i) The Nepalese and the Englishmen are not cycling adjacent to each other. (ii) The Iraqi is not in one of the extreme tracks: (iii) The Chinese is to the left of the Indian. 10. If the Nepalese is in Track 3, the Chinese in Track 1, then the Indian could be in : (a) Track 4 (b) Track 2 (c) Track 2 or 4 (d) None of the above 11. If the Nepalese is in Track 4 and the Indian is in Track 3, then the English man could be in : (a) Track 1 (b) Track 2 (c) Track 1 or 2 (d) None of the above

2

Directions for Question Nos. 15 to 16

E-1, E-2 and E-3 are three engineering students writing their assignments at night. Each of them starts at a different time and completes at a different time. The digit in their name and the order of their starting and completing the assignment is certainly not the same. The last student to start is the rst to complete the assignment. [SNAP- 2010] 15. Who is the rst student to start writing the assignment? (a) E-1

(b)

(c)

(d) cannot be decided

E-3

E-2

16. Who is the last student to complete the assignment? (a) E-1

(b)

(c)

(d) cannot be decided

E-3

E-2

12. If the Iraqi is to the left of the Chinese, then the Iraqi could be in : (a) Track 2 only (b) Track 3 only (c) Track 2 or 3 (d) None of the above 13. Six students are sitting in a row in an examination hall. K is sitting between V and R. V is sitting next to M. M is sitting next to B. B is sitting on the extreme left and Q is sitting next to R. Who are sitting adjacent to V? (a) M and R (b) M and K (c) K and R (d) M and Q 14. Five newly extracted diamonds were weighed by a diamond mine worker. In his report card he pointed out that diamond A is lighter than diamond B. Diamond C is lighter than diamond D. Diamond B is lighter than diamond D but heavier than diamond E. Which diamond is the heaviest among all of them at the time of extraction? (a) E (b) D (c) C (d) none of the above

Q is brighter than R but duller than the Don School student who is brighter than A. The same Don School student is duller than P but is brighter than C. [SNAP- 2010] 17. Who is brightest amongst all? (a) B

(b)

(c)

(d) Cannot be decided

R

P

18. Who is the dullest amongst the three students from Elite School? (a) P

(b)

(c)

(d) cannot be decided

R

Q

19. When Rafael entered the class, there were already 10 students in the class 5 students entered the class between Roger and Rafael. Total 10 students entered after Roger. Exactly how many students are in the class nally?

Directions for Question Nos. 17 to 18 A, B and Care three students from Don School and P, Q and R are three students from Elite School.

[SNAP 2010] (a) 15

(b)

(c)

(d) Cannot be decided

27

25

Linear Arrangement Directions for Question Nos. 20 and 21 A, B, C, D and E sit on a long bench. C does not sit next to A or E. A and E have three persons sitting between them. [SNAP- 2010] 20. Who is sitting in the middle of the bench? (a) B (b) C (c) D (d) None of these 21. Who are sitting at the extreme ends of the bench? (a) A and E (b) B and D (3) C and E (d) None of these 22. There are 6 volumes of books on a rack kept in order (such as, vol. 1, vol.2 and so on). After some readers used them, their order got disturbed. The changes showed as follows: Vol.5 was directly to the right of vol.2. Vol.4 has vol.6 to its left and both were not at Vol.3’s place. Vol.1 has Vol.3 on right and Vol.5 on left. An even numbered volume is at Vol.5’s place. Find the order in which the books are kept now, from the 4 given alternatives: [SNAP- 2010] (a) 6,3,5,1,4,2 (b) 4,6,3,5,1,2 (c) 3,4,1,6,5,3 (d) 2,5,1,3,6,4 Directions for Question Nos. 23 to 27: Seven students Priya, Ankit, Raman, Sunil, Tony, Deepak and Vicky take a series of tests. No two students get similar marks. Vicky always scores more than Priya. Priya always scores more than Ankit. Each time either Raman scores the highest and Tony gets the least, or alternatively Sunil scores the highest and Deepak or Ankit scores the least. [SNAP – 2011] 23. If Sunil is ranked sixth and Ankit is ranked fth, which of the following can be true? (a) Vicky is ranked rst or fourth (b) Raman is ranked second or third (c) Tony is ranked fourth or fth (d) Deepak is ranked third or fourth 24. If Raman gets the highest, Vicky should be ranked not lower than: (a) Second (b) Third (c) Fourth (d) Fifth

5

25. If Raman is ranked second and Ankit is ranked rst, which of the following must be true? (a) Sunil is ranked third (b) Tony is ranked third (c) Priya is ranked sixth (d) none of these 26. If Sunil is ranked second, which of the following can be true? (a) Deepak gets more than Vicky (b) Vicky gets more than Sunil (c) Priya gets more than Raman (d) Priya gets more than Vicky 27. If Vicky is ranked fth, which of the following must be true? (a) Sunil scores the highest (b) Raman is ranked second (c) Tony is ranked third (d) Ankit is ranked second 28. Among Anil, Bibek, Charu, Debu, and Eswar, Eswar is taller than Debu but not as fat as Debu. Charu is taller than Anil but shorter than Bibek. Anil is fatter than Debu but not as fat as Bibek. Eswar is thinner than Charu, who is thinner than Debu. Eswar is shorter than Anil. Who is the thinnest person? [IIFT [2008]] (a) Bibek

(b)

(c)

(d) Eswar

Debu

Charu

29. Ganesh Cultural Centre for promoting arts has appointed 3 instructors for music, dance, and painting. Music instructor takes session from 12 noon to 4:00 pm on Monday, Thursday and Sunday. The sessions of dance instructor are scheduled on Tuesday, Thursday, Wednesday and Sunday between 10:00 am to 2:00 pm. The 9:00 am to 12:00 noon slot on Tuesday, Friday and Thursday and also 2:00 pm to 4:00 pm slot on Wednesday, Saturday and Sunday is lled up by Painting Instructor. On which day(s) of a week the dance and painting sessions are simultaneously held? [IIFT [2009] (a) Sunday and Wednesday (b) Tuesday and Friday (c) Tuesday and Thursday (d) Only on Tuesday

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6 Koncepts of Logical Reasoning Directions for Question Nos. 30 to 34

Directions for Question Nos. 35 to 37

During the cultural week of an institute six competitions were conducted. The cultural week was inaugurated in the morning of 19th October, Wednesday and continued till 26th October. In the span of 8 days six competitions namely debate, folk dance, Fash-p, street play, rock band, and group song, were organized along with various other cultural programs. The information available from the institute is: i. Only one competition was held in a day. ii. Rock band competition was not conducted on the closing day. iii. Fash-p was conducted on the day prior to debate competition. iv. Group song competition was conducted neither on Wednesday nor on Saturday. v. None of the competition was conducted on Thursday and Sunday. vi. Street Play competition was held on Monday. vii. There was gap of two days between debate competition and group song competition.

A parking lot can accommodate only six cars. The six cars are parked in two rows in such a way that the front of the three cars parked in one row is facing the other three cars in the other row. i. Alto is not parked in the beginning of any row ii. Esteem is second to the right of i10 iii. Punto which is the neighbor of Alto is parked diagonally opposite to i10 iv. Swift is parked in front of Alto v. SX4 is parked to the immediate right of Alto

[IIFT 2010] 30. The cultural week started with which competition? (a) Fash-p competition (b) Debate competition (c) Street play competition (d) rock band competition 31. How many days gap is there between rock band competition and group song competition? (a) Two (b) Three (c) four (d) Five 32. Which pair of competition was conducted on Wednesday? (a) Rock band competition and debate competition (b) Debate competition and fash-p competition (c) Rock band competition and folk dance competition (d) None of these 33. Which competition exactly precedes the street play competition? (a) rock band competition (b) Group song competition (c) Debate competition (d) Fash-p competition 34. Fash-p competition follows with which competition? (a) Debate competition (b) Street play competition (c) Rock band competition (d) None of these

[IIFT 2010] 35. If SX4 and Esteem exchange their positions mutually then car(s) adjacent to Esteem is (are)? (a) i10 and Swift

(b)

(c)

(d) Alto and Punto

Only Alto

Only Swift

36. If Alto changes position with i10 and Punto changes position with SX4 and Swift shifts one position to the right to accommodate Beatle then the car(s) parked adjacent to Beatle is (are)? (a) Punto only

(b)

(c)

(d) Alto and Swift

Punto and Alto

i10

37. In the original parking scheme four new cars enter the parking lot such that Wagon-R is second to the right, of i10 and Zen is second to the left of SX4. Jazz is parked second to the left of Wagon-R and Beat is parked to the right of Alto then the cars that moved out are? (a) Esteem and Swift (b) (c)

i10 and Alto

Punto and Alto

(d) Punto and SX4

Directions for Question Nos. 38 and 39 Sampada Apartment is a housing society formed by a group of professors of a University. It has six ats on a oor in two rows facing North and South which are allotted to Prof. Purohit, Prof. Qureshi, rof. Rathor, Prof. Sawant, Prof. Tripathy and Prof. Usman. Prof. Qureshi gets a North facing at and it is not next to Prof. Sawant’s at. Prof. Sawant and prof. Usman get their ats which are diagonally opposite to each other. Prof. Rathor gets a south facing at which is next to Prof. Usman’s at. Prof, Tripathy’s at is North facing. [IIFT 2011]

Linear Arrangement 38. Which of the following professors get South facing ats? (a) Prof. Qureshi, Prof. Tripathy and Prof. Sawant (b) Prof. Usman, Prof. Tripathy and Prof. Purohit (c)

Prof. Usman, Prof. Rathor and Prof. Purohit

(d) none of the above 39. If the ats of Prof. Tripathy and Prof. Purohit are interchanged, whose at will be next to that of Prof. Usman? (a) Prof. Rathor

(b)

(c)

(d) none of the above

Prof. Usman

Prof. Tripathy

40. Gita is older than her cousin Mita. Mita’s brother Bhanu is older than Gita. When Mita and Bhanu are visiting Gita, all three like to play a game of Monopoly. Mita wins more often than Gita does. Which of the following can be concluded from the above? [IIFT 2012] (a) When he plays Monopoly with Mita and Gita. Bhanu often loses. (b) Of the three, Gita is the oldest. (c)

Gita hates to lose at Monopoly.

(d) Of the three, Mita is the youngest. 41. Priya is taller than Tiya and shorter than Siya. Riya is shorter than Siya and taller than Priya. Riya is taller than Diya, who is shorter than Tiya. Arrange them in order of ascending heights.

[IIFT 2012]

(a) Priya –Siya –Riya –Tiya -Diya

Directions for Question Nos. 42 and 43 Era is in charge of seating of the speakers at a table. In addition to the moderator, there will be a pilot, a writer, an attorney, and an explorer. The speaker’s names are Gaj, hema, Jaya, kumar, and Lalit. The moderator must sit in the middle, in seat # 3 The attorney cannot sit next to the explorer Lalit is the pilot The writer and the attorney sit on either side of the moderator Hema, who is not the moderator, sits between Kumar and Jaya. The moderator does not sit next to Jaya or Lalit Gaj, who is attorney, sits in seat # 4. [IIFT 2012] 42. Who is moderator? (a) Lalit (c) Hema

(b) Gaj (d) Kumar

43. Where does Jaya sit? (a) Seat # 1 (b) (c) Seat # 3 (d)

Seat # 2 Seat #

44. In a four day period- Monday through Thursday – each of the following temporary ofce workers worked only one day, each a different day. Jai was scheduled to work on Monday, but he traded with Raj, who was originally scheduled to work on Wednesday. Farid traded with kajal, who was originally scheduled to work on Thursday. Finally, Jai traded with Kajal. After all the switching was done, who worked on Tuesday?

(b) Riya – Siya- Priya – Diya – Tiya (c)

Siya – Riya – Priya – Tiya - Diya

(d) Siya – Priya – Riya – Diya- Tiya

3

Directions for Question Nos. 45 to 49

Sourav’s Fish Salon serves a special Friday night seafood banquet consisting of seven courses - hilsa, pomfret, Indian shrimp, rahu, kingsh, lobster, and bhetki. Diners are free to select the order of the seven courses, according to the following conditions:

7

[IIFT 2012] (a) Jai (c) Raj

(b) Farid (d) Kajal

The kingsh is served some time after rahu. Exactly one course should be served between the pomfret and the Indian shrimp. The lobster is served some time before the pomfret. The kingsh is served either fth or sixth. The hilsa is served second. [XAT 2006]

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8 Koncepts of Logical Reasoning 45. Which one of the following sequences would make for an acceptable banquet? (a) rahu, hilsa, lobster, bhetki, pomfret, kingsh, Indian shrimp (b) rahu, hilsa, bhetki, pomfret, kingsh, Indian shrimp, lobster (c) lobster, hilsa, pomfret. rahu, kingsh, Indian shrimp, bhetki (d) lobster, hilsa, rahu. kingsh. pomfret, bhetki, Indian shrimp 46. If kingsh is the fth course served, then which one of the following MUST BE true? (a) Pomfret is the third course served (b) Indian shrimp is the fourth course served (c) Bhetki is the seventh course served (d) Lobster is the rst course served 47. Which one of the following would make it possible to determine the EXACT ordering of the courses? (a) Pomfret is the fourth course served (b) Indian shrimp is the fth course served (c) Kingsh is the sixth course served (d) Lobster is the rst course served 48. If kingsh is the sixth course served, then which one of the following CANNOT be true? (a) Rahu is the fth course served (b) Indian shrimp is the seventh course served (c) Pomfret is the fth course served (d) Lobster is the third course served 49. If Bhetki is the third course served, which one of the following MUST BE true? (a) Pomfret is the fourth course served (b) Kingsh is the fth course served (c) Rahu is the rst course served (d) Indian shrimp is the seventh course served

50. Which of the following is a proper list of persons in increasing order of prize money won? (a) G,C,F,B,E,D,A

(b)

D,F,C,E,A,B,G

(c)

(d)

A,B,G,C,F,E,D

F,C,D,E,A,B,G

51. If D won more than E, and B and G together won Rs 3.5 lakhs, which of the following MUST be true? (a) D won Rs 3.5 lakhs (b)

A won Rs 1.5 lakhs

(c)

C won Rs 50,000

B won Rs 1.5 lakhs (d)

52. If the difference of prize money between A and C is the minimum, which of the following pairs MUST NOT have won prize money that differs by the minimal amount? (a) B and E

(b)

C and G

(c)

(d)

A and E

D and G

53. If the total money won by A and D is equal to that of G, and the difference between E and D is at least 1 lakh, then which of the following MUST be TRUE? (a) A and B together won Rs. 3 lakhs (b) B and F together won Rs. 3.5 lakhs (c)

Cand E together won Rs. 3 lakhs

(d) B and C together won Rs. 3.5 lakhs Directions for Question Nos. 54 to 58 A company launches eight products – Q, R, S, T, V, W, Y, and Z – in one of the four metros of India. The products were launched one after the other over a period of six months in 2006. The order in which the products were launched is consistent with the following conditions: (a)

V is launched before both Y and Q

Directions for Question Nos. 50 to 53

(b)

Q gets launched after Z

Seven persons A, B, C, D, E, F and G contested in a game show that had total prize money ofRsl4 lakhs. Every contestant won some prize money and the highest prize money was Rs. 3.5 lakhs. No two contestants won the same amount of prize money. For every contestant the difference with the next highest and the next lowest winner is the same. E won Rs.2lakhs.B won more money than A. The difference of prize money between B and A was the minimum. The difference of prize money between D and F was not the least. There was at least one person whose prize money was between that of E and G [XAT 2006]

(c)

T gets launched before V but after R

(d) S gets launched after V (e)

R gets launched before W [XAT 2007]

54. Which one of the following could be true? (a) Y is the second product to be launched. (b) R is the third product to be launched. (c)

Q is the fourth product to be launched

(d) S is the fth product to be launched. (e)

V is the sixth product to be launched.

Linear Arrangement 55. Is Z is the seventh product to be launched, then which one of the following could be true (a) W is the fth product to be launched (b) T is the fourth product to be launched. (c)

R is the second product to be launched.

(d) V is the sixth product to be launched. (e)

(a) Z is the rst product to be launched. (b) T is the second product to be launched. (c)

V is the third product to be launched.

(d) W is the fourth product to be launched. (e)

J is not advertised during a given week unless H is advertised during the immediately preceding week. The product that is advertised twice is advertised during week 4 but is not advertised during week 3. G is not advertised during a given week unless either J or O is also advertised that week. K is advertised during one of the rst two weeks. O is one of the products advertised during week 3.

Y is the eighth product to be launched

56. If Q is the fth product to be launched, then each of the following could be true EXCEPT-

Y is the sixth product to be launched.

57. If R is the second product to be launched, which one of the following MUST be true? (a) S gets launched some time before T.

[XAT 2008] 59. Which one of the following could be the schedule of the advertisements? (a) Week 1: G, J; week 2: K, L; week 3: O, M; week 4: H, L (b) Week 1: H, K; week 2: J, G; week 3: O, L; week 4: M, K (c)

W gets launched some time before V.

(d) Y gets launched some time before Q. (e)

Z gets launched some time before W

58. If V gets launched before Z does, then which one of the following COULD be true? (a) R is the second product to be launched. (b) T is the fourth product to be launched. (c)

Q is the fourth product to be launched.

(d) V is the fth product to be launched (e)

Z is the sixth product to be launched

Directions for Question Nos. 59 to 62 During a four-week period, each one of seven previously unadvertised products–G,H, J, K, L, M and O – will be advertised. A different pair of these products will be advertised each week. Exactly one of the products will be a member of two of these four pairs. None of the other products gets repeated in any pair. Further, the following constraints must be observed:

Week 1: H, K; week 2: J, M; week 3: O, L; week 4: G, M

(d) Week 1: H, L; week 2: J, M; week 3: O, G; week 4: K, L (e)

(b) T gets launched some time before W. (c)

9

Week 1: K, M; week 2: H, J; week 3: O, G; week 4: L, M

60. If L is the product that is advertised during two of the weeks, which one of the following is a product that MUST be advertised during one of the weeks in which L is advertised? (a) G

(b)

H

(c)

J

(d) K

(e)

M

61. Which one of the following is a product that could be advertised in any of the four weeks? (a) H

(b)

J

(c)

K

(d) L

(e)

O

62. Which one of the following is a pair of products that could be advertised during the same week? (a) G and H

(b)

H and J

(c)

H and O

(d)

K and O

(e)

M and O

EBD_7743

10  Koncepts of Logical Reasoning 

4

Concept Deviator

Directions for Question Nos. 63 to 65

Five of India’s leading models are posing for a photograph promoting ‘y know, world peace and understanding’. But then Rakesh Shreshtha, the photographer, is having a tough time getting them to stand in a straight line, because Aishwariya refuses to stand next to Sushmita for Sushmita has said something about her in a leading gossip magazine. Rachel and Anu want to stand together because they are ‘such good friends, ‘y know’. Manpreet on the other hand cannot get along well with Rachel, because there is some talk about Rachel scheming to get a contract already awarded to Manpreet. Anu believes her friendly astrologer who has asked her to stand at the extreme right for all group photographs. Finally, Rakesh managed to pacify the girls and got a beautiful picture of five girls smiling in a straight line, promoting world peace.

[CAT 1994]

63. If Aishwariya is standing to the extreme left, who is standing in the middle? (a) Manpreet (b) Sushmita (c) Rachel (d) cannot say 64. If Aishwariya stands to the extreme left, who stands second from left? (a) Cannot say (b) Sushmita (c) Rachel (d) Manpreet 65. If Anu’s astrologer tells her to stand second from left and Aishwariya decides to stand second from right, then who is standing on the extreme right? (a) Rachel (b) Sushmita (c) cannot say (d) Manpreet Directions for Question Nos. 66 to 69 A, B, C and D are to be seated in a row. But C and D cannot be together. Also B cannot be at the third place. 66.

[CAT 1998]

Which of the following must be false? (a) A is at the first place (b) A is at the second place (c) A is at the third place (d) A is at the fourth place

67. If A is not at the third place, then C has which of the following option?

(a) The first place only



(b) The third place only



(c) The first and third place only



(d) any of the places

68. If A and B are together, then which of the following must be necessarily true?

(a) C is not at the first place



(b) A is at the third place



(c) D is at the first place



(d) C is at the first place

69. P,Q,R and S are four statements. Relation between these statements is as follows

A. If P is true, then Q must be true.



B. If Q is true, then R must be true.



C. If S is true, then either Q is false or R is false.



Which of the following must be true?



(a) If P is true, then S is false



(b) If S is false, then Q must be true



(c) If Q is true, then P must be true



(d) If R is true, then Q must be true Direction for Question No. 70 Read each of the five problems given below and choose the best answer from among the four given choices.

70. Persons X, Y, Z and Q live in red, green, yellow or blue coloured houses placed in a sequence on a street. Z lives in a yellow house. The green house is adjacent to the blue house. X does not live adjacent to Z. The yellow house is in between the green and red houses. The colour of the house X lives in is: [CAT ]

(a) Blue



(b) green



(c) red



(d) not possible to determine

71. At a village mela, the following six nautankis (plays) are scheduled as shown in the table below. No. 1 2 3 4 5 6.









[CAT 2001]

Nautanki

Duration Show Times 9:00 a.m. and Sati-Savitri 1 hour 2:00 p.m. 10:30 a.m. and JorukaGhulam 1 hour 11:30 a.m. 10:00 a.m. and SundarKand 30 minutes 11:00 a.m. 10:00 a.m. and Veer Abhimanyu 1 hour 11:00 a.m. 9:30 a.m., Reshmaaur1 hour 12:00 noon and Shera 2:00 p.m. 11:00 a.m. and Jhansi ki Rani 30 minutes 1:30 p.m.

You wish to see all the six nautankis. Further, you wish to ensure that you get a lunch break from 12:30 p.m. to 1:30 p.m. Which of the following ways can you do this? (a) Sati-Savitri is viewed first; SundarKand is viewed third and Jhansi ki Rani is viewed last. (b) Sati-Savitri is viewed last; Veer Abhimanyu is viewed third and ReshmaaurShera is viewed first. (c) Sati-Savitri is viewed first; SundarKand is viewed third and JorukaGhulam is viewed fourth. (d) Veer Abhimanyu is viewed third; ReshmaaurShera is viewed fourth and Jhansi ki Rani is viewed fifth.







Directions for Question Nos. 73 and 74 Elle is three times older than Yogesh. Zaheer is half the age of Wahida. Yogesh is older than Zaheer. [CAT 2001] 73.



Based on the above clues, which one of the following statements is true? (a) Vaibhav is the oldest, followed by Anshuman who was born in September, and the youngest is Suprita who was born in April.

Which of the following can be inferred? (a) Yogesh is older than Wahida. (b) Elle is older than Wahida. (c) Elle may be younger than Wahida. (d) None of the above.

74. Which of the following information will be sufficient to estimate Elle’s age? (a) Zaheer is 10 years old. (b) Both Yogesh and Wahida are older than Zaheer by the same number of years. (c) Both (a) and (b) above. (d) None of the above. Directions for Question Nos. 75 and 76 Some children were taking free throws at the basketball court in school during lunch break. Below are some facts about how many baskets these children shot. i. Ganesh shot 8 baskets less than Ashish ii. Dhanraj and Ramesh together shot 37 baskets iii. Jugraj shot 8 baskets more than Dhanraj iv. Ashish shot 5 baskets more than Dhanraj v. Ashish and Ganesh together shot 40 baskets.

72. Mrs.Ranga has three children and has difficulty remembering their ages and the months of their birth. The clues below may help her remember. [CAT 2001]  The boy, who was born in June, is 7 years old.  One of the children is 4 years old, but it is not Anshuman.  Vaibhav is older than Suprita.  One of the children was born in September, but it was not Vaibhav.  Suprita’s birthday is in April.  The youngest child is only 2 years old.

  Linear Arrangement 11  (b) Anshuman is the oldest being born in June, followed by Suprita who is 4 years old, and the youngest is Vaibhav who is 2 years old. (c) Vaibhav is the oldest being 7 years old, followed by Suprita who was born in April, and the youngest is Anshuman who was born in September. (d) Suprita is the oldest who was born in April, followed by Vaibhav who was born in June, and Anshuman who was born in September.

[CAT 2003] 75. Which of the following statements is true?

(a) Ramesh shot 18 baskets and Dhanraj shot 19 baskets.



(b) Ganesh shot 24 baskets and Ashish shot 16 baskets.



(c) Jugraj shot 19 baskets and Dhanraj shot 27 baskets.



(d) Dhanraj shot 11 baskets and Ashish shot 16 baskets.

76. Which of the following statements is true? (a) Dhanraj and Jugraj together shot 46 baskets. (b) Ganesh shot 18 baskets and Ramesh shot 21 baskets. (c) Dhanraj shot 3 more baskets than Ramesh. (d) Ramesh and Jugraj together shot 29 baskets. Directions for Question Nos. 77 to 79 Seven varsity basketball players (A, B, C, D, E, F and G) are to be honoured at a special luncheon. The players will be seated on the dais in a row. A and G have to leave the luncheon early and so must be seated at the extreme right. B will receive the most valuable player’s trophy and so must be in the centre to facilitate presentation. C and D are bitter rivals and therefore must be seated as far apart as possible.

77. Which of the following cannot be seated at either end?

(a) C

(b) D



(c) F

(d) G

78. Which of the following pairs cannot be seated together?

(a) B and D

(b) C and F



(c) D and G

(d) E and A

79. Which of the following pairs cannot occupy the seats on either side of B?

(a) F and D

(b) D and E



(c) E and G

(d) C and F

[CAT ]

5

Concept Eliminator

Directions for Question Nos. 80 to 85

In MERI Kolkata, hostel wings are named after famous scientists. In Bose Wing there are six blocks of rooms along a straight corridor with each block containing two rooms facing each other.  Right

Left 

Corridor

There are 12 students who stays in this Bose Wing, these are Chandan, Chandrakant, Gaurav, Dripto, Gopal, Hemant, Chandra, Jayanta, Krishna, Kunal, Saurabh and Raj has occupied some of these rooms. There are a maximum of two people in a room and some rooms may be empty.  Raj and Gopal stay in single rooms two blocks to the left of Gaurav  Chandrakantand his roommate stay two blocks to the right of Jayanta and his roommate Chandra.  Chandan stays alone, three blocks to the left of Hemant and two blocks to the left of Saurabh.  Gaurav stays one block to the left of Jayanta and Chandra.  Dripto stays three blocks to the right of the block on which Krishna and Kunal have single rooms.

80. Which of the following lists the persons in the correct order, going from the left most block to the right?

(a) Raj, Krishna, Gaurav, Jayanta, Chandrakant, Dripto



(b) Rajener, Kunal, Dripto, Saurabh, Chandrakant. Chandra



(c) Gopal, Krishna, Chandan, Chandra, Dripto, Chandrakant



(d) Chandrakant, Saurabh, Jayanta, Gaurav, Kunal, Raj

81. Which of the following pairs must stay on the same block?

I.

Dripto and Saurabh



II. Chandan and Gaurav



III. Jayanta and Chandrakant



(a) I only

(b) III only



(c) I and II only

(d) II and III only

82. Chandrakant’s roommate, assuming that he or she is one of the persons mentioned is

(a) Saurabh

(b) Hemant



(c) Gaurav

(d) Dripto

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12  Koncepts of Logical Reasoning 

Linear Arrangement 13 83. Raj stays on the (a) rst block, and to the left of Krishna or Kunal (b) second block, and to the left of Chandan or Jayanta and Chandra. (c)

third block, and to the right of Gaurav or Saurabh

(d) fourth block as Jayanta and Chandra 84. An empty room or empty rooms may be found in the (a) second block only. (b) fourth block only (c)

third or sixth block, but not both

Directions for Question Nos. 88 to 90 Five 1st year students of MERI are waiting for their punishment. Name of these students are Choudhry, Chitranshi, Oswal, Malhotra and Garg. (i) Choudhry is next to Garg and Malhotra is next to Oswal. (ii) Malhotra is not next to Chitranshi, Chitranshi is on the extreme left hand side and Oswal is on the second position from the right hand side. (iii) Choudhry is on the right hand side of Garg and to the right side of Chitranshi. (iv) Choudhry and Oswal are together.

(d) fourth or sixth block or both 85. Chandan arranges to move into a room two blocks to the left, whose occupant moves into a room one block to the right. In turn, the occupant of this room moves into a room three blocks to the right, whose occupant takes Chandan’s old room. The new occupant of Chandan’ old room is (a) Krishna or Kunal (b) (c)

Gaurav

Dripto or Saurabh

(d) Raj

Directions for Question Nos. 86 and 87 Consider statements (i)−(iv): (i) X was born before Y. (ii) Y was born in the sixteenth century. (iii) Z was born before A. (iv) B was a compatriot of X. S: If either B was a compatriot of X or Z was born before A, then Y was born in the sixteenth century

88. Which of the following statements is redundant in order to nd their correct sequence? (a) Either (i) or (ii) (b) Either (ii) or (iii) (c) Either (iii) or (iv) (d) All of them are required 89. _____ is on the extreme right and _____ is on the extreme left. (a) Chitranshi and Oswal (b) Garg and Chitranshi (c) Malhotra and Chitranshi (d) Chitranshi and Malhotra 90. _____ is in the middle and ____ is on its right. (a) Choudhry and Malhotra (b) Oswal and Malhotra (c) Choudhry and Oswal (d) Garg and Choudhry Directions for Question Nos. 91 to 94

86. To make sentence S true, (a) (ii), (iii) and (iv) should be false. (b) (iii) and (iv) should be true and (ii) should be false. (c)

(iii) should be true and (ii) and (iv) should be false.

(d) (i), (ii), (iii) and (iv) should be true. 87. To make sentence S false, (a) (ii), (iii) and (iv) should be true. (b) (iii) or (iv) should be true and (ii) should be false. (c)

(iii) and (iv) should be true and (ii) should be false.

(d) (iii) and (iv) should be false and (ii) should be true.

In Praxis Business School Kolkata 6 students Umesh, Victor, Wasim, Xavior, Yogita and Zeeshan participated in swimming and running tournament. The results are as follows1. Umesh got better rank in swimming than Victor 2. Umesh got better rank in running than Zeeshan. 3. Victor got better rank in swimming than Yogita 4. Zeeshan got better rank in running than Victor 5. Xavior got better rank in swimming thanWasim. 6. Umesh got better rank in running thanWasim 7. Xavior lost to Yogita in swimming and won in running. 8. Yogita lost to Umesh in swimming and won in running. 9. Zeeshan got better rank in Swimming than Umesh but lost to Wasim in running.

EBD_7743

14 Koncepts of Logical Reasoning 91. How many student won to Xavior in swimming and lost to Yogita in running? (a) 0 (b) 1 (c) 2 (d) 3 92. Which of the following student is/are won in both swimming and running to Victor? (a) Umesh only (b) Wasim only (c) Zeeshan only (d) Umesh and Zeeshan only 93. Which of the following student could be awarded as the 3rd best on the basis of swimming and running if swimming has 60% weightage and running has 40% weightage? (a) Umesh (b) Victor (c) Yogita (d) Zeeshan 94. What should be the minimum weightage on swimming such that Zeeshan be the overall topper? (a) 67% (b) 70% (c) 55% (d) 27% Directions for Question Nos. 95 to 98 5 friends, M, N, O, P, Q of Praxis Business School Kolkata, stay in hostel on 2nd oor and their room is in a row which has 6 rooms. 1. M’s room is either room number 1 or 2. 2. O’s Room is just before or after Q’s room. 3. P stays at room number 3. 4. Q can not stay at room number 1, 2 or 3.

95. If un occupied room are room number 5 and 6, which of the following MUST be true? (a) M stay at room number 2 (b) N stay at room number 2 (c) O stay at room number 4 (d) O and P are room mates 96. If Q stays in room number 4 and no more than two of the friends stay in a room, which of the following MUST be true? (a) lf N stays in room no 2, M stays in room no l. (b) If N stays in room no 3, O stays in room no 5. (c) If N stays in room number 4,O stays in room no 5. (d) If O stays in room no 3, N stays in room no 2 97. If N and O are room mates then in which room they can saty?

I.

Room number 3

II.

Room Number 4

III. Room Number 5 (a) I only

(b)

II and III only

(c)

(d)

I and II only

III only

98. If only un-occupied room is room number 1 then, which of the following cannot be true? (a) N stays just one room before Q (b) P stays just one room before O (c)

M stays at room number 2

(d) N Stays at room number 5. Directions for Question Nos. 99 to 102 The owner of the house has been murdered. The visitors to the house were Aditya, Vijay and Puneet. The following additional information is also given. 1. The murderer who was one of the three visitors, arrived at the house later than at least one of the other two visitors. 2. The driver of the house who was one of the three visitors, arrived at the house earlier than at least one of the two visitors. 3. The driver arrived at the house at midnight. 4. Neither Aditya nor Vijay arrived at the house after midnight. 5. Between Vijay and Puneet, the one who arrived earlier was not the driver. 6. Between Aditya and Puneet, the one who arrived later was not the murderer.

99. Who arrived at the house earliest? (a) Puneet

(b)

(c)

(d) Data insufcient

Vijay

Aditya

100. Who is the murderer? (a) Puneet

(b)

(c)

(d) Data insufcient

Vijay

Aditya

101. Who is the driver? (a) Puneet

(b)

(c)

(d) Data insufcient

Aditya

Vijay

102. Who arrived at the house last? (a) Puneet

(b)

(c)

(d) Data insufcient

Vijay

Aditya

103. Who arrived at midnight? (a) Puneet

(b)

(c)

(d) Data insufcient

Vijay

Aditya

Linear Arrangement 15 Directions for Question Nos. 104 to 110 Seven friends, Susmit, Moumita, Ricky, Zeeshan, Danish, Vivek and Rahul were invited to a party. After party it was found that a necklace is stolen, On being asked by police about their visit to the party, following was their responses Susmit: I went to the party twice, when I returned to the party only Ricky and Vivek were left. Moumita: I went to the party twice and when I returned to the party Susmit was still there and Ricky was the only new person that I met on my second visit. Ricky:- When I was chatting with only friend left at party, I saw Susmit entering the party. Zeeshan: When I arrived at the party I saw Susmit talking to Rahul while Moumita was talking to someone over phone. Danish: When I arrived at the party I met Susmit, Zeeshan and Ricky, After few minutes only I came out with Susmit.. Rahul: I was the second to arrive at the party after Susmit. Since Moumita did not like Zeeshan hence I left the party with Moumita immediately after Zeeshan arrived. 104. Who met with minimum number of friends(a) Only Danish andVivek (b) Only Vivekand Rahul (c) Only Danish or Rahul (d) None of these

Concept Applicator (CA) Solutions (1 to 3) 1. (b) Based on given information we get, D=2E, E=9F/2, F=G/2,G=H/2. Let the weight of H be 8x. Then G=4x,F=2x, E=9x, and D=18x. Hence weight of descending order will be as follows- D (=18x) , E (=9x), H (=8x) , G (=4x) ,F (=2x). 2. (b) Last information i.e(F R Or in other words A/B/C >Q > R It is given that the Don School student who is brighter than Q is also brighter than A. From this we can conclude that A is not the brightest. The same Don School student is brighter than C. Hence it is B who is brighter than Q. But we know that B is duller than P. So the nal sequence should be. P>B>Q>R A and C are duller than B. hence P is the brightest

Linear Arrangement 17 18. (c) From the solution of previous question Among Elite school students P, Q and R, R is the dullest. 19. (d) There are 5 students who entered between Roger and Rafael but it is not mentioned who came 1st hence we cannot determine. Solutions (20 and 21) As there are three persons between A and E, C is not adjacent to A or E hence the nal arrangement is as follows. A/E B/D C D/B E/A 20. (b) C is sitting at the middle of the bench. 21. (a) A and E are sitting at the extreme ends. 22. (d) Initial Position is V1 V2 V3 V4 V5 V6 Now from the 1st condition- V2 V5 From 2nd condition V6 V4 and none of them at position number three hence they cannot be at 1st two places . From 3rd condition V5 V1 V3 At 5th place even numbered book is placed. From these information we can conclude that:V2 V5 V1 V3 V6 V4 hence option (d) is correct. Solutions (23 to 27) Vicky > Priya Priya > Ankit One of them is true- Raman is highest or Tony is least or Sunil got highest and Deepak or Ankit got least. Now for each and every question we have different cases23. (d) Since Sunil got the sixth and Ankit got the fth rank, hence we can conclude that Raman must have got the highest and Tony must have got the least rank. And the order of ranks of Vicky, Priya and Ankit must be Vicky > Priya > Ankit. Vicky must have got the second or third rank and accordingly Priya must have got the third or fourth rank. Deepak must have got any rank among two, three and four. 24. (c) If Raman gets the highest, then Tony will get least and be ranked 7th. Now since Priya and Ankit are always ranked numerically higher than Vicky. Hence, Vicky can never be ranked lower than 4th. 25. (d) Given that Ankit is ranked 1st.But is not possible as he always scores less than Vicky and Priya. Hence this case is not possible and data is inconsistent. 26. (a) If Sunil gets the second rank, then Raman gets the rst rank and Tony gets the last rank. In that case Vicky’s rank can be 3 or 4, and

Priya’s rank can be 4 or 5 and Ankit’s rank can be 5 or 6 then Deepak’s rank can be 3, 4 or 5. 27. (a) If Vicky is ranked 5th, then Priya is ranked 6th and Ankit is ranked 7th. Then only possibility is Sunil is ranked 1st. 28. (d) From the given condition we can form the sequence as Sequence as per height : B>C>A>E>D Sequence as per body build up : B>A>D>C>E 29. (c) From the information , dance and painting will be held simultaneously as dance instructor are scheduled on Tuesday, Thursday, Wednesday and Sunday between 10:00 am to 2:00 pm and the 9:00 am to 12 noon slot on Tuesday, Friday and Thursday and also 2:00 pm to 4:00 pm slot on Wednesday, Saturday and Sunday is lled up by Painting instructor. So on Tuesday and Thursday both sessions will simultaneously held. Tuesday : Dance 10 : 00 am to 2 : 00 pm, Painting 9 : 00 am to 12 : 00 noon.and on Thursday : Dance – 10 : 00 am to 12 : 00 noon,Painting – 9 : 00 am to 12 : 00 noon. Solutions (30 to 34) From the given information: Date for competition from 19th October to 26th October (From Wednesday to Wednesday) From (v), no competition was held on 20th and 23rd (i.e on Thursday & Sunday) From (vi), Street play was held on 24th. From (iii) and (vii), Debate must have been held on 22nd,Flash-P on 21st and Group song on 25th. From (ii), Rock-band was held on 19th. Hence from this we can draw following resultDate

Day

Competition

19

Wednesday

Rock band

20

Thursday

–

21

Friday

Fash - p

22

Saturday

Debate

30. (d) 32. (c) 34. (c)

23

Sunday

–

24

Monday

Street Play

25

Tuesday

Group song

26

Wednesday

Folk dance

31. (d) 33. (c)

Solutions (35 to 37) From given information original arrangement is as following SX4

Alto

Punto

i 10

Swift

Esteem

35. (c) If S´4 and Esteem exchange their positions, then Alto will be adjacent to Esteem.  36. (d) Alto and Swift will be adjacent to Beatle.  37. (d) Under the given conditions, the arrangement of cars will be as follows. Beatle

Alto

Zen

Jazz

Swift

Wagon R

The cars that are moved out are Sx4, Punto, i10 and Esteem.  Solutions (38 to 39) As per the given information arrangements of flats should be as follows North South From the given condition Rathore is facing South and is next to Usman, Usman must be facing South   . Usman is diagonally opposite Sawant, the arrangement can be represented as follows North South

Sawant Usman

Rathore

But it is given that Qureshi is not next to Sawant and Tripathy’s flat is facing North, the final arrangement should beNorth

Qureshi

Tripathi

Sawant

South

Usman

Rathore

Purohit

38. (c) From this final arrangement we can conclude that Prof. Usman, Prof. Rathor and Prof. Purohit staying in south facing flats. 39. (a) The new arrangement would be North

Qureshi

Purohit

Sawant

South

Usman

Rathore

Tripathi



Now from this arrangement Prof Usman is next to Prof Rathore. IIFT 2012 40. (d) From the given condition we can conclude that  Bhanu > Geeta > Meeta.        Hence, Meeta is the youngest.                                  41. (c) The question asked us to arrange in ascending order but options give as per the descending order. From the given condition we can conclude that        Siya>Priya>Tiya        Siya>Riya>Priya Riya>Diya Tiya>Diya        Combining these we get        Siya>Riya>Priya>Tiya>Diya The correct answer is option (c) Solutions (42 and 43) 42. (d) Since the moderator sits in the third position. The writer and the attorney sit on either side of the moderator and Gaj, the attorney sits in seat 4. From these information we will get following arrangement. Jaya

Hema

Kumar

Gaj

abit

Explorer Writer Moderator Attorney Pilot   Hence Kumar is the moderator 43. (a) From the above arrangement Jaya sits in seat 1 44. (a) Raj and Jai traded with each other, Raj is now working on Monday and Jai on Wednesday. Similarly Kajal and Farid work on Tuesday and Thursday respectively. Now, Kajal and Jai traded with each other. Hence Jai works on Tuesday.  Concept Cracker (CC) with XAT Solutions (45 to 49) There are 7 courses- conditions are as follows 1stcondition: The Kingfish is served some time after Rahu. 2ndcondition: Exactly one course should be served between the pomfret and the Indian shrimp. 3rdcondition: The lobster is served some time before the pomfret. 4thcondition: The Kingfish is served either fifth or sixth. 5thcondition: The Hilsa is served second. 45. (a) Option B is wrong as Lobster has to before pomfret, option C is wrong as there are two fishes in between Pomfret and Indian Shrimp and D is wrong because Kingfish should be in 5th and 6th .

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18  Koncepts of Logical Reasoning 

46. (c) From the given condition only possible is lobster serve in first. 47. (b) 48. (a) If Rahu is the 5th course served then pomfret and Indianshrimp on the 1st and 3rd place but lobster should come before pomfret 49. (d) Solutions (50 to 53) Given that total prize money is 14 lacs, Highest prize money 3.5 lacs For every contestant the difference with the next highest and the next lowest winner is the same, hence the amount that they won must be in AP. Let us assume that amount won by them is as follows3.5, 3.5-k, 3.5-2k, 3.5-3k, 3.5-4k, 3.5-5k, 3.5-6k 3.5 + (3.5-k) + (3.5-2k) + (3.5-3k) + (3.5-4k) + (3.5-5k) + (3.5-6k) = 24.5 –21k = 14 or k = 0.5 Hence the prize money must be, 3.5, 3, 2.5, 2.0, 1.5, 1.0, 0.5 Now from given conditions B > A, Since difference between B and A is minimum hence that difference must be 0.5 lacs, D and F is not consecutive. There must be one person between E and G. 50. (c) As B win more money than A so option (A) is not correct. The difference of prize money between D and F was not the least so , Option (B) is ruled out. Option D is also ruled out as all the remaining 4 people will win less than 0.5 lakh. All when add not give 14 lakh. 51. (a) Option B ,C and D are false because as A get 1.5 lakh then B will get 2 lakh which is not possible because E has 2 lakh. 52. (d) Difference of prize money between A and C be 50000/- as B >A as A in between C and B so A and E cannot won prize which differ by the minimal amount. 53. (b) In this case B and C together won 3.5 lakhs Solutions (54 to 58) Assume the notation that x K Hence E> W or Elle is older than Waheeda. 74. (c) We have to nd the value of k, and that can be nd out by both option (A) and option (B) Solutions (75 to 76) According to given condition,we conclude: i) A-G=8 ii)D+R=37 iii)J=D+8 iv)A=D+5 iv) A+G=40 Using i) and iv) we get, A=(40+8)/2=24 and G=4024=16 So, D=A-5=24-5=19, R=37-19=18 , J=19+8=27 75. (a) Ramesh shot 18 baskets and Dhanraj shot 19 baskets. 76. (a) Dhanraj and Jugraj=(19+27)=46. Solutions (77 to 79) A and G must seated Extreme left. So either place 6 or 7. B must seat in the middle means 4th position. C and D not seated together and far apart is possible. So, If any 1 of C and D take 5th position and another take 1st position then only maintain maximum distance. Rest 2 places (2,3) lled by E or F. C/D

77. (c) From above sitting arrangement, either end of row are occupied by C/D or A/G.

E/F

E/F

B

C/D

A/G

A/G

2

3

4

5

6

7

Concept Eliminator (CE) with CAT Guru Gajen From condition 1: Raj/Gopal Gourav Gopal/Raj

From condition 2: Jayanta& Chandra

Chandrakant Room Mate

From condition 3: Chandan (alone)

Saurabh

Hemant

From condition 4 along with condition 1 and 2: Raj/Gopal Gourav

Jayanta & Chandra

Chandrakant & Room Mate

Gopal/ Raj

From condition 5 along with other conditions: Raj/Gopal

Krishna/Kunal

Gourav/ Chandan Jayanta& Chandra

Gopal/ Raj

Kunal/ Chandan/ Krishna Gourav

Dripto Saurabh

Chandrakant& Hemant

80. (c) From the above figure we can say that the correct option is (C) 81. (c) The correct option is (C). 82. (b) From the figure, Hemant is the room-mate of Chandrakant. Hence, the correct option is (B). 83. (a) Raj stays in the first block. Hence, the correct option is (A). 84. (d) An empty room or empty rooms may be found in 4th or 6th block or both. Hence, the correct option is (D). 85. (b) According to the new arrangement, Dripto or Sourabh are the occupants of Chandan’s old room. Hence, the correct option is (B). 86. (a) To make S to be true we have to look for the implications.

(iii) or (ii) ⇒ (ii)



(iv) ⇒ (ii)



(iii) ⇒ (ii)



~ (ii) ⇒ (iv) and ~ (iii).

87. (b) To make S false, we should look for negation.

The negation is,



(iv) or (iii) and ~ (ii).

From statement (i) (Choudhry and Garg) is a pair and (Malhotra and Oswal is a pair) From statement (ii) Chitranshi

Oswal

Malhotra is either at 3rd or 5th (Right most place) From statement (iii) Garg Chitranshi Choudhry From statement (iv) (Choudhry and Oswal are together) 88. (c) If we discard statement (i), or (ii) we can not get a unique sequence,

If we discard statement (iii) then – ChiGarg Choudhry Oswal transhi



Malhotra

Hence statement (iii) can be discarded. If we discard statement (iv) then – Chitranshi Garg Choudhry Oswal Malhotra

89. (c) Malhotra is on the extreme right and Chitranshi is on the extreme left. 90. (c) Choudhry is in the middle and Oswal is on its right.

Consider Swimming result: U>V V>Y X>W Y>X U>Y Z>U If we combine then we will get Z >U>V>Y>X>W Consider Running result: U>Z Z>V U>W X>Y Y>U W>Z If we combine X> Y> U>W>Z>V 91. (d) Students who won to Xavior in swimming are Zeeshan, Umesh, Victor, and Yogita Students lost to Yogita in running are Umesh, Wasim, Zeeshan, and Victor Students who satisfy both the conditions are Zeeshan, Umesh, and Victor. 92. (d) 93. (c) Let topper of any competition get 5 points, then 4 points, and so on the last ranker would get 0 points. Swimming :- Z =5, U = 4, V = 3, Y = 2, X = 1, W=0 Running :- X = 5, Y = 4, U= 3, W = 2, Z = 1, V = 0, Score of U = 4x0.6 + 3x0.4 = 2.4 + 1.2 = 3.6 Score of V = 3x0.6 + 0x0.4 = 1.8 + 0 = 1.8 Score of W = 0x0.6 + 2x0.4 = 0.8 Score of X = 1x0.6 + 5x0.4 = 0.6 + 2 = 2.6 Score of Y = 2x0.6 + 4x0.4 = 1.2 + 1.6 = 2.8 Score of Z = 5x0.6 + 1x0.4 = 3 + 0.4 = 3.4 Hence their ranking and score is U (=3.6) > Z (=3.4) > (Y=2.8) > (X =2.6) > (V = 1.8) > (W = 0.8) Hence Yogita is the 3rd best. 94. (a) From answer of last question we have to see that Zeeshan’s score is more than Umesh’s. let the minimum weightage on swimming is k (here k in fraction) then that on running is (1-k). Score of Umesh 4k + 3(1-k) = k+3 Score of Zeeshan 5k + 1(1-k) = 4k +1 From the given condition 4k +1 > k +3 or 3k > 2 or k > 2/3 = 0.667

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22  Koncepts of Logical Reasoning 

Linear Arrangement 23 M:- Room no 1 or 2 N:O:- Room no 3, 4, 5 or 6 P:- Room no 3 Q: Room no 4, 5 or 6 95. (d) Given that room numbers 5 and 6 are unoccupied hence condition must be:Only option left Q is room number 4 and O stays just before or after Q’s room but room number 5 is un occupied hence O must stay at room number 3. M:- Room no 1 or 2 N:O:- Room no 3, P:- Room no 3 Q: Room no 4, O and P are room mates 96. (b) P stays in room number 3 and if N also stays in room number 3 then, O cannot stay in room number 3, and only option with him is room number 5. 97. (b) Since P stays in room number 3 hence N and O can stay in room number 4, 5 or 6 hence both statement (II) and (III) are correct. 98. (d) Room no 2 :- M Room no 3 :- P Since O and Q will go together hence N can not stay at room number 5

99. (c) From 2, the driver came in rst or second. From 3, he came at midnight. So there were either two people who came after midnight or there was one person who came before midnight and one who came after midnight. From 4, neither Aditya nor Vijay arrived after midnight. So there is only one person who came after midnight. He has to be Puneet. It follows that one of the three arrived before midnight. So we have the following. Before Midnight Aditya/Vijay At Midnight Driver Vijay/Aditya After Midnight Puneet From 6, Puneet was not the murderer. From 1, the murderer came in second or third. As Puneet came third, and was not the murderer, the person who came in second was the murderer. So, Before Midnight Vijay At Midnight DriverAditya - murderer After Midnight Punee Vijay arrived at the house earliest. Hence, Option (c) 100. (b) Aditya is the murderer. 101. (c) Aditya is the driver. 102. (a) Puneet arrived at the house last. 103. (b) Aditya came at midnight.

From Rahul’s 1st statement we can concludeIn

Present

Out

Susmit

----

-----

Rahul

Susmit

From Zeeshan’s Statement In

Present

Out

Zeeshan

Susmit, Rahul, Moumita

-----

If we combine Zeeshan’s statement with Rahul’s then we can conclude thatIn

Present

Out

Susmit

----

-----

Rahul

Susmit

-------

Moumita

Rahul, Susmit

-------

Zeeshan

Susmit, Rahul, Moumita

-----

Susmit, zeeshan

Rahul, Moumita

EBD_7743

24 Koncepts of Logical Reasoning Now from Danish’s statementIn

Present

Out

Susmit

----

-----

Rahul

Susmit

-------

Moumita

Rahul, Susmit

-------

Zeeshan

Susmit, Rahul, Moumita

-----

Susmit, zeeshan

Rahul, Moumita

Ricky

Susmit, zeeshan

Danish

Ricky, Susmit, zeeshan Ricky, Zeeshan

Danish, Susmit

From Susmit’s and along with Moumita,s statement:In

Present

Out

Susmit

----

-----

Rahul

Susmit

-------

Moumita

Rahul, Susmit

-------

Zeeshan

Susmit, Rahul, Moumita

-----

Susmit, zeeshan

Rahul, Moumita

Ricky

Susmit, zeeshan

Danish

Ricky, Susmit, zeeshan Ricky, Zeeshan

Danish, Susmit

Vivek (Here it is not clear whether Vivek came 1st or Zeeshan left 1st)

Ricky

Zeeshan

Susmit

Ricky , Vivek Susmit, Ricky

Moumita

Vivek

Susmit, Ricky Susmit

Moumita

Ricky

Zeeshan

Danish

Vivek

Rahul

Total

Susmit

x

yes

yes

yes

yes

yes

Yes

6

Moumita

Yes

x

Yes

yes

xxxxx

xxxxxx

Yes

4

Ricky

Yes

Yes

x

Yes

Yes

Yes

xxxxx

5

Zeeshan

Yes

Yes

Yes

x

Yes

Uncertain

yes

5 or 6

Danish

Yes

xxxxxx

Yes

Yes

x

xxxxx

xxxx

3

Vivek

Yes

xxxxxx

Yes

Uncertain

xxxxx

X

xxxxx

2 or 3

Rahul

Yes

Yes

xxxxx

Yes

xxxxx

xxxxx

x

3

From this table we can say that if it is Danish and Vivek then it must be Rahul as well. From the nal arrangement Zeeshan was present at both the situations. From the table option (c) is correct. Total number of pair is 7C2 = 21, Since only one pair about whom we could not determine zeeshan and Vivek Required Probabilty is 1/21 108. (c) Maximum number of friends present is 4 109. (c) From the table we have seen that Moumita met with Susmit, Ricky, Zeeshan, and Rahul i.e 4 friends. 110. (a) From the table Moumita came at 3rd position. 104. (d) 105. (b) 106. (c) 107. (a)

Matrix Arrangement 25

2

Matrix Arrangement Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt Exams

Very Important

Other Aptitude Test

Very Important

Matrix arrangement problems are the most common problem types in all aptitude test entrance exams. Unlike linear arrangement problems these are dealt with more than one variable / property of the objects. In linear arrangement, where the objects had only one property – their positioning, objects in matrix arrangement have multiple properties. In general the information that is provided in these question is of two types: 1. Direct information: Information relating an object with its property 2. Indirect information: Information that relates two or more properties of an object. To handle this type of questions 1st we have to note down all the given information, then nd out the parameters about which more information is given and then make a table and co-relate the given parameters to nd out exact match. For example if it is given that four students A, B, C, and D belongs to 4 different cities namely Delhi, Kolkata, Mumbai and Chennai. If it is given that A doesn’t belongs to Chennai and Delhi, B belongs to Kolkata then we can represent this information asA B C D

Delhi X X

Kolkata X √ X X

Mumbai Chennai √ X X X X X

Lets proceed towards questions and see how to deal these types of questions.

EBD_7743

26 Koncepts of Logical Reasoning

1

Directions for Question Nos. 1 to 3

In a business school, four students Dinakaran, Sumit, Tarun and Amul exhibit a very strange mix of hobbies and subject interests. One of them studies Commerce and plays Golf and Lawn Tennis. Dinakaran and Sumit study Psychology. Dinakaran plays Billiards. Both the Psychology students play Chess. Tarun is a student of Physics. The Physics student plays Chess and Badminton. All the friends play two games each and study one subject each. One of the students also does Weight Training. 1.

Who does not play Chess? (a) Dinakaran (b) Sumit (c) Tarun (d) Amul

2.

Who studies Psychology and plays Billiards? (a) Dinakaran (b) Sumit (c) Tarun (d) Amul

3.

How many games are played and subjects studied by all the four students? (a) 2, 1 (b) 3, 2 (c) 6, 3 (d) 5, 4

4.

When is ‘Laugh a While’ scheduled? (a) 9.00 pm to 9.30 pm (b) 9.30 pm to 10.00 pm (c) 10.00 pm to 10.30 pm (d) 10.30 pm to 11.00 pm

5.

Based on the information above which of the following is not implied? (a) ‘Midnight Murders’ is a news based program (b) ‘Laugh a While’ has adult content (c) ‘Detective Dooms’ has adult content (d) ‘HIV and India’ has social message

6.

When is ‘HIV and India’ scheduled? (a) 11.00 pm to 11.30 pm (b) 9.30 pm to 10.00 pm (c) 10.00 pm to 10.30 pm (d) 10.30 pm to 11.00 pm Directions for Question Nos. 7 and 8

Vipin would be visiting ve places: P, Q, R, S and T. The distance between any two of these places in kilometres is as follows:

Directions for Question Nos. 4 to 6 A television channel has scheduled ve half-hour shows between 9.30 pm and midnight. Out of the three family drama shows ‘Main Sati Hoon’ has social message for audience. The family dramas bring in maximum revenues for the channel and they are scheduled one after another. The show ‘Detective Doom’ is a suspense thriller, a family drama and also has social message. Two of the family dramas have social message and one not having social message has adult content. There are two shows which have adult content and they have been scheduled at 10.30 pm and 11.30 pm. The show ‘HIV and India’ is a news based program and also has social message. ‘Midnight Murders’ neither has social message nor is a family drama and is scheduled at 11.30 pm. The two news based programs but are not family dramas have been given two adjacent slots. The show ‘Main Sati Hoon’ has been scheduled for 10.00 to 10.30 pm slot and has a family drama preceding it. ‘Laugh a while’ is the fth program.

P Q R S T

P

Q

R

S

T

0

2

3

5

6

0

2

1

4

0

2

3

0

2 0

7.

Vipin is currently in S and wants to nish his tour at T. If he doesn’t want to visit a city more than once, what is the minimum distance he would have to cover? (a) 7 km (b) 8 km (c) 9 km (d) 10 km

8.

If Vipin starts his tour at Q, and comes back to Q after visiting all other places only once; what is the minimum distance he has to cover? (a) 10 km (b) 11 km (c) 12 km (d) None of these

Matrix Arrangement 27 Directions for Question Nos. 9 to 11 M1, M2, M3 and M4 bought a computer and decided to share the time of each day according to the money they contributed. The price of the computer in rupees as well as the hours of the day, the computer is used (greater than 10), are integers. If M1 contributed Rs 5,000, M2 contributed 25 per cent of the total money, M3 uses 20 per cent of the time of the computer and M4 uses it for 2 hours, then: 9.

M1 uses the computer for: (a) 6 hours 30 minutes (b) 6 hours 45 minutes (c) 6 hours (d) 6 hours 15 minutes

10. M4 contributed: (a) Rs 1,800 (c) Rs 1,500

(b) (d)

Rs 2,000 Rs 1,600

11. If another person M5 wanted to use the remaining free hours at the computer each day by paying them a rent of Rs 100 for certain number of days and after which he could have it free because he would have contributed by then the money had he contributed the same money at the purchase of it for his share of time of computer, then he would have paid the rent for: (a) 48 days (b) 54 days (c) 42 days (d) 45 days Directions for Question Nos. 12 to 14 In a family picnic, Shikhar, Dolly, Snehal, Akash, Swarn and Amar were playing a game. Shikhar’s father, mother and uncle were in the group. Group has only two females. Dolly, the mother of Shikhar got more points than Shikhar’s father. Akash got more points in the game than Swarn but less than that of Amar. Niece of Swarn got the lowest points. Father of Shikhar got more points than Amar but could not win the game. 12. Who was the lady in the group besides Dolly? (a) Snehal (b) Akash (c) Swarn (d) None of these 13. Who stood second in the game? (a) Shikhar (b) Dolly (c) Snehal (d) None of these

14. Who won the game? (a) Shikhar (c) Amar

(b) Dolly (d) Akash

Directions for Question Nos. 15 to 18 In a post-induction training placement of an Export Oriented Unit, ve new recruits Aditya, Aryan, Harish, Puru and Sheetal are to be placed in departments of Corporate Planning, Information Technology, Finance, HR and Exports. Aditya is recommended by Corporate Planning and Exports Departments. Aryan gets recommendations from Information Technology and HR Departments. Harish is recommended by Corporate Planning and Exports Departments. Puru makes it to Information Technology, Finance, HR as well as Exports. Only Sheetal could not get recommended by more than one department and it is also known that every department recommended at least two recruits from this group. As per the placement policy of the company, departments have certain educational preferences while recommending available candidates. The department of: Ø Exports prefers MIBs as most suitable Ø Information Technology stresses more on a qualication of MCA Ø Finance prefers an MFC Ø HR gives top preference to applicants having a PMIR Ø Corporate Planning prefers an MBA 15. Which of the following is an MBA? (a) Aditya alone (b) both Aditya and Aryan (c) both Aditya and Harish (d) None of these 16. For which department is Sheetal shortlisted? (a) Information Technology (b) Finance (c) HR (d) Exports 17. The qualication which is most common among the new recruits: (a) MIB (b) MFC (c) MCA (d) MBA 18. The recruit who is an MCA as well as MFC? (a) Aditya (b) both Aditya and Aryan (c) both Aditya and Harish (d) Puru

EBD_7743

28 Koncepts of Logical Reasoning Directions for Question Nos. 19 to 23 Five experts on Nano-technology involved in an international Research Project hold a Quarterly Review Meeting in Singapore. There are certain limitations on their language skills. Expert R1 knows only Japanese and Hindi; R2 is good at Japanese and English; R3 is good at English and Hindi; R4 knows French and Japanese quite well; and R5, an Indian, knows Hindi, English and French. 19. Besides R5, which of the following can converse with R4 without an interpreter? (a) Only R1 (b) only R2 (c) only R3 (d) both R1 and R2 20. Which of the following cannot converse without an interpreter? (a) R2 and R5 (b) R1 and R2 (c) R1 and R3 (d) R3 and R4 21. Choose the language that is least commonly used at the meeting. (a) English (b) French (c) Japanese (d) none of the above 22. Which of the following can act as an interpreter when R3 and R4 wish to discuss? (a) Only R1 (b) only R2 (c) only R5 (d) none of the above 23. Suppose a sixth Expert R6 joins into chair the session so that maximum number of the earlier ve is able to understand him, he should be uent in: (a) English and French (b) Japanese and Hindi (c) English and Hindi (d) none of the above

at the regional level, lies between the football player and the chess player. Manu is a regional level player and stays in the 3rd at while Tarun is a quality inspector and stays in the 5th at. The football player is a design engineer and stays in the 2nd Flat. Amit is a power engineer and plays Chess while Ambrish is the mechanical engineer and plays Cricket at the national level. 24. Who stays in the 4th at? (a) Ambrish (b) Amit (c) Rohit (d) Manu 25. What does Tarun play? (a) Chess (b) Football (c) Cricket (d) Tennis 26. Who plays football? (a) Ambrish (c) Rohit

(b) Amit (d) Manu

27. Agewise, who among the following lies between Manu and Tarun? (a) Quality inspector (b) Mechanical engineer (c) Power engineer (d) Design engineer 28. Who stay in the same at? (a) Ambrish and Amit (b) Manu and Tarun (c) Amit and Manu (d) Rohit and Tarun 29. The Chess player is a: (a) Power engineer (b) Mechanical engineer (c) Design engineer (d) Quality inspector

Directions for Question Nos. 24 to 29

Directions for Question Nos. 30 to 32

In a Public Sector Undertaking Township, there are ve executives – Ambrish, Amit, Rohit, Manu and Tarun. Two of them play Cricket while the other three play different games viz. Football, Tennis and Chess. One Cricket player and a Chess player stay in the third at, whereas the other three stay in different ats, i.e. 2nd, 4th and 5th. Two of these ve players are mechanical engineers while the other three are quality inspector, design engineer and power engineer respectively. The chess player is the oldest in age while one of the cricket players, who plays at the national level, is the youngest in age. The age of the other cricket player, who plays

P, Q, R, S, T and X are six persons who decided to start a partnership business in computer hardware. Q, R and T are ladies and the rest are men. P, Q, R and X are well versed in hardware but the others do not know much about hardware. Q, S and X know marketing, but the others do not know anything about marketing. 30. The male partner who knows marketing but does not know computer hardware: (a) S (b) P (c) X (d) Q

Matrix Arrangement 29 31. The female partner who is well versed in computer hardware but does not know marketing: (a) X (b) Q (c) T (d) R

2

33. Read the following about the grid given below and answer. [SNAP 2008] The cells in this grid contain the digits 1 to 9 in random order.

1

Ø

Column A contains no odd digits.

3

Ø

Cell C3 minus Cell C2 equals 4.

Ø

The sum of three digits in Row 1 is 17

Ø

Number 7 is in Column B; its left hand neighbor is not 4.

Ø

The digits of Column C add up to 14.

Ø

2 is not in the same horizontal row as 8; and 9 is not immediately below 3.

Ø

A

B

C

2

Which cell holds the number 9? (a)

B1

(b)

B3

(c)

C2

(d) C1

Directions for Question Nos. 34 to 36 Use the information given below to answer. I. there is a group of 5 persons A, B, C, D and E II. In the group there is one badminton player, one chess player and one tennis player. III. A and D are unmarried ladies and they do not play any games IV. No lady is a chess player or a badminton player V. There is a married couple in the group of which E is the husband VI. B is the brother of C and is neither achess player nor a tennis player [SNAP 2008] 34. Which of the following groups has only ladies? (a) ABC (b) BCD (c) CDE (d) none of the above 35. Who is the tennis player? (a) B (b) C (c) D (d) E

32. The female partner who neither knows hardware nor marketing: (a) H (b) Q (c) T (d) X

36. Who is the wife of E? (a) A (b) (c)

D

B

(d) none of the above

Directions for Question Nos. 37 to 42 Study the 10 statements given below and answer the questions. Ø Six businessmen from six different nations are staying in different rooms in succession in the same row in a hotel. Ø Each of them owns a different number of cars and has donated to different number of institutions during the last year. Ø The businessman in Room no. 102 owns twice as many as the number of cars owned by the businessmen who has donated to 8 institutions in the last year. Ø The businessman from Uruguay and the businessman in Room no. 106 together own 40 cars in total. Ø The businessman from Argentina owns 8 cars less than the businessman from England but donated to 10 more instillations in the last year. Ø Four times the number of cars owned by the businessman in Room no. 104 is lesser than the number of institutions to which he has donated in the last year. Ø The businessman in Room No. 103 owns 12 cars and donated to 8 institutions in the last year. Ø The businessman who owns 16 cars donated to 24 institutions in the last year. Ø The businessman in Room no. 105 owns 8 cars and donated to 2 institutions less than those donated by the businessman from Canada in the last year. Ø The Brazilian businessman is staying two rooms ahead of the English businessman who is staying two rooms ahead of the Canadian businessman. [IIFT 2006]

37. In which room is Brazilians businessman staying? (a) Room no. 102 (b) Room no. 103 (c) Room no. 104 (d) Room no. 105 38. What is the number of institutions to which the Argentinean businessman donated in the last year? (a) 8 (b) 3 (c) 18 (d) 24 39. The businessman of which country is staying in room no. 106? (a) Argentina (b) Canada (c) Uruguay (d) Germany 40. The businessman of which country has donated to 24 institutions in the last year? (a) Argentina (b) Uruguay (c) Canada (d) Germany 41. The businessman of which country owns the highest number of cars? (a) Argentina (b) Uruguay (c) Germany (d) Brazil 42. How many cars are owned by the English businessman? (a) 8 (b) 12 (c) 4 (d) 20 Directions for Question Nos. 43 to 45 (i) Five girls – Seema, Reema, Neeta, Mona and Veena have total five tickets of movie theaters – Priya, Chanakya, M2K, PVR Saket, Satyam where movies - Gangster, Khiladi, Hero, Salaam Namaste and Iqbal are currently playing. Each girl has one movie ticket of one of the five theatres. (ii) Movie Gangster is running in Priya theatre whose ticket is not with Veena and Seema. (iii) Mona has ticket of Iqbal movie. (iv) Neeta has ticket for the M2K theatre, Veena has the ticket of Satyam theatre where Khiladi is not running. (v) In PVR Saket theatre Salaam Namaste is running.

[IIFT 2007]

43. Which is the correct combination of the Threate – Girl – Movie? (a) M2K – Neeta – Hero (b) Priya – Mona – Gangster (c) Satyam – Veena – Iqbal (d) PVR Saket – Seema – Salaam Namaste

44. Which movie is running in Chanakya? (a) Gangster (b) Iqbal (c) Hero (d) Data inadequate 45. Who is having the ticket of the movie Hero? (a) Reema (b) Veena (c) Seema (d) Mona 46. Three children won prizes in the ‗Tech India Quiz‘contest. They are from three schools: Lancer, Columbus and Leelavati, which are located in different states. One of the children is named Binod. Lancer school‘s contestant did not come first. Leelavati school‘s contestant‘s name is Rahman. Columbus school is not located in Andhra Pradesh. The contestant from Maharashtra got third place and is not from Leelavati School. The contestant from Karnataka did not secure first position. Columbus school‘s contestant‘s name is not Badal. [IIFT 2008]



Which of the following statements is TRUE? (a) 1st prize: Rahman (Leelavati), 2nd prize: Binod (Columbus), 3rd prize: Badal (Lancer) (b) 1st prize: Binod (Columbus), 2nd prize: Rahman (Leelavati), 3rd prize: Badal (Lancer) (Q. No. 1) (c) 1st prize: Rahman (Leelavati), 2nd prize: Badal (Lancer), 3rd prize: Binod (Columbus) (d) 1st prize: Binod (Columbus), 2nd prize: Badal (Lancer), 3rd prize: Rahman (Leelavati) Directions for Question Nos. 47 and 48

Director of an institute wants to distribute teaching assignments of HRM, Psychology, Development Studies, Trade policy and Finance to five of six newly appointed faculty members. Prof. Fotedar does not want any assignment if Prof. Das gets one of the five. Prof. Chaudhury desires either HRM or Finance or no assignment. Prof. Banik opines that if Prof. Das gets either Psychology or Trade Policy then she must get the other one. Prof. Eswar insists on an assignment if Prof. Acharya gets one.

[IIFT 2008]

47. Which of the following is valid faculty – assignment combination if all the faculty preferences are considered? (a) Prof. Acharya - HRM, Prof. Banik – Psychology, Prof. Chaudhury – Development studies, Prof. Das – Trade Policy, Prof. Eswar – Finance

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30  Koncepts of Logical Reasoning 

Matrix Arrangement 31 (b) Prof. Chaudhury - HRM, Prof. Das – Psychology, Prof. Acharya – Development studies, Prof. Banik – Trade Policy, Prof. Eswar – Finance (c) Prof. Acharya - HRM, Prof. Banik – Psychology, Prof. Eswar – Development studies, Prof. Das – Trade Policy, Prof. Fotedar – Finance (d) Prof. Banik - HRM, Prof. Fotedar – Psychology, Prof. Eswar – Development studies, Prof. Chaudhuri – Trade Policy, Prof. Acharya – Finance 48. If Prof. Acharya gets HRM and Prof. Chaudhury gets Finance, then which of the following is not a correct faculty-assignment combination assuming all faculty preferences are considered? (a) Prof. Das – Development Studies, Prof. Banik – Trade Policy (b) Prof. Fotedar – Development Studies, Prof. Banik – Trade Policy (c) Prof. Banik – Development Studies, Prof. Eswar – Trade Policy (d) Prof. Banik – Development Studies, Prof. Das – Trade Policy Directions for Question Nos. 49 to 51 Five women decided to go for shopping to South Extension, New Delhi. They arrived at the designated meeting place in the following order: 1. Aradhana, 2. Chandrima, 3. Deepika, 4. Heena and 5. Sumitra. Each of them spent at least Rs. 1000. The woman who spent Rs. 2234 arrived before the woman who spent Rs. 1193. One of them spent Rs. 1340 and she was not Deepika. One woman spent Rs. 1378 more than Chandrima. One of them spent Rs. 2517 and she was not Aradhana. Heena spent more than Deepika. Sumitra spent the largest amount and Chandrima the smallest. [IIFT 2008] 49. What was the amount spent by Heena? (a) Rs. 1193 (b) Rs. 1340 (c) Rs. 2234 (d) Rs. 2517 50. Which of the following amount is spent by one of the women? (a) Rs. 1139 (b) Rs. 1378 (c) Rs. 2571 (d) Rs. 2518 51. The lady who spent Rs. 1193 is: (a) Aradhana (b) Chandrima (c) Deepika (d) Heena Directions for Question Nos. 52 to 55 Mr. Mansingh has ve sons – Arun, Mahi, Rohit, Nilesh and Saurav, and three daughters – Tamanna, Kuntala and Janaki. Three sons of Mr. Mansingh

were born rst followed by two daughters. Saurav is the eldest child and Janki is the youngest. Three of the children are studying at Trinity School and three are studying at St. Stefan. Tamanna and Rohit study at St. Stefan School. Kuntala, the eldest daughter, plays chess. Mansoroverschool offers cricket only, while Trinity school offers chess. Beside, these schools offer no other games. The children who are at Mansorover School have been born in succession. Mahi and Nilesh are cricketers while Arun plays football. Rohit who was born just before Janki, plays hockey. [IIFT 2008] 52. Arun is the _________ child of Mr. Mansingh. (a) 2nd (b) 3rd (c) 6th (d) 5th 53. Saurav is a student of which school? (a) Trinity (b) St. Stefan (c) Mansorover (d) Cannot be determined 54. What game does Tamanna play? (a) Cricket (b) Hockey (c) Football (d) Cannot be determined 55. Which of the following pairs was not born in succession (ignore the order)? (a) Mahi and Nilesh (b) Kuntala and Arun (c) Rohit and Janki (d) Arun and Rohit Directions for Question Nos. 56 to 58 Four houses Blue, Green, Red and Yellow are located in a row in the given order. Each of the houses is occupied by a person earning a xed amount of a salary. The four persons are Paul, Krishna, Laxman, and Som. Read the following instruction carefully: I. Paul lives between Som and Krishna II. Laxman does not stay in Blue house III. The person living in Red house earns more than that of person living in Blue IV. Salary of Som is more than that of Paul but lesser than that of Krishna V. One of the person earns Rs. 80, 000 VI. The person earning Rs. 110,000 is not Laxman VII. The salary difference between Laxman and som is Rs. 30,000 VIII.The House in which Krishna lives is located between houses with persons earning salaries of Rs. 30,000 and Rs. 50,000 IX. Krishna does not live in Yellow house, and the person living in yellow house is not earning lowest salary among the four persons. [IIFT 2009]

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32 Koncepts of Logical Reasoning 56. Who lives in Red house? (a) Paul (b) Krishna (c) Laxman (d) Som 57. Which house is occupied by person earning highest salary? (a) Blue (b) Green (c) Red (d) Yellow 58. What is the salary earned by person living in Green house? (a) Rs. 30,000 (b) Rs. 50,000 (c) Rs. 80,000 (d) Rs.110, 000 Directions for Question Nos. 59 to 63 i.

Six friends Rahul, Kabeer, Anup, Raghu, Amit and Alok were engineering graduates. All six of them were placed in six different companies and were posted in six different locations, namely TiscoJamshedpur, Telco-Pune, Wipro-Bangalore, HCLNoida, Mecon-Ranchi and Usha Martin-Kolkata. Each of them has their personal e-mail id with different email providers i.e., Gmail, Indiatimes, Rediffmail, Yahoo, Hotmail, Sancharnet, though not necessarily in the same order. ii. The one having e-mail account with Sancharmet works in Noida and the one having an e-mail account with India times works for Tisco. iii. Amit does not stay in Bangalore and does not work for Mecon, the one who works for Mecon has an e—mail id with Gmail. iv. Rahul has an e-mail id with Rediff mail and works at Pune. v. Alok does not work for Mecon and the one who works for Wipro does not have an e-mail account with Yahoo. vi. Kabeer is posted in Kolkata, and does not have an account with Hotmail. vii. Neither Alok nor Raghu work in Noida. viii. The one who is posted in Ranchi has an e-mail id which is not an account of Rediff mail or Hotmail. ix. Anup is posted in Jamshedpur. [IIFT 2010]

61. Which of the following is true? (a) Amit is posted at Ranchi (b) Raghu is posted at Jamshedpur (c) Kabeer has an e-mail id with Yahoo (d) Rahul has an e-mail id with India times 62. Which of the following sequences of location represent Alok, Kabeer, Anup, Rahul, Raghu and Amit in the same order? (a) Bangalore, Noida, Pune, Jamshedpur, Ranchi, Kolkata (b) Bangalore, Kolkata, Jamshedpur, Pune, Noida, Ranchi, (c) Kolkata, Bangalore, Jamshedpur, Pune, Noida, Ranchi (d) none of these 63. People who have e-mail account with India times, sancharnet and Yahoo work for which companies, in the same sequence as the e-mail accounts mentioned? (a) Usha Martin, HCL, Wipro (b) Tisco, Wipro, Usha Martin (c) HCL, Tisco, Wipro (d) Tisco, HCL, Usha Martin Directions for Question Nos. 64 and 65 Mr.Malhotra’s family is a traditional joint family from Jalandhar having six persons from three generations. Each member of the family has different food preference and they support different sports/ games. Only two couples are there in the family. Rakesh likes continental food and his wife neither likes dry fruits nor supports gymnastics. The person who likes egg supports Rugby and his wife likes traditional food. Mona is mother-in-law of sonalika and she supports Athletics. Varun is grandfather of Tarun and Tarun, who likes Punjabi food, supports Basketbll. Nuri is granddaughter of Mona an she supports Badminton. Nuri’s mother supports horse riding. [IIFT 2011]

59. The man who works in Wipro has a e-mail account with? (a) Sancharnet (b) Yahoo (c) Rediff mail (d) none of these

64. Identify the correct pair of two couples from the following: (a) Mona-Varun and Rakesh-Sonalika (b) Varun-Mona and Rakesh-Nuri (c) Rakesh-Sonalika (d) cannot be determined

60. Which of the following e-mail place of postingperson combination is correct? (a) Kabeer-Kolkata-Rediffmail (b) Alok-bangalore-Indiatimes (c) Amit-Noida-Yahoo (d) Raghu-Ranchi-Gmail

65. Who likes Punjabi food, and what sport/game does he/she support? (a) Nuri and Badminton (b) Sonalika and horse riding (c) Tarun and Basketall (d) none of the above

3

Directions for Question Nos. 66 to 70

Seven instructors - J, K, L, M, N, P and Q – teach management courses at a premier institute in east India. Each instructor teaches during exactly one term: the rst term, the second term, or the third term. The following conditions apply: [XAT 2006] K teaches during the third term. L and M teach during the same term. Q teaches during either the rst term or the second term. Exactly twice as many instructors teach during the third term as teach during the rst term. N and Q teach during different terms. J and P teach during different terms. 66. Which one of the following could be an accurate matching of instructors to terms? (a) M: the rst term; P: the second term; Q: the rst term (b) J: the third term; L: the third term; P: the third term (c) L: the rst term; N: the second term; P: the third term (d) J: the rst term; M: the third term; N: the second term 67. Which one of the following cannot be true? (a) L teaches during the rst term (b) M teaches during the second term (c) M teaches during the third term (d) N teaches during the second term 68. If exactly one instructor teaches during the second term, which one of the following must be true? (a) J teaches during the third term (b) L teaches during the rst term (c) M teaches during the third term (d) P teaches during the second term 69. Each of the following contains a list of instructors who can all teach during the same term EXCEPT: (a) J, K, M (b) J, L, M (c) K, L, P (d) K, P, Q

Matrix Arrangement 33

70. If more instructors teach during the second term than teach during the rst term, then which one of the following instructors must teach during the second term? (a) J (b) M (c) N (d) P Directions for Question Nos. 71 to 75 Analyze the following statements and give an appropriate answer. K.C. Das is preparing special puja sweet packages. Different sweet packages are numbered 1 through 5 from left to right, and K.C. Das is lling them with different sweets. Each package will contain at least one, but not more than two of the following types of sweets: Gulabjamun, Kajubar, Petha, Rasgulla, Sohanhalwa, and Cham cham. Each type of sweet will be placed in at least one sweet package. These sweets will be packed either in a bucket, or a carton or a tin. K.C. Das lls the packages according to the following conditions: At least two packages must contain Rasgulla. Exactly two packages must contain Kajubar, and these packages cannot be adjacent to each -other. Both packages that contain Kajubar must be to the left of any packages that contain Gulabjamun. Package 2, 3, and 4 cannot contain Sohanhalwa: Any package that contains Rasgulla must be packed in a carton. Any package that contains, Kajubar must be packed in a bucket. Package 2 is packed in a carton. [XAT 2006] 71. Which one of the following cannot be true? (a) Package 1 is packed in a tin. (b) Package 2 contains Cham cham. (c) Package 3 is packed in a tin. (d) Package 4 contains Kajubar. 72. If a package containing sweets and packed in a tin is not adjacent to a package packed in a bucket, then which one of the following MUST be true? (a) Package 1 contains Petha (b) Package 4 contains Kajubar. (c) Package 4 contains Rasgulla. (d) Package 5 contains Gulabjamun.

EBD_7743

34 Koncepts of Logical Reasoning 73. If Rasgulla are contained in the maximum number of packages, which one of the following must be true? (a) Package 3 is packed in a bucket. (b) Package 4 is packed in a bucket. (c) A package containing Sohanhalwa is packed in a bucket. (d) A package containing Gulabjamun is packed in a carton. 74. If package 4 contains Petha and Cham cham, which one of the following pairs of sweets must be contained in the same package as each other? (a) Kajubar and Sohanhalwa. (b) Gulabjamun and Petha. (c) Rasgulla and Cham cham. (d) Gulabjamun and Sohanhalwa. 75. If package 3 is packed in a tin, which one of the following could be false? (a) Package 1 contains Sohanhalwa. (b) Package 2 contains Rasgulla. (c) Package 3 contains Cham cham. (d) Package 4 is packed in a bucket. Directions for Question Nos. 76 to 81 Seven lm buffs – Gangadhar, Indara, Lalatakshaya, Maheshwar, Rudra, Vyomkesha and Yogi- attend a showing of classic lms. Three lms are shown: one each directed by Guru Dutt, Stayajit Ray and Ritwik Ghatak. Each of the lm buffs sees exactly one of three lms. The lms are shown only once, one lm at a time. The following restrictions apply: Exactly twice as many of the lm buffs see the Satyajit Ray lm as see the Guru Dutt lm. Gangadhar and Rudra do not see the same lm as each other. Indra and Maheshwar do not see the same lm as each other. Vyomkesha and Yogi see the same lm as each other; Lalatakshaya sees the Satyajit Ray lm. Gangadhar sees either the Guru Dutt lm or the Ritwik Ghatak lm. [XAT 2007] 76. Which one of the following could be an accurate matching of lms buffs to lms? (a) Gangahdar: the Satyajit Ray lm; Indra; the Ritwik Ghatak lm; Maheshwar: the Satyajit Ray lm. (b) Gangadhar: the Ritwik Ghatak lm; Indra; the Guru Dutt lm; Vyomkesha: the Guru Dutt lm. (c) Indra: the Satyajit Ray lm; Rudra: the Ritwik Ghatak lm; Vyomkesha: the Guru Dutt lm.

(d) Maheshwar: the Ritwik Ghatak lm; Rudra: the Ritwik Ghatak lm; Vyomkesha: the Ritwik Ghatak lm. (E) Maheshwar: the Satyajit Ray lm; Rudra: the Satyajit Ray lm: Yogi: the Satyajit Ray lm. 77. Each of the following must be false except: (a) Rudra is the only lm buff to see the Guru Dutt lm. (b) Rudra is the only lm buff to see the Satyajit Ray lm. (c) Yogi is the only lm buff to see the Ritwik Ghatak lm. (d) Exactly two lm buffs see the Ritwik Ghatak lm (e) Exactly three lm buffs see the Satyajit Ray lm. 78. Which one of the following could be a complete and accurate list of the lm buffs who DO NOT see the Satyajit Ray lm? (a) Gangadhar, Maheshwar (b) Gangadhar, Rudra (c) Gangadhar, Indra, Rudra (d) Gangadhar, Maheshwar, Yogi (e) Gangadhar, Vyomkesh, Yogi 79. If exactly one lm buff sees the RitwikGhatak lm, then which one of the following must be true? (a) Vyomkesha sees the Satyajit Ray lm (b) Gangadhar sees the Guru Dutt lm (c) Maheshwar sees the Guru Dutt lm (d) Indra sees the Guru Dutt lm. (e) Rudra sees the Satyajit Ray lm 80. If Vyomkesha and Gangadhar see the same lm, then which one of the following could be true? (a) Gnagadhar sees the Guru Dutt lm. (b) Indra sees the Satyajit Ray lm (c) Rudra sees the RitwikGhatak lm (d) Vyomkesha sees the Satyajit Ray lm (e) Yogi sees the Guru Dutt lm 81. Each of the following could be complete and accurate list of the lm buffs who see the Guru Dutt lm EXCEPT: (a) Gangadhar, Indra (b) Gangadhar, Maheshwar (c) Indra, Rudra (d) Maheshwar, Rudra (e) Vyomkesha, Yogi

Directions for Question Nos. 82 to 87 A famous retail electronics showroom chain has six new mobile phone models – T, V, W, X, Y and Z – each equipped with at least one of the following three options: digital camera, music player, and office document viewer. No mobile has any other option. The following conditions apply: V features both a digital camera and an office document viewer. W has digital camera and music player W and Y have no options in common. X has more options as compared to W. V and Z have exactly one option in common. T has fewer options as compared to Z [XAT 2007] 82. For exactly how many of the six mobile phones is it possible to determine exactly which option each one has? (a) Two (b) Three (c) Four (d) Five (e) Six 83. Which one of the following must be false? (a) Exactly five mobile phones feature a music player. (b) Exactly five mobile phones feature a document viewer. (c) Exactly four mobile phones feature a music player. (d) Exactly four mobile phones feature a digital camera. (e) Exactly four mobile phones feature a document viewer. 84. If Z has no option in common with T but has at least one option in common with every other mobile phone, then which one of the following must be false? (a) T has digital camera (b) Z has document viewer (c) Exactly four of the six mobile phones have digital camera. (d) Exactly four of the six mobile phones have document viewer. (e) Exactly four of the six mobile phones have music player. 85. Suppose no two mobile phone models have exactly the same options as one another. In that case, each of the following could be true except:



(a)



(b)



(c)



(d)



(e)

  Matrix Arrangement 35  Exactly three of the six mobile phones have digital camera. Exactly four of the six mobile phones have digital camera. Exactly three of the six mobile phones have document viewer Exactly five of the six mobile phones have document viewer. Exactly four of the six mobile phones have music player.

86. If exactly four of the six mobile phones have music player, and exactly four of the six mobile phones have digital camera, then each of the following must be true EXCEPT: (a) T and V have no options in common. (b) T and Y have no options in common. (c) T and Z have exactly one option in common. (d) W and Z have exactly one option in common. (e) Y and Z have no option in common. 87. Suppose that the condition “X has more options than W” is replaced by a new condition “X and W have exactly two options in common”. If all of the other original conditions remain in effect, which one of the following must be false? (a) T and X have no options in common. (b) X and Z have no options in common. (c) V and X have exactly two options in common. (d) V and X have exactly one option in common. (e) X and Z have exactly two options in common. Directions for Question Nos. 88 to 92 Krishnapuram’s town council has exactly three members: Arjun, Karn and Bhim. During one week, the council members vote on exactly three bills: a recreation bill, a school bill, and a tax bill. Each council member votes either for or against each bill. The following is known: Each member of the council votes for at least one of the bills and against at least one of the bills. Exactly two members of the council vote for the recreation bill. Exactly one member of the council votes for the school bill. Exactly one member of the council votes of the tax bill. Arjun votes for the recreation bill and against the school bill Karn votes against the recreation bill. Bhim votes against the tax bill. [XAT 2007]

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36 Koncepts of Logical Reasoning 88. Which one of the following statements could be true? (a) Arjun and Karn vote the same way on the tax bill. (b) Karn and Bhim vote the same way on the recreation bill. (c)

Karn and Bhim vote the same way on the school bill.

(d) Arjun votes for one of the bills and Karn votes two of the bills. (e)

Arjun votes for two of the bills and Karn votes for two of the bills.

89. If the set of members of the council who vote against the school bill are the only ones who also vote against the tax bill, then which one of the following statements must be true? (a) Arjun votes for the tax bill (b) Karn votes for the recreation bill (c)

Karn votes against the school bill

(d) Bhim votes against recreation bill (e)

Bhim votes against the school bill.

90. If Karn votes for the tax bill, then which one of the following statements could be true? (a) Arjun and Karn each vote exactly one bill (b) Karn and Bhim each vote for exactly one bill (c)

Arjun votes for exactly two bills

(d) karn votes for the recreation bill (e)

Bhim votes against the recreation bill

91. If Karn votes for exactly two of the three bills, which one of the following statements must be true? (a) Arjun votes for the tax bill. (b) karn votes for the recreation bill (c)

karan votes for the school bill

(d) Karan votes against the tax bill (e)

Bhim votes for the school bill

92. If one of the members of the council votes against exactly the same bills as does another member of the council, then which one of the following statements must be true? (a) Arjun votes for the tax bill (b) Karn votes for the recreation bill (c)

Karn votes against the school bill

(d) Karn votes for exactly one bill (e)

Bhim votes for exactly one bill.

Directions for Question Nos. 93 to 96 Six square states having equal area in a country are located in North-South direction in two columns next to each other. States are located in the given order, State 1, State 3, and State 5 are on the western side and State 2, State 4, and State 6 are on the eastern side. Within the six states, there are exactly four medical institutes, two management institutes, and two technical institutes. These eight institutions are located as follows: No institution is in more than one of the states. None of the states contain more than one management institute, and none contains more than one technical institute. None of the states contain both a management institute and a technical institute. Each management institute is located in a state that contains at least one medical institute. The technical institutes are located in two states that do not share a common boundary. State 3 contains a technical institute and State 6 contains a management institute. [XAT 2008] 93. Which one of the following could be true? (a) State 1 contains exactly one technical institute (b) State 1 contains exactly one medical institute (c) State 2 contains exactly one management institute (d) State 5 contains exactly one technical institute (e) State 6 contains exactly one technical institute 94. A complete and accurate list of the states, any one of which could contain the management institute that is not in State 6, would be _____. (a) 1, 4 (b) 2, 4 (c) 4, 5 (d) 1, 4, 5 (e) 1, 2, 4, 5 95. If each of the six states contains at least one of the eight institutions, then which one of the following must be true? (a) There is a management institute in State 1 (b) There is a medical institute in State 2 (c) There is a medical institute in State 3 (d) there is a medical institute in State 4 (e) There is a management institute in State 4

Matrix Arrangement 37 96. If one of the states contains exactly two medical institutes and exactly one technical institute, then which combination of three states might contain no medical institute? (a) 1, 3, 5 (b) 1, 4, 5 (c) 2, 3, 5 (d) 2, 4, 6 (e) 4, 5, 6 Directions for Question Nos. 97 to 100 There are exactly ten stores and no other buildings on a straight street in BistupurMarket. On the northern side of the street, from West to East, are stores 1, 3, 5, 7, and 9; on the southern side of the street, also from West to East, are stores 2, 4, 6, 8, and 10. The stores on the northern side are located directly across the street from those on the southern side, facing each other in pairs, as follows: 1 and 2; 3 and 4; 5 and 6; 7 and 8; 9 and 10. Each store is decorated with lights in exactly one of the following colours: green, red, and yellow. The stores have been decorated with lights according to the following conditions: No store is decorated with lights of the same colour as those of any store adjacent to it. No store is decorated with lights of the same colour as those of the store directly across the street from it. Yellow lights decorate exactly one store on each side of the street. Red lights decorate store 4. Yellow lights decorate store 5. [XAT 2008] 97. Which one of the following could be an accurate list of the colours of the lights that decorate stores 2,4, 6, 8 and 10, respectively? (a) Green, red, green, red, green (b) Green, red, green, yellow, red (c) Green, red, yellow, red, green (d) Yellow, green, red, green, red (e) Yellow, red, green, red, yellow 98. If green lights decorate store 7, then each of the following statements could be false EXCEPT: (a) Green lights decorate store 2 (b) Green lights decorate store 10 (c) Red lights decorate store 8 (d) Red lights decorate store 9 (e) Yellow lights decorate store 2 99. Which one of the following statements MUST be true? (a) Green lights decorate store 10 (b) Red lights decorate store 1 (c) Red lights decorate store 8 (d) Yellow lights decorate store 8 (e) Yellow lights decorate store 10

100. Suppose that yellow lights decorate exactly two stores on the south side of the street and exactly onestore on the north side. If all other conditions remain the same, then which one of the following statements MUST be true? (a) Green lights decorate store 1 (b) Red lights decorate store 7 (c) Red lights decorate store 10 (d) Yellow lights decorate store 8 (e) Yellow lights decorate store 2 Directions for Question Nos. 101 to 104 Professor Mukhopadhay works only on Mondays, Tuesdays, Wednesdays, Fridays,and Saturdays. She performs four different activities – Lecturing, Conducting quizzes, evaluating quizzes and working on consultancy projects. Each working day she performs exactly one activity in the morning and exactly one activity in the afternoon. During each week her work schedule MUST satisfy the following restrictions: She conducts quizzes on exactly three mornings. If she conducts quizzes on Monday, she does not conduct a quiz on Tuesday. She lectures in the afternoon on exactly two consecutive calendar days. She evaluates quizzes on exactly one morning and three afternoons. She works on consultancy project on exactly one morning. On Saturday, she neither lectures nor conducts quizzes. [XAT 2008] 101. On Wednesdays, the professor could be scheduled to? (a) Work on a consultancy project in the morning and conduct a quiz in the afternoon (b) Lecture in the morning and evaluate quizzes in the afternoon. (c) Conduct a quiz in the morning and lecture in the afternoon (d) Conduct a quiz in the morning and work on consultancy project in the afternoon. (e) Evaluate quizzes in the morning and evaluate quizzes in the afternoon. 102. Which of the following statements must be true? (a) There is one day on which she evaluates quizzes both in the morning and in the afternoon. (b) She works on the consultancy project on one of the days on which lectures. (c) She works on consultancy project on one of the days on which she evaluates quizzes. (d) She lectures on one of the days on which evaluates quizzes. (e) She lectures on one of the days on which she conducts quiz.

103. If the Professor conducts a quiz on Tuesday, then her schedule for evaluating quizzes could be? (a) Monday morning, Monday afternoon, Friday morning, Friday afternoon (b) Monday morning, Friday afternoon, Saturday morning, Saturday afternoon (c) Monday afternoon, Wednesday morning, Wednesday afternoon, Saturday afternoon (d) Wednesday morning, Wednesday afternoon, Friday afternoon, Saturday afternoon (e) Wednesday afternoon, Friday afternoon, Saturday morning, Saturday afternoon 104. Which one of the following must be a day on which professor lectures? (a) Monday (b) Tuesday (c) Wednesday (d) Friday (e) Saturday Directions for Question Nos. 105 The surnames of four professionals are: Bannerji, Chatterji, Mukherji and Pestonji. Their professions are accountant, lawyer, dentist and doctor (not necessarily in this order). The accountant and lawyer work in their offices, while the dentist and doctor work in their nursing homes. The accountant looks after Mukherji’s and Chatterji’s account. Chatterji, does not know Bannerji, although his nursing home is in the same street as Bannerji’s office. Chatterji is not a doctor.

[XAT 2008]

105. What are the occupations of the four people? (a) Bannerji – Doctor, Chatterji – Dentist, Mukherji – Accountant and Pestonji – Lawyer (b) Bannerji – Lawyer, Chatterji – Dentist, Mukherji – Accountant and Pestonji –Doctor (c) Bannerji – Doctor, Chatterji –Accountant, Mukherji – Dentist and Pestonji – Lawyer (d) Bannerji – Lawyer, Chatterji – Dentist, Mukherji – Doctor and Pestonji – Accountant (e) Bannerji – Dentist, Chatterji – Lawyer, Mukherji – Doctor and Pestonji – Accountant Directions for Question Nos. 106 to 108 Four married couples competed in a singing competition. Each couple had a unique team name. Points scored by the teams were 2, 4, 6 and 8. The “Sweet Couple” won 2 points. The “Bindas Singers” won two more points than Laxman’s team. Mukesh’s team won four points more than Lina’s team, but Lina’s team didn’t score the least amount of points. “Just Singing” won 6 points. Waheda wasn’t on the team called “New Singers”. Sanjeev’s team won 4 points. Divya wasn’t on the “Bindas Singers” team. Tapas and Sania were on the same team, but it wasn’t the “Sweet Couple”. 

[XAT 2009]

106. Laxman’s teammate and team’s name were: (a) Divya and Sweet Couple (b) Divya and Just Singing (c) Waheda and Bindas Singers (d) Lina and Just Singing (e) Waheda and Sweet Couple 107. The teams arranged in the ascending order of points are: (a) Bindas Singers, Just Singing, New Singers, Sweet Couple (b) Sweet Couple, New Singers, Just Singing, Bindas Singers (c) New Singers, Sweet Couple, Bindas Singers, Just Singing (d) Sweet Couple, Bindas Singers, Just Singing, New Singers (e) Just Singing, Bindas Singers, Sweet Couple, New Singers 108. The combination which has the couples rightly paired is: (a) Mukesh, Lina (b) Mukesh, Waheda (c) Sanjeev, Divya (d) Sanjeev, Lina (e) Sanjeev, Waheda Directions for Question Nos. 109 to 112 The regular mathematics faculty could not teach because of being sick. As a stop gap arrangement, different visiting faculty taught different topics on 4 different days in a week. The scheduled time for class was 7:00 am with maximum permissible delay of 20 minutes. The monsoon made the city bus schedules erratic and therefore the classes started on different times on different days. Mr. Singh didn’t teach on Thursday. Calculus was taught in the class that started at 7:20 am. Mr.Chatterjee took the class on Wednesday, but he didn’t teach probability. The class on Monday started at 7:00 am, but Mr.Singh didn’t teach it. Mr.Dutta didn’t teach ratio and proportion. Mr.Banerjee, who didn’t teach set theory, taught a class that started five minutes later than the class featuring the teacher who taught probability. The teacher in Friday’s class taught set theory. Wednesday’s class didn’t start at 7:10am. No two classes started at the same time. [XAT 2009] 109. The class on Wednesday started at: (a) 7:05 am and topic was ratio and proportion. (b) 7:20 am and topic was calculus. (c) 7:00 am and topic was calculus. (d) 7:30 am and topic was calculus. (e) 7:05 am and topic was probability.

EBD_7743

38  Koncepts of Logical Reasoning 

110.  The option which gives the correct teacher – subject combination is:

(a) Mr.Chatterjee – ratio and proportion



(b) Mr. Banerjee – calculus 



(c) Mr.Chatterjee – set theory



(d) Mr. Singh – calculus



(e) Mr. Singh – set theory

111. Probability was taught by:

(a) Mr.Dutta on Monday



(b) Mr.Dutta on Thursday



(c) Mr. Singh on Wednesday



(d) Mr. Singh on Monday



(e) None of these

112. The option which gives a possible correct class time – week day combination is:

(a) Wednesday–7:10 am, am, Friday – 7:05 am

Thursday

–

7:20



(b) Wednesday–7:20 am, am, Friday – 7:20 am

Thursday

–

7:15



(c) Wednesday–7:05 am, am, Friday – 7:10 am

Thursday

–

7:20



(d) Wednesday–7:10 am, am, Friday – 7:05 am

Thursday

–

7:15



(E) Wednesday–7:20 am, am, Friday – 7:10 am

Thursday

–

7:05

Directions for Question Nos. 113 to 117 Five people joined different engineering colleges. Their first names were Sarah (Ms.), Swati (Ms.), Jackie, Mohan and Priya (Ms.). The surnames were Reddy, Gupta, Sanyal, Kumar and Chatterjee. Except for one college which was rated 3 star, all other colleges were rated either 4 star or 5 star.  The “Techno Institute” had a higher rating than college where Priya studied. The three-star college was not “Deccan  College.” Mohan’s last name was Gupta but he didn’t study at “Barla College.” Sarah, whose last name wasn’t Sanyal, joined “Techno Institute.” Ms.Kumar and Jackie both studied at four-star colleges. Ms. Reddy studied at the “Anipal Institute,” which wasn’t a five-star college. The “Barla  College” was a five-star college. Swati’s last name wasn’t Chatterjee. The “Chemical  College” was rated with one star less than the college where Sanyal studied. Only one college was rated five star. [XAT 2009]

  Matrix Arrangement 39  113. Which is the correct combination of first names and surnames? (a) Mohan Gupta, Sarah Kumar, Priya Chatterjee (b) Priya Chatterjee, Sarah Sanyal, Jackie Kumar (c) Jackie Sanyal, Swati Reddy, Mohan Gupta (d) Mohan Gupta, Jackie Sanyal, Sarah Reddy (e) Jackie Chatterjee, Priya Reddy, Swati Sanyal 114. Which option gives a possible student – institute combination? (a) Priya – Anipal, Swati – Deccan, Mohan – Chemical (b) Swati – Barla, Priya – Anipal, Jackie – Deccan (c) Jaydeep – Chemical, Priya – Techno, Mohan – Barla (d) Priya – Anipal, Jaydeep – Techno, Sarah – Barla (e) Swati –  Deccan, Priya – Anipal, Sarah –Techno 115. Mohan Gupta may have joined: (a) Techno – Institute which had 5 star rating (b) Deccan College which had 5 star rating  (c) Anipal Institute which had 4 star rating (d) Chemical College which had 4 star rating (e) Techno – Institute which had 4 star rating 116. In which college did Priya study? (a) Anipal Institute (b) Chemical Institute (c) Barla College (d) Deccan College (e) Techno-Institute 117. The person with surname Sanyal was: (a) Sarah studying in Chemical College (b) Swati studying in Barla College (c) Priya studying in Deccan College (d) Jackie studying in Deccan College (e) Sarah studying in Techno-Institute Directions for Question Nos. 118 to 122 Five colleagues pooled their efforts during the office lunch-hour to solve the crossword in the daily paper. Colleagues: Mr. Bineet, Mr. Easwar, Ms. Elsie, Ms. Sheela, Ms. Titli. Answers: Burden, Barely, Baadshah, Rosebud. Silence. Numbers: 4 down, 8 across, 15 across, 15 down, 21 across. Order: First, second, third, fourth, fifth.

EBD_7743

40 Koncepts of Logical Reasoning 1.

2. 3. 4. 5.

6.

7.

Titli produced the answer to 8 across, which had the same number of letters as the previous answer to be inserted, and one more than the subsequent answer which was produced by one of the men. It was not Bineet who solved the clue to ‘Burden’, and Easwar did not solve 4 down. The answers to 15 across and 15 down did not have the same number of letters. ‘Silence’, which was not the third word to be inserted, was the answer to an across clue. ‘Barely’ was the rst word to be entered in the grid, but ‘Baadshah’ was not the second answer to be found. Elsie’s word was longer than Bineet’s; Sheela was neither the rst nor the last lo come up with an answer. Fifth one to be worked out was an answer to an across clue. [XAT 2010]

118. What was Sheela’s word? (a) Baadshah

(b)

Silence

(c)

Rosebud

(d) Barely

(e)

Burden

119. What could be Titli’s answer? (a) Baadshah

(b)

Silence

(c)

Rosebud

(d) Barely

(e)

Burden

120. What was Titli’s order? (a) First

(b)

Second

(c)

Third

(d) Fourth

(e)

Fifth

121. What was Easwar’s number? (a) 4 down

(b)

21 across

(c)

8 across

(d) 15 down

(e)

15 across

122. What was Bineet’s word? (a) Barely

(b)

Burden

(c)

Silence

(d) Rosebud

(e)

Baadshah

Directions for Question Nos. 123 to 125 Seven bands were scheduled to perform during the weeklong music festival at XLRI. The festival began on a Monday evening and ended on the Sunday evening. Each day only one band performed. Each band performed only once. The organizing committee had the task of scheduling the performances of the seven bands - Cactus, Axis, Enigma, Boom, Fish, Dhoom and Bodhi Tree. The festival schedule followed the following conditions: the performance of Bodhi Tree, the home band of XLRI, did not precede the performance of any other band. Among the visiting bands three were rock bands and the other three were fusion bands. All three bands of the same genre were not allowed to perform consecutively. Boom, which was a rock band, refused to perform immediately before or after Fish. Meet, who was a lead vocalist with a rock band, refused to perform after Angelina. Angelina, the only female lead vocalist in the music fest besides Bony, was with the band Enigma. Angelina refused to perform after Thursday citing personal reasons. Ali, who was the lead vocalist of a rock band, was not with the band Dhoom, and did not perform on Saturday. Sid, the lead vocalist of the rock band Cactus, could perform only on Monday. Rupam, the only male among the lead vocalists of the fusion bands, was with Fish and performed on Wednesday. None of the bands performed in absence of their lead vocalist. [XAT 2010] 123. All of the following statements can be true except: (a) If Meet was the lead vocalist of Axis then Ali was the lead vocalist of Boom. (b) If Meet was the lead vocalist of Dhoom then Bony was the lead vocalist of Axis. (c) If Bony was the lead vocalist of Dhoom then Meet was the lead vocalist of Axis. (d) If Ali was the lead vocalist of Boom then Meet was the lead vocalist of Dhoom. (e) If Bony was the lead vocalist of Axis then Meet was the lead vocalist of Boom. 124. Which of the following must be true? (a) Ali performed on Saturday and Enigma performed on Thursday. (b) Dhoom performed on Thursday and Angelina performed on Tuesday. (c) Boom performed on Friday and Meet performed on Tuesday. (d) Ali performed on Friday and Enigma performed on Tuesday. (e) Bony performed on Saturday and Axis performed on Thursday.

Matrix Arrangement 41 125. Which of the following is a plausible performance sequence? (a) Cactus, Enigma, Fish, Dhoom, Boom, Axis (b) Cactus, Dhoom, Fish, Boom, Enigma, Axis (c) Cactus, Axis, Fish, Boom, Enigma, Dhoom (d) Cactus, Axis, Fish, Enigma, Boom, Dhoom (e)

Cactus, Boom, Fish, Axis, Enigma, Dhoom

Directions for Question Nos. 126 to 128 On a certain day six passengers from Chennai, Bengaluru, Kochi, Kolkata, Mumbai, and Hyderabad boarded the New Delhi bound Rajdhani Express from TataNagar. The following facts are known about these six passengers: [XAT 2010] Ø Ø Ø

Ø

Ø Ø

Ø Ø Ø

The persons from Kochi and Chennai are less than 36 years of age. Person Z, the youngest among all is a doctor. The oldest person is from Kolkata and his/her profession is same as that of the person who got down at Mughal Sarai. The pet from Bengaluru, Chennai, Hyderabad and Mumbai got down at four different stations. The eldest among these four got down at Koderma and the youngest at Kanpur. The person who got down at New Delhi is older than the person who got down at Mughal Sarai. The engineer from Bengaluru is older than the engineer from Chennai. While arranging the teachers in increasing order of age it was observed that the middle person is as old as the engineer from Chennai. Person Y who got down at Mughal Sarai is less than 34 years old. The teacher from Kochi is four years older than the 31 year old doctor who is not from Mumbai. In the past, three of the travellers have served in the Indian Army.

4

Directions for Question Nos. 129 to 132

Four people of different nationalities live on the same side of a street in four houses each of different color. Each person has a different favorite drink. The following additional information is also known: The Englishman lives in the red house. The Italian drinks tea.

126. Which of the following options is true? (a) The person from Chennai is older than the person from Kochi. (b) The oldest teacher is from Mumbai. (c) The person from Mumbai is older than at least one of the engineers. (d) The person from Kochi got down at Mughal sarai, and was an engineer. (e) The person who got down at New Delhi is older than Y, who in turn is older than the person from Hyderabad. 127. All six travelers are working in the same organization for at least one year. The organization recruits two categories of employees – fresh graduates and those who have at least ve years of experience in the Indian army. In both cases a 1 recruit should be less than 30 years of age. Among the travelers from same profession, those with military background are at least ve years older than the travelers who joined as fresh graduates. Identify the travelers who joined the organization as fresh graduates. (a) Only Y (b) The person Y and the travelers from Chennai (c) The person y and the travelers from Kochi and Hyderabad (d) The travelers from Kochi and Hyderabad (e) The teacher from Mumbai, the traveller from kochi and the younger engineer 128. If W is neither the youngest nor the oldest among the travelers from her profession, which of the following is true about her? (a) She got down at koderma (b) she is 36 years old (c) She got down at Mughal Sarai (d) She is from kochi (e) None of the above

The Norwegian lives in the rst house on the left. In the second house from the right they drink milk. The Norwegian lives adjacent to the blue house. The Spaniard drinks fruit juice. Tea is drunk in the blue house. The white house is to the right of the red house. [CAT 1993]

EBD_7743

42 Koncepts of Logical Reasoning 129. The color of the Norwegian’s house is (a) yellow

(b)

(c)

(d) red

blue

white

130. Milk is drunk by (a) Norwegian

(b)

(c)

(d) None of the above

Italian

English

131. The Norwegian drinks (a) milk

(b)

(c)

(d) fruit juice

tea

cocoa

132. Which of he following is not true? (a) Milk is drunk in the red house. (b) The Italian lives in the blue house. (c)

The Spaniard lives in a corner house.

(d) The Italian lives next to the Spaniard Directions for Question Nos. 133 and 134 Amar, Akbar and Anthony are three friends. Only three colours are available for their shirts, viz. red, green and blue. Amar does not wear red shirt. Akbar does not wear green shirt. Anthony does not wear blue shirt. [CAT 1998] 133. If Akbar and Anthony wear the same colour of shirts, then which of the following is not true? (a) Amar wears blue and Akbar wears green (b) Amar wears green and Akbar wears red (c)

Amar wears blue and Akbar does not wear blue

(d) Anthony wears red 134. If two of them wear the same colour, then how many of the following must be false? I. II.

Amar wears blue and Akbar does not wear green Amar does not wear blue and Akbar wears blue

III. Amar does not wear blue and Akbar does not wear blue IV. Amar wears green, Akbar does not wear red, Anthony does not wear green (a) None

(b)

(c)

(d) Three

Two

One

Directions for Question Nos. 135 to 138 Mr Bankatlal acted as a judge for the beauty contest. There were four participants, viz. Ms Andhra Pradesh, Ms Uttar Pradesh, Ms West Bengal and Ms Maharashtra. Mrs Bankatlal, who was very anxious about the result, asked him about it as soon as he was back home. Mr Bankatlal just told that the one who was wearing the yellow saree won the contest. When Mrs Bankatlal pressed for further details, he elaborated as follows: All of them were sitting in a row. All of them wore sarees of different colours, viz. green, yellow, white, red. There was only one runner-up and she was sitting beside Ms. Maharashtra. The runner-up was wearing the green saree. Ms West Bengal was not sitting at the ends and was not the runner up. The winner and the runner-up are not sitting adjacent to each other. Ms Maharashtra was wearing white saree. Ms Andhra Pradesh was not wearing the green saree. Participants wearing yellow saree and white saree were at the ends. [CAT 1998] 135. Who wore the red saree? (a) Ms Andhra Pradesh (b) Ms West Bengal (c) Ms Uttar Pradesh (d) Ms Maharashtra 136. Ms. West Bengal was sitting adjacent to (a) Ms Andhra Pradesh and Ms Maharashtra (b) Ms Uttar Pradesh and Ms Maharashtra (c) Ms Andhra Pradesh and Ms Uttar Pradesh (d) Ms Uttar Pradesh 137. Which saree was worn by Ms Andhra Pradesh? (a) Yellow (b) Red (c) Green (d) White 138. Who was the runner-up? (a) Ms Andhra Pradesh (b) Ms West Bengal (c) Ms Uttar Pradesh (d) Ms Maharashtra

Matrix Arrangement 43 Directions for Question Nos. 139 to 140 A, B, C, D, E and F are a group of friends from a club. There are two housewives, one lecturer, one architect, one accountant and one lawyer in the group. There are two married couples in the group. The lawyer is married to D who is a housewife. No lady in the group is either an architect or an accountant. C, the accountant, is married to F who is a lecturer. A is married to D and E is not a housewife. [CAT 1999] 139. What is E? (a) Lawyer (c) Lecturer

(b) Architech (d) Accountant

140. How many members of the group are male? (a) 2 (b) 3 (c) 4 (d) None of these 141. Five persons with names P, M, U, T and X live separately in any one of the following: a palace, a hut, a fort, a house or a hotel. Each one likes two different colours from among the following: blue, black, red, yellow and green. U likes red and blue. T like black. The person living in a palace does not like black or blue. P likes blue and red. M likes yellow. X lives in a hotel. M lives in a: [CAT 2000] (a) Hut (b) palace (c) Fort (d) house 142. Eight people carrying food baskets are going for a picnic on motorcycles. Their names are A, B, C, D, E, F, G and H. They have four motorcycles, M1, M2, M3 and M4 among them. They also have four food baskets O, P, Q and R of different sizes and shapes and each can be carried only on motorcycles M1, M2, M3, or M4, respectively. No more than two persons can travel on a motorcycle and no more than one basket can be carried on a motorcycle. There are two husband-wife pairs in this group of eight people and each pair will ride on a motorcycle together. C cannot travel with A or B. E cannot travel with B or F. G cannot travel with F, or H, or D. The husbandwife pairs must carry baskets O and P. Q is with A and P is with D. F travels on M1 and E travels on M2 motorcycles. G is with Q, B cannot go with R. Who is travelling with H? [CAT 2001] (a) A (c) C

(b) B (d) D

143. Four friends Ashok, Bashir, Chirag and Deepak are out shopping. Ashok has less money than three times the amount that Bashir has. Chirag has more money than Bashir. Deepak has an amount equal to the difference of amounts with Bashir and Chirag. Ashok has three times the money with Deepak. They each have to buy at least one shirt, or one shawl, or one sweater, or one jacket that are priced Rs.200, Rs.400, Rs.600 and Rs.1000 a piece, respectively. Chirag borrows Rs.300 from Ashok and buys a jacket. Bashir buys a sweater after borrowing Rs.100 from Ashok and is left with no money. Ashok buys three shirts. What is the costliest item that Deepak could buy with his own money? [CAT 2001] (a) A shirt (b) A shawl (c) A sweater (d) A jacket 144. In a “keep-t” gymnasium class there are fteen females enrolled in a weight-loss program. They all have been grouped in any one of the ve weight-groups W1, W2, W3, W4, or W5. One instructor is assigned to one weight-group only. Sonali, Shalini, Shubhra, and Shahira belong to the same weight-group. Sonali and Rupa are in one weight-group, Rupali and Renuka are also in one weightgroup. Rupa, Radha, Renuka, Ruchika, and Ritu belong to different weight-groups. Somya cannot be with Ritu, and Tara cannot be with Radha. Komal cannot be with Radha, Somya, or Ritu. Shahira is in W1 and Somya is in W4 with Ruchika. Sweta and Jyotika cannot be with Rupali, but are in a weightgroup with total membership of four. No weight-group can have more than ve or less than one member. Amita, Babita, Chandrika, Deepika and Elina are instructors of weightgroups with membership sizes 5, 4, 3, 2 and 1, respectively. Who is the instructor of Radha? [CAT 2001] (a) Babita (b) Elina (c) Chandrika (d) Deepika 145. A king has uninching loyalty from eight of his ministers M1 to M8, but he has to select only four to make a cabinet committee. He decides to choose these four such that each selected person shares a liking with at least one of the other three selected. The selected persons must also hate at least one of the likings of any of the other three persons selected.



M1 likes fishing and smoking, but hates gambling, M2 likes smoking and drinking, but hates fishing,  M3 likes gambling, but hates smoking, M4 likes mountaineering, but hates drinking, M5 likes drinking, but hates smoking and mountaineering, M6 likes fishing, but hates smoking and mountaineering, M7 likes gambling and mountaineering, but hates fishing, and M8 likes smoking and gambling, but hates mountaineering.

Who are the four people selected by the king?

[CAT 2001] (a) M1, M2, M5, M6 (b) M3, M4, M5, M6 (c) M4, M5, M6, M8 (d) M1, M2, M4, M7

147. Five boys went to a store to buy sweets. One boy had Rs.40. Another boy had Rs.30. Two other boys had Rs.20 each. The remaining boy had Rs.10. Below are some more facts about the initial and final cash positions. (i) Alam started with more than Jugraj. (ii) Sandeep spent Rs. 1.50 more than Daljeet. (iii) Ganesh started with more money than just only one other person. (iv) Daljeet started with 2/3 of what Sandeep started with. (v) Alam spent the most, but did not end with the least. (vi) Jugraj spent the least and ended with more than Alam or Daljeet. (vii) Alam spent 10 times more than what Ganesh did. (viii) Ganesh spent Rs. 3.50

Directions for Question Nos. 146 to 148

[CAT 2002]

146. Four students (Ashish, Dhanraj, Felix and Sameer) sat for the Common Entrance Exam for Management (CEEM). One student got admission offers from three National Institutes of Management (NIM), another in two NIMs, the third in one NIM, while the fourth got none. Below are some of the facts about who got admission offers from how many NIMs and what is their educational background. (i) The one who is an engineer didn’t get as many admissions as Ashish. (ii) The one who got offer for admissions in two NIMs isn’t Dhanraj nor is he a chartered accountant. (iii) Sameer is an economist. (iv) Dhanraj isn’t an engineer and received more admission offers than Ashish. (v) The medical doctor got the most number of admission offers.

Which one of the following statements is necessarily true? (a) Ashish is a chartered accountant and got offer for admission in three NIMs. (b) Dhanraj is a medical doctor and got admission offer in one NIM. (c) Sameer is an economist who got admission offers in two NIMs. (d) Felix who is not an engineer did not get any offer for admission.



In the choices given below, all statements except one are false. Which one of the following statements can be true? (a) Alam started with Rs.40 and ended with Rs.9.50. (b) Sandeep started with Rs.30 and ended with Rs.1.00. (c) Ganesh started with Rs20 and ended with Rs.4.00. (d) Jugraj started with Rs.10 and ended with Rs.7.00.

148. Three children won the prizes in the Bournvita Quiz contest. They are from the schools; Loyola, Convent, Little Flowers, which are located at different cities. Below are some of the facts about the schools, the children the city they are from [CAT 2002] One of the children is Bipin. Loyola School’s contestant did not come first. Little Flower’s contestant was named Riaz. Convent School is not in Hyderabad. The contestant from Pune took third place. The contestant from Pune is not from Loyola School. The contestant from Bangalore did not come first. Convent School’s contestant’s name is not Balbir. Which of the following statements is true? (a) 1st prize: Riaz (Little Flowers), 2nd prize: Bipin (Convent), 3rd prize: Balbir (Loyola). (b) 1st prize: Bipin (Convent), 2nd prize: Riaz (Little Flowers), 3rd prize: Balbir (Loyola). (c) 1st prize: Riaz (Little Flowers), 2nd prize: Balbir (Loyola), 3 rd prize: Bipin (Convent). (d) 1st prize: Bipin (Convent), 2nd prize: Balbir (Loyola), 3rd prize: Riaz (Little Flowers).

EBD_7743

44  Koncepts of Logical Reasoning 

  Matrix Arrangement 45  (vii) Bimal eats two more idlis than Ignesh, but Ignesh eats two more vadas than Bimal.

Directions for Question Nos. 149 and 150 The Head of a newly formed government desires to appoint five of the six elected members A, B, C, D, E and F to portfolios of Home, Power, Defence, Telecom and Finance. F does not want any portfolio if D gets one of the five. C wants either Home or Finance or no portfolio. B says that if D gets either Power or Telecom then she must get the other one. E insists on a portfolio if A gets one.

[CAT 2003]

151. Which one of the following statements is true?

(a) Daljit eats 5 idlis.



(b) Ignesh eats 8 idlis.



(c) Bimal eats 1 idli.



(d) Bimal eats 6 idlis. 

152. Which of the following statements is true?

(a) Sandeep eats 2 vadas.

(a) A-Home, B-Power, C-Defence, D-Telecom, E-Finance.



(b) Mukesh eats 4 vadas



(b) C-Home, D-Power, A-Defence, B-Telecom, E-Finance.



(c) Ignesh eats 6 vadas



(d) Bimal eats 4 vadas.



(c) A-Home, B-Power, E-Defence, D-Telecom, F-Finance  



(d) B-Home, F-Power, C-Telecom, A-Finance. 

149. Which is a valid assignment?

E-Defence,

150. If A gets Home and C gets Finance, then which is NOT A valid assignment for Defence and Telecom?

153. Which of the following statements is true?

(a) Mukesh eats 8 idlis and 4 vadas but no chutney.



(b) The person who eats 5 idlis and 1 vada does not take chutney.



(c) The person who eats equal number of vadas and idlis also takes chutney.



(d) The person who eats 4 idlis and 2 vadas also takes chutney.



(a) D-Defence, B-Telecom



(b) F-Defence, B-Telecom



(c) B-Defence, E-Telecom

Directions for Question Nos. 154 to 156



(d) B-Defence, D-Telecom.

Five women decided to go shopping to M.G. Road, Bangalore. They arrived at the designated meeting place in the following order: 1. Archana, 2. Chellamma, 3. Dhenuka, 4. Helen and5. Shahnaz. Each woman spent at least Rs.1000. Below are some additional facts about how much they spent during their shopping spree. [CAT 2003]

Directions for Question Nos. 151 to 153 Five friends meet every morning at Sree Sagar restaurant for an idli-vada breakfast. Each consumes a different number of idlis and vadas. The number of idlis consumed are 1, 4, 5, 6 and 8, while the number of vadas consumed are 0, 1, 2, 4 and 6. Below are some more facts about who eats what and how much. [CAT 2003] (i)

(ii) (iii) (iv) (v) (vi)

The number of vadas eaten by Ignesh is three times the number of vadas consumed by the person who eats four idlis. Three persons, including the one who eats four vadas, eat without chutney. Sandeep does not take any chutney. The one who eats one idli a day does not eat any vadas or chutney. Further, he is not Mukesh. Daljit eats idli with chutney and also eats vada. Mukesh, who does not take chutey, eats half as many vadas as the person who eats twice as many idlis as he does.

(i)

The woman who spent Rs.2234 arrived before the lady who spent Rs.1193.

(ii) One woman spent Rs.1340 and she was not Dhenuka. (iii) One woman Chellamma.

spent

Rs.1378

more

than

(iv) One woman spent Rs.2517 and she was not Archana. (v) Helen spent more than Dhenuka. (vi) Shahnaz spent the largest Chellamma the smallest.

amount

154. What was the amount spent by Helen?

(a) Rs.1193.

(b) Rs.1340.



(c) Rs.2234.

(d) Rs.2517.

and

EBD_7743

46 Koncepts of Logical Reasoning 155. Which of the following amounts was spent by one of them? (a) Rs.1139.

(b)

Rs.1378.

(c)

(d)

Rs.2718.

Rs.2571.

5

156. The woman who spent Rs.1193 is (a) Archana. (c) Dhenuka.

(b) Chellamma. (d) Helen.

Directions for Question Nos. 157 to 159

Directions for Question Nos. 160 to 163:

These six houses are labeled as P, Q, R, S, T and U. (ii) There are three houses on each side of the road. (iii) The houses are of different colours, namely, Red, Blue, Green, Orange, Yellow and White (iv) The houses are of different heights. (v) T, the tallest house, is exactly opposite to the Red coloured house. (vi) The shortest house is exactly opposite to the Green coloured house. (vii) U, the Orange coloured house, is located between P and S. (viii) R, the Yellow coloured house, is exactly opposite to P (ix) Q, the Green coloured house, is exactly opposite to U. (x) P, the White coloured house, is taller than R, but shorter than S and Q.

Study the following information carefully and answer the questions that follow: Mr Ghosh recently redecorated his house by coordinating orange and three other colours for the walls, carpets and curtains of four different rooms. From the information below, determine the colours of the carpet, walls and curtains for each of the room and answer the following questions:

(i)

157. What is the colour of the tallest house? (a) Red

(b)

Blue

(c)

Green

(d) Yellow

(e)

None of these

158. What is the colour of the house diagonally opposite to the Yellow coloured house? (a) White

(b)

Blue

(c)

Green

(d) Red

(e)

noneofthese

159. Which is the second tallest house? (a) P

(b)

S

(c)

Q

(d) R

(e)

cannot be determined

(a)

Yellow was the only colour used in all the four rooms. It was used at least once for walls, carpets and curtains. (b) Three different colours were used in each room but only the dining room and the bedroom were decorated in the same three colours. (c) The same colour was chosen for the curtains in the bedroom, the carpet in the living room and the walls in the dining room. That colour was not used at all in the study room. (d) The only room with both green and grey in its colour scheme had carpet of the same colour as in the dining room. (e) Grey was the only colour used exactly twice— both times for curtains (f) The study room walls were painted the same colour as the living room walls. 160. Which of the following rooms had orange curtains and green walls? (a) Dining room (b) Living room (c) Bedroom (d) Study 161. Which of the two rooms had green carpets? (a) Dining room and bedroom (b) Study and living room (c) Living room and dining room (d) Study and dining room

Matrix Arrangement 47 162. Which room did not use grey colour at all? (a) Dining room (b) Cannot say (c) Study (d) Living room 163. The dining room had ___ curtains. (a) Green (b) yellow (c) Orange (d) grey Directions for Question Nos. 164 to 168: Four students, Promila, Quadir, Rita and Sridhar, each working under the supervision of one of thefour Professors Anand, Bose, Chandrasekharan and Deshpande made their nal year MBA Project Presentations one by one, one each in the areas of Finance, Marketing, Systems and HumanResource Management (HRM). Each Professor is an expert in only one of the above areas andsupervised exactly one of the above students in his own area. The following clues are provided. (i) (ii) (iii) (iv) (v) (vi)

First presentation was made by Rita. Prof. Bose works in Finance. Prof. Deshpande was Promila’s supervisor. The last presentation was in the Systems area Sridhar’s project was in the HRM area. Prof. Bose’s student’s presentation followed that of Prof. Chandrasekharan’s students.

164. In which area was Rita’s project? (a) Marketing (b) Finance (c) Systems (d) cannot be determined. 165. What is Prof. Deshpande’s area of expertise? (a) Marketing (b) HRM (c) Systems (d) cannot be uniquely determined. 166. In which area was the second presentation? (a) Finance (b) Marketing (c) HRM (d) cannot be uniquely determined. 167. Which student’s project did Prof. Bose supervise? (a) Quadir (b) Rita (c) Sridhar (d) can’t be determined from above 168. What is Prof. Anand’s area of expertise? I. HRM II. Systems III. Marketing (a) Either I or II (b) Either I or III (c) Either II or III (d) Neither I, nor II, nor III

Directions for Question Nos. 169 to 174: A business School with six Professors L, M, N, O, P and Q, has decided to implement a new scheme of course management. Each Professor has to coordinate one course and support another course.This semester; O’s support course is Finance, while three others have it in coordinator’s role. P and Q have marketing as one of their subjects. Q coordinates Operations, which is a support course for both N and P. Finance and IT are L’s subjects. Both L and O have same subjects. Strategy is a support course for only one of the Professors. 169. Who coordinates the Strategy course? (a) M (b) N (c) O (d) none of the six 170. Which course is supported by M? (a) Finance (b) Strategy (c) IT (d) Operations 171. Who coordinates the IT course? (a) L (b) N (c) O (d) None of the six 172. Who all are coordinating the Finance course? (a) L, M and N (b) M, N and O (c) N and O (d) L and N 173. Which course has only one coordinator and only one support Professor? (a) Marketing (b) Operations (c) Finance (d) Strategy Directions for Question Nos. 175 to 178: There are ve friends Amisha, Binaya, Celina, Daisy and Eshaan. Two of them play table tennis while the other three play different games, viz. football, cricket and chess. One table tennis player and the chess player stay on the same oor while the other three stay on oors 2, 4 and 5. Two of the players are industrialists while the other three belong to different occupations viz. teaching, medicine and engineering. The chess player is the oldest while one of the table tennis players, who plays at the national level, is the youngest. The other table tennis player who plays at the regional level is between the football player and the chess player in age. Daisy is a regional player and stays on oor 2. Binaya is an engineer while Amisha is the industrialist and plays tennis at the national level. 174. Who stays on oor 4? (a) Amisha (b) Binaya (c) Celina (d) Eshaan

EBD_7743

48 Koncepts of Logical Reasoning 175. What does Eshaan play?

Directions for Question Nos. 182 to 185:

(a) Chess (b) Football (c) Cricket (d) Table tennis at regional level 176. Age wise, who among the following lies between Daisy and Eshaan? (a) Teacher

(b)

(c)

(d) Doctor

Engineer

Industrialist

177. Who all stay on oor 3? (a) Amisha and Binaya (b) Daisy and Eshaan (c) Binaya and Daisy (d) Celina and Daisy 178. What is the occupation of the chess player? (a) Engineer (b) Industrialist (c)

Doctor

(d) Teacher

Directions for Question Nos. 179 to 181: Five ags, each with a distinct symbol namely Panther, Tiger, Rose, Swan and Quail, have been arranged in the following order: (i) Panther is next to Quail and Swan is next to Rose. (ii) Swan is not next to Tiger, Tiger is on the extreme left hand side and Rose is on the second position from the right hand side. (iii) Panther is on the right hand side of Quail and to the right side of Tiger. (iv) Panther and Rose are together. 179. Which of the following statements is true? (a) (iii) and (iv) are contradicting (b) Either (iii) or (iv) is redundant (c) (iii) is redundant (d) (iv) is redundant 180. Who is on the extreme right and who is on the extreme left. (a) Tiger & Rose (b) Quail & Tiger (c) Swan &Tiger (d) Tiger &Swan 181. Who is in the middle and who is on its right. (a) Panther & Swan (b) Rose &Swan (c) Panther & Rose (d) Quail & Panther

Six friends Abhishek, Deepak, Mridul, Pritam, Ranjan and Salil married within a year in the months of February, April, July, September, November and December and in the cities of Ahmedabad, Bengaluru, Chennai, Delhi, Mumbai and Kolkata, but not necessarily following the above order.The brides’ names were Geetika, Jasmine, Hema, Brinda, Ipsita, and Veena, once again not following any order. The following are some facts about their weddings. Mridul’s wedding took place in Chennai; however he was not married to Geetika or Veena. Abhishek’s wedding took place in Ahmedabad and Ranjan’s in Delhi; however neither of them was married to Jasmine or Brinda. The wedding in Kolkata took place in February. Hema’s wedding took place in April, but not in Ahmedabad. Geetika and Ipsita got married in February and November and in Chennai and Kolkata, but not following the above order. Pritam visited Bengaluru and Kolkata only after his marriage in December. Salil was married to Jasmine in September.

182. Deepak’s wedding took place in (a) Bengaluru (b) Mumbai (c) Kolkata (d) Delhi 183. In Mumbai, the wedding of one of the friends took place in the month of (a) April (b) September (c) November (d) December 184. Ipsita’s wedding took place in (a) Ahmedabad (b) Bengaluru (c) Mumbai (d) Chennai 185. Hema’s husband is (a) Abhishek (c) Ranjan

(b) Deepak (d) Pritam

186. Salil’s wedding was held in (a) Bengaluru (b) Chennai (c) Kolkata (d) Delhi 187. Which one of the following is a correct set of husband/ wife. (a) Abhishek & Geetika (b) Mridul & Brinda (c) Pritam & Jasmine (d) None of these

Matrix Arrangement 49

Concept Applicator (CA) Solutions (1 to 3) Information given is as follows(i)

There are 4 students namely Dinakaran, Sumit, Tarun and Amul

(ii) One of them studies Commerce and plays Golf and Lawn Tennis. (iii) Dinakaran and Sumit study Psychology (iv) Dinakaran plays Billiards (v)

Both the Psychology students play Chess (But from iii Dinakaran and Sumit study Psychology so they play chess)

(vi) Tarun is a student of Physics. (vii) The Physics student plays Chess and Badminton (That means Tarun plays Chess and badminton) (viii) All the friends play two games each and study one subject each. (ix) One of the students also does Weight Training. Based on the information given in the question we conclude the following table: Billiard

Weight Training

Badminton Chess

Yes Yes Yes

Lawn Tennis

Golf

Yes

Commerce

Psychology

Yes

Dinakaran

Yes

Yes

Sumit

Yes

Yes

Tarun Yes

Yes

Amul

Yes Yes

Final result isSTUDENT

SUBJECT

HOBBIES

Dinakaran

Psychology

Billiards, Chess

Sumit

Psychology

Weight Training, Chess

Tarun

Physics

Badminton, Chess

Amul

Commerce

Golf, Lawn Tennis

1.

(d) Amul does not play chess.

2.

(a) Dinakaran studies psychology and plays billiards.

3.

(c) 3 subjects and 6 games.

Solutions (4 to 6) Information given is as follows(i)

There are three family shows.

(ii) ‘‘Main Sati Hoon’ is a family drama show and has social message for audience. (iii) The show ‘Detective Doom’ is a suspense thriller, a family drama and also has social message. So we can complete the table as follows-

Physics

EBD_7743

50 Koncepts of Logical Reasoning From the given information we can draw following conclusionSchedule Time

Show

Adult Content Social msg

5. 6.

Detective Doom

X

√

√

X

10:00-10:30PM

Main Sati hoon

X

√

√

X

10:30-11:00PM

Laugh a while

√

X

√

X

11:00-1:30 PM

HIV & India

√

√

X

√

X

X

X

√

(d) From the above table ‘Laugh a while’ is scheduled at 10:30 to 11:00pm. (c) From the above table ‘Detective Doom’ does not have adult content. (a) From the above table ‘HIV and India’ is scheduled at 11:00 to 11:30pm.

11. (d) M1 + M4 = 55% an and M2 + M = 45% Then Total time taken by M1 + M4 = (6.15 + 2) = 8.15hr and M2 + M3 = (8.15 x45)/55 = 6.45 Total time = 15hrs. Rest time = (24-15) = 9hrs

Solutions (7 to 8) 7

8.

News Based

9:30 -10:00 PM

11:30-Mid Night Mid Night Murder 4.

Family drama

(c) For minimum distance he will take the route S – Q – P – R – T and total distance is 1 + 2 + 3 + 3 = 9 km (b) The minimum distance in this case is given by the root Q – S – T – R – P – Q and total distance is 1 + 2 + 3 + 3 + 2 = 11 km

M1 paid 5000 for 6.15 hrs/15 hrs, then M5 will be paid for 9hrs/24 hrs = 5000 × 9/24 × 15/25 × 4 = Rs.4500 Rent per day = Rs100. No. of days = 4500/100 = 45 days. Solutions (12 to 14)

Solutions (9 to 11)

Given information can be summarized as-

(d) Let the total investment be Rs, x M1 + M4 = (100 – (M2 + M3) = 55% Or, 55x/100 = M1 + M4 Or, 11x/20 = 5000 + M4 This equation satised taking assumption, which is a multiple of 11. M1 + M4 = 6600, = > M4 = 6600 – 5000 = 1600 Using unitary, we can get the time period of M1 = (2 X5000)/1600 = 6.15 hr 10. (d) Above section we already calculated M4 contribution, that is 1600

(i)

9.

Dolly is mother of Shikhar

(ii) Points of Amar >Akash>Swarn (iii) Niece of Swarn got the lowest points (and that must be Shikhar) (iv) Father of Shikhar got more points than Amar but less than Shikhar’s mother. Hence sequence would be Dolly (Shikhar’s mother) >Snehal (Shikhar’s Father) > Amar >Akash>Swarn>Shikhar (Niece of Swarn) 12. (d)

13. (c)

14. (b)

Solutions (15 to 18) We can tabulate the given information as followsCorporate Planning Aditya

Information Technology

HR/Personnel

√

Aryan Harish

Finance

√ √

√

Sheetal 15. (c) From the table, both Aditya and Harish are MBA. 16. (b)

17. (a)

Qualication

√

MIB/MBA

√

√

Puru

Exports

18. (d)

√

MCA/PMIR √

MIB/MBA

√

MIB/MCA/MFC/ PMIR

Matrix Arrangement 51 Solutions (19 to 23)

Concept Builder (CB)

The given information is as follows: R1 – Japanese, Hindi R2 – Japanese, English R3 – English, Hindi R4 – French, Japanese R5 – Hindi, English, French From these we can draw the following conclusion Japanese

Hindi

English

French

R1

√

√

×

×

R2

√

×

√

×

R3

×

√

√

×

R4

√

×

×

√

R5

×

√

√

√

19. (d) 21. (b) 23. (b)

20. (d) 22. (d)

We can conclude the following result:

(3) Chess player

Qualication

Name

Power Engineer

Amit

(3) Cricket (Regional) Mech. Engineer

Manu

(2) Football

Design Engineer Rohit

(5) Tennis

Quality Inspector Tarun

(4) Cricket (National) Mech. Engineer

Ambrish

24. (a) 25. (d) 26. (c) 27. (d) 28. (c) 29. (a) Solutions for 30 – 32: We can summaries the given information as follows. P

M

Hardware

Q

F

Hardware

R

F

Hardware

S

M

T

F

X

M

Hardware

34. (d)

35. (b)

36. (d)

Solutions (37 to 42) From the information given we can draw the following tableRoom no

101

102

103

104

105

106

Country Canada Uruguay England Argentina brazil Germany No of cars No.of car donated

24 x

12

4

8

16

8

18

x-2

24

Let us assume that the number of Canadian institute they have donated is x, then that from Brazil is x-2. On the basis of data above table is drawn. From information 10, residents of Canada, England and Brazil are staying in alternate rooms in that order starting from left.

Solutions (24 to 29) Flat Agewise No. decreasing

33. (b)

Marketing

Though room numbers of residents of Canada, England and Brazil can also be 102, 104 and 106 respectively. 37. (d)

38. (c)

39. (d)

40. (d)

41. (b)

42. (b)

Solutions (43 to 45)

From the given information ticket of Priyatheater is not with Veena and Seema. We can make a table from the given informationPriya Chankaya M2K Seema X X X Reema √ X X Neeta X X √ Mona X √ X Veena X X X Gangster Equbal Khiladi

PVR √ X X X X Salam

Satyam X X X X √ Hero

Marketing

43. (d) From the table we can observe that PVR Saket –Seema – Salam Namaste

Marketing

44. (b) From the table we can observe Eqbalmovie is running in Chanakya

30. (a) From the above result we can say that S is the male who knows marketing but not hardware. 31. (d) 32. (c)

that

45. (b) From the table we can observe that Veena has the ticket of hero.

EBD_7743

52 Koncepts of Logical Reasoning 46. (a) From the above information we can tabulate &(c) the result asLeelavati Lancer Columbus A.P yes no no Karnataka no yes no Maharashtra no no yes Contestant Rahaman Badal Binod Position 1st 3rd 2nd 47. (b) From the data table can be made. Case I when pro. Chaudhary had HRM and pro. Das had psychology. Fotedar Das chaudhary Banik Eswar Acharya

HRM no No yes No No no

Ps no yes no no no no

D.ST no no no no no yes

T.P no no no yes no no

F no no no no yes no

Solutions (49 to 51) According to data table be : Let Rs 1193 be the least amount Aradhana Chandrima Deepika Rs.2234

Rs. 1193

Heena

Susmita

Rs. 2517 Rs. 1340 Rs. 2571

This case does not satisfy because Heena spent more than Deepika. Let Rs 2517 be the maximum amount. Aradhana Chandrima Deepika Heena Susmita Rs.1340 Rs. 1139 Rs. 2234 Rs. 1193 Rs. 2517 Rs.2234

Rs. 1139

Rs1193

Rs. 1340 Rs. 2517

From this table we see that case ii 2nd line gives the answer. 49. (b)

50. (a)

51. (c)

Case II: when pro.Chaudhary had and pro. Das had Trade policy Fotedar Das chaudhary Banik Eswar Acharya

HRM No No No No No Yes

Ps No No No Yes No No

D.ST No No No No yes No

T.P No yes No No No No

F no no yes No No No

48. (d)

Solutions (52 to 55) The nal result Eldest

2nd

3rd

4th

5th

6th

7th

youngest

child

Sourav

M/N

M/N

Kuntala

Tamanna

Arun

Rohit

Janki

School

Trinity

Trinity

St. Stefan St. stefan St. stefan

Chess

football

Mansorover Mansorover

Game 52. (c)

cricketers 53. (a)

cricketers 54. (d)

55. (b)

hockey

Trinity

Matrix Arrangement 53 Solutions (56 to 58) The order of the house is given to us, it is Blue, Green, Red and Yellow, now it is given that Paul stay in between Som and Krishna hence Paul cannot stay at corner houses i.e in the blue and Yellow hose, the same thing applied for Krishna as well. Now from given condition about their salaries we can complete the following tableName

Blue

Green

Red

Yellow

Salary

Paul

X

√

X

X

30000

Krishna

X

X

√

X

110000

Laxman

X

X

X

√

Som

√

X

X

X

Hence Alok must have been placed in WiproBangalore. From (ii), Amit has Sancharnet id and Anup has Indiatimes id. From (iii), Mecon has Gmail id. From (vi), Alok has Hotmail id and Kabeer has Yahoo id. According to these results we can draw the following table 59. (d)

60. (d)

50000

62. (d)

63. (d)

80000

Solutions (64 to 65)

61. (c)

56. (b) Krishna is living in red house. 57. (c) Krishna is earning highest salary. 58. (a) Paul is living in green house and earning low salary.

There are 6 persons in the family from 3 generations. There are two married couples.

Solutions (59 to 63)

Now as there are only 3 generations present.

From the given information we can draw following conclusions

Mona and Varun are married to each other.

From (ix), Anup is placed in Tisco-Jamshedpur. From (iv), Rahul is placed in Telco-Pune and has Rediffmail id. From (vi), Kabeer is placed in Usha Martin – Kolkatta. From (vii), Amit must have been placed in HCL-Noida. From (v), Raghu must have been placed in MeconRanchi.

Mona is the mother-in-law of Sonalika. Varun is the grand Father of Tarun.

Nuri is the grand-daughter of Mona. This means that Tarun and Nuri are siblings. This also means that Rakesh and Sonalika are married to each other. 64. (a) 65. (c) It is given in the question that Tarun likes Punjabi food and supports Basketball.

Concept Cracker (CC) with XAT Solutions (66 to 70) J yes Term 1

K no

×

×

× × Term 2

Term 3

L

M

yes

yes

×

×

×

×

N

P no

no

yes

Q yes ×

×

× yes

×

×

×

×

×

×

×

×

×

×

yes

yes

no

× yes

×

yes no

no

no

yes

×

yes

yes

yes

yes

no

Yes/no

no

no

yes

×

×

yes

66. (d) In option A there will be no instructor in 3rd term. Option B J and P in the same group , not possible. As 2 instructors in the 3rd term so there will be 1 instructor in 1st term, but there is 3 or 4 instructor. So D is only option.

67. (a) If L teaches in the 1st term than M also teaches. So, in 3rd 4 instructors be there but Q doesn’t teach in the 3rd term and only J and P can teach in third term. So 3 instructors in 3rd term. 68. (c) 69. (d) It is clear that k only teach in3rd term but Q in 1st or 2nd so K, P, Q is wrong combination. 70. (b) M must teach in 2nd term so L be in 2nd term Solutions (71 to 75) 71. (a) As package 2 is packed in a cartoon C so can’t have kajuburfi. Packet 2 contain KB have to be in the left of GJ, also the 2 KB are not adjacent so 1st contain KB and packed in bucket. 72. (c) If package 4 contains rasgulla it is packed in a cartoon. 3 and 1 contain KB packed in bucket and package 5 is packed in a tin. 73. (a) 74. (a) 75. (c)

Solutions (82 to 87) Given condition we made the table. Blank space is optional for mobile . Mobile phones T V W X Y Z

Digital Camera

Music Player

Office document

√ √ √ X

X √ √ X √

√ X √ √

82. (c) Hence except for T and Z for other all mobile phones it is possible to determine exactly which option each one has. 83. (a) From the table we can conclude that V and Y don’t have any MP. 84. (e) It is given that Z has no option in common with T but has at least one option in common with every other mobile phone hence Z must have Office document. The table would be -

Solutions (76 to 81) 76. (e) As per given condition, no. of buffs of S.Roy films must be double than G.Dutt films So, no. of SR buffs must be even. Gangadhar sees either GD or RG films. Gangadhar or Rudra and Indra or Maheshwar don’t see films with each other. Now, come to the elimination round of option.A is not possible as Gangadhar must see eithr GD or RG. B is wrong as Indira see S.Roy films. C is also wrong as Vomkesh and Yogi they see together.If they see GD films, then no. of SR fims buff cant double than GD. Option (e) if Rudra see SR fims then no. of SR buffs will be odd, that is also impossible. 77. (a) Among all option B to C are false as we describe earlier question. Rest option is. True that Rudra is only buffs to see GD films. 78. (c) Among total 7 buffs, one pair Vyomkesh and Yogi they see GD fims . There is another buff who only see GD or RG fims, She is Gangadhar. So, 79. (a) If One buff see only RG fims,so rest must see GD and SR fims. It is possible when any pair see SR fims. So, Vyomkesh See Satyajit films. 80. (b) Indira sees the SR films. 81. (e) As we discuss earlier, Yogi and Vyomkesh see SR films.This pair is not possible for GD.

EBD_7743

54  Koncepts of Logical Reasoning 

Mobile phones

Digital Camera

Music Player

Office document

T

√

X

X

V

√

X

√

W

√

√

X

X

√

√

√

Y

X

X

√

Z

X

√

√



From the table we can conclude that exactly 3 phones have Music player hence Option E must be false 85. (d) If no two mobile phone models have exactly the same options as one another, then given chart we can modify according to condition. Mobile phones

Digital Camera

Music Player

Office document

T

√/X

X/√

X

V

√

X

√

W

√

√

X

X

√

√

√

Y

X

X

√

Z

X

√

√



From this table we can conclude that exactly 4 of the 6 mobile phones have office document. 86. (d) If exactly four of the six mobile phones have music player, and exactly four of the six mobile phones have digital camera, then W and Z has more than one condition common false condition is D. 87. (b) If “X and W have exactly two options in common” then table will be – Mobile phones

Digital Camera

Music Player

Office document

T(1 feature) V(2 Features)

√

X

√

W(2 Features)

√

√

X

X

√

√

Y(1 Feature)

X

X

Z (2 Features)

  Matrix Arrangement 55  Option (e) If Arjun voted for 2 of the bills then he voted for Tax bill, then Karn voted for 2 of the bills then he voted for School and Tax bill hence both Arjun and Karn voted for Tax bill, but it is given that only one vote for the Tax bill hence it is ruled out. 89. (e) From the given condition we can draw following table Recreation

School

Tax

Arjun

√

X

X

Karn

X

√

√

Bhim

√

X

X



Now from the table we can find that only option E is true 90. (e) From the given condition we can draw following table

√

√

Name

Name

Recreation

School

Tax

Arjun

√

X

X

Karn

X

X

√

Bhim

√

√

X



From the table we can say that X and Z have one option in common and that is Music player, and hence option B is wrong.

Now from the table we can find that only option A is true 91. (e) Karan must vote for school bills. 92. (a) Karan can exactly vote against A or B.

Solutions (88 to 92)

Solutions (93 to 96)

Since there are two votes for Recreation Bill and Karn voted against it so rest two must have voted for that bill, we can draw following table as per the given information.

Given information are(i) There are 6 states (ii) There are 8 institutes namely Medical institute (MDI) -4, Management Institute (MAI)- 2 and Technical Institute (TI)-2 (ii) None of the states contain both a MAI and TI. (iv) State 3 contain one TI and state 6 contain one MAI i.e it also contain one MDI. (v) The TIs are located in two states that do not share a common boundary, and one of the TI is in state 3 hence the other has to be state 2. Now we can represent the result as follow-

Name

Recreation

School

Arjun

√

X

Karn

X

Bhim

√

Tax

X

88. (d) Now we will eliminate options one by one Option (a) Arjun and Karan can not vote in the same way as Arjun voted for and Karan voted against Recreation Bill. Hence it is eliminated. Option (b) Same as option A. Option (c) Since there was only one vote for the school bill and Arjun voted against that bill hence only one of the Karn or Bhim voted for that bill, hence this option also ruled out. Option (d) If Arjun votes for one of the bill then he will not vote for Tax bill, and since Karn voted for 2 of the bills then he voted for school and Tax bill so tsatisfy all the condition and a possible option.

State 1

State 3

State 5

(TI) State 2 (TI)

State 4

State 6 (MAI & MDI)

93. (b) Now we will eliminate the options one by one Option (a) Already 2 TI got placed in state 3 and 2 hence it is ruled out. Option (b) It may be possible as it doesn’t violate any of the condtion.

EBD_7743

56 Koncepts of Logical Reasoning Option (c) Since TI and MAI can not go together, hence ruled out. Option (d) Already two TIs are in state 2 and 3 hence it cannot be true. Option (e) State 6 contain one MAI so it can not contain TI 94. (d) From the above result State 2 and 3 contain a technical institute, and it is given that TI and MAI cannot be together \ State 2 and 3 cannot have a management institute. But any of States 1, 4 and 5 can have a management institute. 95. (d) Since one remaining MAI can be in anyone of the state 1, 4 or 5 hence Option (A) and (E) are equally probable, but we are not sure.As MAI will must have MDI together hence the remaining three states 1, 4 and 5 must have at least one MDI, given in option (D) 96. (a) Here we will take two cases as we have two states that contain TI Case (I) State 2 contain two MDI and one TI Now start eliminating the options we will nd only option A is the possible option. Case (II) State 3 contain two MDI and one TI Now start eliminating the options we will nd no option satisfy the condition hence this case is ruled out. So from case (I) option (A) is correct. Solutions (97 to 100) Given information are(i) Yellow lights decorate exactly one store on each side of the street. (ii) Store 4- Red light and Store 5- Yellow light. Now consider store 3 it cannot have Red (that is opposite to it) as well as cannot have yellow (that is adjacent to it) hence Store 3 must have Green Now consider store 1 it cannot have green. (as adjacent 3 has green light) and cannot have yellow as well hence store 1- Red, Now we can tabulate the concluded information1 (Red)

3 (Green)

5 (Yellow)

7 Red/ Green

2 Green/ Yellow

4 (Red)

6 (Green)

8 Red/ Yellow

9 Green/ Red 10

97. (b) From the above result we are sure that store 4- Red and store 6- Green hence option C and D ruled out.

Yellow cannot be twice hence option E is ruled out. Yellow should come only in one store hence option A ruled out. Option B satises all the given condition. 98. (d) As per the given condition if Green light decorate store 7 then the condition would be – 1 (Red) 2 Green/ Yellow

3 (Green)

5 (Yellow)

7 Green

9 Red

8 10 Red/ Green/ Yellow Yellow Hence Red light decorate store 9, given in option (D) 4 (Red)

6 (Green)

99. (b) As seen from the table in the set, it is clear that option B is denitely true. 100. (e) We know that Stores 4 and 6 are decorated with Red and Green lights respectively. Now it is given that southern side has 2 yellow lights \ The southern-side stores with Yellow lights have to be from among the stores 2, 8 or 10. \ we also know that no store is decorated with lights of the same colour as those of any store adjacent or opposite to it, \ Yellow lights can decorate only one of stores 8 and 10. \ hence we can conclude that Store 2 has to be decorated with Yellow lights. \ Option E must be true. Solutions (101 to 104) Given informationThere are 5 morning slots and 5 afternoon slots. On exactly 3 mornings she conducts quizzes (CQ) One morning she Evaluate Quizes (EQ) One morning she gives Consultancy (C) On 3 afternoon she Evaluate Quizes (EQ) On two consecutive afternoon she Lectures (L) Hence in morning- 3CQ + 1EQ + 1 C And in afternoon – 3EQ + 2L Given that On Saturday, she neither lectures nor conducts quizzes, hence on that day in morning she will give consultancy (C) and in afternoon EQ. Hence we can tabulate the given information.

Matrix Arrangement 57 Morning

Afternoon

Monday

CQ/ EQ/C

L

EQ

Tuesday

EQ/CQ/C

L

L

Wednesday

CQ

EQ

L

Friday

CQ

EQ

EQ

Saturday

C/EQ

EQ

Since she Lectures on two consecutive afternoon hence we have three cases for afternoon as shown in the table. 101. (c) From the above table we can conclude that on Wednesday she Conduct a quiz in the morning and lecture in the afternoon 102. (e) From the above table we can conclude that she lectures on one of the days on which she conducts quiz. 103. (e) The condition in this case is Morning

Afternoon

Monday

EQ/C

L

EQ

EQ

Tuesday

CQ

L

L

EQ

Wednesday

CQ

EQ

L

L

Friday

CQ

EQ

EQ

L

Saturday

C/EQ

EQ

104. (b) From the table we can observe that on Tuesday she will give lecture. 105. (d) We can complete the table as followsAcct

Law.

Dentist

Doctor

Bannerji

X

√

X

X

Chatterji

X

X

√

X

Mukherji

X

X

X

√

Pestonji √ X X X From the table we nd option (D) is correct Solutions (106 to 108) As per the given information we have four teams, four couples and 4 different points (i) “Sweet Couple” won 2 points. (ii) Mukesh’s team won 4 points more than Lina’s team and Lina’s team has not scored the least points. Therefore, only possible option is Lina has 4 points and Mukesh has 8 points. It also means that Mukesh (with point 8) is not with Lina, and both of them not belongs to “Sweet Couple” (iii) “Just Singing” won 6 points. Hence Mukesh and Lina do not belongs to “Just Singing” (iv) Sanjeev’s team scored 4 points. While from (ii) we have seen that Lina has also 4 points hence Sanjeev and Lina belongs to same group.

(v) Laxman is the only male remaining, who is in the team which scored 2 points i.e. “Sweet Couple” since from (i) we know that “Sweet Couple” has 2 points. (vi) “Bindas Singers” won 2 more points than Laxman’s team hence it has 4 points. (vii) Waheda was not on “New Singers”, so she must belongs to “Sweet Couple” and Divya is in the team “New Singers”. (viii) Tapas &Sania were on the same team but it wasn’t “Sweet Couple”. It cannot be New singer as Divya belongs to the team. So, the only possible team is “Just Singing”. Now we will tabulate this informationTeam

Male

Female

Points

Sweet Couple

Laxman

Waheda

2

Bindas Singers

Sanjeev

Lina

4

Just Singing

Tapas

Sania

6

New Singers

Mukesh

Divya

8

106. (e) From the table we can nd that Laxman’s team mate is Waheda and they belongs to Sweet couple. 107. (d) From the table correct sequence is Sweet Couple, Bindas Singers, Just Singing, New Singers 108. (d) From the table we can nd that Sanjeev and Lina are the paired couple. Solutions (109 to 112) Given information is(i) Mr. Chatterjee → Wednesday. (ii) Class on Monday → 7:00 → (~Mr. Singh ) Hence It has to be taken either by Mr. Dutta or Mr. Banerjee. But it is known that Mr. Banerjee taught a class that started ve minutes later than the class of probability. So he cannot take class on Monday, and will be taken by MrDutta (iii) MrDutta → Monday → 07:00 AM (iv) Mr. Singh → ~ Thursday \ His class has to be on Friday. (v) Mr. Singh → Friday → Set Theory Now we can tabulate the given information. Professor

Day

Mr. Singh

Friday

Mr. Chatterjee

Wednesday

Mr. Dutta

Monday

Mr. Banerjee

Thursday

Timing

Subject Set theory

07:00

EBD_7743

58 Koncepts of Logical Reasoning Further given information is(vi) Calculus → 7:20 am → Chatterjee Since Mr. Singh is teaching set theory, Mr. Dutta’sclass is starting at 7:00 am, and Mr. Banerjee’s class is starting 5 mins after the class on Probability and the timing on probability has to be one from 7: 10 am and 7:20 am. The class on Calculus started at 7:20 am was taken by Mr. Chatterjee. (vii) Mr. Dutta→ ~Ratio and proportion and we can complete the table (viii) Class on ratio and proportion → Mr. Banerjee. Class on probability → Mr. Dutta. Mr. Banerjee → 7:05 am Hence we can complete the table Professor

Day

Timing

Subject

Mr. Singh

Friday

O7:10

Set theory

07:20

Calculus

07:00

Probablity

Mr. Chatterjee Wednesday Mr. Dutta

Monday

Ratio & Proportion 109. (b/d) From the table we can conclude that the class on Monday is scheduled at 7:20. 110. (e) From the table we can conclude that Mr. Singh teaches set theory. 111. (a) From the table we can conclude that Probability was taught by Mr. Dutta on Monday. 112. (e) From the table we can conclude that The correct combination is option (E). Mr. Banerjee

Thursday

Institute and Chemical college has to have three star. Since Sarah studied at Techno institute and Ms. Reddy studied at Anipal Institute. If Chemical college is the one with three star rating then Sanyal has to be from a college with 4 star rating which has to be Deccan since it is the only one remaining. Then the only college remaining for Mohan Gupta is Barla College which is not possible and hence this is ruled out. Anipal Institute has to be the one with three star rating. And since Ms. Reddy studied at Anipal Institute and Priya studied at three star rating college. Hence we can conclude that Priya“s full name has to be Priya Reddy. Sanyal studied at Barla College. Since Sanyal studied at Barla College which is a ve star college. \ It has to be Swati now we have Sarah as the only girl remaining \ She only can be Ms. Kumar

07:05

Solutions (113 to 117) Given information are(i) Only one college was rated ve star and only one college was rated three star. (ii) Barla College → ve star college. Therefore, there are three colleges with four star rating. Since Techno Institute had a higher rating than the college where Priya studied.→ Techno Institute has four star rating and Priya studied in a three star rating college. (iii) Techno Institute – Four star college. (iv) Priya’s college → Three star college. Also Deccan college has four star rating since it is mentioned that it does not have three star rating. (v) Deccan college → four star. \ We know that Barlacollege is a ve star, Techno Institute – four star and Deccan college – four star. Hence one of Anipal

Hence Jackie“s full name has to be Jackie Chatterjee Now we can tabulate the information as. Name

Surname

College

Star

Priya

Reddy

Anipal

3

Jackie

Chatterji

Chemical/ Deccan

4

Sarah

Ms. Kumar

Techno

4

Mohan

Gupta

Deccan / Chemical

4

Swati

Sanyal

Barla

5

113. (e) From the above table we get that Jackie Chatterrjee, Priya Reddy, Swati Sanyal is correct. 114. (b) From the above table we get that Swati – Barla, Priya – Anipal and Jackie – Deccan is correct student – institute combination. 115. (d) From the given options we can conclude that Mohan Gupta may have joined Chemical College which had 4 star rating. 116. (a) From the above table we get that Priya studied in Anipal Institute. 117. (b) From the above table we get that The person with surname Sanyal was Swati studying in Barla

Matrix Arrangement 59 Solutions (118 to 122) From 1st information, Mr.Titli produced the answer to 8 across and it should be a seven lettered word so option we have is either Rosebud or Silence. The next word produced after this by one of the men is a 6 – lettered word. And this can be either ‘Burden’ or ‘Barely’ but given that from 5 ‘Barely’ is the rst word to be entered in the grid, so ‘Burden’ is the word produced by one of the men after a 7 – lettered word. From 5th information, it is clear that ‘Baadshah’ was not the second answer and from 4th information ‘Silence’ was not the third word, so we can conclude sequence them asOrder

Answer

First

Barely

Second

Silence

Third

Rosebud

Fourth

Burden

Fifth

Baadshah

Colleague

Titli

Number

8 across

So, Ms.Titli’s word is Rosebud and it is 8 across. From 4th information., ‘Silence’ was the answer to an across clue then the other two ‘Barely’ and ‘Burden’ were the answers for either of 4 down or 15 down. From 2nd information, it is clear that Mr.Bineet didn’t solve the clue to ‘Burden’ then it is Mr.Easwar who solved it and it should be for 15 down. [Q Easwar didn’t solve 4 down]. Mr.Sheela didn’t solve the rst or the fth word, so she solved the clue for the second word ‘Silence’. From 6, Else’s word was longer than Bineet’s. We can conclude the following. Order

Answer

Colleague

Number

First

Barely

Second

Silence

Third

Rosebud

Fourth

Burden

Fifth

Baadshah

Order First

Answer Barely

Colleague Number Bineet 4 down

Second

Silence

Sheela

across

Third Fourth Fifth

Rosebud Burden Baadshah

Titli Easwar Elsie

8 across 15 down across

119. (c) From the table we can observe that the answer given by Titli was Rosebud. 120. (c) From the table we can observe that Titli gave the answer of the third question. 121. (d) From the table we can observe that Easwar’s number is 15 down. 122. (a) From the table we can observe that Bineet’s word is Barely. Solutions (123 to 125) There are seven bands performing on seven days. Also, there are 3 Rock bands and 3 Fusion Bands. The 3 bands of one genre cannot perform on 3 consecutive days. Given that the festival began on a Monday evening and ended on the Sunday evening. Each day only one band performed. Each band performed only once. The seven bands are namely – Cactus, Axis, Enigma, Boom, Fish, Dhoom and Bodhi Tree. Given information isBodhi Tree’s performance was not preceded by anyone else. This implies that Bodhi Tree was the last to perform i.e. on Sunday. Sid, the lead vocalist of the rock band Cactus, performed on Monday. Rupam, the only male lead vocalist of a fusion band was with Fish and performed on Wednesday. Meet and Ali were lead vocalists of a rock band. Hence, the two females, Angelina and Bony were the lead vocalists with the other two fusion bands. Angelina was with the band Enigma and she could not perform after Thursday. We can tabulate the given information as followsDay Monday

Band

Person

Type

Cactus

Sid

Rock band

Meet

Rock band

Tuesday Titli

8 across

Wednesday Thursday

Fish

Gender

Rupam Fusion band

Enigma Angelina

Friday Saturday

118. (b) Sheela’s word was Silence.

Sunday

Bodhi Tree

Bodhi Tree didn’t precede any performance so it performed on Sunday. Sid a lead vocalist of the rock band Cactus would perform only on Monday. Rupam, of fusion band with Fish performed on Wednesday. Angelina with Enigma refused to perform after Thursday and she didn’t perform before Meet. So Meet of rock band performed on Tuesday and Angelina performed on Thursday. We can tabulate the following results.

EBD_7743

60 Koncepts of Logical Reasoning Rupam is the only male among the lead vocalists of the fusion band, so the other two are females, hence, they should be Angelina and Bony.

The persons from Kochi and Chennai are 35 years old. The person from Bengaluru got down at Koderma, while the person from Hyderabad got down at Kanpur.

Fusion band is performing on Wednesday and Thursday [Rupam, Angelina] so it should be followed by Rock band [since all the three bands of the same genre were not allowed to perform consecutively]. So rock band performs on Friday and Fusion band performs on Saturday [it is by Bony]. And Ali performs on Friday. Boom (a rock band) refused to perform immediately before or after Fish. So it has to perform on Friday. So nally we can tabulate the above information as follows-

The nal condition is as shown in the table.

Day

Band

Person

Type

Male/ Female

Monday

Cactus

Sid

Rock band

Male

Tuesday

Dhoom/ Axis

Meet

Rock band

Male

Wednesday

Fish

Rupam

Fusion band

Male

Fusion band

Female

Thursday

Enigma Angelina

Friday

Boom

Ali

Rock band

Male

Saturday

Axis/ Dhoom

Bony

Fusion band

Female

Sunday

Bodhi Tree

123. (e) Now, from the table we can observe that all options are possible except E, because if E is true then Meet is in Boom and according to the question Boom cannot play before Fish. Hence, the answer is option (E). 124. (c) From the table above, it is obvious that Boom performed on Friday and Meet performed on Tuesday. 125. (d) Again from the table, it is clear that Enigma has to play on Thursday. Hence, the answer is option (d) is correct. Solutions (126 to 128) As per the given information, there are three teachers, two engineers and one doctor. Since the youngest among them is Z, who is a doctor, hence his age must be 31. The persons from Bengaluru and Chennai are engineers, while the persons from Kolkata, Kochi and Mumbai are teachers. Therefore, the person who is the doctor must be from Hyderabad. The person who got down at Mughal Sarai is a teacher and is less than 34 years old.

Age

Profession

>35

Teacher

Boarding point Kolkatta

>35

Engineer

Bengaluru

35

Engineer

Chennai

35

Teacher

Kochi

(31-34)

Teacher

Mumbai

MughalSarai

Y

31

Doctor

Hyderabad

Kanpur

Z

Departure

Person

Koderma

126. (e) From the given options Option E is correct. 127. (e) It is given that the six travelers are working in the same organisation for atleast one year and some are ve years older than the fresh graduates, the present ages of people who are fresh graduates must be between 31 and 35 years. Three persons satisfy the above conditions. They are the teacher from Mumbai, the traveler from Kochi and the young engineer. 128. (d) Here it is given that the W is neither the youngest nor the oldest among the travelers who belong to her profession; she must be the teacher from Kochi. Only Option (D) satises the given condition. Concept Deviator (CD) Solutions (129 to 132) Let us assume that the houses from 1 – 4 is from left to right, then we can summarize the given information as Nationality English

House Colour Red

Favourite drink Milk

House Number 3

Italian

Blue

Tea

2

Norwegain

Yellow

Cocoa

1

Spaniard

White

Fruit Juice

4

129. (a) From the table, we can say that the color of the Norwegian’s house is yellow 130. (b) From the table, we can say that Milk is drunk by the Englishman 131. (b) From the table, we can say that The Norwegian drinks Cocoa 132. (d) From the table, we can say that statement (D) is false.

Matrix Arrangement 61 Solutions (133 to 134) It is given that Amar does not wear red shirt. Akbar does not wear green shirt. Anthony does not wear blue shirt. Red Amar

Green

Blue

Xxxxxx

Akbar

Xxxxxx

Anthony

Xxxxxx

133. (a) Amar does not wear red shirt, hence it has to be worn by either Akbar or Anthony. In that case both of them eitherwear red shirt or one among green or blue shirt. Now sinceAkbar does not wear green and Anthony does notwear blue shirt, it is conrmed that both of them wearred shirts. So Amar wears either blue or green shirt. Hence statement A is false.

134. (b) Given that two of them wear the same colour, then we will have following sixcombinations: Case (i)

Case (ii)

Case (iii)

Case (iv)

Case (v)

Case (vi)

Amar

Green

Blue

Green

Green

Blue

Blue

Akbar

Red

Red

Red

Blue

Blue

Blue

Anthony

Red

Red

Green

Green

Green

Red

From this we can say that statement (i) is true, while statement (ii) may or may not be true, similarly (iii) and (iv) must not be true. Solution from 135 to 138 We can tabulate the given information as follows;Ms Andhra Pradesh Ms Uttar Pradesh Ms West Bengal Ms Maharastra

Saree

Rank

Position

Yellow

1

1 or 4

Green Red White

2

2 or 3 3 or 2 4 or 1

135. (b) From the table we can say that option (B ) is correct 136. (c) From the table we can say that option (C ) is correct 137. (a) From the table we can say that option (A ) is correct 138. (c) From the table we can say that option ( C) is correct Solution from 139 to 140 139. (b) From the given information, Lawyer – married to D, but D married to A hence A –Lawyer. Account C-married to F Lecturer Since E is not a housewife and there are two housewife in the group hence B has to be housewife, then E is an Architect. 140. (b) A –Male husband of D C-male husband of F Out of remaining B is a housewife, and D ia an architect and it is given that architect is no lady in the group is an architect hence there are three males A, C, D total 3

141. (b) Palace Hut P (Blue & Red)

Fort House Hotel

×

M (Yellow) U (Red or Blue)

×

T (Black)

×

X

×

×

×

×

√

From the table we have eliminated all four except M for Palace, hence M will stay in Palace 142. (c) We can tabulate the given information as followsGiven that B cannot go with R and E, so he must go with O, thus C will be with H. 143. (b) Let the amount with Ashok, Bashir, Chirag, and Deepak as A, B, C. and D, then from the given information : A < 3B C>B D=C–B A = 3D Basheer has total amount = Rs. 500 (since he bought a sweater costing Rs. 600 and borrowed Rs. 100 from Ashok and left with no amount hence amount with him = 600-100 = 500). So Ashok has total amount less than Rs. 1500 (since A < 3B) Ashok has atleastRs 1000 amount. So we can write: 333.34 < D < 500. So Deepak can buy ashawl.

144. (b) Radha, Rupa, Renuka, Ruchika and Ritu are in different weight groups. From the given information Rupa is in group W1 with Sonali, Shubra, Shahira and instructor Amita while Kamal and Tara cannot be with Radha. Soumya and Ruchika are in same group so Soumya cannot be in the group as Radha. Renuka and Rupali are in same group so Rupali and Radha cannot be in the same group. So with this above result it is clear that

no any females except Jyotika and Shweta can be with Radha in same group. Since Jyotika and Shweta are in weight group that has total four members so at least one female must be in this group and who cannot be with Radha. So Radha must be alone in her group and her instructor must be Elina. 145. (d) From the given information we can make a table having information about liking and disliking of various persons.

Fishing

Smoking

M1

√

√

M2

×

√

Drinking √

√

M4 ×

M5 M6

√

M7

×

Mountaineering

×

×

M3

×

√

√

×

×

×

√

M8

Here in this table sign (√ ) indicates a particular minister like that and sign (X) denotes a particular minister dislike the same. It is given that each selected minister should share a liking with at least one of the other three ministers.It is also given that the selected persons must also hate at least one of the likings of any of the other three persons selected.Hence after completing the table we will get by table we get the solution that is option no. (4). 146. (c) From Statement (v): doctor got offer from 3 NIMS, hence choices 1 & 2 are wrong. Statement (iv): D > A and D not equal to 2 [from statement (ii)] Hence D = 3, A = 0. Also Engineer = F since he is not D, S or A. This leaves Samir with 2 calls, The final condition would be



Gambling

NIMs

Education

Ashish

1

C.A

Dhanraj

3

Dr.

Felix

0

Engr.

Sameer

2

Eco.

Hence, only the 3rd statement is necessarily true

√

√

√

×

147. (d) boys

alam sandeep ganesh daljeet Jugraj

money

40

30

20

20

10

spent

10x

1.50 + y

x

y

Z

Ganesh spends 3.50, A spends Rs 35. Hence A must start with Rs 40 and G = 20 (statement iii). Also D = 20 and S = 30 (statement iv). Hence J = 10. Now A = 40 - 35 = 5, hence (i) is wrong. G = 20 – 3.50 = 16.50, hence (3) is wrong. Sandeep cannot spend Rs 29 because D cannot spend Rs 27.50, hence (2) is also wrong. Clearly (4) is the only answer that is possible. 148. (a) From the given statements, we know that Pune = 3, Bangalore = 2, Hyderabad = 1. Loyola is not equal to 1, Convent not equal to Hyderabad, hence not equal to 1 which leaves little flowers = 1 (Hyderabad). Now Loyola not equal to 1 not equal to Pune, hence Loyola = Bangalore = 2, We will get the following result Prize

Name

School

City

1st

Riaz

Little Flower Hyderabad

2nd

Balbir

Loyola

Bangalore

3rd

Bipin

Convent

Pune

EBD_7743

62  Koncepts of Logical Reasoning 

Matrix Arrangement 63 Using others condition we get the chart,

Solutions (149 and 150) (i) Either A or D (ii) B/D-power/Telecom. (iii) C-Home/Finance/nothing (iv) If A get any one ,E must also get any one. 149. (b) Option-1,C cant get Defence.It is not his/her choice. Option 3, A and D cant take potfolio at a time. Option 4, C cant handle Telecom. 150. (d) A-Homes , C-Finance. As A is present D cant take any potfolio. B must take Telecom or Power. Option a, b, c is valid, and 'd' is not valid.

2234

1193

1340

1139

2517

ARC

√

×

×

×

×

CHEL

×

×

×

√

×

DHEN

×

√

×

×

×

HELEN

×

×

√

×

×

SHAH

×

×

×

×

√

154. (b) Helen spent Rs.1340. 155. (a) Rs.1139 is spent by Chellamma. 156. (c) Rs.1193 is spent by Dhenuka. Concept Eleminator (CE)

Solutions (151 to 153)

Solutions (157 to 159)

From the given information we can conclude that(a) Number of vadas eaten by Ignesh must be 6, and he has not take 4 idli (b) Sandeep does not take any chutney (c) Mukesh and Sandeepdoesn’t take chutney. (d) Mukesh eats either 1 or 2 vada. And either 2 or 4 vada. (e) Bimal eats 2 more idli than Ignesh so Bimal eats either 6 or 8 idly, and Ignesh eats 4 or 6 idlies. After completing the table we can conclude that

From statements (vii), (viii) and (ix), we can infer that, Orange P U S R Q Yellow Green

Ignesh Sandeep Daljit Mukesh Bimal

Idlis 6 1 5 4 8

Vadas 6 0 1 2 4

Chutney Yes No Yes No No

151. (a) From the table we can conclude that Daljit eats 5 idlis. 152. (c/d) Here both option c and d are correct 153. (c) The person who eats equal number of vadas and idlis also takes chutney is Ignesh and he is taking chutney. Solutions (154 to 156) Given, We rst select the name and RS. But not in same manner. One girl Spent Rs.1378 more than Chellama. Shanaz got highest Means Rs.2517 .Now present data Lowest value is 1193. Now we subtract from height is = (2517-1378) = 1139chellamma. After selecting the position of Shahnaz and Chellamma ,Least value is 1193.that cant be Helen. If we choose Dhenukato 2234 and 1340,it violates conditions.Helen must be 1340.

Then, it is know that T is opposite to S and the colour of S is red. From (x), the colour of house P is white; hence we can infer that the colourof house T is blue. Hence we can conclude that. White Orange Red P U S R Q T Yellow Green Blue Now, from given information (v) and (x), we get T > S, Q > P > R Now frominformation(vi), we will get U is the shortest. Hence we can arrange houses in the descending order of their heightsare T, S/Q, Q/S, P, R, U. 157. (b) From the above result we can conclude that the tallest house is T and its colour is blue. 158. (d) From the above drawn result, R is yellow coloured house and the house diagonally opposite R is S. Then the colour of house S is Red. 159. (e) From the result above, the second tallest house is either S or Q. Solutions (160 to 163) From Statement A :-Yellow was the only colour used in all the four rooms. It was used at least once for walls, carpets and curtains From statement B, Dining and bed room have same set of colours.

EBD_7743

64 Koncepts of Logical Reasoning Walls

Carpet

Curtain

Living Room Study Room From statement C, The same colour was chosen for the curtains in the bedroom, the carpet in the living room and the walls in the dining room. And that colour is not used in Study Room, hence that colour must not be yellow as it is used in all the four rooms. And must not be grey as it is used only twice and that too for curtains (From statement E) Walls

Carpet

Curtain

Case (i) Walls

Carpet

Curtain

Green /Yellow

Yellow/Green

Grey

This combination must belongs to either Living or Study rooms. If it belongs to living rooms then

Living Room

Study Room From statement D:- The only room with both green and grey in its colour scheme had carpet of the same colour as in the dining room. If a room that has green and grey colour then its 3rdcolour must be Yellow as it is used in all the rooms. That room must not be Dining and Bed room as they have same set of colours. The Carpet of this room must not be of grey colour (From statement E).

Carpet

Curtain

Green

Yellow

Grey

But this combination is ruled out as from condition C the common colourcan not be Yellow.

Living Room Living Room

Walls

Walls

Carpet

Curtain

Yellow

Green

Grey

But in this case from statement D Dining room will have Walls and carpet with green colour, but it is not possible as repetition of colour is not allowed. Case (ii) So only possibility is that yellow grey green belongs to study room. And then the common colour mentioned in statement C is Orange colour, hence we have two possibilities-

Living Room

Carpet Curtain Explanation From Dining statement Orange Yellow Grey Room D, Curtains colour is grey Bed Room Orange The 3rd colour in Living Room Green Orange Yellow curtain is Yellow

Study Room

Study Room Green

From statement F:-The study room walls were painted the same colour as the living room walls. Walls

Carpet

Curtain

Dining Room Bed Room

Walls

Yellow

Grey

Now from D, E and F, we have two cases This case is ruled out as Dining and bed room has same combination hence Grey can not be used other than Curtain. OR Walls

Carpet

Curtain

Dining Room

Orange

Green

Yellow

Bed Room

Green

Yellow

Orange In carpet Yellow must be used

Living Room

Yellow

Orange

Grey

Study Room

Yellow

Green

Grey

Carpets Green colour is from Statement D, & Yellow must be present (Step 1)

(As Grey is used at least twice step 2)

Matrix Arrangement 65 160. (c) Using above matrix, we can say Bed room 161. (d) Green carpets: 162. (a) 163. (b) Dining room- yellow curtains. Solutions (164 to 168) Anand

Bose

Chandrashekharan

Deshpande

Promila

Xxxx

xxxx

xxxx

√ (From statement)

Quadir

Xxxx

√

xxxx

Xxxx

Finance

3rd Systems (As it was last)

Rita (1st From statement (i) )

Xxxx (from 6th statement)

Xxxx

Marketting

Sridhar

Xxxx

xxxx

HRM (from statement)

5th

Finance (From 2nd statement) 164. (a) From the table Rita’s project is in marketing. 165. (c) From the table Prof. Deshpande’s area of expertise is Systems. 166. (d) From the table The second presentation could be in the area of Finance or HRM. 167. (a) From the table Prof. Bose supervised Quadir’s project. 168. (b) From the table Prof. Anand’s area of expertise could be HRM or Marketing. Solutions (169 to 173) Given that each Professor has to coordinate one course and support another course L’s subjects are Finance and IT which are also the subjects of O. O coordinates IT. P and Q have marketing as one of their subjects. P coordinates marketing and Q supports it. Three lectures have Finance as their coordinating subjects. M and N are the professors who coordinate nance. Only one professor supports strategy. It has to be M.L and O have same subjects. From the given information we can summarize as follows Faculty

Support

Co- Ordinate

L

IT

Finance

M

Strategy

Finance

N

Operations

Finance

O

Finance

IT

P

Operations

Marketing

Q

Marketing

Operations

169. (d) From the table we can say that no one coordinates strategy course. 170. (b) From the table we can say that Strategy course is supported by M 171. (c) From the table we can say that ‘O’ co-ordinate the IT course. 172. (a) From the table we can say that ‘L, M and N’ co-ordinate the Finance course. 173. (a) From the table we can say that Marketing course has only one coordinator and only one support professor. Solutions (174 to 178) From the given information ;There are ve friends Amisha, Binaya, Celina, Daisy and Eshaan. Games :-table tennis (2 Friends) football (1) , cricket (1) and chess (1) Occupations :-industrialists (2) teaching (1) , medicine (1) and engineering (1) Floors :- 2, 4, 5 and one more room. Folloing table can be made using the available information. Name Amisha (N) Binaya

Game Occupation Floor Table Industrialist Tennis

Youngest (1)

Engineer

Celina Daisy (R) Eshan

Age

2

EBD_7743

66 Koncepts of Logical Reasoning Now in this question we will use question 5 as an hint and then will move ahead, we can eliminate option (B), (C) and (D) because Daisy stays on oor 2, so the answer option will be (A). then nal table would be Name

Game

Occupation Floor

Age

Amisha (N) Tennis Industrialist

3

Youngest (1)

Binaya

Chess

3

oldest (5)

Celina

Football

Daisy (R)

Tennis

Eshan

Cricket

Engineer

3 2

4 2

(N) Represents :- National Level Player (R) Represents :- Regional Level Player Since 1 tennis player and one chess player stay on the same oor. Here, Amisha and Binaya are staying on the same oor. Amisha plays tennis and therefore Binaya will play chess. Next we will use question 4here we can conclude that denately one person is between Daisy and Eshan because none of these is not there in the answer option so, the only possibility will be Age wise

D/E

C

E/D

2

3

4

But it is also given that tennis player is between football and the chess player in age, so only possibility will be Age wise

Football

Tennis

Chess

3

4

3

174. (c/d) From the above table, both options (C) and (D) are correct. 175. (c) From the above table, Eshan plays cricket. 176. (c) From the above table, All the answer options are possible except (C). 177. (a) From the above table, Correct answer is option (A). 178. (a) From the above table, Binya is chess player and she is engineer option (A) is correct. Solutions (179 to 181) Let the ve ags is represented by letters P, T, R, S and Q. From statement (ii) T _ _ R _ Then with statement (i) only two cases are possible: Case (i) T P Q R S Case (ii) T Q P R S 179. (b) Using the statement (iii), the only possible case is T Q P R S . hencestatement (iv) is redundant.

Using statement (iv), the only possible case is T Q P R S . hencestatement (iii) is redundant. Therefore, either (iii) or (iv) is redundant. 180. (c) From the possible case T Q P R S , Swan is on the extreme right and Tiger is on the extreme left. 181. (c) From the possible case T Q P R S , Panther is in the middle and Rose is on its right. Solutions 182 to 185: Given, the following information (1) Mridul belongs to Chennai ;Mridul – didn’t married to (Geetika / Veena) (2) Abhishek belongs to Ahmedabad ; Ranjan belongs to Delhi. (3) In Kolkata marriage happens in February. (4) Hema got married April but not at (Ahmedabad) (5) (Geetika / Ipsita) Married in (February / November) at (Chennai / Kolkata) (6) Pritam married in December but not with (Benguluru / kolkata) (7) Salil married to Jasmine in September. From statement (1), (2) and (6), we can say that Pritam got married in December at Mumbai. From statement (7) and (3), we can say that Salil married to Jasmine in September at Benguluru. And Deepak married in Kolkata in February From (1), (3) and (5), and the above statements, we can say that, Mridul married in Chennai with Ipsita in November Now we can complete the table as followsFriends Wives Venue of wedding Month Abhishek Veena Ahmedabad July Deepak Geetika Kolkata February Mridul Ipsita Chennai November Pritam Brinda Mumbai December Ranjan Hema Delhi April Salil Jasmine Benguluru September 182. (c) From the table we can say that Deepak’s wedding took place inKolkata 183. (d) From the table we can say that In Mumbai, the wedding of Pritam took place in the month of December 184. (d) From the table we can say thatIpsita's wedding took place inChennai 185. (c) From the table we can say thatHema's husband isRanjan 186. (a) From the table we can say that Salil's wedding was held inBenguluru 187. (d) From the table none of the combination is correct.

Circular Arrangement 67

3

Circular Arrangement

Circular arrangement is a type of arrangement of questions with more complexity. In this case, apart from normal arrangement, we have different variables/angles that have to be solved. Left or right information is given in both linear and circular arrangements but in circular arrangement for left or right, you can choose either clockwise or anti-clockwise. When people have to be seated in a row or in a circular table, the best way to determine the left hand and right hand position is to imagine yourself seated in that particular place. In this book, we will take right as anti-clockwise and left as clockwise. To solve these types of questions, we should follow the instructions given below(i)

Draw the diagrammatic representation of the arrangement, e.g. draw a circle for a circular arrangement, draw a dashed line for position on a straight line, draw rectangle for rectangular table, etc.

(ii)

Mark the right hand and left hand position of the people keeping in mind the direction in which the individuals are facing.

(iii)

Mark the direction, i.e. North, South, West or facing towards the table or away from the table.

(iv)

Identify the denite clues and locate them rst on circular/rectangular table whatever is given in the question.

(v)

Identify the relative position and locate the same to get an indicative position.

(vi)

Use denite clues in conjunction with relative position clues to arrive at the nal arrangement.

(vii) If people are seated around a table with no direction mentioned, it has to be read as facing towards the table.

EBD_7743

68 Koncepts of Logical Reasoning

1

8.

Directions for Question Nos. 1 to 5:

Four persons A, B, C and D are sitting around a circular table such that A is not opposite to B, and C is not opposite to D then 1.

If A is sitting right of B then(a) C is left of D (b) D is left of C (c) B is opposite to C (d) None of these

9. Which one of the following is not correct? (a) If A is male then E must be male. (b) If E is male then C must be male. (c) If E is male then F must be male. (d) None of these

2.

If B is opposite to C then(a) B is left of D (b) D is left of B (c) A is opposite to D (d) None of these

3.

If C is equidistant from B and D then(a) A is opposite to D (b) D is left of C (c) B is opposite to C (d) None of these

4.

If A is not sitting opposite to D then(A) C is opposite of D (b) D is opposite of B (c) B is opposite to C (d) None of these

5.

Total number of arrangements is(a) 2 (c) 4

ways

to

make

such

(b) 3 (d) None of these

Directions for Question Nos. 6 to 10: In a family gathering, 3 couples are sitting on a circular dining table such that husband and wife are sitting just opposite to each other, and no 2 females are sitting together. Following further information is known(i) A is sitting adjacent to C and D (ii) B is sitting adjacent to E and F 6.

7.

If E and D is husband wife pair, which one of the following is correct? (a) C and F is husband wife pair. (b) C and B is husband wife pair. (c) C and A is husband wife pair. (d) None of these

If C’s husband is a doctor, which one of the following is correct? (a) E must be a doctor. (b) D’s husband is E. (c) B is sitting adjacent to two males. (d) None of these If E is adjacent to both B and C, which one of the following is not always correct? (a) E and D is husband wife pair. (b) A and B is husband wife pair. (c) A, E and F are males. (d) None of these

10. Which one of the following could be correct pair of husband and wife? (a) (A, B), (C, D) (b) (C, D) (E, F) (c) (C, E) (D, F) (d) None of these Directions for Question Nos. 11 to 15: In a quiz session at Praxis Business School, Kolkata, Zeeshan was the 1st student to enter the classroom. He was guided by the co-ordinator to take a seat from 8 seats around a circular table. Sahana was the 2nd student who has taken a seat such that she is at maximum distance from Zeeshan. Rajesh was the 3rd person who has taken seat such that he is at maximum distance from both Zeeshan and Sahana. This process continues till the 7 th student entered. 11. If Rita is the 5th student who entered the class and she is just opposite of Hena then Hena entered at which place? (a)

6th

(b) 4th (c)

7th

(d) Cannot be uniquely determined 12. If Rita and Hena are sitting opposite to each other, what could be the position of Rita if she didn’t enter at the 5th position? (a)

4th

(b) 6th (c)

7th

(d) Cannot be uniquely determined

Circular Arrangement 69 13. If Rita is sitting right of Zeeshan but left of Hena who is sitting left of Sahana then which of the following seat remains empty? (a) Opposite to Rajesh (b) Opposite to Zeeshan (c) Opposite to Rita (d) None of these 14. If Rita entered just after Hena but just before Rena, which one of the following could be true? (a) Hena is sitting just opposite to Sahana. (b) Hena is sitting just opposite to Rajesh. (c) Rita and Rena are sitting adjacent to each other. (d) none of these 15. If Rita is the 5th student who entered the class and she is just opposite of Raj then Raj entered at which place? (a) 6th (b) 4th (c) 7th (d) Cannot be uniquely determined Directions for Question Nos. 16 to 18: Every day Miss Yadav, Miss Sharma, Miss Toppo and Miss Hussain go to a park for morning walk. One day, they reach the gate of the park at the same time and immediately start walking on the only circular track adjacent to the gate. Miss Yadav, Miss Toppo and Miss Hussain go on a clockwise direction while Miss Sharma goes anti-clockwise. Miss Hussain who is asthmatic, is slowest among the four and soon others move away from her. Like every day, she could walk only one round taking almost the same time as others to complete their morning walk. After her walk, Miss Hussain read the following instruction written at the gate while others join her one after another. “Walkers are requested to use only the 500 m walking track. Plucking of owers and leaves is strictly prohibited. The park will remain closed from 6 pm to 5 am. ”While walking, Miss Yadav overtakes Miss Hussain twice; once near the fountain and the next time at the signature rock. Miss Toppo and Miss Sharma cross her three times. 16. What is the total distance covered by Miss Sharma and Miss Toppo together? (a) 3500 m (b) 4000 m (c) 2500 m (d) 3000 m 17. How many times Miss Yadav and Miss Sharma cross each other on the track? (a) Twice (b) Three times (c) Four times (d) Five times

18. How many times Miss Toppo would overtake Miss Yadav? (a) Never (b) Once (c) Twice (d) Three times Directions for Question Nos. 19 to 21: Seven workmen named Amrita, Bhushan, Ronnie, Deepak, George, Prem and Gunjan are sitting in a circle in a garment fabrication factory. Amrita, Bhushan, Ronnie, Deepak, Prem and Gunjan are sitting at equal distances from one another. Ronnie is sitting two places to the left of Prem, who is sitting one place to the right of Deepak. Amrita forms an angle of 90 degrees from George and an angle of 120 degrees from Bhushan. Bhushan is just opposite to Gunjan, who is sitting to the left of George. 19. Who is the only person sitting between Ronnie and George? (a) Prem (b) Deepak (c) George (d) Amrita 20. George is not sitting at equal distance from: (a) Deepak and Amrita (b) Ronnie and Prem (c) Bhushan and Prem (d) None of these 21. Which of the following statements is not correct? (a) Prem is between Bhushan and Amrita. (b) Bhushan is two places away from Gunjan. (c) George is sitting opposite to Prem. (d) None of these Directions for Question Nos. 22 to 25: There are two circular concentric rings, with 26 sectors on each ring marked A to Z. Sectors on outer and inner circles match in a perfect size t. Whereas the outer ring contains the sectors marked clockwise, the inner ring has sectors marked anticlockwise, such that sectors marked A on both the circles are coinciding at present, B on the outer circle coincides with Z on the inner circle and so on. 22. Apart from alphabet A, how many more alphabets coincide at present? (a)

1

(b)

2

(c)

3

(d) None of these

EBD_7743

70 Koncepts of Logical Reasoning 23. How many vowels on the sectors in the outer circle have vowels on the corresponding sectors on the inner circle? (a)

1

(b)

2

(c)

3

(d) None of these

24. If the inner circle is rotated anti-clockwise by one sector from the original position, how many alphabets will coincide on both the circles? (a) 1

(b)

(c)

(d) None of these

3

2

25. If the outer circle is rotated clockwise by two sectors from the original position, how many alphabets will coincide on both the circles? (a) 1

(b)

(c)

(d) None of these

3

2

Directions for Question Nos. 26 to 28: Arijit, Biplab, Chintan, Debashish, Elangovan, Frederick, Gautam and Himadri are sitting around a circular table. Some information about the order in which they are sitting is available as follows: (i) Debashish is sitting opposite to Himadri and to the immediate right of Gautam. (ii) Elangovan is sitting to the immediate right of Biplab. (iii) Arijit is sitting opposite to Chintan who is not immediately next to Frederick on either side. [SNAP 2010] 26. Who is sitting to the immediate right of Himadri? (a)

Arijit

(b)

Debashish

(c)

Elangovan

(d) Frederick

27. Who is sitting opposite to Biplab? (a)

Arijit

(b)

Debashish

(c)

Frederick

(d) Himadri

28. Who is to the immediate right of Chintan? (a) Arijit

(b)

(c)

(d) Himadri

Elangovan

Biplab

Directions for Question Nos. 29 to 33: This group of questions is based on a set of conditions. In answering some of the questions, it may be useful to draw a rough diagram. Choose the response that most accurately and completely answers each question. A circular eld, with inner radius of 10 meters and outer radius of 20 meters, was divided into ve successive stages for ploughing. The ploughing of each stage was handed over to a different farmer. (i) Farmers are referred to by the following symbols: F1, F2, F3, F4 and F5. (ii) The points between different stages of project are referred to by the following symbols: P1, P2, P3, P4, and P5, not necessarily in the order. (iii) Farmer F5 was given the work of ploughing stage starting at point P4. (iv) The stage from point P5 to point P3 was not the rst stage. (v) Farmer F4 was given the work of the fourth stage. (vi) Stage 3 nished at point P1, and the work of which was not given to farmer F1. (vii) Farmer F3 was given work of stage ending at point P5. [XAT 2009] 29. Which was the nish point for farmer F2? (a) P1 (b) P2 (c) P3 (d) P4 (e) P5 30. Which stage was ploughed by farmer F5? (a) First (b) Second (c) Third (d) Fourth (e) Fifth 31. Which were the starting and nish points of stage 2? (a) P2 and P5 (b) P5 and P3 (c) P3 and P1 (d) P5 and P4 (e) P3 and P2 32. For which farmer was P2 a nishing point? (a) F1 (b) F2 (c) F3 (d) F4 (e) F5 33. Which was the starting point for farmer F3? (a) P2 (b) P3 (c) P4 (d) P1 (e) None of these

Circular Arrangement Directions for Question Nos. 34 to 37:

Directions for Question Nos. 38 to 42:

A leading socialite decided to organize a dinner and invited a few of her friends. Only the host and the hostess were sitting at the opposite ends of a rectangular table, with three persons along each side. The prerequisite for the seating arrangement was that each person must be seated such that at least on one side it has a person of opposite sex. Maqbool is opposite of Shobha, who is not the hostess. Ratan has a woman on his right and sitting opposite of a woman. Monisha is sitting to the hostess’ right, next to Dhirubhai. One person is seated between Madhuri and Urmila, who is not the hostess. The men were Maqbool, Dhirubhai and Jackie, while the women were Madhuri, Urmila, Shobha and Monisha. [CAT 1994]

Twelve people Abhishek, Binit, Chand, Dhiraj, Eshita, Fatima, Garima, Hena, Ishan, Jatin, Kamal and Lalit are sitting around a rectangular table. The following information is knownThe table has 12 chairs numbered from 1 to 12. 6 seats on one side of the table and 6 on the opposite side. The chairs are arranged in such a way that chair number 1 is just opposite to 12, 6 is opposite to 7 and so onAbhishek is sitting opposite to Kamal who is the only person sitting between Chand and Jatin. Eshita is sitting opposite to Ishan who is the only person sitting between Binit and Lalit. Fatima, sitting at chair number 1, is diagonally opposite to Chand who is sitting opposite to Dhiraj.

71

34. The eighth person present, Jackie, must be: (i)

the host

(ii) seated to Shobha’s right (iii) seated opposite of Urmila (a) i only (b) iii only (c) i and ii only (d) ii and iii only 35. Which of the following persons is denitely not seated next to a person of the same sex? (a) Maqbool (b) Madhuri (c) Jackie (d) Shobha 36. If Ratan would have exchanged seats with a person four places to his left, which of the following would have been true after the exchange? (i) No one was seated between two persons of the opposite sex (e.g. no man was seated between two women.) (ii) One side of the table consisted entirely of persons of the same sex. (iii) Either the host or the hostess changed their seats. (a) i only (b) ii only (c) i and ii only (d) ii and iii only 37. If each person is placed directly opposite of his/her spouse, which of the following pairs must be married? (a) Ratan and Monisha (b) Madhuri and Dhirubhai (c) Urmila and Jackie (d) Ratan and Madhuri

38. If Garima is sitting opposite to Fatima then who is sitting opposite to Hena? (a) Lalit (b) Binit (c) Ishan (d) Uniquely not determined. 39. If Lalit is sitting opposite to Hena, then who is sitting opposite to Garima? (a) Eshita or Fatima (b) Jatin or Fatima (c) Jatin or Eshita (d) None of these 40. How many persons are sitting between Binit and Dhiraj, if they are on the same side of the table? (a) 2 or 3 (b) 1 or 2 (c) 1 or 3 (d) None of these 41. Which one of the following is correct? (a) Lalit is sitting at seat number 12 (b) Lalit is sitting at seat number 10 (c) Kamal is sitting at seat number 8 (d) None of these 42. Which one of the following is incorrect? (a) Lalit is opposite to Jatin. (b) Jatin is opposite to Hena. (c) Lalit is adjacent to Chand. (d) None of these Directions for Question Nos. 43 to 45: Six students A, B, C, D, E and F are sitting around a hexagonal conference table, such that A is not opposite to B, C is not opposite to D, and E is not opposite to F.

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72 Koncepts of Logical Reasoning 43. If D is opposite to C, then which one of the following sentences is correct? (a) B must be opposite to F. (b) E must be opposite to B. (c) B must be opposite to either E or F. (d) None of these 44. If A is sitting between E and F, then who is sitting opposite to A? (a) D (b) C (c) C or D (d) None of these 45. If A is not sitting opposite to E and F, then which one of the following is correct? (a) B is sitting opposite to E. (b) B is sitting opposite to F. (c) E is sitting opposite to F. (d) None of these Directions for Question Nos. 46 to 50: Some people enter a conference room and get seated around a circular table. The following is known about them. (i) The rst to enter sits opposite to one who belongs to Durgapur but right of Ranchi. (ii) Teacher belongs to Kolkata and is right to IT professional. (iii) The guy from Durgapur is adjacent to teacher and architect and opposite to doctor. (iv) The guy from Ranchi is opposite to architect who is not from Dharbhanga.v) The doctor belongs to Chennai and sits adjacent to those who came at 3rd and 6th positions.

(vi) The guy from Burdwan is adjacent to guys who came at 3rd and 5th positions. (vii) Professor who didn’t come at 5th position is equidistant from guys from Ranchi and Durgapur. Now answer the following questions.

46. Professor is opposite to(a) Teacher (b) IT professional (c) Dancer (d) None of these 47. Guy from Burdwan is adjacent to(1) Professor and architect (2) Guys from Dharbhanga and Durgapur (3) Architect and IT professional (a) Only 1 (b) Only 2 (c) Only 2 and 3 (d) None of these 48. The guy sitting opposite to guy from Ranchi came at which position? (a) 2nd (b) 4th (c) 5th (d) None of these 49. The dancer is from(a) Ranchi (c) Durgapur

(b) Dharbhanga (d) None of these

50. Which line in the question is redundant. (a) Line 5 (b) Line 7 (c)

Line 6

(d) None of these

Circular Arrangement 73 

Solutions Concept Applicator (CA) Solutions (1 to 5) Since A is not opposite to B so opposite to A can be either C or D. If opposite to A is C then we have two cases: B is left or right of A, and so on we have total 4 cases as shown below. A A

B

D

D

A C/V

D/C

E/F

F/E B

B

Case (i) C Case (i)

C Case (ii)

A

A

B

C

D Case (iii)

C

B E/F

B

C/O

D/C A

D Case (iv)

1. (d) Out of the four cases given condition is represented by case (ii) and case (iv). From this, we can say that none of the given options is correct. Option (a) and (c) are not definitely true. 2. (c) Out of the four cases, given condition is represented by case (iii) and case (iv). From this, we can say that option (c) is correct. 3. (d) Out of the four cases, given condition is represented by case (i) and case (ii). From this, we can say that option (d) is correct. 4. (b) Here case is similar to case (i) or (ii). From this, we can say that option (b) is correct. 5. (c) Solutions (6 to 10) From the given condition, it is known that A is male or female so we have two cases: Case (i) A is a male Case (ii) A is a female These two cases can be shown as followsIn the diagram, rectangle represents male and circle represents female.

F/E

Case (ii) 6. (c) From the given condition case is similar to case (i); and in this case C, B and D are female. Hence, B is sitting adjacent to two males. 7. (c) From the diagram, we can say that option (c) is not always correct. 8. (a) If E and D is a husband wife pair, then C and F is also a husband wife pair. 9. (b) From the diagram, we can say that option (b) is not always correct. 10. (c) From the diagram, we can say that option (c) is not always correct. Solutions (11 to 15) Zeeshan 4th

Rajesh

Sahana

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74 Koncepts of Logical Reasoning

11. (a) If Rita is the 5th student who entered the class and she is just opposite of Hena, then Hena must have entered at the 6th position. 12. (b) From the given condition, Rita’s position is 6th. 13. (c) From the given condition, option (c) is correct. 14. (b) Option (b) that says Hena is sitting just opposite to Rajesh could be true if Hena entered at the 4th position. 15. (a) If Rita is the 5th student who entered the class and she is just opposite of Rajesh, then Rajesh must have entered at the 6th position (This question is same as question 11).

Solutions (19 to 21) According to the information given in question the sitting order will be Left B

Right P

R

A D

19. (b)

20. (c)

George

G

21. (c)

Solutions (22 to 25)

Solutions (16 to 18) It is given that walking track is 500 m and Miss Hussain takes only one round. The given information is as follows: (i)

Miss Yadav overtakes Miss Hussain twice and both are running in the clockwise direction, so Miss Yadav must have walked three times on the same track.

(ii) Miss Toppo crosses Miss Hussain three times and both are running in the same direction, so Miss Toppo must have walked four times on the same track. (iii) Miss Sharma crosses Miss Hussain three times, so she must have walked three times on the same track, as both are running in the opposite directions. Hence from this, we can conclude that the distances covered by all fours are as follows:Miss Hussain 500 m Miss Yadav - 1500 m

22. (a) 23. (a) 24. (d) 25. (b) It can be seen that Z and M coincide on both the circles. Solutions for questions 26 to 28: Let us represent the persons with the rst letter of their names. The nal arrangement should be as follows:

Miss Toppo - 2000 m Miss Sharma - 1500 m 16. (a) From the above result, total distance covered by Miss Sharma and Miss Toppo = 1500 m + 2000 m = 3500 m. 17. (d) From the above result as Miss Yadav and Miss Sharma both covered 1500 meter, but in opposite directions, so both will meet ve times on the track. 18. (c) From the above result, Miss Toppo and Miss Yadav covered 2000 meter and 1500 meter, respectively in the same direction. So, Miss Toppo will overtake Miss Yadav twice.

26. (a) Arijit is to the immediate right of Himadri. 27. (c) Frederick is sitting opposite to Biplab. 28. (b) Biplab is to the immediate right of Chintan.

Circular Arrangement 75 Solutions for questions 29 to 33: Given that there are ve stages of the project from stage 1 to stage 5 on the eld with starting and ending points from amongst P1, P2, P3, P4 and P5. Now given information is as follows: (i)

Stage 3 nished at P1.

(ii) Stage 4 must have started from point P1. (iii) Fourth stage work was given to farmer F4; hence, from all these outcomes we can conclude that 2nd stage is from P5 to P3.

36. (a) If Ratan would have exchanged seat with a person four places to his left, i.e. Sobha (i) will follow. 37. (a) From the options given only Ratan and Monisha are sitting opposite to each other, hence they must be married. Solutions for questions 38 to 42: Let us denote these 12 students by their 1st letter of name, like Abhishek is A and so on.

Stage 2- From point P5 to point P3

From the given information we can conclude that (C) and (D) are at seat numbers 7 and 6, respectively. And (K) is the only person between (C) and (J) while (A) is opposite to (K). Hence, (A), (K) and (J) must be at seat numbers 5, 8 and 9. respectively.

Stage 3- From point P3 to point P1

Then we have following two cases:

Stage 4- From point P1 to point P4

Case I

Hence we can conclude thatStage 1- From point P2 to point P5

Stage 5- From point P4 to point P2

G or H H or G

12

11

E

J

K

C

10

9

8

7

Now it is given thatFarmer F3 → rst stage. Farmer F4 → fourth stage (from 5th information).

1

2

3

4

5

6

Farmer F5 → fth stage (given that work of stage starting at point P4 is given to farmer 5).

F

L or B

I

B or L

A

D

L or B

I

L or B

J

K

C

12

11

10

9

8

7

Farmer F1 → second stage. Stage 1- From point P2 to point P5 → Farmer F3

Case II

Stage 2- From point P5 to point P3 → Farmer F1 Stage 3- From point P3 to point P1 → Farmer F2 Stage 4- From point P1 to point P4 → Farmer F4 Stage 5- From point P4 to point P2 → Farmer F5 29. (a) Point P1 30. (e) From the above result, we get that the fth stage was ploughed by farmer F5. 31. (b) From the above result we get that the starting and ending points of stage 2 are P5 and P3. 32. (e) From the above result, we get that the P2 was the nishing point for farmer F5. 33. (a) From the above result, we get that the starting point for farmer F3 was P2. Solutions for questions 34 to 37: 34. (c) From the information given in the question, it is clear that Jackie is the host and is sitting to Sobha’s right. 35. (d) Sobha is sitting next to Jackie and Dhirubhai. So, she is the only person who is not seated next to a person of the same sex.

1

2

3

4

5

6

F

E

G or H

H or G

A

D

38. (d) From the above 2 cases, it follows case (i) and opposite to Fatima is either Lalit or Binit. 39. (b) From the above 2 cases, In case (i) if Lalit is sitting opposite to Hena then Fatima is sitting opposite to Garima. In case (ii) if Lalit is sitting opposite to Hena then Jatin is sitting opposite to Garima. 40. (c) From the above 2 cases, it follows case (i) and number of persons sitting between Binit and Dhiraj is either 1 or 3. 41. (c) From the given options only option (c) is correct. 42. (c) From the given options option (c) is correct. 43. (c) From the given conditions C and D are opposite to each other and B can’t be opposite to A; hence he must be opposite to either E or F.

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76 Koncepts of Logical Reasoning 44. (c) From the given information opposite to A can be either C or D.

45. (d) From the given condition we cannot be sure about options A, B and C.

Solutions for questions 46 to 50: Durgapur

Ranchi

Kolkata

Darbhanga

Burdwan

Chennai

Xxxx Xxxx

√ Xxxx

Xxxx Xxxx

Xxxx

Architect

Xxxx Xxxx

Xxxx Xxxx

Doctor

Xxxx

IT professional

√

Xxxx Xxxx

Xxxx Xxxx

Xxxx Xxxx

……

Xxxx Xxxx

√ Xxxx

Xxxx Xxxx

Xxxx

Teacher

Professor

46. (a)

47. (b)

48. (d)

√

49. (a)

√ Xxxx Xxxx Xxxx Xxxx

50. (d)

√ Xxxx Xxxx Xxxx

Set Theory

Set Theory

4

77

INTRODUCTION A set is a collection of objects called elements of the set. So, it can be referred to as collection of elements also. In other word we can say that a set is a collection of well-dened objects.

Let’s consider a set P dened as prime numbers less than 10 so P = {2, 3, 5, 7} and number of elements in P is 4. Methods of Describing Sets Roster method or Tabular method or Listing method:- It is a complete or implied listing of all the elements of the set or stating the pattern in case of more elements. Example: A = {a, b, c, d} here number of elements = 4 B = {a, e, i, o, u} here number of elements = 5 Or in cases where listing of all the elements is not possible it is written as: W = {0, 1, 2, 3, …..} E = {2, 4, 6, 8, …….100} here number of elements = 50 Set builder form:Example: A = {x/x N, 10 < x < 100} it is read as x such that x is an element of natural numbers greater than 10 and less than 100. Example: E = {x|x is even} This is read as x such that x is divisible by 2.

Types of Sets Finite and Innite set:- If the elements of a set can be counted or they are denite then the set is called nite set, otherwise innite set. Example: A = {1, 2, 3, 4, 5, ..., 100} natural numbers less than 100, here number of elements in this set is 100, it can also be represented as n(A) = 100

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78 Koncepts of Logical Reasoning If there is no end to the number of members or elements of a set then the set is called innite set. Example: B = {2, 4, 6, 8, ...} multiples of 2, here upper end is not dened hence number of elements in set B is innite. Singleton set: Any set with a single element is called singleton set. Example: A = {0}, B = {odd number between 68 and 70} Empty set or Null set or Void set:- Any set with no elements or members is called empty set or null set or void set. Example: A = {x|x is two digit even prime number}, [actually no such element exists] B = {even number between 42 and 44}, [actually no such element exists] Number of elements is 0 it is also represented as {φ} Equality of Two Sets: Two sets are equal if they have precisely the same members. Example: A = {x|x is even number less than 10 but more than 0} And B = {2, 4, 6, 8} Here each element of set A is also element B and vice versa hence we can say that Set A = Set B. Subsets: A is a subset of B if and only if every element of A is also element of set B. Example: A = {2, 4, 6, 8, 10} and B = { 1, 2, 3, ….. 10} So, we can see that every element of A is present in B. This is written symbolically as A⊆ B. Example: A = {2, 4, 6} and B = {4, 2, 6} Here, A ⊆ B and B ⊆ A and in this case Set A = B

Proper Subset:- If every element of set A is the element of set B and there exists at least one element in B that does not belong to A, then A is said to be the proper subset of B denoted by A ⊂ B. Example: A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6} then A ⊂ B.

Universal Set:- All the sets under study could be the subsets of a xed set, which is called the universal set. Example: A set of natural numbers will be the universal set for the set of even numbers. Equivalent Sets: If 2 sets contain the same number of elements then they are called equivalent sets. Example: A= {1, 2, 3, 4} and B = {a, b, c, d}. Here, the two sets are equivalent sets. Symbolically written as, A ≡ B or n(A) = n(B) Disjoint Sets: If two sets have no elements in common then the two sets are said to be disjoint sets. Example: A = {1, 3, 5, 7, 9} and B = {2, 4, 6, 8} Very Important Relation!!!! Power Set:- The family of all the subsets of any set A is called the power set of A. The power set is denoted by P(A). If the set A has n elements then the power set of A will contain 2n elements. Example: A = {1, 2, 3} then, n = 3 and therefore the number of subsets of A will be 23 = 8. The power set of A is P(A) = { {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}, Φ} Another approach is a subset of A must have either 0 or 1 or 2 or 3 elements and number of ways we can select subset is 3C0 + 3C1 + 3C2 + 3C3 = 23 = 8 Comparable sets:-Two sets A and B are said to be comparable if either A ⊂ B or B ⊂ A.

Example: A = {1, 2, 3} and B = {1, 2, 3, 4} are comparable but the sets C = {d, e, f} and D = {d, e, g, h, i}, here the sets C and D are not comparable.

Set Theory

Operations of Sets

79

Union of Sets:- Union of two or more sets is the set of all elements that belong to any one of the sets or present in both of them. The symbol used for union of sets is ‘∪’, i.e. A∪B = Union of set A and set B. Example: A = {1, 2, 3, 4} and B = {2, 4, 5, 6} then A∪B = {1, 2, 3, 4, 5, 6} Properties of Union: A∪B=B∪A

A∪A=A

A ∪ U = U, where U is the universal set A ∪ Φ =A

A ⊂ A ∪ B and B ⊂ A ∪ B

If B ⊆ A, then A ∪ B = A

A ∪ (B ∪ C) = (A ∪ B) ∪ C, this is known as associative law.

Representation of union of sets:-

Intersection of Sets: It is the set of all the elements, which are common to all the sets. The symbol used for intersection of sets is ‘∩’, i.e. A ∩ B = {x| x ∈A and x ∈B} Example: If A = {1, 2, 3, 4} and B = {2, 4, 5, 6} then A ∩ B = {2, 4}

Properties of Intersection: A∩B=B∩A A∩A=A If A and B are disjoint sets then A ∩ B = Φ A∩Φ=Φ (A ∩ B) ⊆ A and (A ∩ B) ⊆ B

A ∩ (B ∩ C) = (A ∩ B) ∩ C, i.e. associative law Difference of Sets: The difference of set A to B denoted as A – B is the set of those elements that are in the set A but not in the set B, i.e. A – B = {x| x ∈ A and x ∉ B}.

Similarly B – A = {x: x ∈ B and x ∉ A}

Example: If A = {1, 2, 3, 4} and B = {3, 4, 5, 6, 7} then A–B = {1, 2} and B–A = {5, 6, 7}.

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80 Koncepts of Logical Reasoning Properties of Difference of Sets: A–B≠B–A A – B ⊆ A and B – A ⊆ B

A-Φ=A

(A – B) ∪ B = A ∪ B and (A – B) ∩ B = Φ

Complement of a Set:

The complement of a set A is the set of elements which do not belong to A; in other words, U – A. Symbolically, denoted as, Ac or A’ A’ = {x|x∈ U, x ∉ A} = U – A

Approach of Problem with 2 sets:Let the two sets be A and B. Here Set A = a + c Set B = b + c In both set A and set B = c Neither in set A nor in set B = n Universal set U = a + b + c + n Example: If out of 100 students in a school 70 likes Chocolate, 60 likes Tea, while 10 students like neither Tea nor Chocolate then what is the number of students who like both? Solution: Let two sets are A as those who like Chocolate and B those who like Tea Here Set A = a + c = 70 Set B = b+c = 60 In both Set A and set B = c Neither in set A nor in set B = n = 10 Universal set U = a+b+c+n Or 100 = 70 + b+ 10 or b = 20 then c = 40 and a = 30 Hence we can conclude followingThere are 30 students who like only chocolate (means like Chocolate but not tea) There are 20 students who like only tea (means like tea but not Chocolate) There are 40 students who like both. Approach to solve problem with three sets:Lets understand it with an example:Example: In a society 50 families like TOI, 60 Family like Telegraph While 40 family like HT. if 10 family like none of them while the same number of family like all three newspapers. If 15 of them like TOI and Telegraph, 20

Set Theory of them like HT and telegraph while 15 of them like TOI and HT, thenA) How many of them like only TOI B) How many of them like only Telegraph C) How many of them like only HT D) How many of them like TOI but Not HT E) How many of them like TOI and HT but not Telegraph F) How many of them like exactly one news paper G) How many of them like exactly two news papers. H) How many of them don’t like TOI I) How many of them don’t like HT K) How many of them don’t like HT and TOI L) Total number of families in the society. From the given condition Venn Diagram is like as given belowSince 50 students like TOI hence a+ d + e + g = 50 Similarly 60 students like Tel hence b + d + f + g = 60 And 40 students like HT hence c + e + f + g = 40 Since 10 family like none of them hence n = 10 And 10 family like all three hence g = 10 15 of them like TOI & Tel hence d + g = 15 but g =10 hence d = 5 20 of them like HT and Tel hence f + g = 20 but g = 10 hence f = 10 15 of them like TOI and HT hence e + g = 15 but g =10 hence e =5 Since a+d+e+g = 50 or a+5+5+10 =50 or a = 30, similarly b = 35 and c = 15 A) How many of them like only TOI :- =a = 30 B) How many of them like only Telegraph:- b = 35 C) How many of them like only HT :- c = 15 D) How many of them like TOI but not HT :-a+d = 30 +5 = 35 E) How many of them like TOI and HT but not Telegraph :- e = 5 F) How many of them like exactly one news paper:-a+b+c = 30+35+15 = 80 G) How many of them like exactly two news papers :-d+e+f = 5+10+5 = 20 H) How many of them don’t like TOI :- b +f+c+n = 35+10+15+10 = 70 I) How many of them don’t like HT = a + d + b + n = 30 + 5 + 5 + 10 = 50 K) How many of them don’t like HT and TOI:- b + n = 35 + 10 = 45 L) Total number of families in the society :-a + b + c + d + e = f + g + n = 120. Approach to solve problem with four sets:Let us understand with an example:- If there are 4 types of news paper then Venn Diagram will be as followsNumber of persons reading TOI :- a + e + g + k + l + o + i + n Number of persons reading HT = e + b + k + h + o + m + n + j Number of persons reading Hindu = g + k + h + c + l + o + m + f Number of persons reading ET = l + o + m + f + i + n + j + d

81

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82 Koncepts of Logical Reasoning Number of persons reading both TOI and HT = e + k + o + n Number of persons reading both TOI and Hindu = g + k + l + o Number of persons reading both TOI and ET = l + o + i + n Number of persons reading both HT and Hindu = k + h + o + m Number of persons reading both HT and ET = o + m + n + j Number of persons reading both Hindu and ET = l + o + m + f Number of persons reading TOI, HT and Hindu = k + o Number of persons reading TOI, HT and ET = n + o Number of persons reading TOI Hindu and ET = l + o Number of persons reading HT, Hindu and ET = m + o Number of persons reading all the three = o Number of persons reading exactly one news paper = a + b + c + d Number of persons reading exactly two news paper = e + g + i + h + j + f Number of persons reading exactly three news paper = k + l + m + n

Example: (Task for students) If A = {1, 2, 3, 4}, B = {3, 5, 7}, then the set (A – B) È (B – A) would be Consider the sets dened below and answer the questions that follow: Universal set U = {Positive real numbers less than 100} A = {x ЄN I x is divisible by 2} B = {3, 6, 9, 12...} C =The set of all natural numbers divisible by 5 D= The set of all perfect squares. E= The set of all perfect cubes. Then nd the followings:n(A n B)

(B) n(A n D)

(C) n(A U B)

(D) n( C n D/E)

(E) n(A U B/C)

(F) n (A U E’)

Set Theory

1

Directions for Question Nos. 1 to 5

Consider a universal set U = {1, 2, 3, 4…..10} Set P = {x | x is a prime number} Set (E) = {2, 4, 6, 8, 10} Set (O) = { 1, 3, 5, 7, 9} Set (C) = {x | x is a composite number} 1.

How many elements does P’ or Pc has, or nd n(P’) (a) 3 (b) 5 (c) 6 (d) None of these

2.

Find the number of elements in E – C, (a) 3 (b) 2 (c) 4 (d) None of these

3.

Find the number of elements in P U O (a) 3 (b) 2 (c) 4 (d) None of these

4.

Find the number of elements in P n O (a) 3 (b) 2 (c) 4 (d) None of these

5.

Find the number of elements in power set of set E (a) 32 (b) 22 (c) 40 (d) None of these Directions for Question Nos. 6 to 7 If in Praxis Business school 40 students like chocolate while 30 like cake, if there are 10 students who like both and an equal number of students who like none of them.

6.

Find the total number of students at Praxis Business School. (a) 65 (b) 70 (c) 85 (d) None of these

7.

What percentage of students do not like cake? (a) 57.14% (b) 25.12 (c) 55.55 (d) None of these

8.

A dinner party is to be xed for a group of 100 persons. In this party, 50 persons do not prefer sh, 60 prefer chicken and 10 do not prefer either chicken or sh. The number of persons who prefer both sh and chicken is. (a) 20 (c) 25

(b) 22 (d) None of these

9.

83

In an examination, 62% of the candidates failed in English, 42% in Mathematics and 20% in both. The number of those who passed in both the subjects is. (a) 11 (b) 16 (c) 18 (d) None of these

10. In a market research project, 20% opted for ‘Nirma’ detergent whereas 60% opted for ‘Surf blue’ detergent. The remaining individuals were not certain. If the difference between those who opted for ‘Surf blue’ and those who were uncertain was 720, how many respondents were covered in the survey (a) 1100 (b) 1150 (c) 1300 (d) None of these 11. If X and Y are two sets such that (X È Y) has 60 elements, X has 38 elements and Y has 42 elements, how many elements does (X Ç Y) have? (a) 11 (b) 20 (c) 13 (d) None of these 12. In a B School there are 15 teachers who teach marketing or nance. Of these, 8 teach nance and 4 teach both marketing and nance. How many teach marketing but not nance? (a) 15 (c) 11

(b) 20 (d) None of these

13. In a school 50 students play cricket, 70 play football while 30 play none of the games, If total number of students are 100 then how many of them Play only cricket (a) 15 (b) 20 (c) 10 (d) None of these 14. In a school 80 students like chocolate, 40 like coffee if the number of students doesn’t like any of them is equal to the number of students who like both of them then what is the total number of students in the school? (a) 115 (b) 90 (c) 120 (d) None of these 15. There are 200 students in a B school, 120 have been taken in project of nance, 50 in project of marketing and 30 in both the marketing and nance. Find the number of students worked on marketing but not on nance (a) 20 (b) 30 (c) 40 (d) None of these

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84 Koncepts of Logical Reasoning 16. In a school there are 100 students 60 of them don’t like Chocolate and 50 don’t like Biscuit and 10 of them like none then how many of them like both? (a) 20 (b) 30 (c) 40 (d) None of these Directions for Question Nos. 17 to 20 An Airline Company has divided the potential market into three categories through a market survey for the purpose of implementing a Customer Relationship Management Programme. The rectangle given in the diagram below represents Economy Class passengers. The circle represents working Executives and the triangle represents the Indians.

17. A signature hotel is launching a service for Executives who travel Economy Class but are not Indians. Which segments should it concentrate in the diagram above? (a) 7

(b)

(c)

(d) None of the above

5 and 6

6

18. A travel agency wishes to target passengers who are Executives and travel by Economy Class. Which segments in the diagram should it aim? (a) 6 and 5

(b)

6 and 7

(c)

(4)

None of the above

2, 5 and 8

19. The Indian Executives who do not travel by Economy Class are represented by the space numbered : (a) 7

(b)

(c)

(d) None of the above

2

3

20. The Airline Company wishes to target Economy Class, non-Executive passengers for In-ight Duty Free Shopping. Which area must it focus on?

2

(a) 6 and 5

(b)

(c)

(d) None of the above

1 and 2

6 and 2

Directions for Question Nos. 21 to 22

In the diagram below, the circle stands for educated‘, square stands for hard working‘, triangle for urban people ‘and rectangle for honest‘. The different regions of the diagram are numbered from 1 to 12. Study the diagram carefully and answer the questions: [SNAP 2008]

Set Theory 21. Uneducated urban hard-working an honest people are indicated by: (a) 3 (b) 11 (c) 9 (d) 4 22. Non-urban educated people who are neither hard-working nor honest are indicated by: (a) 5 (b) 7 (c) 10 (d) 11 Directions for Question Nos. 23 and 24 The Venn diagram given below shows the estimated readership of 3 daily newspapers (X, Y & Z) in a city. The total r e a d e r s h i p an d a d v e r t i s i n g cost f or ea c h of t h e papers is as below : Newspapers

Readership (lakhs)

Avertising cost (Rs. per sq. cm)

X

8.7

6000

Y

9.1

6500

Z

5.6

5000

The total population of the city is estimated to be 14 million. The common readership (in lakhs) is indicated I the given Venn diagram.

[SNAP 2009] 23. The number of people (in Lakhs) who read at least one newspaper is (a) 4.7 (b) 11.9 (c) 17.4 (d) 23.4 24. The number of people ( in lakhs) who read only one newspaper is (a) 4.7 (b) 11.9 (c) 17.4 (d) 23.4 Directions for Question Nos. 25 and 26 A survey was conducted of 100 people whether they have read recent issues of ‘Golmal’, a monthly magazine. Summarized in formation is presented below: Only September:18 September but not August: 23 September and July: 8 September: 28 July: 48 July and August: 10 None of the three months: 24 [SNAP 2010]

85

25. What is the number of surveyed people who have read exactly for two consecutive months? (a) 7 (b) 9 (c) 12 (d) 14 26. 60 employees in an ofce were asked about their preference for tea and coffee. It was observed that for every 3 people who prefer tea, there are 2 who prefer coffee. For every 6 people who prefer tea, there are 2 who drink both of tea and coffee. The number of people who drink both is the same as those who drink neither. How many people drink both tea and coffee? [SNAP 2011] (a) 10 (b) 12 (c) 14 (d) 16 27. In 2002, according to a news poll, 36% of the voters had leaning towards party “Y”. In 2004, this gure rose to 46%. But in another survey the percentage was down to 40%. Therefore, the party “Z” is likely to win the next election. Which of the following, if true, would seriously weaken the above conclusion? (a) People tend to switch their votes at the last minute. (b) It has been showed that 85% of the voters belonging to the party “Y” vote in an election as compared to 80% of the voters belonging to party ”Z”. (c) 35% of people favour party “Z”. (d) No one can predict how people will vote. Directions for Question Nos. 28 and 29 TT School of Management is a management institute involved in teaching, training and research. Currently it has 37 faculty members. They are involved in three jobs: teaching, training and research. Each faculty member working with TT School of Management has to be involved in atleast one of the three jobs mentioned above: Ø A maximum number of faculty members are involved in training. Among them, a number of faculty members are having additional involvement in the research. Ø The number of faculty members in research alone is double the number of faculty members involved in all the three jobs. Ø 17 faculty members are involved in teaching. Ø The number of faculty members involved in teaching alone is less than the number of faculty members involved in research alone. Ø Ten faculty member involved in the teaching are also involved in at least one more job. [IIFT 2008]

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86 Koncepts of Logical Reasoning 28. After some time, the faculty members who were involved in all the three tasks were asked to withdraw from one task. As a result, one of the faculty member each opted out of teaching and research, while remaining ones involved in all the three tasks opted out of training. Which one of the following statements, then necessarily follows: (a) The least number of faculty members is now involved in teaching. (b) More faculty member are now associated with training as compared to research. (c) More faculty members are now involved in teaching as compared to research. (d) None of the above.

4

Directions for Question Nos. 31 to 35

A professor keeps data on students tabulated by performance and sex of the student. The data is kept on a computer disk, but unfortunately some of it is lost because of a virus. Only the following could be recovered: Performance Average

Good

Male

10

Female Total

Total Excellent 32

30

Panic buttons were pressed but to no avail. An expert committee was formed, which decided that the following facts were self-evident: Half the students were either excellent or good. 40% of the students were females. One third of the male students were average. [CAT 1993] 31. How many students are both female and excellent? (a) 0 (b) 8 (c) 16 (d) 32 32. What proportion of good students are male? (a) 0 (b) 0.73 (c) 0.4 (d) 1.0 33. What proportion of female students are good? (a) 0 (b) 0.25 (c) 0.5 (d) 1.0

29. Based on the information given above, the minimum number of faculty members involved in both training and teaching, but not in research is: (a) 1 (b) 3 (c) 4 (d) 5 30. A survey shows that 61 %, 46% and 29% of the people watched “3 idiots”, “Rajneeti” and “Avatar” respectively. 25% people watched exactly two of the three movies and 3% watched none. What percentage of people watched all the three movies? [IIFT 2010] (a) 39% (b) 11% (c)

14%

(d) 7%

34. How many students are both male and good? (a) 10 (b) 16 (c) 22 (d) 48 35. Among average students, what is the ratio of male to female? [CAT 2002] (a) 1:2 (b) 2:1 (c) 3:2 (d) 2:3 36. In a hospital there were 200 Diabetes, 150 Hyperglycaemia and 150 Gastro-enteritis patients. Of these, 80 patients were treated for both Diabetic and Hyperglycaemia. Sixty patients were treated for Gastro-enterities and Hyperglycaemia, while 70 were treated for Diabetes and Gastro-enteritis. Some of these patients have all the three diseases. Doctor Dennis treats patients with only Diabetes. Doctor Hormis treats patients with only Hyperglycaemia and Doctor Gerard treats patients with only Gastro-enterities. Doctor Paul is a generalist. Therefore, he can treat patients with multiple diseases. Patients always prefer a specialist for their disease. If Dr. Dennis had 80 patients the other three doctors can be arranged in terms of the number of patients treated as: (a) Paul > Gerard > Hormis (b) Paul > Hormis > Gerard (c) Gerard >Paul > Hormi (d) none of these.

[CAT 2003]

Set Theory Directions for Question Nos. 37 to 38 New Age Consultants have three consultants Gyani, Medha and Buddhi, The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved. All three consultants are involved together in 6 projects. Gyani works with Medha in 14 projects. Buddhi has 2 projects with Medha but without Gyani, and 3 projects with Gyani but without Medha. The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved. 37. What is the number of projects in which Gyani alone is involved? (a) Uniquely equal to zero. (b) Uniquely equal to 1. (c) Uniquely equal to 4. (d) Cannot be determined uniquely. 38. What is the number of projects in which Medha alone is involved? (a) Uniquely equal to zero (b) Uniquely equal to 1 (c) Uniquely equal to 4 (d) Cannot be determined uniquely. Directions for Question Nos. 39 to 42 Help Distress (HD) is an NGO involved in providing assistance to people suffering from natural disasters. Currently, it has 37 volunteers. They are involved in three projects: Tsunami Relief (TR) in Tamil Nadu, Flood Relief (FR) in Maharashtra, and Earthquake (ER) in Gujarat. Each volunteer working with Help Distress has to be involved in at least one relief work project. A Maximum number of volunteers are involved in the FR project. Among them, the number of volunteers involved in FR project alone is equal to the volunteers having additional involvement in the ER project. The number of volunteers involved in the ER project alone is double the number of volunteers involved in all the three projects. 17 volunteers are involved in the TR project. The number of volunteers involved in the TR project alone is one less than the number of volunteers involved in ER project alone. Ten volunteers involved in the TR project are also involved in at least one more project. [CAT 2005]

87

39. Based on the information given above, the minimum number of volunteers involved in both FR and TR projects, but not in the ER project is: (a) 1

(b)

(c)

(d) 5

4

3

40. Which of the following additional information would enable to nd the exact number of volunteers involved in various projects? (a) Twenty volunteers are involved in FR. (b) Four volunteers are involved in all the three projects. (c)

Twenty three volunteers are involved in exactly one project.

(d) No need for any additional information. 41. After some time, the volunteers who were involved in all the three projects were asked to withdraw from one project. As a result, one of the volunteers opted out of the TR project, and one opted out of the ER project, while the remaining ones involved in all the three projects opted out of the FR project. Which of the following statements, then, necessarily follows? (a) The lowest number of volunteers is now in TR project. (b) More volunteers are now in FR project as compared to ER project. (c)

More volunteers are now in TR project as compared to ER project.

(d) None of the above. 42. After the withdrawal of volunteers, as indicated in Question 42, some new volunteers joined the NGO. Each one of them was allotted only one project in a manner such that, the number of volunteers working in one project alone for each of the three projects became identical. At that point, it was also found that the number of volunteers involved in FR and ER projects was the same as the number of volunteers involved in TR and ER projects. Which of the projects now has the highest number of volunteers? (a) ER (b) FR (c)

TR

(d) cannot be determined

EBD_7743

88 Koncepts of Logical Reasoning Directions for Question Nos. 43 to 46 The proportion of male students and the proportion of vegetarian students in a school are given below. The school has a total of 800 students, 80% of whom are in the Secondary Section and rest equally divided between Class 11 and 12. Male (M) Class 12

0.60

Class 11

0.55

Vegetarian (V)

0.55 0.475

0.53 [CAT 2007]

43. What is the percentage of vegetarian students in Class 12? (a) 40 (b) 45 (c) 50 (d) 55 (e) 60 44. In Class 12, twenty ve per cent of vegetarians are male. What is the difference between the number of female vegetarians and male nonvegetarians?

5

Directions for Question Nos. 47 to 51

In Praxis Business School Kolkata 60 1st year students participated either in Marketing quiz (MQ) or in Finance quiz (FQ) some information is as followsNo girl student participated in both MQ and FQ 40 % students are girls out of which 75% has work experience. 60 % students have work experience of which one third participated in only MQ. The number of boys students with work experience participated in only MQ is 8. 8 number of girl students participated in only MQ 12 fresher students (without work experience) participated in only FQ. 47. What is the maximum percentage of boys participated in only FQ? (a) 66.67%

(b)

(c)

(d) 75%

50%

(b) (4)

10 14

45. What is the percentage of male students in the secondary section? (a) 40

(b)

(c)

(d) 55

50

45

(5) 60 0.5

Secondary section Total

(a) Less than 8 (3) 12 (5) 16

55.56%

46. In the Secondary Section, 50 % of the students are vegetarian males. Which of the following statements is correct? (a) Except vegetarian males, all other groups have same number of students. (b) Except non-vegetarian males, all other groups have same number of students. (c)

Except vegetarian females, all other groups have same number of students.

(d) Except non-vegetarian females, all other groups have same number of students. (e)

All of the above groups have the same number of students.

48. If K is the ratio of students participated in only MQ to that participated in only FQ then nd the range of K (a) (4/9, 12/13)

(b)

(c)

(d) None of these

(7/11, 17/18)

(8/11, 18/17)

49. Find the minimum and maximum number of students who participated in only MQ. (a) 16 and 24

(b)

(c)

(d) None of these

18 & 26

16 and 28

50. What could be the number of fresher boys participated only in FQ (a) 12

(b)

(c)

(d) None of these

29

11

51. What could be the ratio of number of boys with work experience to the number of girls without work experience participated in only FQ? (a) 9/2 or 7/3

(b)

(c)

(d) None of these

13/2 or 13/3

11/2 or 17/2

Set Theory Directions for Question Nos. 52 to 56 In a society 60 family read Times Of India (TOI), 70 read Hindustan Times HT), and 40 read Telegraph (Tel). 10 family read both HT and Tel but not TOI, 18 family read HT & TOI, number of family who read only TOI & Tel but not HT is 10 less than the number of family who read all the three news paper. 52. What could be the total number of family in the society assuming that each family read at least one news paper? (a) 114 (b) 126 (c) 129 (d) None of these 53. If number of family who read both TOI and HT but not Tel is more than the number of family who read both TOI and Tel but not HT then what could be the number of family who read only Tel? (a) 15 (b) 10 (c) 16 (d) None of these 54. If on a particular day HT was not available so those who read only HT shifted to either TOI or Tel such that number of family who read only TOI and only Tel, become equal then nd the ratio of number of families who shifted to TOI to the number of family who shifted to Tel. (a) 31:9 (b) 9 : 31 (c) 7 ; 36 (d) none of these 55. If T, H and L is the number of family who read only TOI, HT and Tel respectively then which one of the following is correct relation (a) H > T > L (b) H > L > T (c) T > H > L (d) None of these 56. If P is the number of family who read exactly two news paper then which of the following is true about P. (a) 17 < P < 21 (b) 21 < P < 29 (c) 5 < P < 17 (d) None of these Directions for Question Nos. 57 to 61 In Praxis Business School Kolkata, 8 friends A, B, C, D, E, F, G & H have interest in one or more of the three subjects Marketing, Finance and HR. No two students like the same set of subjects. Further known facts are as follows – A like all the three subjects. Pair of students (B, C) , (D, E) (G- H) (B, D) and (F, G) have no common subjects i.e B and C have not common subjects and so on. F likes only HR. H like none of the subjects.

89

57. Which of the following pair of students don’t like Mkt. (a) B & D (b) C & E (c) F & C (d) F & H 58. Which of the following could be students who like Finance? (a) C, E, A, G (b) B, D, A, G (c) B, E, A, G (d) None of these 59. Which of the following student passed in exactly two subjects? (a) B (b) D (c) C (d) None of these 60. Which of the following statement require to nd liking and disliking of each and every subjects. (i) There is 1 student who don’t like a single subject. (ii) B and E have a common subject in liking. (iii) D and C have a common subject in liking. (a) Only (i) & (ii) (b) Only (ii) & (iii) (c) All three (d) None of these 61. If G don’t like Mktany more and F likes Mkt and no other change happen then which of the following is set of students like same subject? (a) G, D, A, C (b) A,B,E,F (c) A,C,E,F (d) None of these Directions for Question Nos. 62 to 66 In a school students at Pioneer career Kolkata wrote Mock test which has three subjects DI, VA and QA, here is the result of these students. 80 students cleared cut off in DI, 70 in VA and 60 in QA. Only 40 students cleared all the three subjects. 10 students failed to clear cut off even in one subjects. 50 students cleared cut off in VA and QA. 5 students cleared in cut off in only QA.

62. What is the minimum number of students who appeared in the Mock test. (a) 105 (b) 110 (c) 115 (d) None of these 63. What is the minimum number of students who did not clear cut off in exactly two subjects? (a) 5 (b) 15 (c) 20 (d) None of these

EBD_7743

90 Koncepts of Logical Reasoning 64. If number of students who didn’t clear cut off in at least two subjects is maximum possible then nd the number of students who failed in exactly one subject? (a) 5

(b)

(c)

(d) None of these

20

15

65. What is the ratio of number of students who didn’t clear cut off DI, but cleared cut off in QA to the number of students who didn’t clear cut off in exactly 1 subject. (a) 3:1

(b)

(c)

(d) None of these

1: 3

3:2

66. What is the maximum number of students who didn’t clear cut off in QA (a) 5

(b)

(c)

(d) None of these

20

15

Directions for Question Nos. 67 to 71 Out of 400 cadets at MERI Kolkata, 50% students participated in exactly one of the three activities, Swimming, Kick Boxing and Marching, total 160 students participated in swimming, 150 in Kick Boxing and 120 in Marching. Number of cadets who enrolled for swimming and kick boxing only is 20 more than the number of students who enrolled for kick boxing and marching only. 67. If all those who didn’t participated in any of the three said activity got an order from authority to participate in Social Activity, what is the ratio of maximum to minimum number of cadets participated in social Activity? (a) 216:185

(b)

(c)

(d) None of these

206:185

216:195

68. What is the maximum number of cadets who participated in exactly two activities? (a) 125

(b)

(c)

(d) None of these

105

115

69. Due to increasing popularity of Social Activity all those who earlier participated in only one activity joined Social Activity, then what is the maximum possible number of cadets who like only one activity? [Use data of previous question if require] (a) 112

(b)

(c)

(d) None of these

82

102

70. Due to increasing popularity of Social Activity Kick boxing program is dissolve and all the cadets shifted to Social Activity, then what is the Minimum possible number of cadets who like only one activity? [Use data of previous question if require] (a) 282

(b)

(c)

(d) None of these

182

285

71. Due to increasing popularity of Social Activity all the cadets at MERI participated in Social Activity then what could be the number of cadets who like only 2 activities. (a) 112

(b)

(c)

(d) None of these

122

116

Directions for Question Nos. 72 to 76 At Pioneer Career a leading IIT JEE training institute, a survey is conducted on 80 students of a batch and it was found that 20% students don’t like a single subject from three subjects Phy, Che and maths. 50% students like Phy and Maths. Numer of students who like Phy and Che both is equal to number of students who like Maths and Che both, Total 10% students like exactly 1 subjects. Number of students who like Phy is same as that of students who like Che and in turn equal to number of students who like Maths which is equal to 50. After 1st mock test each of the students changed their liking (i.e if a student who liked maths and Phy only now like Che, the one who like all the three now like none, if a student who like only one subject will now like 2 subjects, if a student earlier like 2 subjects will now like only 1 subject and so on) 72. How many students who like Phy now had liking for Maths before mock test? (a) 8

(b)

(c)

(d) None of these

10

6

73. How many students don’t like Phy before and after Mock. (a) 8

(b)

(c)

(d) None of these

10

6

74. What is the difference between total number of students who like Phy before Mock and like Che after Mock. (a) 20

(b)

(c)

(d) None of these

24

16

Set Theory 75. What is difference between total number of students who like all the three subject before and after Mock test. (a) 20

(b)

(c)

(d) None of these

24

14

76. Total number of students who didn’t like Phy or Che after Mock test is (a) 20

(b)

(c)

(d) None of these

24

38

Directions for Question Nos. 77 to 81 In society only three types of news paper TOI, HT or Tel are supplied, number of families who read TOI, HT and Tel is 43, 65, 37. Given that total number of family in the society is 100 and each family read at least one news paper. 77. If number of family who read HT only is minimum possible then what is the maximum possible number of family who read TOI and Tel but not HT. (a) 6

(b)

(c)

(d) None of these

12

8

78. If number of family who read all the three news paper is maximum possible then what is the maximum possible number of family who read only TOI (a) 18

(b)

(c)

(d) None of these

32

21

79. If number of family who read TOI only is more than the number of family who read HT only then what is the maximum number of family who read all the three news paper. (a) 17

(b)

(c)

(d) None of these

12

11

80. What is the minimum number of family who read at least two news paper. (a) 18

(b)

(c)

(d) None of these

32

21

81. What is the Sum of maximum to minimum number of students who read only Tel? (a) 18

(b)

(c)

(d) None of these

32

21

91

Directions for Question Nos. 82 to 86 In a school when a survey is conducted on liking of subjects out of three subjects physic, chemistry, mathematics, then outcome is as follows :Total number of students who like physics, chemistry and mathematics is 30, 40, 50 respectively.10 students shown interest in all three subjects, Number of students who like physics and mathematics both is equal to number of students who like chemistry and mathematics both and is equal to ¾ times number of students who like physics and chemistry both15 students like only chemistry. When class teacher saw this survey report he told head master that at least 20 and at max 40% of those who told they like physics are liars about physics liking. Same percentage for chemistry is 40% to 50%. Further class teacher added that the liars must belong to those who like only 1 subject and if numbers of liars is more then it belongs to those who like 2 subjects and probability that the who likes all three subjects is a liar is negligible.

82. According to survey what is the ratio of number of students who like only Phy to those who like only Che to those who like only Maths. (a) 1:3:4

(b)

(c)

(d) None of these

1:3:5

2:3:3

83. As per the observation of class teacher what could be the number of students who don’t like a single subject? (a) 12

(b)

(c)

(d) None of these

20

15

84. As per the observation of class teacher what could be the number of students who like only one subject? (a) 12

(b)

(c)

(d) None of these

20

15

85. As per the observation of class teacher what is the maximum number of students who like Phy and Che both. (a) 12

(b)

(c)

(d) None of these

16

15

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92 Koncepts of Logical Reasoning 86. As per the observation of class teacher if number of liars from maths group is maximum then how many students like only one subject? (a) 22

(b)

(c)

(d) None of these

26

In MERI kolkata 400 cadets have to participate in exactly one activity at Morning, One Activity at day Time and one activity at evening. Morning activities are Yoga and PT, Day time activity is Mess Club and Cooperative society and evening activity is sports and marching. Following facts are known further:Number of students who participate in PT and sports both is 100 Number of students who participate in yoga and co-operative both is 90 Number of students who participate in marching and mess actives both is 110 Number of students who participate in PT and morning but not in mess and sports is 60

87. If 190 students do not participate in mess activity then what is the number of students who neither participate in yoga in morning, nor in mess society in day time? (a) 100

(b)

(c)

(d) none of these

115

88. If number of students who like yoga is 190 then what is the number of students who participate in PT and also in Marching? (a) 102

(b)

(c)

(d) none of these

90

110

89. If k is the number of students who participate in yoga in morning and mess in day time and PT in evening then which of the following is correct about K? (a) 0 ≤ K ≤ 110

(b)

(c)

(d) None of these

40 ≤ d ≤ 120

10 ≤ d ≤ 100

90. What is the maximum number of students who participated in Yoga? (a) 190

(b)

(c)

(d) None of these

210

(i)

25

Directions for Question Nos. 87 to 91

78

91. Which of the following statement is sufcient enough to nd the exact number of students in each activity?

230

Number of students who participate in yoga and Sports both is 60

(ii) Number of students who participate in PT and Sports but not in mess activity is 80 (iii) Number of students who participate in yoga and Mess both is 70 (a) Only (i) & (ii)

(b)

(c)

(d) None of these

All three

Only (ii) & (iii)

Directions for Question Nos. 92 to 96 4 types of news paper namely HT, TOI, ET and Hindu are available in a society. Number of readers of news paper HT, TOI Et and Hindu is in AP with common difference 10. Number of readers who read exactly 1 news paper is 80, those who read exactly 2 news paper is 70, those who read exactly 3 news paper is 60. 20 families don’t read any of these news paper. 92. What is the minimum total number of families reading Hindu news paper. (a) 105

(b)

(c)

(d) None of these

120

115

93. If number of families who read ET is 115 then how many families read at least two news paper? (a) 120

(b)

(c)

(d) None of these

140

125

94. If number of families who read at least two news paper is equal to number of families who read at most two news paper then maximum how many families are reading TOI? (a) 120

(b)

(c)

(d) None of these

140

135

95. If on a particular day Hindu news paper (Whose readership is 130) was not available then what could be the maximum number of families reading 3 news paper? (a) 45

(b)

(c)

(d) None of these

75

15

96. What could be the total number of families in the society? (a) 220

(b)

(c)

(d) None of these

235

225

Set Theory Directions for Question Nos. 97 to 101 In a college student can opt for any one or more available sports, these are Foot Ball (FB), Cricket (Cr), Chess (Ch), and Volley Ball (VB), number of students who play FB and any one more game is 10, (I.e FB and Ch is 10, FB and Cr is 10 and so on), similarly number of students who play Cr and any one more game (Except FB as it is already dened as 10) is 8 and number of students who play FB and any two more games is 12. Total count for each of four Game is 100. 97. How many student play Cricket and exactly one more game? (a) 26 (b) 28 (c) 32 (d) None of these 98. If number of students who play Ch and Exactly one more game is maximum possible then what is the number of students who play only Cr. (a) 25 (b) 50 (c) 46 (d) None of these 99. What is maximum number of students who play Ch and VB? (a) 59 (b) 70 (c) 88 (d) None of these 100. Which of the following statement/s are correct? (i) Number of students who play Cricket and Chess is equal to those who play Ch and VB. (ii) Number of students who play FB and VB and Chess is equal to those who play Cr andCh and VB. (iii) Number of students who play FB and VB and Chess is equal to those who play Cr and Ch and F.B. (a) Only (i) (b) Only (i) & (iii) (b) Only (i) & (ii) (d) None of these 101. If number of students who play all four games is maximum possible then number of students who play only FB is(a) 6 (b) 8 (c) 12 (d) None of these Directions for Question Nos. 102 and 103 In a paramilitary college each of the students has to participate in one of the two available sports i.e Cricket or Foot ball, One of the club from Social or Media, one of the subject from History or Science and one of the Committee from mess or placement. Following additional information is as givenNumber of students who play Cricket, represents Social club, study Fin and a part of Mess Committee is 10.

93

Number of students who play Cricket, represents Media club, study Fin and a part of Placement Committee is 12. Number of students who play Foot Ball, represents Media club, study Mkt and a part of Placement Committee is 12. Number of students who play Foot Ball, represents Social club, study Mkt and a part of Placement Committee is 12. Number of students who play Cricket, represents Media club, study Mkt and a part of Placement Committee is 9. Number of students who play Cricket, represents Media club, study Mkt and a part of Mess Committee is 11. Number of students who play Cricket, represents Media club, is 40. Number of students who play Cricket, represents Mess committee, is 40. Number of students who study Mkt and a part of Mess Committee is 30. Number of students who play Cricket, and a part of Social club is 40. Number of students who play Foot Ball, study Fin and represents Mess committee, is 20. 102. If number of students play cricket but don’t study Fin is maximum then what is the number of students who study Fin but not a part of Mess Committee. (a) 40 (b) 32 (c) 24 (d) None of these 103. If number of students who study Fin is less than the number of students who participated in social activity then what can be the maximum number of students who neither play Cricket nor in Social activity but in Mess Committee? (a) 21 (b) 23 (c) 24 (d) None of these 104. What could be the total number of students participated in Media club. (a) 100 (b) 57 (c) 92 (d) None of these 105. What could be the maximum number of students who play Foot Ball and in Media Club? (a) 60 (b) 58 (c) 92 (d) None of these 106. What is the maximum difference between number of students who Study Fin to those who is involved in Social group? (a) 25 (b) 22 (c) 27 (d) None of these

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94 Koncepts of Logical Reasoning

Concept Applicator (CA)

8.

(a) Total number of persons = a + b + c + n = 100

Solutions (1 to 5) Consider a universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Set P = {2, 3, 5, 7} Set (E) = {2, 4, 6, 8, 10} Set (O) = { 1, 3, 5, 7, 9} Set (C) = { 4, 6, 8, 9, 10} 1.

(c) P’ = U –P = { 1, 4, 6, 8, 9, 10} so number of elements in P’ = 6

2.

(d) since E – C = { 2, 1}A

Do not prefer sh b+n = 50 60 prefer chicken hence b+c = 60

hence number of elements = 1 3.

Do not like sh and chicken is n = 10

(d) Since Set P = {2, 3, 5, 7}

On solving these equations we will get

And Set (O) = { 1, 3, 5, 7, 9}

a = 30, b = 40, c = 20

Then P U O = { 1, 2, 3, 5, 7, 9} = 6 4.

The number of persons who prefer both sh and chicken is = c = 20

(a) Since Set P = {2, 3, 5, 7} And Set (O) = { 1, 3, 5, 7, 9} Then P Ç O = { 3, 5, 7}

9.

(b)

= number of elements = 3 5.

(a) Since number of elements in set e is 5 hence number of elements is 2^5 = 32

Solutions (6 to 7) Since 40 students like Chocolate hence a+c = 40 Since 30 students like Cake hence b+c = 30 Since 10 students like both hence c = 10 = n or a = 30, b = 20 From the given condition – a + c = 62k b + c = 42k and c = 20k Hence a= 42k, and b = 22k Number of students who failed in none means passed in both = n = 100k – (a + b + c) 6.

(b) Total number of students is a + b + c + n = 30+ 20+10+10 = 70

7.

(a) Number of students who do not like cake is a+n = 30+10 = 40 Required percentage value is 40/70 × 100= 57.14%

= 100k – (42k+22k+20k) = 16k. or 16% 10. (d) Let those who opted for Nirma = a and those who opted Surf Blue = b and those who opted for none is n. Hence a = 20k and b = 60k, then n = 100k – 20k – 60k = 20k

Set Theory

11. (b)

12. (d)

13. (d)

14. (c)

15. (a)

16. (d)

The difference between those who opted for 'Surf blue' and those who were uncertain = 60k – 20k = 40k = 720 hence k = 18, Hence total number of persons covered in survey = 100k = 1800 Since (X È Y) has 60 elements, X has 38 elements and Y has 42 elements. We know that (X È Y) = X + Y – X Ç Y or 60 = 38 + 42 – (X Ç Y) or (X Ç Y) = 80 – 60 = 20 From the given condition (M È F) = 15, (F) = 8, (M Ç F) = 4 hence M = 15 + 4 – 8 = 11, M – F = 11 – 4 = 7 From the given information (C) = 50, (F) = 70, (n) = 30, (U) = 100 Since (U) = (C) + (F) – (C Ç F) + (n) Or 100 = 50 + 70 – (C Ç F) + 30 Or (C Ç F) = 50 Play only cricket = (C) – ( C Ç F) = 50 – 50 = 0 From the given information (C) = 80, (F) = 40, and (C Ç F) = (n) Hence (U) = (C) + (F) – (C Ç F) + (n) Or 80 + 40 – x + x = 120 From the given information (U) = 200, (F) = 120, (M) = 50, (F Ç M) = 30 Marketing but not on Finance = 50 – 30 = 20 If number of students who like chocolate =a+c Number of students who like Biscuit = b + c Number of students who like Both = c Number of students who like none = n = 10 From the given condition 100 = a + b + c + n Since 60 of them don’t like Chocolate, hence b + n = 60 or b = 50 And 50 of them don’t like Biscuit hence a + n = 50, a = 40 Hence 100 = 40 + 50 + c + 10 or c = 0

Solutions (17 to 20) The given diagram is

95

In the given diagram, Circle ® working executives Rectangle ® economy class passengers Triangle ® Indians 17. (b) Executives who travel by Economy class but not are Indians is represented by the area common for circle and rectangle but not triangle, which is 6. 18. (a) Executives who travel by Economy class are represented by circle and rectangle i.e., 6 and 5. 19. (d) Indian executives who do not travel by Economy class are represented by the area common for circle and triangle but not rectangle i.e. 4. 20. (c) The Economy class non-executive passengers is represented by the area which is in rectangle but not in circle i.e., 1 and 2. Concept Builder (CB) 21. (d) The uneducated urban hard-working and honest people are represented by the area common to the rectangle, the triangle and the square and excluding the common area of circle. This area is marked by 4. 22. (b) The non-urban educated people who are neither hardworking nor honest are represented by the part of thecircle, which is not common to any of the other 3gures. This part of circle is marked by 7. 23. (c) Total readership of X is 8.7 lakhs. Hence total readership of X alone is 8.7 – 2.5 – 0.5-1 = 4.7 Also, total readership of Y is 9.1 lakhs total readership of Y. alone is 9.1 – 2.5 – 0.5-1.5 = 4.6 Lacs Also, total readership of Z is 5.6 lakhs total readership of Z alone is 5.6 – 1 – 0.5 – 1.5 = 2.6 Lacs The number of people who read at least one newspaper = 4.7 + 2.5 + 0.5 + 1.0 + 4.6 + 1.5 + 2.6 = 17.4 lakhs 24. (b) The number of people who read only one newspaper = 4.7 + 4.6 + 2.6 = 11.9 lakhs 25. (b) The given information can be represented as follows.



Now from the given information we can frame following equations c = 18, f + c = 23, f + g = 8, c + f + g + e = 28, a + d + f + g = 48, d + g = 10, n = 24 People who read exactly two consecutive months is represented by d and e. f + c = 23 and c = 18 ∴ f = 5 f + g = 8 and f = 5     ∴ g = 3 d + g = 10 and g = 3  ∴ d = 7 c + f + g + e = 28, c = 18, f = 5 and g = 3 ∴ e = 2 or ∴d + e = 9 26. (b) Let the number of people who prefer tea = 3K ∴ Those who prefer coffee = 2K Those who like both = K Those who like neither = K This can be represented as follows. µ = 60 Tea = 3K

Coffee = 2K

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96  Koncepts of Logical Reasoning 

29. (a) 30. (d) The given condition is as followsRajneeti

3 idiot a

db

e g

f

c



Avatar



We know that {(a + d + e + g) + ( b + d + f + g) + (c + e + f + g) } – ( d + e + f) – 2g



= a+b+c+d+e+f+g



or 61x + 46x + 29x – 25x – 2g = 97x



or 2g = 14x or g = 7x Concept Deviator (CD)

2K

K

Solutions (31 to 35)

K

nK ∴ 5K = 60 or K = 12. 27. (c) As per the question according to a news poll in 2002party Y had 36% of voter’s support,. In 2004, one poll says that the support for Y has increased to 46% while another says it is 40%. And the conclusion is that party Z is likely to win the next election. Option 3 that says that only 35% of the people favour party Z (35% is less than 40% or 46% estimated as support for Y by two different polls) weakens the conclusion most.

Solutions (28 and 29) teaching

17

training

7

a b

d

x c 2x

Females students = 32 which is 40% of total students. Hence, total number of students 32 = 80 . Hence, 0.4 males = (80 – 32) = 48. It is further given, that half the students were either excellent or good and one third of the male students were average. Hence, the table can be completed as under: Performance Average Good Male 16 22 Female 24 8 Total 40 30





When x = 4, a ≥ 2.

Total 48 32 80

31. (a) There is no female excellent student in the class. 32. (c) Proportion 22 = 0.73 = 30

of

good

male

students

33. (d) Proportion of good female students =

research 28. (d) Since, a + x + b = 10 x >3.5, a + b + c + d + 3x = 30, c + d + 2x = 20

Excellent 10 – 10

8 = 0.26 30

34. (b) There are 22 male students who are good 35. (b) Ratio of average male to average female

= 6:24 = 2:3

Set Theory

97

From the rst condition:

36. (a)

G + B = M + 16 => G – M + B =16……(1) No. of projects in which more than one consultant are involved is 6+3+2+8 =19; From condition 5: Total no of projects = 19*2-1 =37 Also, G+M+B+19 = 37; => G+M+B =18 ………(2) Subtracting 1 from 2, we get: 2M=2 => M=1; 37. (d) 38. (b) Solutions (39 to 42) From the above diagram we get the relation, A+D+G+F = 200 A = 80 (given)

G+F = 80

B+D+G+E = 150

G+E = 60

C+F+G+E = 150

D+G = 70

39. (c) FR project has the maximum number of volunteers. Let number of volunteers involved in all the three projects be g. Then the number of people involved in ER project alone will be 2g, and number of volunteers in TR alone will be 2g –1.

by solving we get, B= 50, C= 40 D=40, E =30, F=50 , G=30 Paul treat = 150, gerad = 50 and hormis = 40 Solutions (37 and 38) Let G =>Gyani working alone; M =>Medha working alone; and B =>Buddhi working alone Given conditions are: 1.

The sum of the number of projects handled by Gyani and Buddhi individually is equal to the number of projects in which Medha is involved.

2.

All three consultants are involved together in 6 projects.

3.

Gyani works with Medha in 14 projects.

4.

Buddhi has 2 projects will Medha but without Gyani, and 3 projects with Gyani but without Medha.

5.

The total number of projects for New Age Consultants is one less than twice the number of projects in which more than one consultant is involved.

Based on the above information the following diagram is drawn.

Since 17 volunteers are involved in the TR project and 10 of them are involved in at least one other project, 7 of them are involved in TR project alone. Hence 2g – 1 =7 or g = 4. Now it is given that the number of volunteers in FR alone is the same as those who have additional involvement in ER hence b=d+4 Now total of TR is 17 hence a+c = 6 Total volunteer are 37 hence 8+7+4+a+b+c+d =37 Or a+b+c+d = 18, but a+c =6 hence b+d =12, but b =d+4, hence 2d+4 =12 or d=4, b=8 Since FR= 16+a has a maximum number of volunteers hence it must be more than ER=16 + c or TR=17. Hence 16 + c < 16+a, or c < a but a+c =6 Hence minimum possible value of a = 4 40. (a) If it is known that twenty volunteers are involved in FRi.e 16+a , then we can nd the value of 'a' and 'b' and hence the number of volunteers in each project.

41. (b) The new venn diagram becomes FR

TR = 17

2g – 1 = 7

a+1

b=8

g=4 d= c+2 4+1

=5

2g = 8

ER If one volunteer withdraw of the TR project among the four who are involved in all the projects, he will be working only on FR and ER, So FR + ER becomes 4 + 1 = 5. Similarly when one opts out of ER project, FR + TR becomes a + 1 and when two opt out of FR project among the four, TR + ER becomes c + 2. From the previous question 3 ER 42. (a) Since the addition of volunteers was such that they were allotted to only one projects each and the number of volunteers working in one project alone for each of the three projects became equal, let us take it as x. Here, the number of volunteers in (FR+ ER) and (TR + ER) will not be affected .Given (FR + ER) = (TR + ER) Hence (4 + 1 + 0) = c + 2 (from the figure in solution 89) or c = 3 and hence a = 3 Also FR = (a + 1) + x + (4 + 1) = x+ 9 ER = (c + 2) + x + (4 + 1) = x + 10 TR = (c + 2) + x + (a + 1) = x + 9 Hence ER has the highest number of volunteers. 43. (a) Total no of male students = 0.475 × 800 =380, Total no of veg. students = 0.53 × 800 = 424 Total number of students in secondary section = 0.80 × 800=640, out of these 0.55 × 640 = 352 are vegetarian, so we can re-write and complete the table as follows

Class 12 Class 11 Secondary section Total

Male (M) 48 44 288 380

Vegetarian (V) 32 40 352 424



Hence % of vegetarian students in class 12



= 32/80 × 100 = 40%

44. (a) The number of vegetarian male in class 12 = 0.25 × 32 =8, since total number of males in class 12 is 48, hence the number of nonvegetarian males in this class = 48 – 8 =40.

The number of vegetarian female



= 32 – 8 =24



Hence required difference = 40 – 24 =16

45 . (b) total number of male students in secondary section = 288, and total number of students in this section is 640, hence required % = 288/640 × 100 =45% 46. (b) The data given in this question is not consistent with the table provided in the common information. The official answer key released by IIM –A has not given any answer key to this question, however if we interpret the question as “ In the secondary section, 50% of the vegetarian are males” then we can solve the question.

Total veg students = 352



Veg males= veg females = 176



Total number of males = 288, total no of nonveg males = 288 – 176 = 112



Hence except non-veg males all other groups are equal. Then option (b) is true. Concept Eliminator (CE)

Solutions (47 to 51) Number of Boys students –36, with work experience – 18, fresher students - 18 Number of Girl students – 24, with work experience18, fresher students- 6 Number of students with work experience – 36 Number of fresher students -24 Boys with WE Fresher Boys Girls with WE Fresher Girl Total

18 18 18 6 60

MQ 12– X= 8 Y X=4 4 16+Y

MQ & FQ 10 – Z 8–Y 0 0 18 – Y – Z

FQ Z 10 14 2 26+Z

From the table maximum value of Z = 10, minimum value of Z = 0 Maximum value of Y = 8 and minimum value of Y = 0

EBD_7743

98  Koncepts of Logical Reasoning 

Set Theory

99

47. (b) Number of boys participated in FQ is 10 + Z, here maximum value of Z is 10 then 10 + Z = 20, so required maximum percentage is 20/36 × 100 = 55.56%

53. (c) From the given condition 18 – x > x – 10 or 28 > 2x or x < 14

As per the given condition K = (16+Y)/(26+Z)

Number of family who read only Tel is 40-2x whose minimum value is 12 and maximum value is 20 with even number.

48. (a) For maximum value of K, Y should be maximum and Z should be minimum then K = (16+8)/(26+0) = 24/26 = 12/13

But we have seen that x≥ 10 hence range of x is 10 ≤ x < 14

54. (b)

For minimum value of K, Y should be minimum and Z should be maximum then K = (16+0)/(26+10) = 16/36 = 4/9 49. (a) From the given table number of students participated only in MQ is (16 + Y) So maximum value is 16 + 8 = 24 and minimum value is 16 – 0 = 16. 50. (c) From the table number of fresher boys participated only in FQ = 26 + Z from maximum and minimum value of Z we will get its range as 26 to 36, hence only possible value is 29 51. (b) Here we have to nd the ratio of Z and 2 where Z is between 10 and 18. So required ratio could be any one of 10/2, 11/2, 12/2, …. 18/2 Solutions (52 to 56)

Here k is the number of family out of 42 family who shifted to Tel and remaining 42-k shifted to TOI. From the given condition 50-2x +k = 112 -2x –k or 2k = 62 or k =31 So required ratio is (42-k) : k = 9:31 55. (a) Here H = 42, T = 52-x and L = 40 -2x as we have seen 10 ≤ x ≤ 18 so minimum and maximum value of T is 34 and 42 while that of L is 4 and 20, hence H must be more than T and L. Now consider T and L we will nd T> L, hence required sequence is H > T > L 56. (a) From the venn diagram (18 – x) + (10) + (x –10) = 18 Solutions (57 to 61)

Let us assume that the number of family who read all the news paper is x, then remaining is as given the venn diagram. 52

(b) Total number of family is 152 – 2x From venn diagram 10 ≤ x ≤ 18 So minimum total number of family 152 – 2 × 18 = 152 – 36 = 116 And maximum number of family = 152 – 2 × 10 = 152 – 20 = 132 So total number of family must be between 132 and 116 and an even number hence 126 is a possible option.

From the given condition no pair of students like same subject hence condition must be like 1 students like all the three subjects (A in this case), 3 students who like exactly 2 subjects, 3students like only one subjects and 1 student like none of the subjects. B has no common subjects with C and D so B must like only 1 subject and one of C or D must like 2 subjects and one like only 1 subject. But D has not common liking with E and B so D must like only one subject then C and E must like 2 subjects each. Similarly G likes two subjects. Students who like 2 subjects :- C, E and G Students who like 1 subject :- B, D and F Since F like only HR so G must like both Mkt and Fin. B and D like only one subject either Mkt or Fin

EBD_7743

100 Koncepts of Logical Reasoning C and E like two subjects these pair of subjects must be either (Fin & HR) or (Mkt& HR) A B C D E F G H

Mkt √ Y/N

Fin √ N/Y

N/Y

Y/N

X √ X

X √ X

HR √ X

64. (b) From the given condition 70-2x is maximum when x = 0, then required value is 10 + 5+ x = 15+0 =15 65. (a) From the given condition (10+5)/5 = 15/5 = 3/1 66. (d) From the venn diagram required value is 65-x and maximum value is 65.

√ X X

Solutions (69 to 71)

57. (d) From the venn diagram we can see that F and H denitely don’t like Mkt. 58. (c) From Venn diagram option (c) could be true although not must be. 59. (c) From table option (C) is correct. 60. (d) All the three statements are all ready derived hence these are redundant information. 61. (c) From the venn diagram A, C, E, and F like HR Solutions (62 to 66) Since 40 students cleared cut off in all the three subject, while 50 students cleared cut off in VA and QA hence number of students who cleared cut off in only VA and QA is 50-40 =10 5 students cleared cut off in only QA hence number of students who cleared cut off in only DI and QA is 6040-10-5 =5. Let number of students who cleared cut off in only DI and VA is x the venn diagram is as shown. Draw a similar venn diagram based on the facts about the subjects in which students failed to clear the cut off.

From the given condition- a + b + c = 200 d = f + 20 Let d + e + f = P = 2f + 20 + e or P ≥ 20 a + b + c + d + e + f + g + n = 400 a + d + e + g = 160 b + d + f + g = 150 c + e + f + g = 120 Adding these three equations we will get (a + b + c) + 2(d + e + f) + 3g = 430 200 + 2(d + e + f) +3g = 430 or 2(d + e + f) + 3g = 230 Or 2P + 3g = 230 or 2(P + g) = 230 – g or P + g = 115 – g/2 67. (a) Number of students who didn’t participated in a single activity is ‘n’ hence we have to nd the ratio of maximum to minimum value of ‘n’. Since total number of students is 400 hence a + b + c + d + e + f + g + n = 400

62. (a) From the venn diagram range x ≤ 20 Total number of students = 125-x , for minimum value of this x should be maximum. Minimum total number of students = 125-20 = 105. 63. (c) From the 2ndvenn diagram number of students who didn’t clear cut off in exactly 2 subjects is 5 + (35-x) + (20-x) = 60-2x, for minimum value x = 20 So required minimum value = 60 -2x20 = 20

or 200 + P + g + n = 400 or n = 200 – (P + g) For maximum value of n condition is P + g is minimum, We know that P + g = 115 – g/2 Or P + 3g/2 = 115 for minimum value of P + g, g should be maximum, and since minimum value of P is 20 so maximum value of g is 62, when P is 2 So minimum value of P + g = 62 + 22 = 84

Set Theory or nmax = 200 – 84 = 116 For minimum value of n, P + g should be maximum, and it will happen when g = 0 or P = 115 So nmin = 200 –115 = 85 Hence required ratio is 116 : 85 68. (b) Number of cadets who participated in exactly two subjects is d + e + f = P

101

Before Mock b = those who like only Che = after Mock those who like both Phy and maths Before Mock c = those who like only Maths = after Mock those who like both Che and Phy Before Mock d = those who like both Che and Phy = after Mock those who like only maths And so on-

From the previous solution maximum value of P is 115 when g =0 69. (d) Total number of cadets who like Social Activity is n + a + b + c , then number of cadets who like only one activity that must be social activity is given by n + (a + b + c) Since maximum value of n is 116 and a + b + c = 200 hence required maximum value is 116 + 200 = 316 70. (b) Total number of cadets who like Social Activity is n + a + b + c , and its minimum value is 185 + 100 = 285 71. (d) All those who liked only one will like exactly 2 activity, hence required value is 200. Solutions (72 to 77) From the given conditions20% of the students do not like a single subjects hence n =16 Then a + b + c + d + e + f + g = 80 – 16 = 64 50% students like Phy and Maths so f+g = 40

72. (c) After mock those who like Phy now is c, b, e and n out of these e and c like Maths before Mock hence required value is e+c = 2+8 = 10 73. (d) Those who didn’t like Phy after Mock is a, d, f and those who didn’t like it before Mock is b, c, e and n hence required number is 0. 74. (a) Required value is 50 – 30 =20 75. (b) Required value is given by g – n = 30–16 = 14

And a + b + c =8

76. (b) From the 2nd Venn diagram required value is d + g = 8 + 30 = 38

Since a + b + c + d + e + f + g = 64

Solutions (77 to 81)

Also d + g = e + g then d =e

or 8 + 40 + d + e = 16 or d = e = 8, a+d+f+g = 50 or a + 8+40 = 50 or a =2 Similarly c= 2, b =4, g = 30

Since total number of family is 100 hence a + e + c + 65 = 100 After Mock test distribution will be as follows-

Or 43 – (d+e+g) + e + 37 – (e+f+g) + 65 = 100

Before Mock a = those who like only Phy = after Mock those who like both Che and maths

Or d+e+f + 2g = 45

EBD_7743

102 Koncepts of Logical Reasoning 77. (d) Since 65 – (d + f + g) is minimum possible hence d + f + g should be maximum possible. Maximum value of d + f + g is 45 when e =0 and g=0, Hence required minimum value is 0 78. (b) Since d + e + f + 2g = 45 hence maximum possible value of g = 22 when d + e + f = 1 For maximum value of a = 43 – (d + e + g) ,d + e + g should be minimum that is possible when d =e = 0 Hence minimum value of a = 43 – (0 + 0 + 22) = 21 79. (b) From the given condition 43 – (d + e + g) > 65 – (d + f + g) or f > e + 22 Since d + e + f + 2g = 45 since f = e + 22 hence d + 2e + 2g = 23, maximum value of g is 11 80. (d) At least two news paper is read by d+e+f+g, since d+e+f+2g = 45, for given condition g = 22, then minimum value of d+e+f+g = 22+1 = 23 81. (d) Number of family who read only Tel = c = 37 – (e + f + g) Maximum value is cmax when e + f + g is minimum possible it is when d = 43, e = 0, g = 0, a = 0, f = 2 so minimum value of e + f + g = 2 or cmax = 37 – 2 = 35

As per the observation of class teacher actual number of students who like Phy is 18 – 24 that means number of students those who lied about their liking of Phy is 6 to 12 out of these 6-12 students 5 students must be those who liked only Phy. Remaining 1 to 7 students must be from d or f (as g =10 students assumed that they are not liar) Let x = actual number of students who like only Phy and Che not Maths. And y = actual number of students who like only Phy and Maths not Che. Similarly z = actual number of students who like only Maths and Che not Phy. Hence from the given condition 1≤ x + y ≤ 7 And 1 ≤ y + z ≤ 5

Minimum value of cmin0 when e + f + g = 37. So required summation = 35 + 0 = 35 Solutions (82 to 86) From the given condition g = 10, Number of students who like both Phy and Maths = 10 + f Number of students who like both Che and Maths = 10 + e From given condition 10 + f = 10 + e or e =f Similarly from 3rd condition ¾ (10 +e) = 10 + d Or 30 + 3e = 40 + 4d

82. (a) From the survey report a = 5, b = 15, c = 20 and required ratio is 1:3:4

3e - 10 4 Since 15 students like only chemistry hence b = 15

83. (c) Students who don’t like a single subject must be all the liars who said they like only Phy or only Che hence required number is 5+ 15 = 20

Number of students who like only Chemistry is 40

84. (c) From the venn diagram made by the observation of class teacher number of students who only one subject is 0 + 0 + 20 = 20

Or d =

Hence e + g + d + b = 40 or e + 10 + (3e – 10)/4 + 15 = 40 or (7e – 10 )/ 4 = 15 or 7e = 70 or e = 10 Hence a = 5, c = 20, f = e = 10, d = 5 Hence the distribution is as follows

85. (b) From the Venn diagram of teacher’s observation we have to nd maximum value of 10+x, maximum possible value of x is 5, hence required maximum value is 10+5 = 15

Set Theory 103 86. (d) As per the given condition y and z should be maximum hence y = 7 and z =5, Since earlier f = 10 but now same value is y = 7 hence the difference is 10 – 7 = 3 those who liked only maths but replied for both Mahs and Phy, similarly 10 – z = 10 – 5 = 5 so remaining 5 liked only maths. Total number of students who liked only one subject = 20 + 3 + 5 = 28. Solutions (87 to 91) Try to understand the Venn diagram. Number of students who like yoga is a + d + f + g Number of students who like PT is b + c + e + n Number of students who like mess is b + d + e + g Number of students who like co-operative is a + c + f + n Number of students who like sports is c + e + f + g Number of students who like marching is a + b + d + n From the given condition Number of students who like PT and sports = c + e = 100 Similarly a + f = 90 And b + d = 110 Then number of students who perform PT in morning but not in mess activity as sports is = n = 60 Since total number of students = 400 hence a + b + c + d + e + f + g + n = 400 or 100 + 90 + 110 + 60 + g = 400 or g = 40 87. (a) From the given condition those who do not participate in mess activity is a + f +c +n =190 90 + 60 + c Or 40 Required number of students = c + n or 40 + 60 = 100 88. (b) From the given condition a + d + f + g = 190 90 + 40 + d =190 or d = 60 Since b + d =110 , b = 50 The number of students who participate in PT and also in Marching = b + n = 50 + 60 = 110 89. (a) Students who Participate in Yoga = a+d+f+g Students who Participate in Mess = b + d+e+g Students who Participate in PT = a+d+b+n Common to all these is d, hence K = d Since b+d = 110 hence 0 ≤ d ≤ 110 90. (b) Here we have to nd the value of a+d+f+g = 90 d+40 = 130 +d Since maximum value of d is 110 hence maximum value of 120 + d = 12 + 110 = 230 91. (c) From statement (i) we can nd f+g = 60 or f= 20 From statement (ii) we can nd c+n = 80 or c= 20 From statement (iii) we can nd d + g = 70 or d= 30 Hence we require all the three statements

Solutions (92 to 96) Total number of families is = T0 + T1 + T2 + T3 + T4 here Ti is the number of families reading exactly ‘i’ news papers. Let number of readers of HT is ‘a’ then that of TOI, ET and Hindu is ‘a+10’, ‘a+20’ and ‘a+30’ HT +TOI + ET + Hindu = 4a + 60= T1 + 2T2 +3T3 + 4T4 = 80+ 140+180 + 4T4 or 4a = 340 + 4T4 or a = 85 + T4 92. (b) So minimum number of families who like Hindu is 85+30= 115 93. (c) At least two news paper = T2 + T3 + T4 Since ET =115= 85+20+ T4 or T4 = 10 Hence T2 + T3 + T4 = 70+60+10 = 140 94. (b) From the given condition T0 + T1 + T2 = T2 + T3 + T4 or T0 + T1 = T3 + T4 or T4 = 20 + 80 – 60 = 40 TOI = a+10+40 = 85+10+40 =135 95. (c) Since 130 = 85 + 30+ T4 or T4 = 15 In new condition the number of families who read three news paper is those who were reading all for plus those who were reading TOI + HT + ET , required maximum value is 15 +60 = 75 96. (c) Total number of families is = T0 + T1 + T2 + T3 + T4 = 20 + 80 + 70 + 60 + T4 = 230 + T4 Solutions (97 to 101) From the given condition number of students who play FB and any one more game is 10 we can conclude e = g = i = 10, From the given condition number of students who play Cr and any one more game is 8 we can conclude, h= j = 8. From the given condition number of students who play FB and any two more games is 12 we can conclude k = n = l =12 Hence the Venn diagram will be as follows Since total number of students who play FB is 100 hence a + e + g + k + l + o + i + n = 100 or

a + o = 34

Similarly o + m + b = 50 O + m+ c + f = 58 O + m+ d+ f = 58 Or c + f = 8 +b

EBD_7743

104 Koncepts of Logical Reasoning From statement (ix) From this we conclude that n + i + j + d = 30 or j + d = 8 From statement (x) From this we conclude that k + o + e + n = 40 or k + 10 + 11 + e =40 or k + e = 19 From statement (xi) From this we conclude that m + f= 20.

97. (a) From the Venn Diagram we have to nd the value of Cr + FB, Cr + Ch , and Cr + VB or we have to nd the value of e+ h + j = 10 + 8 + 8 = 26 98. (b) Since number of students who play Chess and exactly 1 more game is g + h + f = 18+f is maximum when f is maximum and since c + f = 8 + b hence for f is maximum b has to be maximum and maximum value of b is 50 when o = m = 0. 99. (b) Student who play Ch and VB is l + o + m + f = 12 + o + m + f = 70 when c =0 100. (d) From the Venn diagram none of the statement is correct. 101. (d) Since a+o = 34, since o is maximum hence a should be minimum i.e 0. Solutions (102 to 106) From Venn Diagram those who Play Cricket:-a + e + g +k+l+o+i+n Number of students who play Foot Ball is :- b + h + m +j+c+f+d+N And so on Number of students who play Cricket, represents Social club, study Fin and a part of Mess Committee is 10 or o = 10. Number of students who play Cricket, represents Media club, study Fin and a part of Placement Committee is 12 i.gg =12 Number of students who play Foot Ball, represents Media club, study Mkt and a part of Placement Committee is 12 i.e N =12 Number of students who play Foot Ball, represents Social club, study Mkt and a part of Placement Committee is 12 i.e b =12 From statement (v) a =9 From statement (vi) i= 11 From statement (vii) From this we conclude that a + g + l + i = 40 or 9 + 12 + l + 11= 40 or l = 8 From statement (viii) From this we conclude that l + o + i + n = 40 or 8 + 10 + 11 + n =40 or n = 11

102. (b) From the given condition e+n is maximum when e is maximum i.ee = 19 when k =0 Then the number of students who study Fin but not a part of Mess Committee is g+k+h+c = 12+0+12+8 =32 103. (a) Here we have to nd maximum value of d + f It is given that m + f = 20 and j + d = 8 hence m + j + f + d = 28, for maximum value of d + f, the value of m + j must be minimum. Number of students who study Fin is 12 + k + 12 + 8 + 8 + 10 + m + f = 70 + k Number of students who participate in Social activity e + k + 10 + 11 + j + m + 12 + 12 = 64 +m+j or 70 + k < 64 + m + j or 6+k < m +j or minimum value of m + j is 7. Hence maximum value of d+ f = 28-7 = 21 104. (d) Those who participated in Media Club is a + g + l + i + c + f + d + N = 9 + 12 + 8 + 11 + 8 + f + d + 12 = 60 + f + d So minimum value of this is 60 and maximum is 60 + 20 + 8 = 88. 105. (b) Required number is given by c + f + d + N for maximum value f = 20 and d = 8, hence required maximum value is 8 + 20 + 8 + 22 = 58 106. (a) From the Venn Diagram number of students who study Fin is 70 + K so its minimum and maximum value is 70 and 89. Number of students who involved in Social club is 64 + m + j and its minim value is 64 and maximum value is 64 + 28 = 92 So maximum value of difference is 89 – 64 = 25

Group/ Team Formation

5

105

Group/ Team Formation Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

SSC

Very Important

IIFT

Very Important

CSAT

Very Important

SNAP

Very Important

Other Govt Exams

Very Important

NMAT

Very Important

Other Aptitude Test

Very Important

In this chapter we will solve the questions related to group /Team formation. In these type of questions we have to form one or more group/teams from the given condition with the given members. Information provided are of two types(i)

Positive information :- e.g A and B is together

(ii) Negative information:- e.g A and B are not together (iii) Conditional information:- If A is with B then C is with D (iv) Restrictive Information :- Number of members in a group can not be more/less than 3. With the use of these information we have to form group/team. In certain questions the main data has only few information and then in each and every question some more condition is given, in these type of questions we have to form team/group for each and every question.

EBD_7743

106 Koncepts of Logical Reasoning

1

Directions for Question Nos. 5 to 8

Directions for Question Nos. 1 to 4

A scientist is trying to nd a cure for the common cold using four ingredients. He can choose from the stable chemicals A, B and C and the unstable chemicals W, X, Y and Z. In order for the formula not to explode, there must be two stable chemicals in it. Also, certain chemicals cannot be mixed because of their reaction together. Chemical B cannot be mixed with chemical W. Chemical C cannot be mixed with Chemical Y. Chemical Y cannot be mixed with Chemical Z. 1.

If the scientist calculated that Y is the most important chemical and must be used in the formula, which other ingredients must be a part of the cure? (a) (c)

2.

(b) (d)

B, C and X A, B and X

engineers:

The scientist rejected chemical B because of its possible side effects but decided to use chemical Z. Which is a possible combination of the four ingredients in the formula? (a) (c)

3.

A, B and Z A, B and W

A company wants to select a team of four mechanical engineers from its South Indian Factory for transfer to North India, where they are going to set up a new plant. The company is managed by professional managers and is very particular about human resources and industrial relations. There are seven engineers of equal ability; X, Y and Z (who are in Senior Executive Cadre) and A, B, C and D (who are in Junior Executive Cadre). The company requires that there should be two Senior Executives and two Junior Executives in each team. It is also necessary that all the engineers in a particular team are friendly with each other, in order to have a real team spirit and avoid any industrial relations problem in the new factory being set up in the North. Following is the situation of relations between the seven

A, W, Y and Z A, W, X and Z

(b) (d)

A, C, W and Z A, X, Y and Z

Using chemical Y and W together.

II.

Using chemical B and C together.

5.

6.

III. Using chemical W, X and Z together. (a) (c) 4.

III only I only

(b) (d)

I and III only II only

Y and A are not friendly.

II.

Z and C are not friendly.

III. A and B are not friendly

Which of the following combinations of chemicals is impossible? I.

I.

(a)

X, Z, C and B

(b)

Z, C, D and B

(c)

X, Z, A and B

(d)

X, Z, D and B

If A is on the team then which other engineers must be on the team as well? (a)

X, Z and B

(b)

X, Z and C

(c)

X, Y and D

(d)

X, Z and D

Which of the following can never be true?

If both Y and Z are selected, which of the other engineers must be on the team with them?

I.

(a) Both B and A

(b) Both B and D

(c) Both C and D

(d) Only D

II.

If chemical C is used, chemical Z is added. If chemical B is not used, chemical Y is added.

III. If chemical C is used, chemical W is added. (a) (c)

7.

If B is selected and Y is rejected, the team will consist of

(III) only (I), (II) and (III)

(b) (d)

(I) and II only II only

8.

Which statement (s) must be false? I.

Y and C are never selected together.

II.

Z and B are never selected together.

III. Z and D are never selected together. (a)

III only

(b)

I, II and III

(c)

I only

(d)

II only

Group/ Team Formation 107 driving the blue car, which of the following can be true? (a) Suman is a man (b) Manjit is a woman (c) Both (a) and (b) are correct (d) (a) is correct but (b) is not

Directions for Question Nos. 9 to 10 Three men and three women are travelling in two cars (red and blue). Each car has exactly three persons. The cars cannot have all women or all men passengers. Mala, Ajit and Suman know how to drive a car. Mala and Sapna are women. Manjit and Sarat are not in the same car. Ajit and Sarat are men.

9.

If the red car can have either Manjit or Sarat but not both while Ajit drives it and Mala is

2

Directions for Question Nos. 11 to 12

The Vice Chancellor of a University wants to select a team of ve member organizing committee for the next convocation of the University to be held in March 2012. The committee members are to be selected from ve shortlisted professors (Prof. Ahuja, Prof. Banerjee, Prof. Chakravarty, Prof. Das and Prof. Equbal) and four short listed students (Prakash, Queen, Ravi and Sushil). Some conditions for selection of the committee members are given below: [IIFT 2011] i

Prof. Ahuja and Sushil have to be together

ii.

Prakash cannot be put with Ravi

iii.

Prof. Das and Queen cannot go together

iv.

Prof. Chakravarty and Prof. Equbal have to be selected

v.

Ravi cannot be selected with Prof. Banerjee.

11. If two members of the committee are students and Prof. Das is one of the members of the committee, who are the other committee members? (a) Prof. Banerjee, Prof. Chakravarty, Prakashand Queen (b) Prof. Ahuja, Prof. Banerjee, Sushil and Prakash (c) Prof. Chakravarty, Prof. Equbal, Prakash and Sushil (d) None of the above

10. If Suman and Manjit are the two passengers in the red car, who can be passengers of the blue car? [Multiple correct options] (a) Sapna and Sarat (b) Mala and Ajit (c) Sapna and Ajit (d) Mala and Sarat

12. In case Prof. Ahuja and Prof. Chakravarty are members, who are the other members who cannot be selected for the committee? (a) (b) (c) (d)

Prof. Banerjee, Prof. Equbal and Sushil Prof. Equbal, Sushil and Prakash Prof. Equbal, Prakash and Queen None of the above

Directions for Question Nos. 13 to 14 A weekly television show routinely stars six actors, J, K, L, M, N, and O. Since the show has been on the air for a long time, some of the actors are good friends and some do not get along at all. In an effort to keep peace, the director sees to it that friends work together and enemies do not. Also, as the actors have become more popular, some of them need time off to do other projects. To keep the schedule working, the director has a few things she must be aware of: [IIFT 2012] Ø

J will only work on episodes on which M is working.

Ø

N will not work with K under any circumstances.

Ø

M can only work every other week, in order to be free to lm a movie

Ø

At least three of the actors must appear in every weekly episode.

13. In a show about L getting a job at the same company J already works for and K used to work for, all three actors will appear. Which of the following is true about the other actors who may appear?

14.

(a) M, N, and O must all appear. (b) M may appear and N must appear. (c) M must appear and O may appear. (d) O may appear and N may appear. (e) Only O may appear not there. Next week, the show involves N's new car and O's new refrigerator. Which of the

3

Concept Cracker

Directions for Question Nos.15 to 20

During one week, a human resource director conducts five interviews for a new job, one interview per day, Monday through Friday. There are six candidates for the job - Ram, Shyam, Trilochan, Usha, Veena, and Kishore. No more than two candidates are interviewed more than once. Neither Shyam nor Usha nor Veena is interviewed more than once, and no other candidate is interviewed more than twice. The schedule of interviews is subject to the following conditions:



following is true about the other actors who may appear? (a) M, J, L, and K all may appear. (b) J, L, and K must appear. (c) L and K must appear. (d) Only L may appear. (e) Only K may appear not there.



(a) Monday: Shyam: Tuesday: Usha; Wednesday:Ram; Thursday: Kishore; Friday: Ram;



(b) Monday: Shyam; Tuesday: Kishore; Wednesday: Ram; Thursday: Kishore; Friday: Usha;



(c) Monday: Trilochan; Tuesday: Ram; Wednesday: Shyam; Thursday: Ram; Friday: Trilochan;



(d) Monday: Trilochan; Wednesday: Kishore; Friday: Trilochan;

[XAT 2006]

If Trilochan is interviewed, then Trilochan must be interviewed on both Monday and Friday. If Shyam is interviewed, then Usha is also interviewed, with Shyam's interview taking place earlier than Usha's interview. If Ram is interviewed twice, then Ram's second interview takes place exactly two days after Ram's first interview. If Veena is interviewed, then Kishore is interviewed twice, with Veena's interview taking place after Kishore's first interview and before Kishore's second interview. If Usha is interviewed, then Ram is also interviewed, with Usha's interview taking place on a day either immediately before or immediately after a day on which Ram is interviewed. 15. Which of the following could be a complete and accurate list of candidates the human resources director interviews and the days on which those interviews take place?

Tuesday: Thursday:

Ram; Veena;

16. If Veena is interviewed on Tuesday, then which one of the following MUST BE true?

(a) Trilochan is interviewed on Friday



(b) Usha is interviewed on Thursday



(c) Ram is not interviewed



(d) Shyam is not interviewed

17. If Kishore is not interviewed, then which one of the following MUST BE true?

(a) Ram is interviewed on Thursday



(b) Shyam is interviewed on Tuesday



(c) Trilochan is interviewed on Monday



(d) Usha is interviewed on Wednesday

18. If Shyam is interviewed, then which one of the following could be true?

(a) Kishore is interviewed on both Tuesday and Wednesday



(b) Usha is interviewed on Monday.



(c) Veena is interviewed on Tuesday



(d) Shyam is interviewed on Thursday

EBD_7743

108  Koncepts of Logical Reasoning 

19. If neither Usha nor Trilochan is interviewed, then each of the following MUST BE true EXCEPT: (a) Ram is interviewed on Monday (b) Ram is interviwed on Thursday (c) Veena is interviewed on Tuesday (d) Kishore is interviewed on Wednesday 20. If both Usha and Veena are interviewed, then which one of the following is a complete and accurate list of the days on which Kishore could be interviewed? (a) Monday, Friday (b) Tuesday, Thursday (c) Monday, Wednesday,Friday (d) Tuesday, Wednesday, Thursday Directions for Question Nos. 21 to 23: In a local pet store, seven puppies wait to be introduced to their new owners. The puppies, named Ashlen, Blakely, Custard, Daffy, Earl, Fala and Gabino, are all kept in two available pens. Pen 1 holds three puppies, and pen 2 holds four puppies. [XAT 2010]

4

24. There are ten animals–two each of lions, panthers, bison, bears, and deer–in a zoo. The enclosures in the zoo are named X, Y, Z, P and Q and each enclosure is allotted to one of the following attendants: Jack, Mohan, Shalini, Suman and Rita. Two animals of different species are housed in each enclosure. A lion and a deer cannot be together. A panther cannot be with either a deer or a bison. Suman attends to animals from among bison, deer, bear and panther only. Mohan attends to a lion and a panther. Jack does not attend to deer, lion or bison. X, Y, and Z are allotted to Mohan, Jack and Rita respectively. X and Q enclosures have one animal of the same species. Z and P have the same pair of animals. The animals attended by Shalini are: [CAT 2000]

(a)

bear and bison

(b)

bison and deer

(c)

bear and lion

(d)

bear and Panther

Group/ Team Formation 109 If Gabino is kept in pen 1, then Daffy is not kept in pen 2. If Daffy is not kept in pen 2, then Gabino is kept in pen 1. If Ashlen is kept in pen 2, then Blakely is not kept in pen 2. If Blakely is kept in pen 1, then Ashlen is not kept in pen 1. 21. Which of the following groups of puppies could be in pen 2? (a) Gabino, Daffy, Custard, Earl. (b) Blakely, Gabino, Ashlen, Daffy. (c) Ashlen, Gabino, Earl, Custard. (d) Blakely, Custard, Earl, Fala. (e) Gabino, Ashlen, Fala, Earl. 22. If Earl shares a pen with Fala, then which of the following MUST be true? (a) Gabino is in pen 1 with Daffy. (b) Custard is in pen 2. (c) Blakely is in pen 2 and Fala is in pen 1. (d) Earl is in pen 1. (e) Gabino shares a pen with Blakely. 23. If Earl and Fala are in different pens, then which of the following must NOT be true? (a) Fala shares a pen with Custar(d) (b) Gabino shares a pen with Ashlen. (c) Earl is in a higher-numbered pen than Blakely. (d) Blakely shares pen 2 with Earl and Daffy. (e) Custard is in a higher-numbered pen than Fala.

Directions for Question Nos. 25 to 27 A group of three or four has to be selected from seven persons. Among the seven are two women: Fiza and Kavita, and ve men: Ram, Shyam, David, Peter and Rahim. Ram would not like to be in the group if Shyam is also selected. Shyam and Rahim want to be selected together in the group. Kavita would like to be in the group only if David is also there. David, if selected, would not like Peter in the group. Ram would like to be in the group only if Peter is also there. David insists that Fiza be selected in case he is there in the group. [CAT 2001] 25. Which of the following is a feasible group of three? (a) David, Ram, Rahim (b) Peter, Shyam, Rahim (c) Kavita, David, Shyam (d) Fiza, David, Ram

26. Which of the following is a feasible group of four? (a) Ram, Peter, Fiza, Rahim (b) Shyam, Rahim, Kavita, David (c) Shyam, Rahim, Fiza, David (d) Fiza, David, Ram, Peter 27. Which of the following statements is true? (a) Kavita and Ram can be part of a group of four. (b) A group of four can have two women. (c) A group of four can have all four men. (d) None of the above. Directions for Question Nos. 28 to 30 Rang Barsey Paint Company (RBPC) is in the business of manufacturing paints. RBPC buys RED, YELLOW, WHITE, ORANGE, and PINK paints. ORANGE paint can also be produced by mixing RED and YELLOW paints in equal proportions. Similarly, PINK paint can also be produced by mixing equal amounts of RED and WHITE paints. Among other paints, RBPC sells CREAM paint, (formed by mixing WHITE and YELLOW in the ratio 70 : 30) AVOCADO paint (formed by mixing equal amounts of ORANGE and PINK paint) and WASHEDORANGE paint (formed by mixing equal amounts of ORANGE and WHITE paint). The following table provides the prices at which RBPC buys paints. [CAT 2003] Color

Rs./litre

RED

20.00

YELLOW

25.00

WHITE

15.00

ORANGE

22.00

PINK

18.00

28. The cheapest way to manufacture AVOCADO paint would cost (a)

Rs.19.50 per litre. (b)

Rs.19.75 per litre.

(c)

Rs.20.00 per litre. (d)

Rs.20.25 per litre.

29. WASHEDORANGE can be manufactured by mixing (a)

CREAM and RED in the ratio 14 : 10.

(b)

CREAM and RED in the ratio 3 : 1.

(c)

YELLOW and PINK in the ratio 1 : 1.

(d) RED, YELLOW, and WHITE in the ratio 1 : 1 : 2.

30.

Assume the AVOCADO, CREAM, and WASHEDORANGE each sells for the same price. Which of the three is the most protable to manufacture? (a) AVOCADO. (b) CREAM. (c) WASHEDORANGE (d) Sufcient data is not available. Directions for Question Nos. 31 to 35

K, L, M, N, P, Q, R, S, U and W are the only ten members in a department. There is a proposal to form a team from within the members of the department, subject to the following conditions: [CAT 2006] (i) (ii) (iii) (iv) (v) (vi)

A team must include exactly one among P, R, andS. A team must include either M or Q, but not both. If a team includes K, then it must also include L, and vice versa. If a team includes one among S, U, and W, then it must also include the other two L and N cannot be members of the same team. L and U cannot be members of the same team The size of a team is dened as the number of members in the team.

31. What could be the size of a team that includes K? (a) 2 or 3 (b) 2 or 4 (c) 3 or 4 (d) Only 2 (e) Only 4 32. In how many ways a team can be constituted so that the team includes N? (a) 2 (b) 3 (c) 4 (d) 5 (e) 6 33. What would be the size of the largest possible team? (a) 8 (b) 7 (c) 6 (d) 5 (e) Cannot be determined 34. Who can be a member of a team of size 5? (a) K (b) L (c) M (d) P (e) R 35. Who cannot be a member of a team of size 3? (a) L (b) M (c) N (d) P (e) Q

EBD_7743

110 Koncepts of Logical Reasoning

Group/ Team Formation

5

Directions for Question Nos. 36 to 40

At Pioneer Career Kolkata six students A, B, C, D, E and F aspire to be a member of organizing team of New Year Party The team should have at least two people with following conditions. (i)

If the size of the team is less than 4 then D and F cannot be selected together. (ii) If B is selected then D should also be selected while E should be rejected. (iii) If A is selected then exactly one from B and C should also be selected. (iv) If C is selected then E should also be selected while F should be rejected. 36. If the team selected is of the maximum possible size, then who is/are denitely not selected in the team? (a) 37.

38.

39.

40.

A

(b) D & B

(c) B & E (d) None of these If B and F are not selected in the team then who should denitely be selected? (a) C (b) E (c) D (d) A What could be the strength of team if C and E are always part of the team? (a) 2 (b) 3 or 4 (c) 2 or 3 (d) None of these Which of the following could never be part of same team (a) A D (b) B & C (c) D & F (d) None of these If team size is 3 then which one of the following can not be a part of the team? (a) B (c) A (c) F (d) None of these Directions for Question Nos. 41 to 45

Prof. Das wants to make his team for his next project. The conditions are as follows(i)

The team should have 5 students

(ii) The team must have two boys and 3 girls (iii) Name of boys is Amit, Bimal and Chandan

111

(iv) Name of girls is Deepika, Eshita, Fatima, Garima and Hena (v) Bimal and Hena can not go together. (vi) Fatima and Garima can not be selected together. (vii) Eshita and Hena can not go together (viii) Amar will always be selected in the team 41. Who will be denitely selected in the team? (a) Bimal (b) Fatima (c) Deepika (d) None of these 42. How many different Teams can be formed? (a) 6 (b) 21 (c) 24 (d) None of these 43. If Hena is selected then Who will be denitely selected in the team? (a) Bimal (b) Fatima (c) Chandan (d) None of these 44. If Hena is not selected then Who will be denitely selected in the team? (a) Bimal (b) Eshita (c) Chandan (d) None of these 45. If Chandan and Garima is selected then who all will be denitely selected in the team? (a) Amar, Chandan, Deepika, Eshita and Garima (b) Amar, Chandan, Deepika, Fatima and Garima (c) Amar, Chandan, Deepika, Hena and Garima (d) None of these Directions for Question Nos. 46 to 50 Prof Das has 8 students for his next project research, on three different shopping Malls, namely South City, Forum, and Manisquare at Kolkata. Following further information is known (i) Name of boys is Amit, Bimal and Chandan (ii) Name of girls is Deepika, Eshita, Fatima, Garima and Hena (iii) All these 8 students must have to visit 1 mall. (iv) At least 2 students will visit a Mall. (v) Chandan don’t want to go to Manisquare because of distance problem. (vi) Deepika and Fatima do not like each other hence they will never go together. (vii) If Amar goes to Manisquare then Bimal and Eshita will also go with him.

Koncepts of Logical Reasoning

(viii) Garima and Hena are very good friends hence will always be together. (ix) If Fatima goes to Southcity, then Bimal also goes with him. 46. Which one of the following is true if Fatima and Hena will go to Southcity? (a) Amit and Chandan are together (b) Deepika visit Southcity (c) Garima and Bimal are not together (d) None of these 47. If two more students with Bimal and Deepika went to Forum Mall then which one of the following is true? (a) Amit and Chandan is not in the same group (b) Bimal and Fatima is not in the same group (c) Deepika and Chandancan not be in the same group (d) None of these 48. If two more students with Bimal and Deepika went to Forum Mall then which one of the following is correct combination –

Concept Applicator (CA) 1.

(b) If B is rejected so W can be mixed and Z is present ,Y cant mixed. Again we 2 stable ingredients. So, A &C will be there.

3.

4.

5.

(4) According to the given condition, only one of them can be selected-Y/A, Z/C, A/B If B is selected A is rejected is also rejected then 2 senior member must be X, Z And 2 junior will be B, D.

6.

(4) If A is selected Y and B are rejected. Only Senior member rest X, Z.If Z is selected then C is ruled out. Only rest member be D.

7.

(b) Both Y and Z are selected then A and C are ruled out respectively.

(d) Stable chemicals: A, B and C Unstable chemicals W, X, Y and Z. Chemical cannot be mixed with another chemical – B/W, C/Y,Y/Z. If Y is present then Z&C can’t be with it. Among 4 ingredients there must be 2 stable ingredients. So, they are A and B and only unstable is x because due to presence of B.

2.

(a) Forum:-Bimal, Deepika, Fatima, Eshita (b) ManiSquare:- Amit and Chandan (c) South City:-Amit and Chandan (d) None of these 49. If two students Garima and Hena went to Forum Mall then which one of the following is correct. (a) deepika and Eshita can not be together (b) Fatima and Amit must be together (c) Garima and Chandan may be together (d) None of these 50. If only two students Garima and Hena went to Forum Mall then which one of the following is correct combination? (a) ManiSquare:- Fatima, Amit, Bimal and Eshita (b) ManiSquare:- Fatima, Amit, Bimal and Chandan (c) South City:-Amit, Bimal Fatima and Deepika (d) None of these

(a) According to the given Option iii) said 3 unstable chemical so, there must be 1 stable, which is also impossible. (d) If B is not used and Y is added then W is selected and C is rejected respectively. Again problem appear in fulll the no. stable of ingredients.

There must be both B and D. 8.

(b) All 3 conditions are false. Because none of them violating the selection procedure.

Solution (9 to 10) As per the given conditions are – (i)

Mala, Ajit and Suman- know how to drive a car

(ii) Females are Mala and Sapna and Males are Ajit and Sarat.

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112

  Group/ Team Formation

Condition-1: At least one person should be in each car who knows driving.



Condition-2 Manjit and Sarat are not in the same car.



Condition-3 Each car has three persons and the cars cannot have all women or all men passengers.



Condition-4 Each car has exactly three persons.

9. (c) If Manjit is in red car, Sarat will be in blue car, then

Red car ⇒ Manjit, Ajit (M)



Blue car ⇒ Sarat (M), Mala (F)



As per the given condition since at least one man (Sarat) and one woman (Mala) in blue car. So, one man and one woman will certainly go in red car. Therefore, ‘Manjit is a woman’ can be a true statement. Similarly, ‘Suman is a man’ can be a true statement.

10.



It is given that Suman and Manjit are the two passengers in the red car. And Suman can drive the car. And Manjit is in red car, so Sarat must go in the blue car. Also, only one out of Mala and Ajit can go in the red car (as three persons in one car) or Sapna can go in the red car. It satisfy all the conditions. Here all the four given options are correct.

K1

Tues

Wed

Thrus

S

T1

R1

K1

T2

U

V

Fri

K2

R2

This is the case when Usha interviewed after a day when Ram interviewed Mon T1 S



K2

This is the case when Usha interviewed before a day when Ram interviewed. Mon



V

113 

Tues U K1

Wed R1 V

Thrus

Fri T2

K2

R2

This is the case when Shayam and Usha are not interviewed on same day.

15. (a) 16. (d) 17. (c) 18. (a) 19. (d) 20. (c)

Concept Builder (CB)

Solution (21 to 23)

Solution (11 to 12) 11. (d) No team can be formed with the given conditiont

21. (d) Considering all the options one by one

Option 1: Since one among Ashlen and Blakely has to be in Pen 2, which is not in the given option

Solution (13 to 14)



So, Option 1 is incorrect.

13. (c) J, L and K work on an episode, as J works M must also work with him. Hence, J K L M work together.                               



Option 2: Ashlen and Blakely has to be opposite Pens.(Condition 2)



So, Option 2 is incorrect.



Option 3: According to the condition 2, if Daffy is not kept in pen 2, then Gabino is kept in pen 1.



So, Option 3 is also incorrect.



Option 5: We know that if Daffy is not kept in pen 2 then Gabino is kept in pen 1.



Option 5 is incorrect.

12. (d) Option (1) and (2) satisfy all the conditions. Option (3) does not satisfy                                                                     

14. (d) In the next week M cannot work as he works on the alternate weeks. Similarly J also cannot work. As N works K must not work.So, the remaining N O L must work. Concept Cracker (CC) with XAT Solution (15 to 20) Mon T1

R1

Tues

Wed

Thrus

S U

R2

Fri T2

Option 4 satisfies all the given conditions. 22. (b) According to the given condition: Earl shares a pen with Fala.

Two conditions arises here, either they are in Pen 1 or they are in pen 2. When they are in Pen 1, one among Ashlen and Blakely has to be in pen 1. This implies Custard has to be in pen 2. When they are in Pen 2, one among Ashlen and Blakely has to be in pen 1 along with Gabino and Daffy. This implies Custard has to be pen 2. In both the condition Custard is in pen 2. 23. (e) Case 1: When Earl is in pen 1 and Fala is in Pen 2. Also, there is one among Ashlen and Blakely in pen 1, this will mean Gabino and Daffy will have to be in Pen 2 and Custard will have to be in Pen 1. Case 2: When Fala is in pen 1 and Earl is in Pen 2. The same situation arises and custard will have to be in pen 1. So the last option which says Custard is in a higher-numbered pen than Fala is always incorrect. Concept Deviator (CD) 24. (c) Write down the information given L→ -D (Lion and Dear cannot be together) P→ -D & -Bs (A Panthor cannot be with a Dear or a Bison) Suman → -L Jack → P or B Hence we can short out following table

Animal L + P (given) B+P B+L D+B D+B

Solution (25 to 27) 25. (b) Refer to the given information and eliminate the options we will get only option (2) satises all the given conditions. 26. (c) Refer to the given information and eliminate the options, David and Peter cannot be in the same team so option (4) must be eliminated. If Rahim is selected then Shyam must be selected so option (1) must be eliminated. David and Fiza must be in the same team so option (2) must be eliminated. 27. (d) Refer to the given information and eliminate the options we will get no option that satises all the given conditions. Solution (28 to 30) Main colours = RED, YELLOW, WHITE, ORANGE, and PINK paints. ORANGE= RED and YELLOW (50 :50) PINK= RED and WHITE (50 :50) CREAMS= WHITE and YELLOW (70:30) AVOCADO = ORANGE & PINK (50 :50) WASHEDORANGE =ORANGE & WHITE (50 :50) 28. (b) Avocado-[Pink +Orange] It can be prepared directly or other different way to calculate least the cost/liter Avocado-[Pink +Orange]=(22+28)/2=20 Avocado-Pink +[Red +Yellow ] = [18 + (20 + 25) / 2] / 2 = 20.25

Mohan →L + P

Keeper Mohan Jack Shalini Suman Rita

Jack gets Bear and Panther as he doesn’t handle lion (given) that leaves. Q with Lion and Bear; which goes to Shalini as Suman doesn’t handle lion (given).

Enclosure X (given) Y (given) Q P Z (given)

Since Z & P have the same pair of animals, the other 3 enclosures viz., X, Y and Q must contain the other 3 species within themselves. So apart from Lion and Panther, Bear must be third species in X, Y, Q.

Avocado - [Red + white] + Orange = [(20+15) / 2 + 22] / 2 = 19.5 Avacado - [Red + white] + [Red + Yellow ] = [(20+25) / 2 + (20 + 15) / 2] / 2 = 20 Least cost /Lt = 19.5 29. (d) Was hedorange – Orange + White (50 : 50) = 1 : 1 = >[Red +Yellow] +White=1 : 1 : 2 [When we mix any quantity of liquid with other Total volume must be change] Now we take assumption that Orange is produce with Red and Yellow(1:1) Quantity of Orange is 2 .To maintain this we can have Orange : White 1:1 ,White must be 2 in the new mixture. Hence Red : Yellow: White = 1:1:2

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114 Koncepts of Logical Reasoning

30. (b) We already calculated Cheapest Price of Avocado-19.5/lt. Cream-White : Yellow-(7 :3)=(7 x15 +3 x 25)/10=18/lt Washed orange - (1:1) = (22 x1+15 x1)/2 = 18.5 31. (e) It is given that K is included hence L is included. If L included then so, N and U cannot be included. As U is not included, S and W are not included. One out of M and Q and one out of P and R will be included. Thus, the team will include: K, L, (M / Q) and (P /R). Hence total 4 members 32. (e) If the team includes N, then it does not include L and K. One out of M and Q can be included and one out of P, S and R can be included. If S is a member, so are U and W will also be member. So we can select one out of M and Q in 2 ways And we can select one out of P, S and R in 3 ways Hence total number of ways = 2x3=6 ways 33. (d) Since S, U and W come together but there is no restriction with S hence if S is included, the team will have S, U, W, M or Q, N. This is the largest possible team with 5 members. 34. (c) From the above answer M is the one of the options given in question 35. (a) A team of size 3 has to have one of the M and Q and one of the P& R. The only other member that can be selected all alone is N. L cannot be selected as K has to be selected with him and then the total member will increase to 4. Solution (36 to 40) Lets consider the different cases here Case (i) Teams with 2 members: A cannot be a member of any such team as otherwise at least one of B and C must be selected (From condition (iii) and if B is selected then D also and if C is selected then E also hence in this case number of teams would be more than Possible teams are: (B, D), ( C, E), (D, E) and (E, F)

Group/ Team Formation 115 Case (ii) Teams with 3 members: Possible teams with 3 members are (A, C, E), (A, B, D), (C, E, D) Case (iii) Teams with 4 members: Possible teams with 4 members are (A, B, D, F), (A, C, E, D) Team with 5 or 6 members is not possible. Concept Eliminator (CE) with CAT Guru Gajen 36. (d) Maximum possible size is 4 then from case (iii) we can say that all of them could be a member of this team. 37. (b) Possible teams without B and F are: From Case (i) (C, E), (D, E) From case (ii) (A, C, E), (C, E, D) From case (iii) A, C, E, D 38. (d) From all the three available cases A and C can be a member of the team that has strength 2 or 3 or 4. 39. (b) From the available 3 cases we can say that B and C can never be together. 40. (c) From the case (ii) we can say that F can not be a part of the team. Solution (41 to 45) It is given that Amar is always selected in the committee. Then Possible committees are as follows: Case 1. Amar, Bimal, Deepika, Eshita and Fatima Case 2. Amar, Bimal, Deepika, Eshita and Garima Case 3. Amar, Chandan, Deepika, Eshita and Fatima Case 4. Amar, Chandan, Deepika, Eshitaand Garima Case 5. Amar, Chandan, Deepika, Fatima and Hena Case 6. Amar, Chandan, Deepika, Garima and Hena 41. (c) From the above 6 cases we can say that Deepika will always be selected. 42. (a) From the above chart we have seen that total 6 such ways are possible. 43. (c) If Heena is always selected then case 5 and 6 and Chandan is always selected. 44. (b) If Heena is not selected then case 1, 2, 3 and 4 and Eshita is always selected.

45. (a) The case is similar to that of case (4) hence option (A) is correct. Solution (46 to 50) 46. (a) In this case the combination would beSouth City:-Fatima, Bimal, Garimaand Hena ManiSquare:-Deepika, Eshita Forum : - Amit, Chandan, Hence option (A) is correct. 47. (b) In this case the combination have 2 casesCase (1) South City:- Garima and Hena ManiSquare:Fatima, Eshita Forum:- Bimal, Deepika, Amit and Chandan Case (2) South City:- Amit and Chandan ManiSquare:Fatima, Eshita

Forum:- Bimal, Deepika, Garima and Hena From the options option(B) is correct. 48. (c) From solution of previous question only option (C) is correct. 49. (c) In this case the combination has 2 cases Case (1) South City:- Chandan and Deepika ManiSquare:- Fatima, Amit, Bimal and Eshita Forum:- Garima and Hena Case (2) South City:- Amit, Bimal Chandan and Fatima ManiSquare:-, Deepika and Eshita Forum:- Garima and Hena In both the cases Fatima and Amit together option (C) is correct. 50. (a) From solution of previous question Option (A) is correct

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116 Koncepts of Logical Reasoning

Syllogism

Syllogism Exam

Importance

Exam

6

117

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt Exams

Very Important

Other Aptitude Test

Very Important

INTRODUCTION Syllogism is one of the very important chapters for any aptitude test exam. In these types of questions premise has generally two statements on the basis of which a deduction has to be made for conclusion. And then that conclusion we have to select from the given options We may have a case where from the given premise no conclusion can be drawn There are two methods to solve these types of questions:(i)

Venn Diagram

(ii) Rules of deduction.

Now we will see how to derive conclusion from the given premise from these two methods but before that let’s have a look at the different components of the premise and for that take two example of premiseAll Rats are Hats……………….. (i) All Hats are Pats……………….. (ii)

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118 Koncepts of Logical Reasoning (i)

The premises normally start with qualiers or quantiers, e.g. the word All, No, some and Some – Not. The word “All” has its synonyms as – Every, Any, Each, whereas the word “Some” can also be replaced by Many, Few, A little, Most of, Much of, More, etc.

(ii) A premise consists of a subject and a predicate wherein the rst term [e.g. “Rats” in statement (i)] is the subject and the second term [e.g. “Hats” in statement (i)] the predicate. Similarly, in statement (ii), ‘Hats” is called the subject and “Pats” is the predicate. (iii) The word that occurs in both the premises is known as the ‘middle term’ (in this example since “Hats” is in both the premise hence it is called middle term). (iv) The “conclusion” of the premise middle term should not appear and conclusion should consist of the other two words (“Rats” and “Pats” in the above example) and the. The premises can be divided into 2 types (Based on qualier) (A) Universal statements [ if the qualier used in the premise is “All”, “Every”, “Any”, “Each”] (B) Particular statements [if the qualier used in the premise is “Some”, Many, Few, A little, Most of, Much of, More, etc] The premises can be divided into 2 types (Based on type of statement)(A) Positive (afrmative) statements [ if premise has no negation] (B) Negative statements [If premise has a negative term like “not” or “no] The combination of the two different categories of classications leads to four different premises as given in Table below. Universal/ Particular

Afrmative/ Negative

“All”, “Every”, “Any”, “Each”

Universal

Afrmative

“No” , “Not” “None”

Universal

Negative

Some, Many

Particular

Afrmative

Some not, Many not

Particular

Negative

The subject or the predicate can be either distributed or not distributed in the given premise. The subject and the predicate are either distributed (indicated as yes) or not distributed (indicated as no) depending on what kind of a statement it is. Table below shows the distribution pattern of the subject and the predicate. Universal afrmative Universal negative Particular afrmative Particular negative

Example

Subject

Predicate

“All”, “Every”, “Any”, “Each”

Yes

No

“No” , “Not” “None”

Yes

Yes

Some, Many

No

No

Some not, Many not

No

Yes

Please note that:(i)

Subject is distributed only in Universal statements.

(ii) Predicate is distributed in Negative statement.

RULES FOR DEDUCTIONS 1.

Every deduction should contain three and only three distinct terms.

2.

The middle term must be distributed at least once in the premises.

3.

If one premise is negative, then the conclusion must be negative.

4.

If one premise is particular, then the conclusion must be particular.

5.

If both the premises are negative, no conclusion can be drawn.

Syllogism 6.

If both the premises are particular, no conclusion can be drawn.

7.

No term can be distributed in the conclusion, if it is not distributed in the premises.

119

Now let’s take few examples to understand thisExample – 1) Find the conclusion of (i)

All Rats are Pats

(ii) All Pats are Cats Solution : Now look at the minute details of each premise(i)

Here the rst statement starts with “All” which is Universal afrmative hence it is a universal afrmative statement, and the subject (Rats) is distributed but the predicate (Pats) is not distributed.

(ii) The second statement is also Universal afrmative, the subject Pats is distributed and the predicate Cats is not distributed. (iii) Here the middle term is Pats as it occurs in both the premises. (iv) Middle term is Pats is distributed once in the premises (In this example Premise ii) hence it satises Rule [2] hence we can nd a conclusion. (v)

Conclusion will have two terms and these terms are “Rats” and “Cats”

(vi) As “Rats” is distributed in the 1st premises and “Cats” is not distributed, (vii) In nal conclusion “Rats” is distributed but “Cats” is not distributed. Conclusion: All Rats are Cats Note of Caution: The conclusion cannot be All Cats are Rats as in this case we have distributed the Venn diagram approach:(i)

All Rats are Pats: Can be represented asPats Rats

(ii) All Pats are Cats: Can be represented asPats

Cats

Overall conclusion is: Cats Pats Rats

Hence nal conclusion is all rats are cats. Example –2) Find the conclusion of (i) All Rats are Pats (ii) Some Rats are Cats

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120 Koncepts of Logical Reasoning Solution: Now look at the minute details of each premise(i)

Here the rst statement starts with “All” which is Universal afrmative hence it is a universal afrmative statement, and the subject (Rats) is distributed but the predicate (Pats) is not distributed.

(ii) Here the 2ndstatement starts with “Some” which is Particular afrmative hence it is a Particular afrmative statement, and the subject (Rats) is not distributed and the predicate (Pats) is not distributed. (iii) Here the middle term is Rats as it occurs in both the premises. (iv) Middle term is Rats is distributed once in the premises (In this example Premise i) hence it satises Rule [2] hence we can nd a conclusion. (v)

Conclusion will have two terms and these terms are “Pats” and “Cats”

(vi) In premise neither “Pats” nor “Cats” are distributed; so in conclusion “they should not be distributed. Conclusion: Some Pats are Cats or some Cats are Pats

Venn diagram approach (i)

All Rats are Pats: Can be represented asPats

(ii) Some Rats are Cats: Can be represented asCats

Rats

Rats

Overall conclusion is: Pats Rats

Cats

Pats

Rats

Cats

Or Hence nal conclusion is Some Pats are Cats or some Cats are Pats. Example – 3) Find the conclusion of (i)

All Rats are Pats

(ii) No Pats are Cats

Solution: Now look at the minute details of each premise(i)

Here the rst statement starts with “All” which is Universal afrmative hence it is a universal afrmative statement, and the subject (Rats) is distributed but the predicate (Pats) is not distributed.

(ii) Here the 2ndstatement starts with “No” which is Universal negative hence both subject (Rats) and the predicate (Pats) is distributed. (iii) Here the middle term is Pats as it occurs in both the premises. (iv) Middle term is Pats is distributed once in the premises (In this example Premise ii) hence it satises Rule [2] hence we can nd a conclusion. (v)

Conclusion will have two terms and these terms are “Rats” and “Cats”

(vi) Since one of the premises is negative hence conclusion must be negative. (vii) In premise both “Rats” and “Cats” is distributed, so in nal conclusion they should be distributed. Conclusion: No Rats are cats or No Cats are Rats

1

Syllogism

4.

Directions for Question Nos. 1 to 10

Each of these questions contains six statements followed by 4 sets of combinations of 3.Choose the set in which the statements are most logically [CAT 1990] related. 1.

A. Some of my closest friends disapprove of me. B. Some of my aardvarks.

closest

friends

5.

All roses are majestic.

C.

All roses are plants.

D.

All roses need air.

E.

All plants need air.

A. All men are men of scientic ability. B. Some women are women of scientic ability.

E.

C. Some men are men of artistic genius.

Some who aardvarks.

disapprove

of

me

are

D. Some men and women are of scientic ability. E. All men of artistic genius are men of scientic ability. F. Some women of artistic genius are women of scientic ability. (a) ACD (b) ACE (c) DEF (d) ABC

A. All those who achieve great ends are happy. B. All young people are happy. C. All young people achieve great ends. D. No young people achieve great ends.

6.

E. No young people are happy, F. Some young people are happy. (a) ADE (b) ABF (c) ACB (d) ADF 3.

B.

D. All who disapprove of me are aardvarks.

F. Some of my closest friends are no aardvarks. (a) BCD (b) ABD (c) BCE (d) ABE 2.

All roses are fragrant.

F. All plants need water. (a) CED (b) ACB (c) BDC (d) CFE

are

C. All of my closest friends disapprove of me.

A.

121

A. All candid men are persons acknowledge merit in a rival.

who

No shes breathe through lungs.

B.

All shes have scales.

C.

Some shes breed up stream.

D.

All whales breathe through lungs.

E.

No whales are shes.

F. All whales are mammals. (a) ABC (b) BCD (c) ADE (d) DEF

B. Some learned men are very candid. C. Some learned men are not persons who acknowledge merit in a rival.

A.

A.

Some mammals are carnivores.

B.

All whales are mammals.

C.

All whales are aquatic animals.

E. Some learned men are not candid.

D.

All whales are carnivores.

F. Some persons who recognize merit in a rival are learned. (a) ABE (b) ACF (c) ADE (d) BAF

E.

Some aquatic animals are mammals.

D. Some learned men are persons who are very candid.

7.

F. Some mammals are whales. (a) ADF (b) ABC (c) AEF (d) BCE

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122 Koncepts of Logical Reasoning 8.

A. First-year students of this college like to enter for the prize. B. All students of this college rank as University students.

Mary is John’s wife.

B.

Mary and John danced together.

C.

Mary wears John’s ring.

C. First-year students of this college are entitled to enter for the prize.

D. Husband and wives danced the last waltz.

D. Some who rank as University students are First-year students.

E.

E. All University students are eligible to enter for the prize. F. All those who like to are entitled to enter for the prize. (a) AEF (b) ABC (c) BEC (d) CDF 9.

12. A.

John loves Mary.

F. John danced last with Mary. (a) ADF (b) ABD (c) ACE (d) AEF 13. A.

All roses are fragrant.

B.

All roses are majestic.

C.

All roses are plants.

A.

Some beliefs are uncertain.

D.

All plants need air.

B.

Nothing uncertain is worth dying for.

E.

All roses need air.

C.

Some belief is worth dying for.

D.

All beliefs are uncertain.

E.

Some beliefs are certain.

F. All plants need water. (a) ABC (b) BCD (c) CDE (d) CEF

F. No belief is worth dying for. (a) ABF (b) BCD (c) BEF (d) BDF

14. A.

Laxman is a man.

B.

Meera is Laxman’s wife.

C.

Some women are islands.

D.

No man is an island.

B. Everyone who is sane can do logic.

E.

Meera is not an island.

C. None of your sons can do logic.

F. Laxman is not an island. (a) ADE (b) ABE (c) ADF (d) CDE

10. A. No lunatics are t to serve on a jury.

D. Some who can do logic are t to serve on a jury. E. All who can do logic are t to serve on a jury. F. Everyone who is sane is t to serve on a jury. (a) BDE (b) BEF (c) BDF (d) ADE Directions for Question Nos. 11 to 25 Each of these questions contains six statements followed by four sets of combinations of three. Choose the set in which the statements are logically [CAT 1991] related. 11. A.

No attendants are qualied.

15. A.

College students are intelligent.

B.

Intelligence is a collegian’s attribute.

C.

Ram’s sister is a college student.

D.

Ram is a college student.

E.

All intelligent persons go to college.

F. Ram is an intelligent person. (a) ADF (b) BCD (c) ABF (d) ABD 16. A. B.

Smoking causes cancer. All cigarettes are hazardous to health.

B.

Some nurses are qualied.

C. Smoking doesn’t sometimes.

C.

Some nurses are not qualied.

D.

One brand of cigarettes is Cham–Cham.

D.

All nurses are attendants.

E.

Brand X causes cancer.

E.

All attendants are qualied.

F. Cham–Cham is bad for health. (a) ABE (b) BDF (c) ABD (d) ABC

F. Some attendants are qualied. (a) ABF (b) CDF (c) BDF (d) BDE

cause

cancer

Syllogism 123 17. A.

All good bridge players play good chess.

B. Many good chess players are not bridge players. C.

Goren is a good bridge player.

D.

Goren plays chess well.

E.

Spassky plays chess well.

F. Spassky plays bridge badly. (a) ABD (b) BEF (c) ACE (d) ACD 18. A.

All snakes are reptiles.

B.

All reptiles are not snakes.

C.

All reptiles are cold blooded.

D.

All snakes lay eggs.

E.

All reptiles lay eggs.

F. Snakes are cold blooded. (a) ADE (b) BDE (c) ABE (d) ACF 19. A.

All leaves are green.

B.

All leaves have chlorophyll.

C.

Chlorophyll is green.

D.

All plants have leaves.

E.

All plants have chlorophyll.

F. Only leaves have chlorophyll. (a) BDE (b) BEF (c) BDF (d) AEF 20. A.

Some men are bald.

B.

Bald people are intelligent.

C.

Raman is a man.

D.

Raman is bald.

E.

Raman is intelligent.

E.

Some barbarians are not rude.

F. All barbarians are rude. (a) ABE (b) BCE (c) ADF (d) BDE 22. A.

Metal is good material for desks.

B.

Desks are made of metal.

C.

This object is not a desk.

D.

This object is a desk.

E.

This object is not made of metal.

F. This is made of metal. (a) ADF (b) BCE (c) ABD (d) BDF 23. A.

Mathew and Paul are brothers.

B.

Siblings are known to quarrel often.

C.

Mathew and Paul don’t quarrel.

D.

All those who quarrel are siblings.

E.

Paul and Mathew quarrel often.

F. Mathew and Paul cannot be siblings. (a) BDE (b) ADF (c) CDE (d) ABE 24. A.

Painting and music is art.

B.

Art is symptom of culture.

C.

Culture and art are complementary.

D.

Music is a form of art.

E.

Painting is a form of art.

F. Music shows culture. (a) BDF (b) AEF (c) ACE (d) CEF 25. A. Different hues primary colours.

are

obtained

from

B.

A rainbow consists of several hues.

C.

Blue and red can give different hues.

D.

Red is a primary color.

No barbarian is gentleman.

E.

Blue can give different hues.

B.

Some gentlemen are barbarians.

C.

Some gentlemen are rude.

D.

No gentlemen are rude.

F. Red can give different hues. (a) ACE (b) AEF (c) ADF (d) CDF

F. All men are intelligent. (a) ABF (b) BDE (c) BCD (d) BEF 21. A.

EBD_7743

124 Koncepts of Logical Reasoning

2

Directions for Question Nos. 26 to 34

Each question contains six statements followed by four sets of combinations of three. Choose the set in which the statements are logically related. [CAT 1992] 26. A.

All boys are good.

B.

Some girls are bad.

C.

Good people are educated.

D.

Boys are educated.

E.

Ram is an educated boy.

F. Lata is an educated girl. (a) BCF (b) ACD (c) DEF (d) ADF 27. A.

All who are sincere are graduates.

(a) BCE (c) ACF 30. A.

(b) ABD (d) ADF

Some college athletes are professionals.

B.

No college athlete is a professional.

C.

Some professionals are well paid.

D.

All professionals are well paid.

E. All well – paid persons are professionals. F. No well – paid person is a college athlete. (a) BEF (b) ABF (c) BDF (d) ACF 31. A. Some intolerant are poor – thinkers. B. Some poor – thinker is intolerant.

B.

Some graduates are not sincere.

C. All people intolerant.

C.

All who are sincere are dull.

D. No poor thinker is intolerant.

D.

All graduates are dull.

E. No poor thinker has high ideals.

E.

Some who are dull are graduates.

F. Some people with high ideals are not poor thinkers. (a) CDE (b) CDF (c) ABD (d) BCF

F. No one who is dull is sincere. (a) BEF (b) ADF (c) ABF (d) ACD 28. A. Sham won a lottery. B. Sham lost in a chess game. C. Sham is not intelligent. D. One need not be intelligent to win a lottery. E. One need not be intelligent to win a chess game. F. Sham plays chess. (a) BEF (b) ACD (c) BDE (d) BDF

32. A.

with

high

ideals

All engineers can sing.

B.

No music lover can sing.

C.

All who can sing are music lovers.

D.

All music lovers can sing.

E.

Some who can sing are engineers.

F. All engineers are music lovers. (a) ACE (b) ACF (c) ABF (d) ACD 33. A.

Some well-dressed people are sociable.

B.

All sociable people are well-dressed.

Good managers are intuitive.

C.

Some well dressed people are dull.

B.

Some managers are women.

D.

No dull person is well-dressed.

C.

Supriya is a good manager.

E.

Some sociable people are dull.

D.

Supriya is a woman.

E.

Some women are intuitive.

F.

Supriya is intuitive.

F. Some dull ones are well-dressed. (a) ACE (b) BCE (c) ADE (d) BEF

29. A.

are

Syllogism 125 34. A.

Iran and Iraq are members of the UN.

E.

B.

Iran and Iraq are not friends.

C.

Iran and Iraq are neighbours.

D.

Some UN members are friends.

F. All falcons are yellow. (a) ABC (b) CDF (c) DEF (d) BCA

E.

Not all members of the UN are friends.

F. All neighbours are not friends. (a) ABE (b) ABD (c) CDF (d) AEF Directions for Question Nos. 35 to 44 Each question contains six statements followed by four sets of combinations of three. Choose the set in which the combinations are logically related. [CAT 1994] 35. A. All vegetarians eat meat. B.

All those who vegetarians.

eat

meat

are

not

C. All those who eat meat are herbivorous. D. All vegetarians are carnivorous. E. All those who eat meat are carnivorous. F. Vegetarians are herbivorous. (a) BCE (b) ABE (c) ACD (d) ACF 36. A.

All roses have thorns.

B.

All roses have nectar.

C.

All plants with nectar have thorns.

D.

All shrubs have roses.

E.

All shrubs have nectar.

F. Some roses have thorns. (a) BEF (b) BCF (c) BDE (d) ACF 37. A.

No spring is a season.

B.

Some seasons are springs.

C.

Some seasons are autumns.

D.

No seasons are autumns.

E.

Some springs are not autumns.

F. All springs are autumns. (a) DFA (b) BEF (c) CEB (d) DEB 38. A.

All falcons y high.

B.

All falcons are blind.

C.

All falcons are birds.

D.

All birds are yellow.

39. A.

All birds are thirsty.

No wires are hooks.

B.

Some springs are hooks.

C.

All springs are wires.

D.

Some hooks are not wires.

E.

No hook is a spring.

F. All wires are springs. (a) AED (b) BCF (c) BEF (d) ACE 40. A.

Some abra are dabra.

B.

All abra are cabra.

C.

All dabra are abra.

D.

All dabra are not abra.

E.

Some cabra are abra.

F. Some cabra are dabra (a) AEF (b) BCF (c) ABD (d) BCE 41. A.

No plane is a chain.

B.

All manes are chains.

C.

No mane is a plane.

D.

Some manes are not planes.

E.

Some planes are manes.

F. Some chains are not planes. (a) ACD (b) ADF (c) ABC (d) CDF 42. A.

All dolls are nice.

B.

All toys are nice.

C.

All toys are dolls.

D.

Some toys are nice.

E.

Some nice things are dolls.

F. No doll is nice. (a) CDE (c) ACD

(b) CEF (d) BEF

43. A.

Some buildings are not sky-scrappers.

B.

Some sky-scrappers are not buildings.

C.

No structure is a sky-scrapper.

D.

All sky-scrappers are structures.

E.

Some sky-scrappers are buildings.

F.

Some structures are not buildings.

EBD_7743

126 Koncepts of Logical Reasoning (a) ACE (c) CDE 44. A.

(b) BDF (d) ACF

All bins are buckets.

5.

Some sad things are men

6. Some sad things are bad (a) 165 (b) 236 (c) 241 (d) 235

B.

No bucket is a basket.

C.

No bin is a basket.

D.

Some baskets are buckets.

2.

No bright Toms are Dicks

E.

Some bins are baskets.

3.

Some Toms are Dicks

F. No basket is a bin. (a) BDE (b) ACB (c) CDF (d) ABF

4.

Some Dicks are bright

Directions for Question Nos. 45 to 49 Each of the questions contains six statements followed by four sets of combinations of three. You have to choose that set in which the statements are logically related. [CAT 1995] 45. 1.

Some bubbies are not dubbles

2.

Some dubbles are not bubbles

3.

No one who is rubbles is dubbles

4.

All dubbles are rubbles

5.

Some dubbles are bubbles

6. Some who are rubbles are not bubbles (a) 136 (b) 456 (c) 123 (d) 246 46. 1.

Some men are bad

2.

All men are sad

3.

All bad things are men

4.

All bad things are sad

3

Directions for Question Nos. 50 to 58

Given below are six statements followed by sets of three. You are to mark the option in which the statements are most logically related. [CAT 1996]

47. 1.

All Toms are bright

5. No Tom is a Dick 6. No Dick is a Tom (a) 123 (b) 256 (c) 126 (d) 341 48. 1.

All witches are nasty

2.

Some devils are nasty

3.

All witches are devils

4.

All devils are nasty

5.

Some nasty are devils

6. No witch is nasty (a) 234 (b) 341 (c) 453 (d) 653 49. 1.

No tingo is a bingo

2.

All jingoes are bingoes

3.

No jingo is a tingo

4.

Some jingoes are not tingoes

5.

Some tingoes are jingoes

6. Some bingoes are not tingoes (a) 123 (b) 132 (c) 461 (d) 241

6. Some copper is used for tin (a) 123 (b) 356 (c) 341 (d) 125 51. 1.

An ostrich lays eggs

2.

All birds lay eggs

Some pins are made of tin

3.

Some birds can y

2.

All tin is made of copper

4.

An ostrich cannot y

3.

All copper is used for pins

5.

An ostrich is a bird

4.

Some tin is copper

5.

Some pins are used for tin

6. An ostrich cannot swim (a) 251 (b) 125 (c) 453 (d) 532

50. 1.

Syllogism 127 Some paper is wood

3.

Some ve is ten

2.

All wood is good

4.

Some six is twelve

3.

All that is good is wood

4.

All wood is paper

5.

All paper is good

5. Some twelve is ve 6. Some ten is four (a) 145 (b) 123 (c) 156 (d) 543

52. 1.

58. 1.

Poor girls want to marry rich boys

2.

Rich girls want to marry rich boys

3.

Poor girls want to marry rich girls

All bricks are tricks

4.

Rich boys want to marry rich girls

2.

Some tricks are shrieks

5.

Poor girls want to marry rich girls

3.

Some that are shrieks are bricks

4.

Some tricks are not bricks

5.

All tricks are shrieks

6. Rich boys want to marry poor girls (a) 145 (b) 123 (c) 234 (d) 456

6. Some paper is good (a) 254 (b) 246 (c) 612 (d) 621 53. 1.

6. No tricks are shrieks (a) 513 (b) 234 (c) 123 (d) 543 54. 1.

Some sand is band

Directions for Question Nos. 59 to 63 Each of the following questions contains six statements followed by four sets of combinations of three. You have to choose that set in which the third statement logically follows from the rst two. [CAT 1997]

2.

All sandal is band

3.

All band is sandal

59. A.

4.

No sand is sandal

B.

All mammals are viviparous.

5.

No band is sand

C.

Bats are viviparous.

D.

No bat is a bird.

E.

No bird is a mammal.

6. Some band is sandal (a) 231 (b) 165 (c) 453 (d) 354 55. 1.

No wife is a life

No bird is viviparous.

F. All bats are mammals. (a) ADC (b) ABE (c) FBA (d) AFC

2.

All life is strife

3.

Some wife is strife

4.

All that is wife is life

B.

Some nurses like to work.

5.

All wife is strife

C.

No woman is a prude.

D.

Some prudes are nurses.

E.

Some nurses are women.

6. No wife is strife (a) 256 (b) 632 (c) 126 (d) 245 56. 1.

Some crows are ies

60. A.

No mother is a nurse.

F. All women like to work. (a) ABE (b) CED (c) FEB (d) BEF

2.

Some ies are mosquitoes

3.

All mosquitoes are ies

4.

Some owls are ies

B.

All oranges are apples.

5.

All owls are mosquitoes

C.

Some sweet are apples.

D.

Some oranges are apples.

E.

All sweet are sour.

6. Some mosquitoes are not owls (a) 123 (b) 356 (c) 145 (d) 542 57. 1. 2.

Six is ve Five is not four

61. A.

Oranges are sweet.

F. Some apples are sour. (a) DAC (b) CDA (c) BCA (d) FEC

EBD_7743

128 Koncepts of Logical Reasoning 62. A.

Zens are Marutis.

B.

Zens are fragile.

C.

Marutis are fragile.

D.

Opels are fragile.

E.

Marutis are Opels.

F. Opels are stable. (a) ACB (b) EFD (c) CEA (d) ABC

IV. No stone is a diamond. (a) Only I and II follow (b) Only II and III follow (c) Either II or III follows (d) Either I or IV follows 66. Select the alternative that logically follows the two given statements: [SNAP 2010] Some rocks are not tables Some rocks are balloons

Dogs sleep in the open.

(a)

Some tables are not balloons

B.

Sheep sleep indoors.

(b)

Some tables are balloons

C.

Dogs are like sheep.

(c)

Some balloons are not tables

D.

All indoors are sheep.

(d)

None of the above

E.

Some dogs are not sheep.

63. A.

F. Some open are not sheep. (a) AFE (b) DCA (c) ABE (d) FBD. Directions for Question Nos. 64 to 65 In each question below are given two statements, followed by four conclusions numbered I, II, III and IV. You have to take everything given in the statements to be true although it may seem at variance with commonly accepted facts. Then decide which of the conclusions follows from the statements. Mark the right answer from (a), (b), (c), and (d) 64. Statements : 1.

All children are adults.

2.

All adults are fat.

Conclusions: I.

All fat persons are children.

II.

All children are fat

III. Only some children are fat. IV. (a) (b) (c) (d)

Some fat persons are adults. Only I and II follow Only III and IV follow Only II and IV follow Only I and III follow

65. Statements: 1.

All stones are marbles.

2.

All marbles are diamonds.

Conclusions: I.

Some diamonds are stones.

II.

Some diamonds are not marbles.

III. Every diamond is either a marble or a stone.

67. All good athletes who want to win are disciplined and have a well balanced diet. Therefore, athletes who do not have well balanced diets are bad athletes. Based on the sentence above which of the statement below strongly supports the view: (a) No bad athlete wants to win. (b) No athlete who does not eat a well balanced diet is good athlete. (c) Every athlete who eats a well balanced diet is good athlete. (d) All athletes who want to win are good athletes. Directions for Question Nos. 68 to 71 Each question consists of ve statements followed by options consisting of three statements put together in a specic order. Choose the option which indicates a valid argument, that is, where the third statement is a conclusion drawn from the preceding two statements. [IIFT 2007] 68. A. All universities appoint experienced teachers. B. Kahsi Vidyapeeth appoints experienced teachers. C. Kashi Vidyapeeth is a university. D. Some universities employ experienced teachers. E. Kashi Vidyapeeth only experienced teachers. (a) ABC (b) CDB (c) ACB (d) ACE

appoints

69. A. Migration of people augments housing problem in urban areas. B. Increase in housing problem in urban areas in detrimental to economic growth.

C. Migration of people is detrimental to economic growth.

D. Some migration does not cause increase in urban housing problem.



E. Some migration is not detrimental to economic growth. (a) CBA (b) BDE  (c) CDE   (d) BAC



70.  (A) Some drivers are drug addicted.

(B) All drug addicted drivers should be terminated.



(C) Driver Balbeer should be terminated.



(D) Driver Balbeer is drug addicted.

(E) Some drivers should be terminated.  (a)  BAE (b) BDE (c)  CDE (d) BAC 71. (A) No officer is a teacher.

(B) Mr. Rangachary is not a teacher.



(C) Mr. Rangachary is an officer.



(D) Dr. Nandi is not an officer.



(E) Dr. Nandi is a teacher. (a) ABE  (b) ABC     (c)  ADE     (d) ACB



72. Statements: I. Some drivers are technicians II. All technicians are engineers

III. Some engineers are lecturers

Conclusions:

1.

Some technicians are lecturers



2.

Some lecturers are drivers



3.

All engineers are technicians

Some engineers are drivers Only 3 follows Only 4 follows Only 3 and 4 follows None of the above

  Syllogism

129 

73. Statements:

I.



II. No fashion designers are businessmen



III. Some businessmen are traders

Some barbers are fashion designers

Conclusions:

1.

No Fashion designers are traders



2.

Some traders are not fashion designers



3.

Some fashion designers are traders



4. Some barbers are not businessmen (a) Either 1, 2 and 4 or 3, 2 and 4 follow (b) Either 1 and 4 or 3 and 4 follow (c) Either 1 and 2 or 3 and 2 follow (d) None of the above Directions for Question Nos. 74 to 75

Each of the questions below starts with a few statements, followed by four conclusions numbered 1, 2, 3 and 4. You have to consider every given statement as true, even if it does not conform to the accepted facts. Read the conclusions carefully and then decide which of the conclusion(s) logically follow(s) from the given statements, disregarding commonly known facts. [IIFT 2011]

Directions for Question Nos. 72 to 73

In each question below three statements (I, II, III) are given followed by four conclusions numbered 1, 2, 3 and 4. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements, disregarding commonly known facts. Choose the correct options (A to D) presented below. [IIFT 2008]

4. (a) (b) (c) (d)

74. Statements:

I. Some boys are scholars



II. Some teachers are boys



III. All scholars are observers

Conclusions:

1. Some scholars are boys



2. Some scholars are not boys



3. Some observers are boys



4. (a) (b) (c) (d)

Some teachers are scholars 1 and 3 follow 1, 3, and 4 follow Either 1 or 2 and 3 follow None of the above

75. Statements:

I. All teachers are professors



II. All professors are researchers



III. All researchers are consultants

EBD_7743

130 Koncepts of Logical Reasoning (a) (b) (c) (d)

Conclusions: 1.

Some consultants are teachers

2.

All professors are consultants

3.

Some researchers are teachers

4. (a) (b) (c) (d)

All professors are teachers Answer: Only 1 and 2 follow Only 1 and 3 follow Either 1 or 4 follow None of the above

77. Statement 1: Pictures can tell a story.

Statement 2: Some chickens are hens. Statement 3: Female birds lay eggs. If the above statements are facts, then which of the following must also be a fact? All birds lay eggs.

II.

Hens are birds.

IIFT 2012

III. Some chickens are not hens.

4

Directions for Question Nos. 78 to 87

Each question consists of ve statements followed by options consisting of three statements put together in a specic order. Choose the option which indicates a valid argument, that is, where the third statement is a conclusion drawn from the preceding two statements. [CAT 1999] Example: A All cigarettes are hazardous to health. B Brand X is a cigarette C Brand X is hazardous to health ABC is a valid option, where statement C can be concluded from statement A and B. 78. A. All software companies knowledge workers.

employ

B. Tara Tech employs knowledge workers. C. Tara Tech is a software company. D. Some software companies knowledge workers.

employ

Statement 3: Some storybooks have words. If the above statements are facts, then which of the following must also be a fact?

carbon

Pictures can tell a story better than words.

II. The stories simple.

in

storybook

are

very

III. Some storybooks have both words and pictures. (a) I only (b) II only (c) III only (d) None of the statement is known fact

B. Increase in carbon monoxide is hazardous to health. C. Trafc health.

congestion

is

hazardous

to

D. Some trafc congestion does not cause increased carbon monoxide. E. Some trafc congestion is not hazardous to health. (a) CBA (b) BDE (c) CDE (d) BAC 80. A.

Apples are not sweets.

B.

Some apples are sweet.

C.

All sweets are tasty.

D.

Some apples are not tasty.

E. No apple is tasty. (a) CEA (b) BDC (c) CBD (d) EAC 81. A.

E. Tara Tech employs only knowledge workers. (a) ABC (b) ACB (c) CDB (d) ACE 79. A. Trafc congestion increases monoxide in the environment.

Statement 2: All storybooks have pictures.

I.

76. Statement 1: All chickens are birds.

I.

II only II and III only I, II and III None of the statement is a known fact.

Some towns in India are polluted.

B.

All polluted towns should be destroyed.

C.

Town Meghana should be destroyed.

D.

Town Meghana is polluted.

E. Some towns destroyed.

in

India

should

be

Syllogism (a) (c) 82. A.

BDE ADE

(b) BAE (d) CDB

No patriot is a criminal.

B.

Bundledas is not a criminal.

C.

Bundledas is a patriot.

D.

Bogusdas is not a patriot.

E. Bogusdas is a criminal. (a) ACB (b) ABC (c) ADE (d) ABE 83. A.

Ant eaters like ants.

B.

Boys are ant eaters.

C.

Balaram is an ant eater.

D.

Balaram likes ants.

E. Balaram may eat ants. (a) DCA (b) ADC (c) ABE (d) ACD 84. A.

All actors are handsome.

B.

Some actors are popular.

C.

Ram is handsome.

D.

Ram is a popular actor.

E. Some popular people are handsome. (a) ACD (b) ABE (c) DCA (d) EDC 85. A.

Modern industry is technology driven.

B.

BTI is a modern industry.

C.

BTI is technology driven.

D.

BTI may be technology driven.

E. Technology driven industry is modern. (a) ABC (b) ABD (c) BCA (d) EBC 86. A. All Golmal islanders are blue coloured people. B. Some smart people are not blue coloured people. C. Some babies are blue coloured. D. Some babies are smart. E. Some smart people are not Golmal islanders. (a) BCD (b) ABE (c) CBD (d) None of the above 87.

A. MBAs are in great demand. B. Ram and Sita are in great demand. C. Ram is in great demand. D. Sita is in great demand. E. Ram and Sita are MBAs. (a) ABE (b) ECD (c) AEB (d) EBA

131

Directions for Question Nos. 88 to 97 [CAT 1999] Each question has a set of four statements. Each statement has three segments. Choose the alternative where the third segment in the statement can be logically deduced using both the preceding two, but not just from one of them. 88. A. No cowboys laugh. Some who laugh are sphinxes. Some sphinxes are not cowboys. B. All ghosts are uorescent. Some ghosts do not sing. Some singers are not uorescent. C. Cricketers indulge in swearing. Those who swear are hanged. Some who are hanged are not cricketers. D. Some crazy people are pianists. All crazy people are whistlers. Some whistlers are pianists. (a) A and B (b) C only (c) A and D (d) D only 89. A. All good people are knights. All warriors are good people. All knights are warriors. B. No foot ballers are ministers. All foot ballers are tough. Some ministers are players. C. All pizzas are snacks. Some meals are pizzas. Some meals are snacks. D. Some barkers are musk-deer. All barkers are sloth bears. Some sloth bears are musk-deer. (a) C and D (b) B and C (c) A only (d) C only 90. A. Dinosaurs are pre-historic creatures. Water-buffaloes are not dinosaurs. Waterbuffaloes are not pre-historic creatures. B. All politicians are frank. No frank people are crocodiles. No crocodiles are politicians. C. No diamond is quartz. No opal is quartz. Diamonds are opals. D. All monkeys like bananas. Some GI Joes like bananas. Some GI Joes are monkeys. (a) C only (b) B only (c) A and D (d) B and C 91. A. All earthquakes cause havoc. Some landslides cause havoc. Some earthquakes cause landslides.

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132 Koncepts of Logical Reasoning B. All glass things are transparent. Some curios are glass things. Some curios are transparent. C. All clay objects are brittle. All XY are clay objects. Some XY are brittle. D. No criminal is a patriot. Ram is not a patriot. Ram is a criminal. (a) D only (b) B only (c) C and B (d) A only 92. A. MD is an actor. Some actors are pretty. MD is pretty. B. Some men are cops. All cops are brave. Some brave people are cops. C. All cops are brave. Some men are cops. Some men are brave. D. All actors are pretty. MD is not an actor. MD is not pretty. (a) D only (b) C only (c) A only (d) B and C 93. A.

All IIMs are in India. No BIMs are in India. No IIMs are BIMs.

B.

All IIMs are in India. No BIMs are in India. No BIMs are IIMs.

C. Some IIMs are not in India. Some BIMs are not in India. Some IIMs are BIMs. D. Some IIMs are not in India. Some BIMs are not in India. Some BIMs are IIMs. (a) A and B (b) C and D (c) A only (d) B only 94. A.

Citizens of Yes Islands speak only the truth. Citizens of Yes Islands are young people. Young peoplespeak only the truth.

B. Citizens of Yes Islands speak only the truth. Some Yes Islands are in the Atlantic. Some citizens of YesIslands are in the Atlantic. C. Citizens of Yes Islands speak only the truth. Some young people are citizens of Yes Islands. Some young people speak only the truth. D. Some people speak only the truth. Some citizens of Yes Islands speak only the truth. Some people who speak only the truth are citizens of Yes Islands.

(a) A only (c) C only

(b) B only (d) D only

95. A. All mammals are viviparous. Some sh are viviparous. Some sh are mammals. B. All birds are oviparous. Some sh are not oviparous. Some sh are birds. C. No mammal is oviparous. Some creatures are oviparous and some are not. Some creatures are not mammals. D. Some creatures are mammals. Some creatures are viviparous. Some mammals are viviparous. (a) A only (b) B only (c) C only (d) D only 96. A. Many singers are not writers. All poets are singers. Some poets are not writers. B. Giants climb beanstalks. Some chicken do not climb beanstalks. Some chicken are not giants. C. All explorers live in snowdrifts. Some penguins live in snowdrifts. Some penguins are explorers. D. Amar is taller than Akbar. Anthony is shorter than Amar. Akbar is shorter than Anthony. (a) A only (b) B only (c) B and C (d) D only 97. A. A few farmers are rocket scientists. Some rocket scientists catch snakes. A few farmers catch snakes. B. Poonam is a kangaroo. Some kangaroos are made of teak. Poonam is made of teak. C. No bulls eat grass. All matadors eat grass. No matadors are bulls. D. Some skunks drive Cadillacs. All skunks are polar bears. Some polar bears drive Cadillacs. (a) B only (b) A and C (c) C only (d) C and D

5

Directions for Question Nos. 98 to 101

These questions are presented with three true statements: Fact 1, Fact 2 and Fact 3. Then, you are given three more statements (labeled I, II and III),and you must determine which of these, if any, is also a fact. 98. Fact 1: A project team consisting of males and females has four members. Fact 2: Two of the members are procient in mathematics and the other two are procient in computer programming. Fact 3: Half the members are female. If the rst three statements are true, which of the following statements must also be a fact? I.

At least one female member is procient in mathematics.

II. Two of the members are male. III. The male members are procient in computer programming. (a) II only (b) I and III only (c) II and III only (d) None 99. Fact 1: Manoj said. “Anush and I both went to a movie last night.” Fact 2: Anush said, “I was only studying last night.” Fact 3: Manoj always tells the truth, but Anush sometimes lies. If the rst three statements are true, which of the following statements must also be a fact? I.

Anush went to a movie last night.

II.

Manoj went to a movie last night.

III. Anush was studying last night. (a) II only (b) I only (c) I and II only (d) I, II and III 100. Fact 1: Chairs cost between Rs. 200 to Rs. 2000. Fact 2: Some chairs are made of aluminum. Fact 3: Some chairs are made of plastic.

Syllogism 133

If the rst three statements are true, which of the following statements must also be a fact? I.

Aluminum chairs cost more than plastic chairs.

II. Expensive chairs last longer than cheap chairs. III. Plastic chairs costs around Rs. 200 and aluminum chairs costs around Rs. 2000. (a) I only (b) II only (c) I and III only (d) None 101. Fact 1: All metros have ring roads. Fact 2: Delhi is a metro. Fact 3: Delhi has a population of more than 5 million. If the rst three statements are true, which of the following statements must also be a fact? I.

Delhi has a ring road.

II. All metros have a population more than 5 million. III. All cities with a ring road are metros. (a) I only (c) I and II only (c) I and III only (d) I, II and IIII Directions for Question Nos. 102 to 104 On the basis of the given two facts, determine which of the conclusions marked A, B, C, or D can be most logically drawn. 102. Fact 1: Some musicians play tabla. Fact 2: All the tabla players need to be trained for at least 10 years. (a) Children of tabla players may require less than 10 years of training. (b) All the musicians who have trained for at least 10 years are tabla players. (c) Some of the musicians may have been trained for at least 10 years. (d) All the musicians are tabla players. 103. Fact 1: Cloudy days tend to be windier than sunny days. Fact 2: Foggy days tend to be less windy than cloudy days.

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134 Koncepts of Logical Reasoning (a) Sunny days tend to be less windy than foggy days. (b) Sunny days tend to be windier than foggy days. (c) Foggy days and cloudy days tend to be windier than sunny days. (d) Foggy and sunny days tend to be less windy than cloudy days. 104. Fact 1: At a parking lot, a car is parked to the right of a truck and to the left of a van. Fact 2: A jeep is parked to the right of the truck. (a) The car is to the left of the jeep. (b) The jeep is to the right of the van. (c) The jeep is parked between the car and the truck. (d) The truck is to the left of the jeep. 105. He must be an IIT student; he is wearing a shirt with the IIT logo on it. This conclusion is valid only if it is true that (a) All IIT students wear shirts with logo on it. (b) IIT students never wear any shirt without IIT logo on it. (c) IIT students are required to wear shirts with IIT logo on it. (d) Only IIT students wear shirts with IIT logo on it. Directions for Question Nos. 106 to 107 There are statements followed by four conclusions numbered I, II, III, IV. You have to take the given statements to be true even if they seem to be at variance with commonly known facts and then decide which of the given conclusion logically follows from the given statements. 106. Statements: (i)

All Roses are owers.

(ii) Some owers are Vegetables. Conclusions I.

Some Vegetables are Roses.

II.

All Vegetables are Roses.

III. Some Roses are Vegetables. IV. All Roses are Vegetables.

107. Statements: (i)

Some Players are Athletes.

(ii) All Athletes are Humans. (iii) No Human is an Ant. Conclusions I.

Some Players are not Ants.

II.

Some Humans are Players.

III. No Athlete is an Ant. IV. Some Ants are not Athletes. (a) I, II and III follow. (b) II, III and IV follow. (c) I, II and IV follow. (d) All follow. 108. Manisha will eat the orange if Rajesh does not cook. Based on the information above which of the following must be true (a) Manisha will not eat the orange if Rajesh cooks. (b) If Manisha did not eat the orange, then Rajesh did cook. (c) If Manisha ate the orange, then Rajesh did not cook. (d) If Rajesh does not cook, Manisha will eat the orange. 109. Look at the sentences given below (i)

If the contract is valid, then X is liable.

(ii) If X is liable, he will be bankrupt. (iii) If the bank loans him money, he will not go bankrupt. Select the statement that is consistent with the above statements (a) The contract is valid and the bank will loan him money. (b) The contract is valid and the bank will not loan him money. (c) The contract is not valid and he will go bankrupt. (d) The contract is not valid and he is liable Directions for Question Nos. 110 to 111 Each question has two statements. You have to take the statements t be facts even if they seem to be at variance with commonly known facts. Which of the given three conclusions can then be logically concluded from the given facts?

(a)

All conclusions follow.

(b)

Only I and II follow.

(c)

None of the conclusions follow.

·

Some medical doctors are professors.

(d)

Only II and III follow.

·

Some professors perform surgery.

110. Statements:

Syllogism 135 Conclusions:

iii. No football player is a basketball player.

I.

All medical doctors perform surgery.

iv. No basketball player is a hockey player.

II.

Some medical doctors perform surgery.

Conclusions:

III. (a) (b) (c) (d)

All surgeons are medical doctors. I and III II only I, II and III None of the above conclusions

I.

Some hockey players are football players.

II.

No basketball player is a cricketer.

111. If all the three statements, marked (i), (ii) and (iii) are true, then which one of the following deductions, marked (1), (2), (3) and (4) is logically most weakly supported. i.

Some rich people are philanthropists.

ii.

No thief is a philanthropist.

iii. No good person is a thief. (a) No good people who are philanthropists are thieves. (b) No rich person who is good is a thief. (c) All rich people are good and philanthropists. (d) No rich people who are thieves are philanthropist. 112. There are four statements followed by four conclusions. You have to take the given statements to be true even if they seem to be at variance from commonly known facts, and decide which of the given conclusions logically follows from the given statements. Statements: i.

All cricketers are football players.

ii. Some hockey players are also cricketers.

III. Some hockey players are basketball players. IV. Some football players are cricketers. (a) I only (b) I, II and IV only (c) I and IV only (d) I and II only. Directions for Question Nos. 113 Question consists of a set of numbered Statements. Among them there is only one statement which logically follows from the rest. Choose this “conclusion” Statement from the given options. Example: With the four statements. (i) None of the paper setters can do logic. (ii) Insane people are not t to serve in a jury. (iii) None of the paper setters are t to be in a jury. (iv) Anyone who is sane can do logic. The correct answer is (iii), which is a logical consequence of the rest. 113. (i)

No soup, that is cold, has Croutons in it.

(ii) The soup in the cup is hot. (iii) No soup that does not have Croutons in it is t for drinking. (iv) The soup in the cup is t for drinking. (a) (i) (b) (ii) (c) (iii) (d) (iv)

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136 Koncepts of Logical Reasoning

Concept Applicator (CA) 1.

2.

3.

4.

5.

6.

7.

8.

9.

(c) If some of my closest friends are aardvarks (Statement B), but all of them disapprove of me Statement C), it implies that some of those who disapprove of me are aardvarks (Statement C). Hence BCE is the correct set. (c) All those who achieve good ends are happy (Statement A) and all young people achieve good ends (Statement C). Therefore all young people are also happy (Statement B). Hence ACB is the correct set. (d) If some learned men are candid (Statement B) but all candid men recognize merit in a rival (Statement A), it follows that some men recognize merit in a rival are learned (Statement F). Hence BAF is the correct set. (a) If All roses are plants (Statement C) and all plants need air (Statement E), it follows that roses need air (Statement D). Hence CED is the correct set. (b) If all men are men of scientic ability (Statement A) and some men are men of artistic genius (Statement C), it follows that all the men of artistic genius are men of scientic ability (Statement E). Hence ACE is the correct set. (c) If no shes breathe through lungs (Statement A) and all whales breathe through lungs (Statement D) t implies that no whales are shes (Statement E) . Hence ADE is the correct set. (d) Since all whales are mammals (Statement B) and some of them are aquatic animals (Statement C), it follows that some of the aquatic animals are mammals (Statement E). Hence BCE is the correct set. (c) If all students of this college rank as university students (Statement B), and all university students are entitled for the prize (Statement C), it follows that even the rst year students can enter for the prize (Statement E). Hence BEC is the correct set. (d) If nothing that is uncertain is worth dying for (Statement B)and all beliefs are uncertain (Statement D)then it implies that no belief is worth dying for (Statement F). Hence BDF is the correct set.

10. (b) Since everyone who is sane can do logic (Statement B), and all who can do logic is t to serve on a jury (Statement E). Therefore anyone who is sane is t to serve on a jury (Statement F). Hence BEF is the correct set. 11. (c) Some nurses are qualied (Statement B) and all nurses are attendants (Statement D). This implies that some of the attendants are qualied (Statement F). Hence BDF is the correct set. 12. (a) If Mary is John’s wife(Statement A) and the last waltz was danced by husbands and wives (Statement D), it follows that John danced last with Mary. Hence ADF is the correct set. 13. (c) If all roses are plants (Statement A) and all plants need air (Statement D), then all roses will also need air (Statement E). Hence CDE is the correct set. 14. (c) Laxman is a man (Statement A)and no man is an island (Statement D), so Laxman cannot be an island (Statement F). Hence ADF is the correct set. 15. (a) If college students are intelligent (Statement A) and Ram is a college student (Statement D), it follows that Ram is intelligent (Statement F). Hence ADF is the correct set. 16. (b) If all cigarettes are hazardous to health (Statement B) and cham-cham is a brand of cigarette (Statement D), then cham-cham would also be hazardous to health (Statement D). Hence BDF is the correct set. 17. (d) If all good bridge players play good chess (Statement A), then Goran being a good bridge player (Statement C) should also play good chess (Statement D). Hence ACD is the correct set. 18. (d) If all snakes are reptiles (Statement A) and all reptiles are cold blooded (Statement C), it implies that all snakes are cold blooded (Statement F). Hence ACF is the correct set. 19. (a) If all leaves have chlorophyll (Statement B) and all plants have leaves (Statement D), hence we can conclude that all plants have chlorophyll (Statement E),. Hence BDE is the correct set.

  Syllogism 137  This conclusion is supported by the statement given in option (2). The given statements talk about the athletes with respect to the kind of die they have. 68. (c) Option A- ABC the middle term “experienced teachers” is not distributed, hence it is ruled out. Option B- CDB the middle term “University” is not distributed, hence it is ruled out. Option C- ACB it follows all rules of Syllogism hence it is correct. Option D- ACE – There are four terms, hence ruled out. 69. (d) Option A- CBA the middle term “detrimental to economic growth” is not distributed, hence it is ruled out. Option B- BDE- The term “detrimental to economic growth” is not distributed, in the premises but it is in conclusion hence it is ruled out. Option C- CDE-the middle term “Migration” is not distributed in the premises , hence it is ruled out. Option D- BAC - it follows all rules of Syllogism hence it is correct. 70. (a) Option A- BAE - it follows all rules of Syllogism hence it is correct. Option B- BDE- There are four terms. Option C – There are four terms. Concept Builder (CB) Option D- CDB- The term “drug addict” is 26. (b) 27. (b) 28. (b) 29. (c) 30. (a) not distributed, in the premises but it is in conclusion hence it is ruled out. 31. (a) 32. (b) 33. (d) 34. (a) 35. (d) 71. (d) Option A- ABE- There is two negative terms 36. (c) 37. (a) 38. (b) 39. (d) 40. (b) in the premises and hence ruled out. 41. (c) 42. (a) 43. (b) 44. (d) 45. (d) Option B- ABC- There is two negative terms 46. (b) 47. (c) 48. (b) 49. (a) in the premises and hence ruled out. Option C- ADE- There is two negative terms Concept Cracker (CC) with XAT in the premises and hence ruled out. 50. (c) 51. (a) 52. (b) 53. (a) 54. (a) Option D- ACB- IT follows all rules of 55. (d) 56. (d) 57. (a) 58. (a) 59. (b) Syllogism hence it is correct. 60. (c) 61. (a) 62. (a) 63. (a) 72. (b) Drivers technicians 64. (a) Here in the given 2 statements middle term engineers lectures “adults” are distributed hence we can find some 73 (d) Case i Case ii conclusions. Trader Businessmen Barber Conclusion I- It follows all the rules of syllogism Fashion designer Trader Conclusion II- It follows all the rules of Barber Businessmen syllogism Case i follows 1 and 2 and case ii follows 3 Conclusion III- It is incorrect since it is not only and 2 some children but all children. 74. (a) Some boys are scholars: Conclusion IV- The term “Adult” is the middle There are two possible deductions from this term and should not come in the conclusion. statement. Hence only conclusion I and II are correct. All scholars are boys (i.e. the set of scholars is a subset of boys) 65. (a) Statement I and II can be the conclusion Some scholars are boys (i.e. the intersection 66. (d) The two given statements have particular of the sets of scholars and boys is not a null and none of them are distributed. Hence, no set) conclusion can be derived. Some teachers are boys 67. (b) The conclusion is :- athletes who do not have There are two possible deductions from this well balanced diet are bad athletes. statement. 20. (b) If bald people are intelligent (Statement B) and Raman is bald (Statement D), hence we can conclude that Raman is intelligent. BDE is the correct set. 21. (d) Some gentlemen are barbarians (Statement B) and no gentlemen are rude (Statement D). Hence we can conclude that Some barbarians are not rude. BDE is the correct set. 22. (d) Desks are made of metals (Statement B). So if an object is a desk (Statement D) it should be made of metal (Statement F). BDF is the correct set. 23. (d) Mathew and Paul are siblings (Statement A) and siblings are known to quarrel often (Statement B). Therefore we can conclude that Mathew and Paul quarrel often (Statement E). ABE is the correct set. 24. (a) Art is a symptom of culture (Statement B) and music is a form of art (Statement D), therefore music also shows culture (Statement F). BDF is the correct set. 25. (c) If Different hues are obtained from primary colours, (Statement A)and red is a primary colour, (Statement D)it implies that red also gives different hues. (Statement F) ADF is the correct set.

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138 Koncepts of Logical Reasoning All boys are teachers (i.e. the set of boys is a subset of teachers) Some boys are teachers (i.e. the intersection of the sets of teachers and boys is not a null set) All scholars are observers. There are two possible deductions from this statement. All observers are scholars (i.e. the set of observers and scholars are identical) Some observers are scholars (i.e. the set of observers is a subset of scholars) Let us analyze the conclusions: Some scholars are boys. This is true for both the cases that have been mentioned above. Some scholars are not boys. This may or may not be true because there are chances that all scholars may also be boys. Some observers are boys. We know that either all or some observers are scholars (i.e. the set of observers is denitely a subset of the set of scholars) and the intersection of the sets of scholars and boys is not a null set. This means that some observers will always be boys. Hence conclusion 3 is true. Some teachers are scholars. The intersection of set of boys and scholars is not a null set and the intersection of set of boys and teachers is not a null set. But this doesn’t mean that the intersection of boys and teachers will also be a null set. There may or may not be some teachers who are scholars. Thus this statement is incorrect. 75. (d) All teachers are professors. This means that either the set of teachers is identical to the set of professors or the set of teachers is a subset of the set of professors. All professors are researchers. This means that either the set of researchers is identical to the set of professors or the set of professors is a subset of the set of researchers. All researchers are consultants. This means that either the set of researchers is identical to the set of consultants or the set of researchers is a subset of the set of consultants. Let us analyze the conclusions now; Some consultants are teachers This is true as the set of teachers will at least be a subset of professors which at least will be a subset of researchers which will at least be a subset of consultants.

All professors are consultants. Set of professors will always be at least a subset of researchers which will always be at least a subset of consultants. Some researchers are teachers From the analysis of statement 1, we know that this statement is also true. All professors are teachers. This is not true because the set of professors is not a subset of teachers. 76. (d) If we draw the Venn diagram from the given informationBirds Hens Females Eggs

Chicken

Hence, none of the above statements is a known fact. Alternately:Or else from elimination methodStatement (i) has “all birds” i.e. “birds” is distributed but it is not distributed in any of the premises hence this one cannot be the conclusion. Statement (ii) has “hens” i.e. “hens” is distributed but it is not distributed in any of the premises hence this one cannot be the conclusion Statement (iii) from afrmative premises we cannot conclude negative conclusion hence this conclusion is also not possible. Hence, none of the above statements is a known fact. 77. (c) If we draw the Venn diagram from the given information. Story

Pictures

Words

Story Books

Hence, some story books have both words and pictures. Alternately:Or else from elimination methodStatement (i) we cannot compare from the given information. Statement (ii) “simple “ is a word that is not dened in the premise. Statement (iii) we can conclude from the premise.

Concept Deviator (CD) 78. (b) Consider Option (1) In the sequence ABC, the term “knowledge worker” is in all the statement A,B and C hence ruled out. Option (2)-It satisfy all the conditions, middle term “software company” distributed once. And statement B can be a logical deduction of A & C. Option (3) CDB- Middle term “software company “ is not distributed, note the word “some” before it in D. Option (4) ACE- This option is a bit tricky, at 1sr glance it looks correct but “knowledge worker” in the conclusion is distributed but it is not distributed in any of the premises. 79. (d) Option (1) – CBA- Middle term is used as a predicate in each of the premises, hence it is ruled out. Option (2) – BDE- the term “hazardous to health” is distributed in conclusion but not distributed to any of the premises, hence it is ruled out. Option (3) CDE- same as above. Option (4) It satisfy all the condition. 80. (a) Option (1)- It satisfy all the condition of syllogism. Option (2) – BDC- Middle term “apple’ is not distributed in any of the premises. Option (3) – CBD- the term ‘apple” is distribute in conclusion but not distributed in any of the premises. Option (4) – EAC – Same as above. 81. (b) There is so many faults in this sequence, The word “Meghna “ is .used only once. Option (B)- BAE, satisfy all the condition of syllogism. Option ADE – the term “Meghna“ and “town” came only once. Option (D) – the term “polluted” is distributed in the conclusion but not distributed in any of the premises hence this option is ruled out. 82. (a) Option (A) – ACB, It satisfy all the condition of syllogism. Option (B) – ABC- Middle term “criminal’’ is not distributed in any of the premises. Option (C) – ADE- There is no middle term. Option (D) – ABE- Middle term appears in conclusion. 83. (d) Option (A) – DCA – The term “ant eater” is not distributed in premises but it is distributed in conclusion. Option (B) – ADC - Middle term “ants’’ is not distributed in any of the premises hence ruled out. Option (C) – ABE-There are four terms. Option (D) – ACD -, It satisfy all the condition of syllogism

  Syllogism 139  84. (b) Option (A)-ACD- There are four terms. Option (B)- ABE -, It satisfy all the condition of syllogism. Option (C) – DCA- The term “actor” is not distributed in premises but it is distributed in conclusion. Option (D)- EDC- Middle term “popular actor’’ is not distributed in any of the premises hence ruled out. 85. (a) Option (A) - It satisfy all the condition of syllogism. Option (B) ABD- the term “may be” make it less appropriate than option (1). Option (C) BCA- There are four terms. Option (D) EBC- Middle term “modern’’ is not distributed in any of the premises hence ruled out. 86. (b) Option (A) BCD- The term “some babies” came thrice in the sequence. Option (B)- ABE- The term “Golmal Islanders” is negated in the conclusion but not in the premises. Option (C) – CBD- The middle term is not distributed. Hence none of the above is true. 87. (c) Option (A) ABE- Middle term “great demand’’ is not distributed in any of the premises hence ruled out. Option (B)- ECD-There are four terms. Option (C)- AEB- It satisfy all the condition of syllogism. Option (D)- EBA – The term “MBAs” are distributed in conclusion but not in any premises. 88. (c) Consider each of the statement one by one Statement A- Satisfy all rules of syllogism. Statement B- negation is not in the premise but in the conclusion. Statement C- negation is not in the premise but in the conclusion. Statement D- Satisfy all rules of syllogism. 89. (a) Statement A- the term “ Knights” distributed in conclusion but not in premise. Statement B- There are four terms. Statement C- Satisfy all rules of syllogism. Statement D- Satisfy all rules of syllogism. 90. (b) Statement A- The term “pre-historic creatures” distributed in conclusion but not in premise. Statement B- Satisfy all rules of syllogism. Statement C- There are two negations in premises. Statement D- Middle term not distributed.

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140 Koncepts of Logical Reasoning 91. (c) Statement A - Middle term not distributed. Statement B - Satisfy all rules of syllogism. Statement C- Satisfy all rules of syllogism. Statement D- There are two negations in premises. 92. (b) Statement A - Middle term not distributed. Statement B- The term “cops” came thrice. Statement C- Satisfy all rules of syllogism. Statement D- The term “MD” distributed in conclusion but not in premises. 93. (a) Statement A- Satisfy all rules of syllogism. Statement B- Satisfy all rules of syllogism. Statement C- There are two negations in premises. Statement D- There are two negations in premises. 94. (c) Statement A- The term “young” distributed in conclusion but not in premises. Statement B - Middle term not distributed. Statement C- Satisfy all rules of syllogism. Statement D- Middle term is not distributed. 95. (c) Statement A- Middle term is not distributed. Statement B- There is negation in premises but not in conclusion. Statement C- Satisfy all rules of syllogism. StatementD - Middle term is not distributed. 96. (b) Statement A - Middle term is not distributed. Statement B- Satisfy all rules of syllogism. Statement C - Middle term is not distributed. Statement D- id doesn’t satisfy. 97. (d) Statement A - Middle term is not distributed. Statement B - Middle term is not distributed. Statement C- Satisfy all rules of syllogism. Statement D- Satisfy all rules of syllogism.

102. (c) Given, two facts can be represented by Venndiagram The shaded portion represents the statement C. So, only statement C can be logically drawn. Hence, the correct option is (c). 103. (d) According to fact 1, cloudy days are windier then sunny days. By fact 2 foggy days tend to be less windy than cloudy days. We don’t have any fact regarding the relation between foggy and sunny days. So, statement D can be most logically drawn from the facts. Hence, the correct option is (d). 104. (d) By fact 1 and 2, we can say that the car is between the truck and the van. The position of jeep is to right side of truck. So, only statement (D) can be logically drawn. Hence, the correct option is (d).

Concept Eliminator (CE) with CAT Guru Gajen

106. (c) Since middle term owers is not distributed hence we cannot conclude anything. 107. (d) From (i) and (ii) we can conclude that some players are humans. Similarly we can conclude that Some Humans are Players, No Athlete is an Ant and Some Ants are not Athletes. Hence all statements follow. 108. (b, d)Option B and D are correct. 109. (b) From (i) and (ii) we get, If the contract is valid then he will bankrupt. From the above statement and (iii) If he goes bankrupt then the bank does not loan him money. Hence option B is correct. 110. (d) 111. (c) cannot be concluded. 112. (b) Statement I, II and iv are correct. 113. (b) From (iv), the soup in the cup is t drinking. By comparing the above with statement (iii), it can be concluded that the soup in the cup has croutons in it. Now with statement (i), it can be concluded that the soup in the cup is not cold i.e. the soup in the cup is hot, which is statement (ii)

98. (a) From the facts 1 and 3 we can say that there are 2 male and 2 female members in the team. But from facts 2 it is not clear that who is procient in mathematics and in computer programming. So, only statement II must also be a fact. Hence, the correct option is (a). 99. (c) According to fact 3: Manoj always tells the truth but Anush sometimes tells lies. So, Manoj and Anush both went to a movie by fact 1. So, statement I and II only must be facts. Hence, the correct option is (c). 100. (d) In fact 1, the cost of chairs is given but we don’t know what kind of chairs i.e. aluminum or plastic. So none of the statements must be true. Hence, the correct option is (d). 101. (a) By fact 1, fact 2 and fact 3 we can draw the Venn diagram So from the Venn Diagram given above we can say that Only I is a fact. Hence the correct option is (a).

105. (d) The conclusion co - relates wearing the T Shirt with the logo and being an IIT student. This will be true only if IIT students were to wear such T - shirts, otherwise reaching a denite conclusion would not be possible. Now eliminate the optionsOption (A) does not specify that non - IIT students may or may not wear such T shirts. Option (B) is a contradiction to the given statement. Option (C) says that the students are required to wear the T - shirts but this does not mean that they will necessarily wear them Option (d) is logically correct.

Logical Connectivity

7

141

Logical Connectivity

Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt Exams

Very Important

Other Aptitude Test

Very Important

INTRODUCTION Questions based on logical statements and logical connectives have a wide scope and questions related to these concepts are very frequent in different aptitude test exams. These questions can be handled easily if the basic funda is clear. Logical connectives are a statement which states that an event depends on another event. The name ‘logical’ is derived from the fact that the occurrence of the second event depends only on the occurrence or non-occurrence of the rst means these two statements are logically connected. In these types of logical connectivity situations, we deal with statements that are essentially sentences in the English language. However, in logic we are not interested about the factual correctness of the sentence. We have to only check the logical ‘truthfulness’ of the statements.

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142 Koncepts of Logical Reasoning There are different types of questions we will see each type one by oneType 1: “If X the Y” or “Y if X” or “When X then Y” or “Whenever X then Y” This statement implies that:(i)

If event X has occurred then event Y has to occur. ( X → Y)

(ii) If event Y has not occur then Even X has not occur. (~ Y → ~X) (iii) If event Y has occurred then event X may or may not occur. (No denite conclusion) Type 2: “Y only if X” This statement implies that :(i)

If event Y has occurred then event X has to occur. (Y → X)

(ii) If event X has not occur then event Y has not occur. (~ X → ~Y) (iii) If event X has occurred then event Y may or may not occur. (No denite conclusion) Type 3: “Unless X, Y” or “Y unless X” or “X otherwise Y” or “Either X or Y” This statement implies that :(i)

If event X has not occurred then Even Y has to occur. (~ X → Y)

(ii) If event Y has not occurred then Event X has to occur. (~ Y → X) (iii) If event X has occurred then event Y may or may not occur. (No denite conclusion) (iv) If event Y has occurred then event X may or may not occur. (No denite conclusion)

COMPOUND STATEMENT Type 1: “If X then Y and Z” This statement implies that :(i)

(X → Y & Z)

(ii) (~ Y or/and ~Z → ~ X) Type 2: “If X then Y or Z” or “Whenever X then Y or Z” This statement implies that :(i)

(X → Y or Z)

(ii) (~ Y & ~Z → ~ X) (iii) ( X & ~Y → Z) (iv) (X & ~Z → Y) Type 3: “Unless X, Y & Z” or “Either X or Y & Z” This statement implies that: (i)

(~X → Y or Z)

(ii) (~ Y or/and ~Z → X) Type 4:- “Only if X then Y & Z” This statement implies that :(i)

(Y & Z → X)

(ii) (~ X → ~ Y &/or ~Z)

1.

2.

3.

1

If Rajesh studies, he will pass his exam. Rajesh fails the exam. Which one of the following can be concluded from the above statements? (a) Rajesh studied for the exam (b) Rajesh did not study for the exam (c) Rajesh may or may not have studied for the exam (No denite conclusion) (d) None of these Whenever I go in sun I feel headache, which one of the following can be concluded from t his? (a) I went in sun so I have headache now. (b) I didn’t go in sun so I don’t have headache now. (c) I m feeling headache that means I went in sun. (d) None of these Ram will either buy a Car or a Flat which one of the following can be concluded from this? (a) Ram bought a Flat so he will not buy a car. (b) Ram bought a Car so he will buy a Flat. (c) Ram did not bought Car so he must have bought at. (d) None of these

Logical Connectivity 143

6.

4.

5.

The teacher gives a break, only if students are exhausted. (i) The teacher gives break (ii) The teacher did not give a break (iii) Students are not exhausted (iv) Students are exhausted (a) i and iii (b) i and iv (c) ii and iv (d) None of these Either CAT is tough or IIT JEE is easy (i) CAT is tough (ii) CAT is Easy (iii) IIT JEE is easy (iv) IIT JEE is tough.

(b)

ii and iv

(c)

(d)

None of these

iii and iv

Unless you study, you cannot crack CAT (i) you cracked CAT (ii) You could not crack CAT (iii) You did study (iv) you didn’t study

7.

(a) i and iv

(b)

(c)

(d) None of these

iv and i

ii and iii

Mobile is charged, if power is off. (i) mobile is charged (ii) mobile is not charged (iii) power is off (iv) power is not off

8.

(a) i and iv

(b)

(c)

(d) None of these

iv and i

ii and iii

Either Susmit is happy or he met his girlfriend. (i) Susmit is happy (ii) Susmit is sad (iii) Susmit met his girlfriend (iv) Susmit didn’t meet his girlfriend. (a) i and iv (b) ii and iv

Directions for Question Nos. 4 to 10 In each of the following sentences, the main statement is followed by four sentences each. Select a pair of sentences that relate logically to the given statement.

(a) ii and iii

(c) 9.

iv and i

(d) None of these

Unless boss is present employees don’t work (i) Boss is present (ii) Boss is not present (iii) Employee work (iv) Employee don’t work (a) i and iv (b) (c)

iv and i

ii and iv

(d) None of these

10. If you don’t save money you will not get reward. (i)

You saved money

(ii) you didn’t save money (iii) you got reward (iv) you didn’t get reward (a) i and iv (b)

ii and iii

(c)

None of these

iii and i

(d)

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144  Koncepts of Logical Reasoning 

2

Concept Builder

Directions for Question Nos. 11 to 15

Directions for Question Nos. 16 to 20

In each of the following sentences, the main statement is followed by four sentences each. Select a pair of sentences that relate logically to the given

In each question, the main statement is followed by four sentences. Select the pair of sentences that relates logically to the given statement.

statement. 11.

[CAT 1998]

[CAT 1997]

Either Sam is ill, or he is drunk. A. Sam is ill. B. Sam is not ill. C. Sam is drunk. D. Sam is not drunk. (a) AB (b) DA (c) AC (d) CD

12. Whenever Ram hears of a tragedy, he loses sleep. A. Ram heard of a tragedy. B. Ram did not hear of a tragedy. C. Ram lost sleep. D. Ram did not lose sleep. (a) CA (b) BD (c) DB (d) AD 13.

Either the train is late, or it has derailed. A. The train is late. B. The train is not late. C. The train is derailed. D. The train is not derailed. (a) AB (b) DB (c) CA (d) BC

14.

When I read a horror story I have a nightmare. A. I read a story. B. I did not read a horror story. C. I did not have a nightmare. D. I had a nightmare. (a) CB (b) AD (c) BC (d) AC

15.

When I eat berries I get rashes. A. I ate berries. B. I did not get rashes. C. I did not eat berries. D. I got rashes. (a) DA (b) BC (c) CB (d) None of these

16.

Either Sita is sick or she is careless. A. Sita is not sick. B. Sita is not careless. C. Sita is sick. D. Sita is careless. (a) AB (b) AD (c) BA (d) DA

17. Ram gets a swollen nose whenever he eats hamburgers. A. Ram gets a swollen nose. B. Ram does not eat hamburgers. C. Ram does not get a swollen nose. D. Ram eats hamburgers. (a) AB (b) DC (c) AC (d) BC 18. Either the employees have no confidence in the management or they are hostile by nature. A. They are not hostile by nature. B. They are hostile by nature. C. They have confidence in the management. D. They have no confidence in the management. (a) BA (b) CB (c) DA (d) BD 19. Whenever Ram reads late into the night, his father beats him. A. His father does not beat Ram. B. Ram reads late into the night. C. Ram reads early in the morning. D. Ram’s father beats him in the morning. (a) CD (b) BD (c) AB (d) None of these 20. All irresponsible parents shout if their children do not cavort. A. All irresponsible parents do not shout. B. Children cavort. C. Children do not cavort. D. All irresponsible parents shout. (a) AB (b) BA (c) CA (d) All of these

Logical Connectivity 145 Directions for Question Nos. 21 to 24 Each question has a main statement followed by four statements labelled A, B, C and D. Choose the ordered pair of statements where the rst statement implies the second, and the two statements are logically consistent with the main statement. [CAT 1999] 21. Either the orangutan is not angry, or he frowns upon the world. A. The orangutan frowns upon the world. B. The orangutan is not angry. C. The orangutan does not frown upon the world. D. The orangutan is angry. (a) CB only (b) DA only (c) AB only (d) CB and DA 22. Either Ravana is a demon, or he is a hero. A. Ravana is a hero. B. Ravana is a demon. C. Ravana is not a demon. D. Ravana is not a hero.

3

Directions for Question Nos. 25 to 26

Each question consists of a set of numbered statements. Assume that each one of these statements is individually true. Each of the four choices consists of a subset of these statements. Choose the subset as your answer where the statements therein are logically consistent among themselves:

25. A. B. C. D. E.

[SNAP 2011]

Only if the water level in the coastal areas rises, then the people change their lifestyle. People change their lifestyle only if they are rewarded. If people are rewarded, then they will not change their lifestyle. If the temperature rises, then the water level in the coastal areas rises. Whenever the water level in the coastal area rises, then the temperature rises.

(a) CD only

(b)

BA only

(c)

(d)

DB and CA

CD and BA

23. Whenever Rajeev uses the internet, he dreams about spiders. A.

Rajeev did not dream about spiders.

B.

Rajeev used the internet.

C.

Rajeev dreamt about spiders.

D.

Rajeev did not use the internet.

(a) AD

(b)

(c)

(d) DA

CB

DC

24. If I talk to my professors, then I do not need to take a pill for headache. A.

I talked to my professors.

B.

I did not need to take a pill for headache.

C.

I needed to take a pill for headache.

D.

I did not talk to my professors.

(a) AB only

(b)

(c)

(d) AB and CD

F.

CD only

DC only

Unless the people change their lifestyle, temperature rises. G. People are rewarded. H. Water level in the coastal areas does not rise. (a) C, D, F, G and H (b) G, F, D, B and H (c) E, F, G, H and B (d) None of the above 26. A. If Kumar sings, then the audiences sleep. B. If Kumar sings, then the audiences dance. C. Unless audience do not dance, the concert will be successful. D. Only if the audience dance, the concert will be successful. E. If Vina dances, then Kumar sings. F. Kumar sings only if Vina dances. G. Vina dances H. The concert is successful. (a) C, F, G, B and H (b) A, C, F, G and H (c) E, C, G, B and H (d) Both (b) and (c)

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146 Koncepts of Logical Reasoning 27. Manisha will eat the orange if Rajesh does not cook.

(a) If an automatic alarm is set off at the re department that means there must be a re.

Based on the information above which of the following must be true

(b) If the sprinklers do not start, the automatic alai-in at the re department is not set off.

(a) Manisha will not eat the orange if Rajesh cooks.

(c)

(b) If Manisha did not eat the orange, then Rajesh did cook.

(d) If there is no re, no automatic alarm is set off in the re department.

(c)

If Manisha ate the orange, then Rajesh did not cook.

(d) If Rajesh does not cook, Manisha will eat the orange 28. Look at the sentences given below (i)

If the contract is valid, then X is liable.

(ii) If X is liable, he will be bankrupt.

31. (i)

Whenever there is a re, an automatic alarm is set off in the re department.

Doing well in CAT implies doing well in JMET.

(ii) Good JMET results ensure that you get into one of the IITs or IISc. (iii) Poor CAT results do not get you an admission into any of the IIMs. (a) Doing poorly in CAT always implies doing poorly in JMET.

(iii) If the bank loans him money, he will not go bankrupt.

(b) Good CAT result ensures that one gets an admission into the IIMs.

Select the statement that is consistent with the above statements

(c)

(a) The contract is valid and the bank will loan him money.

(d) Anyone getting admission in one of the IIMs is guaranteed to get admission in one of the IITs or IISc.

(b) The contract is valid and the bank will not loan him money. (c)

The contract is not valid and he will go bankrupt.

(d) The contract is not valid and he is liable. Directions for Question Nos. 29 to 31 If all the three statements, marked (i), (ii) and (iii) are true, then which one of the following deductions, marked (1), (2), (3) and (4) can be MOST LOGICALLY deduced: 29. (i)

Whenever milk is kept in front of a child, he/ she starts crying.

(ii) Children cry if they are hungry. (iii) Unhappy children are hungry. (a) When hungry, a child likes milk.

Admissions to the IITs or IISc may mean that one has done well in CAT.

32. In all family photos, if Geeta is present, her two sisters are also present. A certain photo features ve members of his family. Among the people in that photo are Geeta’s parents with their son. Which of these statements is necessarily true? (a) Geeta is present in the photo (b) Geeta is absent from the photo (c)

Geeta may or may not be present in the photo

(d) The two sisters of Geeta are present in the photo 33. If a student sees a teacher in the class he would not sleep in the class, One day, a student does not see a teacher in the class. Which of the following statements is true?

(b) A child crying means he/she is unhappy.

(a) The student will sleep in the class.

(c)

(b) The student will not sleep in the class.

A happy child does not cry.

(d) An unhappy child usually cries. 30. (i)

Whenever there is a re, the re alarm goes off.

(ii) If the sprinklers do not start, the re alarm does not go off. (iii) If the sprinklers start, an automatic alarm is set off at the re department.

(c)

Cannot conclude anything.

(d) None of these Each question consists of a main statement followed by 4 statements in the answer options. From the given options select the one that logically follow the main statement.

Logical Connectivity 147 34. Only if it is a national holiday, Pioneer career is not open and employee enjoy together.

(b) The class will not continue or the institute is closed implies that students are agree.

(a) Today is not a national holiday it implies that Pioneer career is not open and employee enjoy together.

(c)

(b) Employees are not enjoying together and Pioneer Career is open that implies it is not a national holiday.

(d) None of these

(c)

Today is not a national holiday it implies that Pioneer career is open or employees enjoy together.

(d) None of these 35. If it is not raining then I will not walk slowly but I will walk at least 4 km. (a) If it is raining then I will walk slow but will not walk 4 km. (b) If it is raining then I will not walk slow but will not walk 4km. (c)

If I am walking slow or not walking for 4 km then it is raining.

(d) None of these 36. If Ricky Singh is not at Pioneer Career then he is at his sleeping or watching movie. (a) Ricky Singh is not sleeping and not watching movie implies that he is at Pioneer Career. (b) Ricky Singh is not sleeping or not watching movie implies that he is at Pioneer Career. (c)

Ricky Singh is at Pioneer Career it implies that he is not sleeping but watching movie.

(d) None of these 37. Unless students are agree, the class will continue and Institute will remain open. (a) Students are agree, it implies that either class will not continue or institute will not close.

Students are not agree and the class will not continue implies that institute will remain closed.

38. If you study then you will pass the exam and will get a good girlfriend. (a) You did not study then you will not pass the exam or will not get a good girlfriend. (b) You did not pass the exam and did not get a good girlfriend” implies that you did not study. (c)

You passed the exam and also got a good girlfriend implies that you have studied.

(d) None of these 39. If it is a holiday then Suvrojyoti will play cricket or will watch movie. (a) It is not a holiday then Suvrojyoti will not play cricket and will not watch movie. (b) If it is a holiday and Suvrojyoti is not playing cricket then he will watch movie. (c)

It is a holiday then Suvrojyoti will not play cricket and will not watch movie.

(d) None of these 40. Unless teacher is agree, the class will continue and there will be a test. (a) Teacher is agree it implies that either class will not continue or there will not be a test. (b) The class will not continue or there is a test implies that teacher is agree. (c)

Teacher is not agree and the class will not continue implies that there will be a test.

(d) None of these

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148 Koncepts of Logical Reasoning

Concept Applicator (CA) 1.

(b) Let X the event that Rajesh studies for the exam. And Y the event that Rajesh passes the exam. We have been given the information that X Y. Thus ~ Y ~ X means if Ram failed the exam, he did not study for it.

2.

(a) The given statement is: Whenever I go in sun I feel headache. Here let us assume that X is “I go in sun” and Y is “I have headache”

4.

6.

Boss is not present means employees don’t work. [ii & iv] Employee work means boss is present [iii & i] 10. (c) this is the case of If X then Y the conclusion should beYou don’t saved money means you will not get reward [ii & iv]. You got reward means you saved money [iii & i].

11. (b) Sam is not drunk, so he must be ill.

Option (A) I went in sun so I have headache now. (X→ Y) its right conclusion.

12. (c) As Ram did not lose sleep, means he did not hear of the tragedy.

Option (B) I didn’t go in sun so I don’t have headache now. (~ X → Y) it cannot be concluded

13. (d) The train is not late, so it must have derailed.

(c) It is the case of “Either X or Y” hence (~ X → Y) and (~ Y → X) hence option (C) is correct. (b) The conclusion can be: The teacher gives break means Students are exhausted (as given in i and iv). Students are not exhausted hence teacher will not give a break (as given in ii and iii)

5.

(b) it is the case of “Unless X, Y” the conclusion is-

Then situation is “Whenever X then Y” and implications in this is (X→ Y) and (~ Y → X). Now let’s check options one by one-

Option (C) I m feeling headache that means I went in sun. (Y → X), it cannot be concluded. 3.

9.

(a) This is the case of either X or Y the conclusion should be- CAT is not tough (means Easy) then IIT JEE is Easy [as given in ii and iii) IIT JEE is tough then CAT is tough [as given in iv and i] (d) this is the case of “unless X, Y” so in this case negation of one statement will be followed by other statement. (iv) & (ii) follow.

7.

(d) correct combination is i & iii and also iv & ii

8.

(c) The correct conclusion should be – Susmit is sad that means he met with his girlfriend [ii & iii]. Susmit didn’t meet his girlfriend so he must be happy [iv & i]

Concept Builder (CB)

14. (a) I did not have a nightmare, so I must not have read a horror story. 15. (b) I did not get rashes, which means I did not eat berries. 16. (b) If Sita is not sick, means she is careless. 17. (d) Ram does not eat hamburgers, means he does not get a swollen nose. 18. (b) If the employees have condence in the management, it follows that they are hostile. 19. (d) None of the given options relates logically to the given statements. 20. (a) As all irresponsible parents do not shout, it follows that the children cavort. Concept Cracker (CC) 21. (d) It is an example of either a or b typeConsider CB it implies –b → a, that is logically correct. DA- it implies –a → b that is also logically correct. AB it implies –a → -b that is logically incorrect. Hence both CB and DA are correct.

22. (d) It is an example of either a or b type



CD- -a →-b that is logically incorrect.





BA a → b that is also logically incorrect.





DB- -b → a, that is also logically correct





CA –a → b that is also logically correct





30. (c) Lets eliminate options one by one



Option A: Cannot be true since we cannot infer anything if we know that an automatic alarm is set off at the fire department.





Option B: If sprinklers do not start then the fire alarm doesn’t go off. Further we cannot say anything about the automatic alarm at the fire department.





Option C: When there is a fire the fire alarm goes off so sprinklers start. Hence an automatic alarm is set off at the fire department.

23. (a) Same as above AD is logically correct 24. (d) Same as above AB and CD are logically correct Concept Deviator (CD) 25. (c) (H) → Water level in the coast area does not rise

  Logical Connectivity 149  Option D: Unhappy children are hungry and hungry children cry. So choice (D) is most logically supported.

       

From (D) → Temperature will not rise





From (F) → People change their lifestyle







From (B) → People will be rewarded, which is given in G.  

31. (d) Admission in one of the IIM’s ensures good performance in CAT. This implies doing well in JMET. This ensures that you get into one of the IITS or IISC. So option D is true.



Hence Option (b) is correct.

26. (c) (G)  → Vina dances



From (E) →        Kumar sings





From (B) →        Audience dance





From (C) → Concert will be successful, which is given in H.   



28. (b) From first two statements we can conclude: ‘if the contract is valid then X would be bankrupt.’

Statement 3 is ‘If he goes bankrupt, then the bank will not loan him the money.’ From above two statements, we have: If the contract is valid, then the bank will not loan him the money. This is reflected in option B.

29. (d) Lets eliminate options one by one











32. (b) Let X be the event that Geeta is present in the photo and event Y that hersisters are present. Given that X Y.

Hence option C is correct.

27. (b) The conclusion should be if Manisha ate the orange, then Rajesh did not cook,

Option A: Child is hungry. So, from (ii) child is crying, but we cannot say anything about the child liking milk or not. Option B: A child is crying, but from this we cannot say anything regarding the child being hungry or unhappy. Option C: The statements are not speaking about the happy child.

Option (D) is similarly ruled out.

As per the given information in the photo already has three people (Geeta’s parents with their son 3 persons so remaining can be only 2 persons). Thus, Geeta cannot be in the photo.

33. (c) Here let X be the event that the student sees a teacher, and event Y be the event that he is sleeping. We have been given that X ~Y. However, we do not know anything about ~X, and the question asks us what Y will be if ~X.   We cannot conclude anything. 34. (c) This is the situation of “Only if X then Y and Z” it implies that



(Y and Z → X)





(~ X → ~Y or/and ~Z) given in option (C)

35. (c) This is the situation of “If X then Y and Z” it implies that



(X → Y and Z)





(~ Y or/and ~Z → ~X) given in option (C)

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150 Koncepts of Logical Reasoning 36. (a) This is the situation of “If X then Y or Z” it implies that-

39. (b) This is the situation of “If X then Y or Z” it implies that-

(X → Y or Z)

(X → Y or Z)

(~ Y and ~Z → ~X)

(~ Y and ~Z → ~X)

( X and ~Z → Y)

( X and ~Z → Y)

(X and ~Y → Z) given in option (A)

(X and ~Y → Z) given in option (B)

37. (b) This belongs to Type 3:- “Unless X, Y & Z”

40. (b) This belongs to Type 3:- “Unless X, Y & Z”

This statement implies that:

This statement implies that:

(~X → Y or Z)

(~X → Y or Z)

(~ Y or/and ~Z → X) Given in option (B)

(~ Y or/and ~Z → X) Given in option (B)

38. (b) This is of type 1:- “If X then Y and Z” This statement implies that: (X → Y & Z) (~ Y or/and ~Z → ~ X) given in option B

Input Output

8

151

Input Output

Exam

Importance

Exam

Importance

CAT

Important

IBPS/Bank PO

Very Important

XAT

Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt Exams

Very Important

To solved Input / Output question we have to observe the pattern looking at output and input. There is no general rule of pattern but in general we have following patterns. Ø

Arrangement of words in ascending or descending order based on the number of letters in each of the words.

Ø

Arrangement of words alphabetically in ascending or descending order.

Ø

Arrangement of words in ascending or descending order based on the number of vowels in the words.

Ø

Check for position of the word in the input for clues on the pattern.

Ø

Check for forward and backward number position of the alphabets in the given input.

Ø

Check if a certain alphabet is being replaced or position of a particular alphabet in the given set of words.

Ø

Check for common series such as squares, cube root, common ratio, common difference etc.

Ø

Check for position of the number in the input for clues on the pattern.

Ø

Check for arrangement (ascending or descending) based on the sum of the digits of each number.

EBD_7743

152 Koncepts of Logical Reasoning

1

3.

Directions for Question Nos. 1 to 5

When an input line of words is given to a word arrangement machine, it rearranges them following a particular rule in each step. Input :- India is a largest democratic country Step 1:- democratic India is a largest country Step 2:- democratic country India is a largest Step 3:- democratic country largest India is a Here 3rd step is the output step. 1.

2.

After how many steps we will get output if Input is :-Reasoning is a good and interesting subject (a) 2 (b) 3 (c) 4 (d) None of these

4.

5.

What will be the 3rd step if Input is:- Desktop is faster than tab. (a) Desktop faster than tab is (b) Desktop faster than is tab (c) Desktop faster is than tab (d) None of these

What will be the output if input is :- mobile phone is very important for businessmen (a) businessmen important phone mobile very for is (b) businessmen important mobile phone very for is (c) businessmen important mobile very phone for is (d) None of these Which step will be the output step if Input is :- mobile phone is very important for businessmen (a) 4th (b) 3rd (c) 2nd (d) None of these What would be the 2nd step if Input is :- mobile phone is very important for businessmen? (a) businessmen important mobile phone is very for (b) businessmen important mobile phone very for is (c) businessmen mobile phone is very important for (d) None of these

Directions for Question Nos. 6 to 10 When an input line of words is given to a word arrangement machine, it rearranges them following a particular rule in each step. Input: 1 12 123 1234 12345 123456 1234567 Step I: 1 123 12 1234 12345 123456 1234567 Step II: 1 123 12345 1234 12 123456 1234567 Step III: 1 123 12345 1234567 12 123456 1234 Step IV: 1 123 12345 1234567 123456 12 1234 Step V: 1 123 12345 1234567 123456 1234 12 Here step V is the output 6.

7.

What will be the output if input is 98765 98 765 6 456789 (a) 6 765 98765 98 9876 (b) 6 765 456789 98 9876 (c) 6 98 765 98765 9876 (d) None of these

9876

8.

What will be the 3rd step if 2nd step is 5 535 32 4568 25315 124578 (a) 5 32 535 4568 25315 124578 (b) 5 535 25315 32 4568 124578 (c) 5 535 25315 4568 32 124578 (d) None of these

9.

What could be the 3rd step if 2nd step is:- 2 135 12 12456 158665 1235684 (a) 2 135 12456 12 1235684 158665 (b) 2 135 12456 15866512 1235684 (c) 2 135 12456 12 158665 1235684 (d) None of these

456789 98765 456789

How many steps are required if input is:21 213 3241 56423 123456 to get the output. (a) 2 (b) 3 (c) 4 (d) None of these

10. How many steps are required to get the output if input is: 3 135 13527 258658 2486 24 (a) 2 (b) 3 (c) 4 (d) None of these

  Input Output 153  15. How many steps will be required for getting the final output if the input is Input: 11 33 55 77 99 22 44 66 88 (a) 5 (b) 6 (c) 4 (d) None of these

Directions for Question Nos. 11 to 15

Directions for Question Nos. 16 to 20

A number sorting machine when given an in-put of numbers, rearranges the numbers in a particular manner step by step as indicated below till all the numbers are arranged in a particular order. Given below is an illustration of this arrangement: Input : 41 121 62 21 81 119 15 45 63 96 Step I : 15 41 121 62 21 81 119 45 63 96 Step II: 15 41 62 21 81 119 45 63 96 121 Step III: 15 21 41 62 81 119 45 63 96 121 Step IV: 15 21 41 62 81 45 63 96 119 121 Step V: 15 21 41 45 62 81 63 96 119 121 Step VI: 15 21 41 45 62 63 81 96 119 121 (This is the final arrangement and Step VI is the last

A word arrangement machine, when given a particular input, re-arranges it following a particular rule. The following is the illustration of the input and the steps of arrangement: Input: 99 if 77 my work Step l : if 99 77 my work Step II : if 77 99 my work Step III : if 77 my 99 work. And it is the output. Now, study the logic given above and answer the questions that follow:

step for this input)

16. What will be the output if input is:- “11 22 33 what is your name”

11. What will be the Step II for the following input? Input: 47 62 19 99 88 42 25 79 (a) 19 25 47 62 88 42 79 99 (b) 19 47 62 88 42 25 79 99 (c) 19 47 62 88 25 42 79 99 (d) None of these



(a) Is 11 name 22 your what 33



(b) Is 11 name 22 your 33

what



(b) Is 33 name 22 your 11

what



(d) None of these

12. Which of the following will be the third step for the following input ? Input : 45 77 95 132 28 16 150 54 99 125 (a) 16 28 45 77 95 150 54 99 125 132 (b) 16 28 45 95 77 54 99 125 132 150 (c) 16 28 45 77 95 132 54 99 125 150 (d) None of these 13. If the second step for an input is as given below, what will be the fourth step for the same input? Step II: 11 51 32 88 99 132 101 200 (a) 11 32 88 51 99 101 132 200 (b) 11 32 99 51 88 101 132 200 (c) 11 32 51 88 99 101 132 200 (d) None of these 14. If following is the fifth step of an input, what will be the third step ? Step V: 11 32 44 82 69 91 49 55 99 115 (a) 11 32 44 82 69 91 49 55 99 115 (b) 11 32 82 69 44 91 49 55 99 115 (c) 11 32 82 69 91 44 49 55 99 115 (d) None of these

17. After how many steps we will get output if input is :- 3 5 7 9 a e i o u

(a) 2

(b) 3



(c) 4

(d) None of these

18. What will be the output if input is: 3 5 7 9 a e I o u

(a) a

3

e

5

I

7

o 9

u



(b) a

3

e

5

o

7

I

9

u



(c) a

3

e

5

I

7

u 9

o



(d) None of these

19. If output is following then the input is : a 3 e 5 I 7 o 9 u

(a) 3

5

7

9

a

e

I

o

u



(b) 3

5

7

a

e

I

o

u

9



(c) 3

5

7

e

I

7

o

u

9



(d) None of these

20. After how many steps we will get output if input is :-11 22 33 what is your name

(a) 2

(b) 3



(c) 4

(d) None of these

EBD_7743

154 Koncepts of Logical Reasoning

2

Directions for Question Nos. 21 to 25

An electronic device rearranges numbers step-by-step in a particular order according to a set of rules. The device stops when the nal result is obtained. In this case the device stops at Step V. Input: 85 16 36 04 19 97 63 09 Step I 97 85 16 36 04 19 63 09 Step II 97 85 63 16 36 04 19 09 Step III 97 85 63 36 16 04 19 09 Step IV 97 85 63 36 19 16 04 09 Step V 97 85 63 36 19 16 09 04 [SNAP 2006] 21. Which of the following will be Step III for the input below? Input: 09 25 16 30 32 18 17 06 (a) 32 09 25 16 30 18 17 06 (b) 32 30 09 25 16 19 17 06 (c) 32 30 09 25 16 18 17 06 (d) 32 30 25 09 16 18 17 06

22. What is last step for the input below? Input: 16 09 25 27 06 05 (a) Step II (b) Step III (c) Step IV (d) None of the above 23. What is the output of Step V for the input below? Input: 25 08 35 11 88 67 23 (a) 88 67 35 25 23 11 08 (b) 88 67 35 25 23 08 11 (c) 08 11 23 25 35 67 88 (d) None of the above 24. Which one of the following would be last step for the input below? Input: 03 31 43 22 11 09 (a) Step II (b) Step III (c) Step IV (d) None of the above 25. If the output of Step IV is as given below, what was the input? Step IV: 92 86 71 69 15 19 06 63 58 (a) 86 92 69 71 15 19 06 63 58 (b) 15 86 19 92 06 69 63 58 71 (c) 15 19 06 63 58 86 92 69 71 (d) None of the above

Directions for Question Nos. 26 to 29

t

A word arrangement machine, when given a particular input, rearranges it using a particular rule. The following is the illustration and the steps of the arrangement in each question nd the options that are incorrect. [Multiple correct] INPUT : lemon apple Choco college girl dreams rooms book calf STEP 1: choco apple Lemn college girl dream calf book room STEP 2 : lemon apple Choco dream girl college room book calf STEP 3 : calf lemon Apple book choco college room girl dream STEP 4 : apple calf Lemn book choco college dream room girl STEP 5 : lemon calf Apple college choco book girl room dream STEP 6 : dream lemon calf room apple book girl choco college [IIFT 2006]

26. Which will not be Step 10 for the given input (a) Calf lemon dream room apple (b) apple calf lemon book choco (c) lemon college dream choco calf (d) dream college lemon room calf

book college room choco

college dream book apple

choco room girl girl

girl girl apple book

27. Indicate all the step numbers for which the following will not be an output dream lemon calf book apple room girl choco college (a) Step 7 (b) Step 8 (c) Step 9 (d) Step 12

Input Output 155 28. Mark all the arrangements that do not fall between step numbers 11 and 15 (a) choco book dream calf college lemon apple girl (b) book dream college girl lemon calf apple room (c) book dream college room lemon girl apple calf (d) college dream book girl lemon calf choco room 29. Mark two arrangement which will fall as consecutive steps at any time. (a) Calf lemon dream room apple book college choco (b) Choco book dream calf college lemon apple girl (c) Book dream college girl lemon room apple calf (d) College dream book girl lemon room choco calf

room choco choco apple girl room choco apple

Directions for Question Nos. 30 to 33

Directions for Question Nos. 34 to 36

A number arrangement machine, when given a particular input, rearranges it following a particular rule. Illustrations of the input and the steps of arrangement is given below. Input: 245, 316, 436, 519, 868, 710, 689 Step 1: 710, 316, 436, 519, 868, 245, 689 Step 2: 710, 316, 245, 519, 868, 436, 689 Step 3: 710, 316, 245, 436, 868, 519, 689 Step 4: 710, 316, 245, 436, 519, 868, 689 Step 4 is the last step for the given input

A wood arrangement machine, when given a particular input, rearranges it following a particular rule. Following is the illustration of the input and the steps of arrangement:

[IIFT 2008:] 30. If the input is given as ― 655, 436, 764, 799, 977, 572, 333 , which of the following step will be ―333, 436, 572, 655, 977, 764, 799 ? (a)

Step Third

(b)

Step Second

(c)

Step Fourth

(d)

None of the above

31. How many steps will be required to get the nal output from the following input? Input: 544, 653, 325, 688, 461, 231, 857 (a)

6

(b)

5

(c)

4

(d)

None of the above

32. Step third for an input is ―432, 433, 542, 666, 734, 355, 574 What will be the rst step for the input? (a)

666, 542, 432, 734, 433, 574, 355

(b)

542, 666, 734, 432, 433, 574, 355

(c)

355, 574, 433, 432, 734, 666, 542

(d) Cannot be determined 33. What will be the third step for the following input? Input: 653, 963, 754, 345, 364, 861, 541 (a)

541, 345, 754, 963, 364, 816, 653

(b)

541, 345, 364, 653, 963, 754, 861

(c)

541, 345, 364, 963, 754, 861, 653

(d)

541, 345, 364, 653, 861, 754, 963

Input: She was interested in doing art lm Step 1: art she was interested in doing lm Step 2: art was she interested in doing lm Step 3: art was in she interested doing lm Step 4: art was in lm she interested doing Step 5: art was in lm doing she interested Step 5 is the last step of the given input. Now study the logic and rules followed in the above steps, nd out appropriate step for the question given below for the given input.

[IIFT 2008:]

34. Which of the following will be the last step for the input given below? Input: he is going out to search air (a)

out is air to going search he

(b)

out is air to search going he

(c)

search he out is air to going

(d) None of the above 35. If step 2 of an input is “not is the casino considering legal action”, which step is: ―not is casino action legal the considering ? (a)

Step: 3

(b)

Step: 6

(c)

Step: 4

(d)

None of the above

36. How many steps will be required to get the nal output from the following input? Input: Father needs to check on the boy (a)

Four

(b)

Five

(c)

Six

(d)

None of the above

EBD_7743

156 Koncepts of Logical Reasoning Directions for Question Nos. 37 to 39 A word arrangement machine, when given a particular input, rearranges it using a particular rule. The following is the illustration and the steps of the arrangement INPUT: Smile Nile Style Mile Shine Wine Mine Swine Bovine Feline STEP 1: Smile Nile Style Mile Shine Wine Bovine Feline Mine Swine STEP 2: Style Mile Smile Nile Shine Wine Bovine Feline Mine Swine STEP 3: Style Mile Smile Nile Wine Shine Bovine Feline Mine Swine STEP 4: Mile Style Nile Smile Wine Shine Feline Bovine Swine Mine STEP 5: Nile Smile Mile Style Wine Shine Swine Mine Feline ovine STEP 6: Nile Smile Mile Style Wine Shine Feline Bovine Swine Mine STEP 7: Mile Style Nile Smile Wine Shine Feline Bovine Swine Mine [IIFT 2009] 37. Which of the following will be step 14 for the given input: (a)

Style

Mile

Smile

Nile

Wine

Shine

Bovine

Feline

Mine

wine

(b)

Smile

Nile

Style

Mile

Shine

Wine

Mine

Swine

Bovine

Feline

(c)

Mile

Style

Nile

Smile

Shine

Wine

Feline

Bovine

Swine

Mine

(d)

Style

Mile

Smile

Nile

Shine

Wine

Bovine

Feline

Mine

Wine

38. Mark the arrangement that does not fall between step numbers 12 and 14. (a)

Style

Mile

Smile

Nile

Wine

Shine

Bovine

Feline

Mine

Swine

(b)

Mile

Style

Nile

Smile

Wine

Shine

Feline

Bovine

Swine

Mine

(c)

Style

Mile

Smile

Nile

Shine

Wine

Bovine

Feline

Mine

Swine

(d) Smile Nile Style Mile Shine Wine Bovine Feline Mine Swine 39. If the arrangement is repeated which of the steps given below is same as the INPUT row? (a)

Step 9

(b)

Step 11

(c)

Step 20

(d)

Step 14

Directions for Question Nos. 40 to 41 A number arrangement machine, when given a particular input, rearranges it using a particular rule. The following is the illustration and steps of the arrangement. Input 105 241 67 347 150 742 292 589 Step I: 67 105 241 347 150 742 292 589 Step II: 67 742 105 241 347 150 292 589 Step III: 67 742 105 589 241 347 150 292 Step IV: 67 742 105 589 150 241 347 292 Step V: 67 742 105 589 150 347 241 292 Arrangement at Step V is the last for the given input.

[IIFT 2012]

40. What should be the fourth step of the following input? 64

326

187

87

118

432

219

348

(a) 64

432

87

326

118

187

219

348

(b) 64

432

87

348

326

187

118

219

(c) 64

432

87

348

118

326

187

219

(d) None of the above. 41. How many steps will be required to get the nal output from the following input? 319 318 746 123 15 320 78 426 (a)

Four

(b)

Five

(c)

Six

(d)

Seven

Input Output 157

3

Directions for Question Nos. 42 to 46

In a game, "words" (meaningful or meaningless) consist of any combination of at least ve letters of the English alphabets. A "sentence" consists of exactly six words and satises the following conditions: The six words are written from left to right on a single line in alphabetical order. The sentence can start with any word, and successive word is formed by applying exactly one of three operations to the preceding word: delete one letter; add a letter; replace a one letter with another. At the most three of the six words can begin with the same letter. Except for the rst word, each word is formed by a different operation used for the preceding word.

[XAT 2008]

42. Which one of the following could be a sentence in the word game? (a) Bzaeak blaeak laeak paeak paea paean (b) Crobek croeek roeek soeek sxoeek xoeek (c) Doteam goleam golean olean omean omman (d) Feted freted reted seted setega seteg (e) Forod forol forols forpls orpls morpls 43. The last letter of the English alphabet that the rst word of a sentence in the word game can begin withis (a) t (b) w (c) x (d) y (e) z 44. If the rst word in a sentence is “illicit” and the fourth word is “licit”, then the third word can be (a) Implicit (b) Explicit (c) Enlist (d) Inlist (e) Elicit 45. If “clean” is the rst word in a sentence and “learn” is another word in the sentence, then which one of the following is a complete and accurate list of the positions “learn” could occupy? (a) Third (b) Second, third, fourth (c) Third, fourth (d) Third, fourth, fth (e) Third, fourth, fth, sixth

46. If the rst word in a sentence consists of ve letters, then the maximum number of letters that the fth word in the sentence could contain is (a)

Four

(b)

Five

(c)

Six

(d)

Seven

(e)

Eight

Directions for Question Nos. 47 to 50 A computer program converts a two-digit number into another number in ve steps. The following example illustrates the operations on six two-digit numbers. Input:

20 25 31 11 07

72

Step-1:

02 07 04 02 07

09

Step-2:

24 74 47 15 56

153

Step-3:

44 99 78 26 63

225

Step 4:

07 17 14 07 08

08

Step-5:

09 24 18 09 15

17

47. If the output in step-2 of a given input is 02, what would be the nal output of that input? (a)

3

(b)

7

(c)

13

(d)

0

48. If the input number is 17, what is the output of step-5? (a)

189

(b)

271

(c)

24

(d)

39

49. What is the input for the output 8 in step-5? (a)

0

(b)

1

(c)

2

(d)

3

50. If the input number is 11, what is the output of step-3? (a)

9

(b)

8

(c)

26

(d)

11

EBD_7743

158 Koncepts of Logical Reasoning

4

Directions for Question Nos. 51 to 54

Directions for Question Nos. 59 to 60

A number arrangement machine, when given a particular input, rearranges it by following a particular rule.The following is the illustration of the input and the steps of arrangements: Input: 22, 37, 77, 89, 16, 21, 47 Step I : 4, 1, 5, 8, 7, 3, 2 Step II : 12, –3, 21, 60, 45, 5, 0 Step III : 3, –3, 3, 6, 9, 5, 0 Step IV: 5, 0, 8, 13, 20, 18, 17 Step V : 5, 0, 8, 4, 2, 9, 8 Understand the rule by following the patterns carefully and then answer the questions that follow:

A number arrangement machine, when given a particular input, rearranges it by following a particular rule. The following is the illustration of the input and the steps of arrangements: Input:2, 4, 5, 7, 8, 11 Step 1:- 4, 16, 25, 49, 64, 121 Step 2:- 4, 7, 7, 4, 1 4 Step 3:- 16, 49, 49, 16, 1, 16 Step 4:- 7, 4, 4, 7, 1, 7 Step 5:- 49, 16, 16, 49, 1, 49 Step 6:- 4, 7, 7, 4, 1, 4 and so on…

51. If step III is What could be the output? :“1 2 3 4 5 6” then (a) 3 5 8 2 7 1 (b) 2 5 8 2 7 1 (c) 3 5 8 2 7 5 (d) None of these 52. Which of the following statement is correct? (a) If step II is known then we can calculate Input (b) If step IV is known then we can nd Step II (c) If step II is known we can calculate step I. (d) None of these 53. If step II is 5, 12, 45, 77, 60, 32 then what is the step I? (a) 3, 4, 7, 9, 4, 6, (b) 3, 4, 7, 9, 8, 6, (c) 3, 4, 7, 9, 8, 5, (d) None of these 54. If input is 11, 22, 33, 44, 55, 66 then what is the output. (a) 2 6 1 4 5 9 (b) 2 6 3 4 5 9 (c) 2 6 1 4 5 5 (d) None of these 55. Which one of the following is not correct? (a) If step IV is known then we can not nd input. (b) If output is given then we can not nd step II. (c) If step IV is known then we can not nd Step III (d) None of these

56. If input is 2, 3, 4, 5, 6, 7, 8, 9 then in 100th step will have how many terms as zeros? (a) 1 (b) 2 (c) 3 (d) None of these 57. Which of the following statement is not correct? (a) Irrespective of the value in Input number of Zeros in 99th step is same as that in 100th step. (b) If 48th step has 4 zeros then 54th step will also have 4 zeros. (c) If 48th step has 4 zeros then 44th step will also have 2 zeros. (d) None of these 58. If input has all the prime numbers less than 100 then 100th step will have how many zeros as an element. (a) 1 (b) 2 (c) 4 (d) None of these 59. if input is 11, 22, 33, 44, 55 then what will be 100th step? (a) 4 7 0 1 1 (b) 16 49 0 1 1 (c) 16 49 1 1 1 (d) None of these 60. Which of the following statement is not correct? (a) Step 3 can have a 2 digit prime number (b) Step 5 must have single digit numbers (c) Step 5 can not have zero as an element (d) None of these

Input Output 159 Directions for Question Nos. 61 to 65

Directions for Question Nos. 66 to 70

A word – number arrangement machine, when given a particular input, rearrange them in the dictionary order by following a particular rule. The following is the illustration of the input and the steps of are rearranged by the machine.

A number arrangement machine, when given a particular input, rearranges it by following a particular rule. The following is the illustration of the input and the steps of arrangements: Input:- 2,

3,

5,

7,

9,

11

Input : cool, don, cook, cent, cart, chit, cinema

Step 1:- 4,

6,

10,

14,

18,

22

Step i: cool, cinema, cook, cent, cart, chit, don

Step 2:- 12,

18,

30,

42,

54,

66

Step ii: chit, cinema, cook, cent, cart, cool, don

Step 3:- 48,

72,

120, 168,

216,

264

Step iii: chit, cinema, cart, cent, cook, cool, don

Step 4:- 240, 360, 600, 840, so on

1080,

1320 and

Step iv: chit, cent, cart, cinema, cook, cool, don Step v: cart, cent, chit, cinema, cook, cool, don 61. if input is “ I want freedom to learn and to earn” then what will be the output (a)

and earn learn freedom I to to want

(b)

and earn freedom I learn to to want

(c)

earn freedom I learn to to want and

(d) None of these 62. if output is “dog, doll, dome,don, dope, door” and this output came after maximum number of steps then what is the total number of steps? (a)

5

(b)

6

(c)

7

(d)

None of these

63. Which one of the following is not correct? (a)

“Bat, get, net, let, set, war, wet” must be 3rd step of any input with these words.

(b)

“Bat, get, net, let, set, war, wet” can not be 4thstep of any input with these words.

(c)

“Bat, get, net, let, set, war, wet” must be in maximum 3rd step of any input with these words.

(d) None of these 64. If 2nd step of an input is “Bat, get, net, let, set, war, wet” then how many possibility exist for input? (a)

24

(b)

30

(c)

74

(d)

None of these

65. If input has ‘k’ number of words then what is the maximum to minimum number of steps? (a)

(k-1):1

(b)

K:1

(c)

(K+1) : 1

(d)

None of these

66. If input is 7 consecutive integer then after how many steps each term will be divisible by 343 (a) 6 (b) 7 (c) 8 (d) None of these 67. If input is 7 consecutive integer then after how many steps at least one term will be divisible by 343 (a) 6 (b) 7 (c) 13 (d) None of these 68. After how many steps digital sum of all the terms will be 0. (a) 3 (b) 6 (c) 9 (d) None of these 69. After how many steps each term will be divisible by 1000? (a) 14 (b) 11 (c) 7 (d) None of these 70. Which one of the following is not correct? (a) After 4th step each term will be divisible by 10 (b) After 6th step each term will have at least four digits. (c) In Kth step each term will be divisible by 2K. (d) None of these Directions for Question Nos. 71 to 75 Study the following information carefully to answer the questions given below. In a toy exhibition, a machine processes a given input by the following rule. Participants are shown one by one till it reaches its last step. Following is an illustration of the working of this machine. Input :tea am drink free when u I Step I :free tea am I drink when u Step II :I free tea u am drink when Step III: u I free when tea am drink Step IV: when u I drink free tea am Step V :drink when u am I free tea and so on.

71. If step 3 is “ dil wale dulhaniyan le jayenge part two” then which one of the following is the 1st step? (a) “dulhaniyan jayenge dil wale two le part” (b) “wale dulhaniyan jayenge dil part two le ” (c) “wale dulhaniyan jayenge part two le dil” (d) None of these 72. If input is “ you are not as good as Amitabh” then which of the following statement is correct? (a) “Amitabh as you as are not good” is in step 50

(a)

“Amitabh as you as are not good” is in step 50

(b)

“good as Amitabh not as you are” is in step 102

(c)

“are not Amitabh good you as as” is in step 4534

(d) None of these 74. If 11th step is “ you are not as good as Amitabh” then which of the following statement is correct? (a)

“Amitabh as you as are not good” is in step 50

(b)

“good as Amitabh not as you are” is in step 102

(b)

“good as Amitabh not as you are” is in step 102

(c)

“are not Amitabh good you as as” is in step 4534

(c)

“are not Amitabh good you as as” is in step 4534

(d) None of these

(d) None of these 73. If 15th step is “ you are not as good as Amitabh” then which of the following statement is correct?

75. If 100th step is given to us then we can nd which one of the following step. (a)

93 and 107

(b)

107 and 114

(c)

106 and 94

(d)

All the above

EBD_7743

160 Koncepts of Logical Reasoning

  Input Output

161 

Solutions Concept Applicator (CA) Solutions (1 to 5) Here if we look at carefully then we will find that in final output words are arranged in decreasing order of number of letters that the word has. In each step word with largest no. of letters are placed at left place. 1. (c) Here in this case-

6. (d) 7. (c) From the given condition

Input:- 21

213

3241

56423 123456



Step 1:- 213 21

3241

56423 123456



Step 2:- 213 56423 3241



Step 3:- 213 56423 123456 21

21



Input :-Reasoning is a good and interesting subject



Step 4:- 213 56423 123456 3241



Step 1:- interesting Reasoning is a good and subject.



Here Step 4 is output

8. (c) Since step 2 is :-



Step 2:- interesting Reasoning subject is a good and



Step 3:- interesting Reasoning subject good is a and



Step 4:- interesting Reasoning subject good and is a



And step 4 is the output hence we will get output after 4 steps.

2. (a) Here in this case

Input :-Desktop is faster than tab



Step 1:- Desktop faster is than tab



Step 2:- Desktop faster than is tab



Step 3:- Desktop faster than tab is

3. (b) Here option (b) is correct. 4. (a) Input:-mobile phone is very important for businessmen

Step 1:- businessmen mobile phone is very important for



Step 2:- businessmen important mobile phone is very for



Step 3:- businessmen important mobile phone very is for



Step 4:- businessmen important mobile phone very for is And 4th is output

5. (a) From solution of previous question answer is option (a) Solutions (6 to 10) Here numbers are arranged such that 1st number with odd digits are arranged in increasing order of number of digits and then number with even digits are arranged in decreasing order of number of digits.

5 535 32

123456 3241

4568 25315 124578

Step 3:-

5 535 25315 4568 32

1245

9. (c) Given that 2nd step is:- 2 135 12456 158665 1235684 then

12

3rd step:- 2 1235684

158665

135

12456 12

10. (d) The given format is already in output format hence number of steps required is 0. Solutions (11 to 15) Here the rule followed is: Odd numbered step:- The smallest number becomes first and the remaining numbers shift one position rightward. [In case the first number is small then the next number just larger than it will become second and the rest will shift one position rightward and follow this process so on.] Even numbered step:- the largest number among givennumbers becomes last and the remaining numbers shift one position leftward. [In case the largest number is first from the right end, the second largest number re-places the second number from the right end, and so on.] These steps are repeated alternately till all the numbers get arranged in ascending or-der and that will be the last step for that particular input. 11. (b) As per the given condition

Input:-47 62 19 99 88 42 25 79



Step 1:- 19 47 62 99 88 42 25 79



Step 2:- 19 47 62 88 42 25 79 99

EBD_7743

162 Koncepts of Logical Reasoning 16. (d) From above rule the nal output will be

12. (c) Input :45 77 95 132 28 16 150 54 99 125

Is

Step 1 :- 16 45 77 95 132 28 150 54 99 125 Step 2 :- 16 45 77 95 132 28 54 99 125 150

17. (c) Input

Step 3 :- 16 28 45 77 95 132 54 99 125 150 13. (c) Since step II is given hence follow the following stepsStep II:-11 51 32 88 99 132 101 200 Step III:-11 32 51 88 99 101 132 200

14. (d) In these type of questions backward calculation can not give us a unique value. 15. (c) Input: 11 33 55 77 99 22 44 66 88

22

what

33

your.

3

5

7

9

a

e

i

o

u

Step 1

a

3

5

7

9

e

i

o

u

Step 2

a

3

e

5

7

9

i

o

u

Step 3

a

3

e

5

i

7

9

o

u

Step 4

a

3

e

5

i

7

o

9

u

18. (a) From solution of the previous question output isOutput

a

3

e

5

i

7

o

9

20. (b)

Step 3:- 11 22 33 44 55 77 66 88 99

Solutions (21 to 25)

66 77 88 99

Solutions (16 to 20) From the last step it can be concluded that words and numbers are arranged alternately, word with least number of letters shifts to the left most position followed by the least number among the given numbers. In case of two words with same number of letters, words are arranged as per their dic-tionary order. For getting arranged they are interchanged with the word/number whose place it occupies

In this case numbers are arranged in decreasing order with largest number is kept right most place and remaining numbers are shifted one place right. Concept Builder (CB) 21. (d)

22.

(a)

23. (a)

24.

(d)

25. (d)

Solutions (29 to 29) INPUT

u

19. (d) In these type of question we can not be sure about the input.

44 66 88

Step 2:- 11 22 33 55 77 44 66 88 99 Step 4:- 11 22 33 44 55

name

Hence step 4 is the output.

Step IV:-11 32 51 88 99 101 132 200

Step 1:- 11 22 33 55 77 99

11

lemon

apple

choco

college

girl

dream

room

book

calf

1

2

3

4

5

6

7

8

9

rules

Step 1

3

1

2

4

5

6

9

8

7

1

Step2

1

2

3

6

5

4

7

8

9

2

Step 3

9

1

2

8

3

4

7

5

6

3

Step 4

2

9

1

8

3

4

6

7

5

4

Step 5

1

9

2

4

3

8

5

7

6

2

Step 6

6

1

9

8

2

7

5

3

4

3

Step 7

9

1

6

7

2

8

4

3

5

1

Step 8

6

1

9

8

2

7

5

3

4

2

Step 9

4

6

1

3

9

7

5

2

8

3

Step10

1

4

6

3

9

7

8

5

2

4

Step11

6

4

1

7

9

3

2

5

8

2

Step12

8

6

4

5

1

3

2

9

7

3

Step13

4

6

8

5

1

3

7

9

2

1

Step14

8

6

4

3

1

3

7

9

2

2

Step15

7

8

6

9

4

5

2

1

3

3

26.

Option C is correct, hence option A,B and D is not possible. Option A, B and D

27.

From the table we can observe that Option B is output in 8th step. Option A, C and D

28.

29.

  Input Output 163  From steps 11 to 15 no such option fall in the arrangement, so all options, (A, B, C, D) Options, (A, B, C, D) From the table we can observe that Rule 1 is applied in option C and D Option C and D

Solutions (30 to 33) 30. (a) This type of question we should solve by observing the pattern in which steps are changing, after a close look we can observe that it is the sum of the digits that plays main role here rather than the number itself. The given input and steps we can describe asInput

245(11)

316(10)

436(13)

519(15)

868(22)

710(8)

689(23)

Step 1

710(8)

316(10)

436(13)

519(15)

868(22)

245(11)

689(23)

Step 2

710(8)

316(10)

245(11)

519(15)

868(22)

436(13)

689(23)

Step 3

710(8)

316(10)

245(11)

436(13)

868(22)

519(15)

689(23)

Step 4

710(8)

316(10)

245(11)

436(13)

519(15)

868(22)

689(23)



From this given data we can observe the pattern that the number that has lowest sum of digits shifts its position towards left and this process continues.



Now in this case the table will be as follows-



Input

655(16)

436(13)

764(17)

799(23)

977(23)

572(14)

333(9)

Step 1

333(9)

436(13)

764(17)

799(23)

977(23)

572(14)

655(16)

Step 2

333(9)

436(13)

572(14)

799(23)

977(23)

764(17)

655(16)

Step 3

333(9)

436(13)

572(14)

655(16)

977(23)

764(17)

799(23)

So the given sequence is the 3rd step.

31. (b) Following the similar logic as above we will find that it takes 5 steps.

36. (d) Here the steps will be

Input: Father needs to check on the boy

32. (d) In this type of question backward progress is not a right approach as the number whose sum is smallest interchange its position, and there is only one shift per step hence given any step we can not determine the input.



Step I: Boy father needs to check on the



Step II: Boy needs father to check on the



Step III: boy needs father to on the check.



Step III will be the final output of the given input. Hence machine will need three steps to make it.

33. (c) Solutions (34 to 36)

Solutions (37 to 39)

Observe the given pattern closely and we will find that neither the number of alphabets nor the 1st letter plays the main role but it is the last letter that plays the important role, and we can observe that last letter is in reversing order,

Continuation of the given information is as follows-

34. (b) The last step can be written directly following the pattern discussed above. So for the input “He is going out to search air”, the last step would be “out is air to search going he”. 35. (d) Steps is output

Step 9

→ 3

4

1

2

5

6

9

10

7

8

Step 10 → 1

2

3

4

5

6

7

8

9

10

Step 11 → 1

2

3

4

5

6

9

10

7

8

We observed that step 5 is repeated 37. (c) 38. (d) Option D is not between step 12 and 14. 39. (c) As step 10 is same as input then step 20 will be same.

EBD_7743

164  Koncepts of Logical Reasoning  Solutions (40 to 41)

40. (c) In this the numbers are arranged as follows. The least number followed by the highest number followed by the second least number followed by the second highest number and so on. This is done through the process of shifting the number.

The given input is hence subsequent output isIn Put

:

64

326

187

87

118

432

219

348

Step I

:

64

432

326

187

87

118

219

348

Step II

:

64

432

87

326

187

118

219

348

Step III

:

64

432

87

348

326

187

118

219

Step IV

:

64

432

87

348

118

326

187

219

41. (d) Similar to the previous question.      



Input

319

318

746

123

15

320

78

426

Step I

15

319

318

746

123

320

78

429

Step II

15

746

319

318

123

320

78

426

Step III

15

746

78

319

318

123

320

426

Step IV

15

746

78

426

319

318

123

320

Step V

15

746

78

426

123

319

318

320

Step VI

15

746

78

426

123

320

319

318

Step VII

15

746

78

426

123

320

318

319

Hence, in this case we require total VII steps to get the final output.          Concept Cracker (CC) with XAT



As per the given information 6 words are to be written from left to right on a single line in alphabetical order. It is also known that at most 3 of the 6 words can begin with the same letter.



Hence “z” is eliminated.



Now start with “y” we can write only 3 words with the letter “y’’. Then we can write the next 2 words with the letter „z’’. then we left with no letter for 6th word, hence it is also ruled out. On the same logic we if we start with “x” we find that it satisfy all the condition.

42. (b) We will eliminate the options one by one.

Option A, the first word “Bz…” is changed to “bl…”. It is not in alphabetical order.Hence eliminated.



Option B – It satisfy all the condition.



Option C, “Doteam” cannot be changed to “goleam” here “d” changed to “g” and “t” changed to “l” since only one letter should change hence this is eliminated.



Option D, the third word “reted” is changed to “seted” by operation C, and “seted” is then changed to “seteg” again using operation C. Since each word has to be formed by a different operation than that used for the preceding word, this option is also eliminated.



Option E, Here in the first four words 1st letter is “f". We know that at most 3 of the 6 words can begin with the same letter, therefore option E is eliminated.

43. (d) It is a pure logical question and we will start with eliminating the options-

44. (e) Here we should make a note that the first word contains 7 letters i.e. “illicit”. And the fourth word is “licit” that has 5 letters.

Since word “Implicit” has 8 letters hence it cannot be third word.



On the same logic “Explicit” also eliminated.



Also the third word cannot be “Enlist” because we have to remove three letters “ens” from “



Option C and D eliminated on the similar logic.



Word “Elicit” can be third word operation A, C and C.

by the

Input Output 165 45. (e) Here given that “clean” is the rst word. And we need to nd the place of word “learn” Hence we need to convert “CLEAN” to “LEARN” Now identify the changes these are C deleted and R added, and there are so many ways to achieve this. These are

Step 2- 02 Step 3- 01+ 02=03 Step 4- 03-1 = 02 Step 5- 01+ 02= 03 48. (c) The given input number is 17. Following the above rules

clean Þ lean Þ learn (Operation A, B)- 3rd position

Step 1- 1+7= 8

clean Þ clear Þ lear Þ learn (Operation C A, B) 4th position

Step 3- 17 +81= 98

clean Þ dlean Þ lean Þ lear Þ learn (Operation C A C, B) 5th position

Step 5- 8+16 = 24

clean Þ dlean Þ dleapn Þ eleapn Þ leapn Þ learn (Operation C, A, C, A, C) 6th position 46. (d) Given that the rst word contains 5 letters. And we need to nd the maximum number of letters in the fth word, we can use maximum 2 operations i.e. add the letter and then replace the letter (or vice-versa). Then, the second word can contain 6 letters. Also, the third word also contains 6 letters. Now, fourth word contains 7 letters and thus fth word can contain only 7 letters. Solutions (47 to 50) To understand the logic, let’s take the rst number of the input series; i.e. 20 and explain how it changes in every stepWe have assumed Input: 20 Step-1: Sum of the digits of given number in Input = 2 + 0 = 02 Step-2: Input + square of the resultant number of step 1 = 20 + (2)2 = 20 + 4 = 24 Step-3: Input + the resultant number of step 2 = 20 + 24 = 44

Step 2- 17+ 82 = 81 Step 4- 9+8-1= 16 49. (d) The given output number is 8. To nd the input number we will have to check all options. As per the rules we will nd that If input is 3, then output is 8. 50. (c) The given input number is 11. Following the above rules Step 1- 1 + 1 = 2 Step 2- 1 + 22 = 5 Step 3- 1 + 52 = 26

Concept Deviator (CD) Solutions (51 to 54) Input: 22, 37, 77, 89, 16, 21, 47 Step I: 4, 1, 5, 8, 7, 3, 2 [This step is digital summation**] Step II : 12, -3, 21, 60, 45, 5, 0 [ in this step number k is written as ] Step III: 3, -3, 3, 6, 9, 5, 0 [This step is digital summation] Step IV: 5, 0, 8, 13, 20, 18, 17

Step-4: Sum of the digits of the resultant number of step 3 — 1 = 4 + 4 — 1 = 7

[in this step prime numbers 2, 3, 5, 7, 11, 13 are added]

Step-5: Sum of the results of step-1 and step-4 = 7 + 2 =9

[This step is digital summation]

47. (a) It is given that Step-2 is 02, so the resultant number at step-1 should be 01 and input should be 01. Now, following the same procedure

Step V: 5, 0, 8, 4, 2, 9, 8 [** Digital summation means sum of digits or remainder when number divided by 9] Step V is the nal output step.

EBD_7743

166 Koncepts of Logical Reasoning 51. (a) In step IV we add prime numbers hence step III

1

2

3

4

5

6

Step IV

3

5

8

11

16

19

Step V

3

5

8

2

7

1 [this is the nal output]

52. (c) Since step II is hence if step II is known to us we can calculate step I 53. (b) Since step II is hence if Step II is 5,

12,

45,

77,

60,

3,

4,

7,

9,

8,

6,

11,

22,

33,

44,

77,

66

Step I

2

4

6

8

5

3

Step II

0

12

32

60

21

5

Step III

0

3

5

6

Step IV

2

6

10

13

14

18

Step V

2

6

1

4

5

9

Hence Step I is 54. (a) Input

32

3 5

55. (c) From solution of previous question option (c) is correct. Solutions (56 to 60) In this case all the odd numbered step i.e, 1st, 3rd, 5thetc are square of previous steps. All the even numbered steps are digital summation of previous steps. 56. (c) All the multiples of 3 will have value 0 in 100th step hence 3 terms will have value 0. 57. (c) Option (c) is correct. 58. (a) Only 3 will give us a term as zero hence only one term will have zero. 59. (b) 59. Input:-

11,

22,

33,

44,

55

Step 1

121,

484

1089

1936

3025

Step 3

4

7

0

1

1

Step 4

16

49

0

1

1

Step 5

7

4

0

1

1

Step 6

49

16

0

1

1

Step 7

4

7

0

1

1

Step 8

16

49

0

1

1

and so on…from cyclicity 100th step will be similar to 4th step. 60. (b) must be correct. 61. (b) From dictionary arrangement option (b) is correct. 62. (a) Since total number of elements is 6 hence maximum number of elements is 5 63. (a) Since last three words are arranged, hence it will come in maximum 3rd step. Option B and C is correct.

Input Output 167

64. (c) Since step 2 is “Bat, get, net, let, set, war, wet”

In 3rd step each number is multiplied with 4!, and so on.

So step 1 must end with ‘wet’ so possible values of step 1 is

66. (d) In this case we are searching for numbers divisible by 343 = 7x7x7 or highest power of 7 should be 3 i.e in 21! Hence in 20th step each number will be multiplied with 21! And each term will be divisible by 343.

Case (1) :- if word “war” is shifted in step 2 then step 1 will have “Bat, get, net, let, set, wet” and word ‘war’ is may be at 1st, 2nd, 3rd,4th or 5th place so in total there are 5 possibility. If we list down these possibility then – “war, Bat, get, net, let, set, wet” “Bat, war, get, net, let, set, wet” “Bat, get, war, net, let, set, wet” “Bat, get, net,war, let, set, wet” “Bat, get, net, let, war, set, wet” ‘wet’ is shifted in 1st step then input has 6 possibilities so total possible value for input is 6x5 = 30 Case (2) :- if word “set” is shifted in step 2 then step 1 will have “Bat, get, net, let, war, wet” and word ‘set’ is may be at 1st, 2nd, 3rd,or 4th place so in total there are 4 possibility. If we list down these possibility then – “set, Bat, get, net, let, war, wet” “Bat, set, get, net, let, war, wet” “Bat, get, set, net, let, war, wet” “Bat, get, net,set, let, war, wet” If word ‘wet’ is shifted in step 1 then we have 6 possibilities for input and if word ‘war’ is shifted in step 1 then we have 5 possibilities

67. (a) Out of 7 consecutive integers at least one would be divisible by 7, hence after 13th step when each of the term will be multiplied by 14! Then at least one term will be divisible by 7x7x7 or 343 68. (d) Digital sum is zero that means number should be divisible by 9. Since 6! Is divisible by 9 hence in 5th step each term will be multiplied with 6! Hence in 5th step digital summation of all the terms is 0. 69. (a) Divisible by 1000 means highest power of 10 is 3 or highest power of 5 is 3 and that is 15! And it will be in step 14. 70. (c) only option C can be false.

Solutions (71 to 75) Here the rule followed is: In each step the fourth word becomes rst word and the last word becomes fourth word and all other words shift one place right wards except the third, which shifts two place right wards. 71. (a) Given that step 3:- “ dil wale dulhaniyan le jayenge part two ”

So total number of possible ways for Input is 4x6 +4x5 = 44

step 2:- “wale dulhaniyan jayengedil part two le ”

Total values of Input is 30+44 = 74

Step 1:- “dulhaniyan jayengedil wale two le part”

65. (a) Required ratio is (K-1) :1

Solutions (66 to 70)

72. (b) Here Input:- you are not as good as Amitabh Step 1:- as you are Amitabh not good as Step 2:- Amitabh as you as are not good

This question is similar to factorial and the number processing machine is just calculating the factorials.

Step 3:- as Amitabh as good you are not

In 1st step each number is multiplied with 2!

Step 4:- good as Amitabh not as you are

In 2nd step each number is multiplied with 3!,

Step 5:-not good as are Amitabh as you

Step 6:-are not good you as Amitabh as Step 7:-you are not as good as Amitabh (Or Input) so after every 7 steps it will repeat e.g step 2 is same as step 9, step 4 is same as step 11 and so on. Step 1 will repeat in every (7k + 1)th step Step 2 will repeat in every (7k + 2)th step and so on… Option A is in every step 2, 9, 16 or in 7k+2 step hence it is incorrect. Option (b) is correct Option (c) is not possible.

73. (b) In this question situation is shifted by 14 steps so cycle will remain same and similar to that of previous question option (b) is correct. Option (b) 74. (d) Here 11th step is input then 1st step will be 12th step of arrangement given in previous question and so on so each term will shift by 11 place, Consider each option one by one. Option (a) 50th step is same as 50 + 11 = 61st or 6th step in previous question and so on none of them are correct. 75. (d) If any one step is known to us then we can nd the remaining steps.

EBD_7743

168 Koncepts of Logical Reasoning

Games and Tournaments 169

9

Games and Tournaments Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Important

XAT

Important

BANK Clerk

Important

IIFT

Very Important

SSC

Important

SNAP

Very Important

CSAT

Important

NMAT

Very Important

Other Govt. Exams

Important

Other Aptitude Test

Very Important

Questions based on games & tournament is frequently asked in CAT and other aptitude test examination. In general there are three types of questions on this topic:(i)

Questions based on Seed or Rank (Knockout tournament)

(ii) Questions based on scheduling of tournament or who won/lost against whom (iii) Questions based on goals for /goals against etc. Let us understand these types of questions and how to handle these type of questions. Knockout Tournament: In these type of tournament generally (not necessarily) teams or players are seeded, i.e. ranked from top to bottom. Let us suppose 16 players are participating a knockout tournament and these players are seeded or ranked from 1 to 16, then tournament is scheduled in 4 stages:-

EBD_7743

170 Koncepts of Logical Reasoning In stage 1: Seed 1 will play with seed 16 Seed 2 will play with seed 15 Seed 3 will play with seed 14 Seed 4 will play with seed 13 Seed 5 will play with seed 12 Seed 6 will play with seed 11 Seed 7 will play with seed 10 Seed 8 will play with seed 9 i.e. total 16/2 = 8 matches at this stage. In general if a higher ranked (seeded) player play against a lower ranked player then higher ranked player should win but if lower ranked player wins the match then we call it as an upset caused by the lower ranked player. E.g. in a match between seed 4 and seed 13, ideally seed 4 should win but if seed 13 wins the match, we call it as an Upset caused by seed 13. Now assume that there is no upset in stage 1 then stage 2 matches will be scheduled as below:Stage 2: Seed 1 will play with seed 8 Seed 2 will play with seed 7 Seed 3 will play with seed 6 Seed 4 will play with seed 5 Number of matches at this stage is 8/2 =4 Now assume that there is no upset in stage 2 then stage 3 matches will be scheduled as below:Stage 3 (Seminal): Seed 1 will play with seed 4 Seed 2 will play with seed 3 Number of matches at this stage is 4/2 =2 Then stage 4 or nal match between seed 1 and seed 2. So total number of matches is 16/2 + 8/2 + 4/2+2/2 = 15 or in other ways since only one winner remaining 15 are looser and one match can give only one looser so we should have 15 matches (1 less than the total number of players). In this case if you look at carefully then you will nd that in stage 1 sum of seed is always 17 so if I ask you seed 6 played with whom you just have to nd 17-6 =11 so seed 6 played with seed 11 in stage 1. Round Robin Tournament :- In this case each team plays with other team exactly once. Let’s take an example of 6 players namely A, B, C, D, E, F in this case we have to make a table. In this case total number of matches is 6C2 = 15 Lets A won against D so we will represent it in 1st row as W. the same match we can also represent as D lost against A, and we will represent that in 1st column 4th row that shows D lost against A A A

F

D

E

F

W XXXX

C E

C

XXXX

B D

B

XXXX L

XXXX XXXX XXXX

In this way from the given information we have to ll up the table then solve the questions based on this information.

Games and Tournaments

1

Directions for Question Nos. 1 to 5

In a knockout tournament 64 players participated. These 64 players are seeded from 1 to 64 with seed 1 being the top seed and seed 64 being the bottom seed. The tournament is conducted in different stages. In stage 1 seed 1 played with seed 64 and that match is named as match 1 of stage 1, seed 2 played with seed 63 and that match is named as match 2 of stage 1, and so on. In stage 2, winner of match 1 and match 32 of stage 1 played against each other and that match is named as Match 1 of stage 2, then winner of match 2 and match 31 of stage 1 played against each other and that match is named as Match 2 of stage 2. And so on The same procedure is followed in further stages Now answer the following questions. 1.

2.

3.

4.

5.

171

If seed 15 won the tournament then what is the minimum number of upsets caused by him? (a) 1 (b) 2 (c)

3

(d)

None of these

Directions for Question Nos. 6 to 10 8 teams namely A, B, C, D, E, F, G & H participated in a tournament whose 1st stage is a round robin stage where each team play with other team exactly once. Following further information is known to us(i) A won against B, C and E (ii) Number of matches won by A, B and D is 3 each no other team won 3 matches (iii) C won against B and D but lost to E (iv) H won all the matches. (v) G won against B but lost to E (vi) D lost to F and C won only 2 matches

How many stages are in the tournament? (a) 5

(b)

(c)

(d) None of these

7

6

Now answer the following questions 6.

What is the total number of matches in the tournament? (a) 63

(b)

(c)

(d) None of these

127

36

7.

If seed 9 reached nal then which one of the following could play with him in nal? (a) 56

(b)

(c)

(d) None of these

11

24

Which lowest seeded player can win the tournament without causing an upset by him? (a) 32

(b)

(c)

(d) None of these

34

33

8.

9.

How many matches F won (a) 2

(b)

(c)

(d) None of these

4

3

How many matches G lost? (a) 2

(b)

(c)

(d) None of these

4

3

E won against which all teams? (a) C and B

(b)

(c)

(d) None of these

C and F

C and G

Which team won minimum number matches ? (a) C and B (c) C and F

(b) C and G (d) None of these

10. How many teams won 4 matches? (a) 2 (b) 3 (c) 4 (d) None of these

of

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172 Koncepts of Logical Reasoning

2

Directions for Question Nos. 11to 15

Directions for Question Nos. 16 to 20

1st stage of the TT tournament is a round robin stage where 8 players played against each other, after this stage it is found that 5 players won 5 matches each. After this 2nd stage is conducted in the similar fashion where again 5 players won 5 matches each. Only two players won 10 matches each. There is no tie in any match.

The world cup tournament is arranged as per the following rules: In the beginning 16 teams enter and are divided into 2 groups of 8 teams each, where the team in any group plays exactly once with all the teams in the same group. At the end of this round top four teams from each group advance to the next round in which two teams play each other and the losing team goes out of the tournament. The winner in the rst round gains one point from the win and the loser gets zero. In case of tie on a position there is a complex rule to decide the position.

11. What is the minimum number of matches won by any player. (a) 0 (b) 5 (c) 6 (d) None of these 12. If a player won 0 matches in stage 1 then which one of the following is correct? (a) There must be a player who won 3 matches. (b) There must be a player who won all the matches. (c) There must be a player who won 1 match. (d) None of these 13. If a player Amit won against Bimal but lost to Chand then Chand has won how many matches? (a) 1 (b) 3 (c) 5 (d) Cannot be determined 14. How many players in can win 3 matches in stage 1 and 2 together? (a) 0 (b) 5 (c) 6 (d) None of these 15. If ‘K’ represents number of matches won by a particular player then which one of the following best represents the range of ‘K’? (a) 5 < K < 10 (b) 6 ≤ K < 11 (c) 5 ≤K ≤10 (d) None of these

16. What is the total number of matches played in the tournament? (a) 63 (b) 56 (c) 64 (d) None of these 17. The maximum number of matches that a team going out of the tournament in the rst round itself can win is (a) 3 (b) 5 (c) 4 (d) None of these 18. The minimum number of matches that a team must win in order to qualify for the second round is (a) 3 (b) 5 (c) 6 (d) None of these 19. The minimum number of matches that a team who advanced to the next stage has won? (a) 2 (b) 5 (c) 3 (d) None of these 20. What is the minimum number of matches won by the winner of the tournament? (a) 8 (b) 6 (c) 5 (d) None of these

Games and Tournaments 173

3

21. There are four teams in the Indian Professional Volleyball league: Karnataka, Punjab, Jharkhand and Gujarat. Karnataka has 19 points and would be playing 1, 2 and 1 matches against Punjab, Jharkhand and Gujarat respectively. Punjab has 29 points and would be playing 2 and 1 matches against Jharkhand and Gujarat respectively. Jharkhand has 32 points and has 2 matches left against Gujarat. Gujarat is currently trailing the point table with 18 points. A win in a match fetches 2 points and loses 0 point. In the nal points table the team(s) which score(s) the lowest point is eliminated from the league. Based on the data above which team would surely be eliminated? (a) Karnataka (c) Punjab

(b) Gujarat (d) None of these

22. The football league of a certain country is played according to the following rules: Ø Each team plays exactly one game against each of the other teams. Ø The winning team of each game is awarded I point and the losing team gets 0 point. Ø If a match ends in a draw, both the teams get ½ point. After the league was over, the teams were ranked according to the points that they earned at the end of the tournament. Analysis of the points table revealed the following: Ø Exactly half of the points earned by each team were earned in games against the ten teams which nished at the bottom of the table. Ø Each of the bottom ten teams earned half of their total points against the other nine teams in the bottom ten. How many teams participated in the league? [XAT 2011] (a) 16 (c) 19 (e) 30

(b) 18 (d) 25

Directions for Question Nos. 23 to 24 There are ve teams – Paraguay, Qatar, Russia, Spain and Turkey playing in a tournament where each team plays against every other team only once. These are the following possibilities: each match can result in a draw where each team scores two points; or a team can win where it scores three points, while the losing team scores one point 23. If Paraguay has won all the matches and Turkey has lost all the matches and all the remaining three teams score equal points, how many points have each of the three remaining teams scored? (a) 5 (b) 7 (c) 8 (d) None of these 24. If all the ve teams have an equal score, what is thenumber of points scored by each team? (a) 5 (b) 6 (c) 7 (d) None of these Directions for Question Nos. 25 to 29 Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament is conducted in two stages. In the rst stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of the rst stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprises of several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated. The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup. The tournament rules are such that each match results in a winner and a loser with no possibility of a tie. In the rst stage a team earns one point for each win and no points for a loss. At the end of the rst stage teams in each group are ranked on the basis of total points to determine the qualiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams from each group advance to the next stage. [CAT 2000]

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174 Koncepts of Logical Reasoning 25. What is the total number of matches played in the tournament? (a) 28

(b)

(c)

(d) 35

63

55

26. The minimum number of wins needed for a team in the rst stage to guarantee is advancement to the next stage is: (a) 5

(b)

(c)

(d) 4

7

6

27. What is the highest number of wins for a team in the rst stage in spite of which it would be eliminated at the end of rst stage? (a) 5 (c) 3

(b) 2 (d) 4

4

28. What is the number of rounds in the second stage of the tournament? (a) 1 (b) 2 (c) 3 (d) 4 29. Which of the following statements is true? (a) The winner will have more wins than any other team in the tournament. (b) At the end of the rst stage, no team eliminated from the tournament will have more wins than any of the teams qualifying for the second stage. (c) It is possible that the winner will have the same number of wins in the entire tournament as a team eliminated at the end of the rst stage. (d) The number of teams with exactly one win in the second stage of the tournament is 4

In the opening match, Spain lost to Germany. After the second round (after each team played two matches), the pool table looked as shown below:

Directions for Question Nos. 30 to 33

The year was 2006. All six teams in pool A of World Cup hockey, play each other only once. Each win earns a team three points, a draw earns one point and loss earns zero points. The two teams with the highest points qualify for the semi-nals. In case of a tie, the team with the highest goal difference (Goal For - Goals Against) qualies. [CAT 2004]

Pool A Teams

Games Played

Won

Drawn

Lost

Goals For

Goals Against

Points

Germany

2

2

0

0

3

1

6

Argentina

2

2

0

0

2

0

6

Spain

2

1

0

1

5

2

3

Pakistan

2

1

0

1

2

1

3

New Zealand

2

0

0

2

1

6

0

South Africa

2

0

0

2

1

4

0

In the third round, Spain played Pakistan, Argentina played Germany, and New Zealand played South Africa. All the third round matches were drawn. The following are some results from the fourth and fth round matches. (a)

Spain won both the fourth and fth round matches.

(b)

Both Argentina and Germany won their fth round matches by 3 goals to 0.

(c)

Pakistan won both the fourth and fth round matches by 1 goal to 0.

Games and Tournaments 175 30. Which one of the following statements is true about matches played in the rst two rounds? (a) Germany beat New Zealand by 1 goal to 0. (b) Spain beat New Zealand by 4 goals to 0. (c) Spain beat South Africa by 2 goals to 0. (d) Germany beat South Africa by 2 goals to 1. 31. Which one of the following statements is true about matches played in the rst two rounds? (a) Pakistan beat South Africa by 2 goals to 1. (b) Argentina beat Pakistan by 1 goal to 0. (c) Germany beat Pakistan by 2 goals to 1. (d) Germany beat Spain by 2 goals to 1. 32. Which team nished at the top of the pool after ve rounds of matches? (a) Argentina (b) Germany (c) Spain (d) Cannot be determined 33. If Pakistan qualied as one of the two teams from Pool A, which was the other team that qualied? (a) Argentina (b) Germany (c) Spain (d) Cannot be determined Answer Questions 34 to 37 on the basis of the information given below: [CAT 2005] In the table below is the listing of players, seeded from highest (#1) to lowest (#32), who are due to play in an Association of Tennis Players (ATP) tournament for women. This tournament has four knockout rounds before the nal, i.e., rst round, second round, quarternals, and semi-nals. In the rst round, the highest seeded player plays the lowest seeded player (seed # 32) which is designated match No.1 of rst round; the 2nd seeded player plays the 31st seeded player which is designated match No.2 of the rst round, and so on. Thus, for instance, match No. 16 of rst round is to be played between 16th seeded player and the 17th seeded player. In the second round, the winner of match No. 1 of rst round plays the winner of match No.16 of rst round and is designated match No. 1 of second round. Similarly, the winner of match No. 2 of rst round plays the winner of match No. 15 of rst round, and is designated match No. 2 of second round. Thus, for instance, match No. 8 of the second round is to be played between the winner of match No. 8 of rst round and the win-

ner of match No. 9 of rst round. The same pattern is followed for alter rounds as well. Seed #

Name of player

1

Maria Sharapova

2

Lindsay Davenport

3

Amelie Mauresmo

4

Kim Clijsters

5

Svetlana Kuznetsova

6

Elena Dementieva

7

Justine Henin

8

Serena Williams

9

Nadia Petrova

10

Venus William

11

Patty Schnyder

12

Mary Pierce

13

Anastasia Myskina

14

Alicia Molik

15

Nathalie Dechy

16

Elena Bovina

17

Jelena Jankovic

18

Ana Ivanovic

19

Vera Zvonareva

20

Elena Likhovtseva

21

Daniela Hantuchova

22

Dinara Sana

23

Silva Farina Elia

24

Tatiana Golovin

25

Shinobu Asagoe

26

Francesca Schiavone

27

Nicole Vaidisova

28

Gisela Dulko

29

Flavia Pennetta

30

Anna Chakvetadze

31

Ai Sugiyama

32

Anna-lena Groenefeld

34. If there are no upsets (a lower seeded player beating a higher seeded player) in the rst round, and only match Nos. 6, 7 and 8 of the second round result in upsets, then who would meet Lindsay Davenport in quarter nals, in case Devenport reaches quarter nals? (a) Justine Henin (b) Nadia Petrova (c) Patty Schnyder (d) Venus Williams

35. If Elena Dementieva and Serena William lose in the second round, while Justine Henin and Nadia Petrova make it to the semi-finals, then who would play Maria Sharapova in the quarterfinals, in the event Sharapova reaches quarterfinals?

(a) DinaraSafina

(b) JustineHenin



(c) Nadia Petrova

(d) Patty Schnyder

36. If, in the first round, all even numbered matches (and none of the odd numbered ones) result in upsets, and there are no upsets in the second round, then who could be the lowest seeded player facing Maria Sharapova in semi-finals?

(a) Anastasia Myskina



(b) Flavia Pennetta



(c) Nadia Petrova



(d) Svetlana Kuznetsova

37. If the top eight seeds make it to the quarterfinals, then who, amongst the players listed below, would definitely not play against Maria Sharapova in the final, in case Sharapova reaches the final?

(a) Amelie Mauresmo (b) Elena Dementieva



(c) Kim Clijsters

(d) Lindsay Davenport

Directions for Question Nos. 38 to 41 In a sports event, six teams (A, B, C, D, E and F) are competing against each other. Matches are scheduled in two stages. Each team plays three matches in Stage-I and two matches in Stage-II. No team plays against the same team more than once in the event. Noties are permitted in any of the matches. The observations after the completion of Stage-I and Stage-II are as given below. [CAT 2008]

Stage I: One team won all the three matches. Two teams lost all the matches. D lost to A but won against C and F. E lost to B but won against C and F. B lost at least one match. F did not play against the top team of Stage-I. Stage II: The leader of Stage-I lost the next two matches. Of the two teams at the bottom after Stage-I, one team won both matches, while the other lost both matches.

One more team lost both matches in Stage-II.

38. The team (s) with the most wins in the event is (are):

(a) A

(b) A & C



(c) F

(d) E



(e) B & E

39. The two teams that defeated the leader of Stage-I are:

(a) F & D

(b) E & F



(c) B & D

(d) E & D



(e) F & D

40. The only team(s) that won both the matches in Stage-II is (are):

(a) B

(b) E & F



(c) A, E & F

(d) B, E & F



(e) B & F

41. The teams that won exactly two matches in the event are:

(a) A, D & F

(b) D & E



(c) E & F

(e) D, E & F



(e) D & F

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176  Koncepts of Logical Reasoning 

Games and Tournaments 177

5

Directions for Question Nos. 42 to 46

Kolkata Premier League has 16 players seeded from 1 to 16. These players are divided in 4 groups namely South Kolkata, North Kolkata, Central Kolkata and Salt lake. 16 players are divided in these groups as followsSouth Kolkata:- 1, 8, 9, 16 North Kolkata:- 2, 7, 10, 15 Central Kolkata:- 3, 6, 11, 14 Salt lake:- 4, 5, 12,13 In stage I, each player in a group plays with all the other members in their group exactly once. Two 2 from each group is selected for stage 2 based on their number of points. A player qualifying for stage II carries forward only those points that he gained stage I against the another player who qualies for stage 2. In stage II:- total 8 players (2 from each group) advanced to stage II, where each player plays every other player except the player from same group. At the end of stage II, the top 4 players on the basis of points would qualify for the 3rd stage The Semi-Final stage. Stage III (Semi-Final) :- winners of the Semi-Final advanced to the nal round and losers play for 3rd place. The following more rules of tournament is as followsPoints in stage I and stage II are awarded as follows:winner 2 points, looser 0 points. An ‘upset’ is caused when, in any match a lower seeded player beats a higher seeded player. There is no tie at any stage. If points of the two players at the end of any stage are same then there is a complex rule to calculate the tie breaker. Seed 1 is the highest seed and seed 16 is the lowest rank seed. 42. If Ricky Singh a lucky player from South Kolkata reached semi nal stage after getting the minimum possible points, then what is the point scored by Mr. Ricky Singh. (a) 4

(b)

(c)

(d) None of these

7

5

43. If Ricky Singh a lucky player from South Kolkata aspire to play semi nal stage what is the minimum points he should aim to guarantee a seat in seminal? (a) 10 (b) 6 (c) 12 (d) None of these 44. Find the ratio of number of matches at Stage I and Stage II. (a) 4:7 (b) 3:4 (c) 4:5 (d) None of these 45. What is the total number of points (From all the stages including nal) in the tournament? (a) 52 (b) 88 (c) 104 (d) None of these 46. If there is no upset in stage 1 then the lowest seeded player who can win the tournament is(a) 11 (b) 12 (c) 15 (d) None of these Directions for Question Nos. 47 to 51 In sports week at MERI Kolkata 8 students participated in table tennis tournament. The following additional information is known. (i) The tournament was conducted on Round Robin Basis (i.e. each player plays with every other player exactly once) (ii) At the end of a match winner awarded 1 point while looser 0 point. There is no tie in the tournament. (iii) At the end of the tournament it was found that no two player has same points. (iv) 8 players are seeded from 1 to 8 on the basis of previous year tournament. (v) Chandra got 2 points and Eshan got either 5 or 6 or 7 (vi) Amit won over Dripto (vii) Bhanu lost to Dripto but won over Firoz (viii) Chandra lost to Eshan but won over Hemant (ix) Eshan lost to Gandhi (x) Firoz lost to Hemant 47. Who won maximum number of matches? (a) Amit (b) Gandhi (c) Dripto (d) None of these

EBD_7743

178 Koncepts of Logical Reasoning 48. Who won least number of matches? (a) Chandra

(b)

(c)

(d) None of these

Firoz

Hemant

49. Eshan lost to which all players? (a) Chandra & Gandhi (b) (c)

Amit& Gandhi

Hemant & Amit

(d) None of these

50. How many players got more than Dripto? (a) 2

(b)

(c)

(d) None of these

4

3

51. Which one of the following statement is incorrect? (a) Dripto won over Bhanu but lost to Eshan (b) Hemant lost to both Firoz and Gandhi. (c)

Eshan won 5 matches.

(d) None of these Directions for Question Nos. 52 to 56 At Praxis Business School, 8 students participated in a carom tournament which is held during annual sports meet. The following additional information is known to usThe tournament is in 3 stages The 1st stage of the tournament is double round robin stage, i.e. every player plays exactly 2 matches against each other. In 1st stage winner gets 3 points while looser gets 0 points, in case of tie both the player gets 1 point. At the end of stage 1 if two players has same points then there is complex rule to break tie. At the end of stage 1, top 4 players advanced to stage 2 which is seminal. The winner of seminal advanced to 3rd stage, i.e. nal There is no tie in seminal or nal round.

52. If a player Rajesh could not advanced to seminal round then what is the maximum points scored by him? (a) 29

(b)

(c)

(d) None of these

31

30

53. If a player Rajesh could not advanced to seminal round even after scoring a minimum possible points, what is the minimum points scored by him? (a) 9

(b)

(c)

(d) None of these

8

15

54. If at the end of stage 1 each player has distinct number of points then what is the minimum point scored by the highest scorer? (a) 19 (b) 15 (c) 8 (d) None of these 55. If at the end of stage 1 each player has distinct number of points then what is the maximum point scored by the lowest scorer? (a) 19 (b) 15 (c) 8 (d) None of these 56. If at the end of stage 1 each player has distinct number of points and number of draws is maximum then what is the ratio of number of decisive matches to indecisive matches? (a) 6:1 (b) 1:6 (c) 8:3 (d) None of these Directions for Question Nos. 57 to 61 8 teams namely A, B, C, D, E, F, G, H participated in a, tournament. In 1st stage or qualifying stage these 8 teams are divided into 2 groups namely P and Q having 4 teams each. In qualifying match each team played 2 matches against each other in its own group. The rules are designed in such a way that there cannot be a tie in any match. Top 2 teams from each group advanced to next stage that is seminal, in each group, teams own a different number of matches, i.e. there is no tie in calculating rank of 4 teams in any group in 1st stage. In the 1st stage F lost both the matches against G, Number of matches won by A and G is the same and C won only 1 match and that is against D which is a team that lost maximum 1 match against other teams except 1 team. A won 1 match against top scorer of its group. The teams who played seminal round is B, F, G and H. 57. Which one of the following is correct? (a) B, H, A, E are in the same group (b) B, H, C, E are in the same group (c) F, G, C, A are in the same group (d) F, G, D, A are in the same group 58. Which team won the 2nd highest number of matches in the 1st stage? (a) B (b) H (c) G (d) Cant be determined 59. Which team won the least number of matches? (a) C (b) D (c) E (d) A

Games and Tournaments 179 60. A lost both the matches to which team. (a) B

(b)

(c)

(d) None of these

B and H

H

61. If in seminal round top scorer of one group played with 2nd scorer of another group and vice versa then which one of the following statement is correct? (a) If F played with B in seminal round then B lost only one match against H in stage 1 (b) If G played with B in seminal round then H lost both the matches against B in stage 1 (c)

If F played with H in seminal round then H lost 2 matches in stage 1

(d) None of these Directions for Question Nos. 62 to 66 16 players participated in world Chess Championship. These 16 players are seeded from seed 1 to seed 16 with seed 1 as the best rank. These 16 players are divided in two groups such that all the odd numbered seed are in group A and all the even numbered seed are in group B. In each group each team plays with each other exactly once and no match ended in a tie. For a win winner awarded 2 points while looser 0 points. From each group top two players based on the points scored are advanced to the next stage, i.e. seminal stage. In seminal stage, top scorer of one group plays with 2nd best scorer of the other group. Winners of the semi-nal play for the nal while loosers play for the 3rd place. An upset is when a lower seeded player beat a higher seeded player. In case of same number of points at the end of the 1st stage there is a complex tie breaker rule which is used to determine the rank. 62. What is the total number of matches in the tournament? (a) 59

(b)

(c)

(d) None of these

61

60

63. If seed 9 won the tournament then what is the minimum points scored by him in stage 1 given that minimum number of upsets in the tournament? (a) 5

(b)

3

(c)

(d)

None of these

6

64. If more than 75% of the matches are upset then which highest seed can win the tournament? (a) 4 (b) 8 (c)

7

(d) None of these

65. If upset caused by a winner is 3 then which lowest seed can win the tournament? (a) 14 (c) 8

(b) 13 (d) None of these

66. Who is the lowest seeded player won the tournament without causing an upset? (a) 10 (b) 9 (c) 8 (d) none of these Directions for Question Nos. 67 to 71 At MERI Kolkata students stay in different sadans namely Bose, Diesel, Edison, Faraday, Marconi, and Rankin. Students who stay in these sadans represent their sadan in Football tournament. 1st stage of the tournament is conducted in 5 days from Monday to Friday such that everyday each team played exactly 1 match. As per the rule 3 points are awarded for a win, 0 for a loss and each team gets 1 point for a tie. On Wednesday evening due to some un avoidable reason tournament is postponed. When Ofcer In charge sports Mr. A B Halder asked the update of tournament he got the following information. Maximum number of goals scored by a team in a match is 4 No team lost with a goal difference more than 2 In each match at least 1 goal is scored. Total number of goals made by the teams are Bose (11), Diesel (9) , Edison (=5), Faraday (=1), Marconi (=7) , and Rankin (= 4) Total number of goals against the teams are Bose (5), Diesel (9) , Edison (=7), Faraday (=4), Rankin (= 5) Faraday won match while Marconi and Rankin got 4 and 3 points respectively. On Monday Bose played against Diesel On Tuesday Diesel won a match against Edison with a goal difference of 2 On Wednesday Edison won a match in which total number of goals scored is 2. 67. What percentage of total match end up with a result where winner won the match with a goal difference of 2. (a) 66.67 % (b) 77.77 % (c) 55.55 % (d) None of these

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180 Koncepts of Logical Reasoning 68. What is the total number of points given to

Ø

Dripto wins against Chandan and Eshan.

these teams in these 3 days?

Ø

Faisal loses to Ahaskar and Bhanu.

(a) 28

(b)

Stage 2

(c)

(d) None of these

26

27

69. Diesel team won against which team? (a) Edison

(b)

(c)

(d) None of these

Faraday

Marconi

70. How many teams got more than 10% of total points but less than 20%? (a) 2

(b)

(c)

(d) None of these

4

3

71. If result of all the matches are interchanged, i.e. goals for will become goals conceded and vice versa then how many teams will have same number of goals scored as before? (a) 1

(b)

2

(c)

(d)

None of these

3

Directions for Question Nos. 72 to 76 Seven players Ahaskar, Bhanu, Chandan, Dripto, Eshan, Faisal and Gaurav participate in a Carrom tournament in which each player plays exactly once against each of the other six players. The tournament was held on three stages such that an equal number of matches is played on all the three stages. As per the rules of the tournament winner will get 2 points and looser will get 0 point while none of the matches ends in a draw. Points scored by all the players is distinct. Further known about the tournament isStage 1 •

Gaurav loses to Bhanu but wins against Ahaskar and Chandan.

Ø

Faisal loses to Chandan, Dripto, Eshan and Gaurav.

Ø

Only one player, wins more than one match in stage 2.

Ø

Chandan won at most 1 match in this stage

Stage 3 Ø

Bhanu wins against Chandan, Dripto and Eshan.

Ø

Eshan loses to Ahaskar and Chandan.

Ø

Gaurav wins exactly two matches in this stage.

72. Who own highest number of match? (a) Ahaskar (b) Bhanu (c) Chandan (d) None of these 73. Result of minimum how many matches has to change in order to make Faisal a highest scorer keeping all the rules of the tournament valid? (a) 6 (b) 4 (c) 3 (d) None of these 74. Dripto lost how many matches (a) 2 (b) 4 (c) 3 (d) None of these 75. If result of stage 2 is interchanged, i.e. winner becomes looser and vice versa then who won the maximum points in the tournament? (a) Ahaskar (b) Bhanu (c) Chandan (d) None of these 76. If result of stage 2 is interchanged, i.e. winner becomes looser and vice versa and as a result three players got same points then Dripto got how many points? (a) 2 (b) 4 (c) 3 (d) None of these

181

Games and Tournaments

Solution (6 to 10)

Concept Applicator (CA)

A

Solution (1 to 5)

A XXXX

Assuming there is no upset Stage 1

Stage 2

Stage 3

Stage 4

Stage 5

Stage 6

1-64

1-16

1-16

1-8

1-4

1-2

2-63

2-15

2-15

2-7

2-3

3-62

3-14

3-14

3-6

4-13

4-5

5-12 6-11 7-10 31-34

15-16

32 -33

16-17

1.

2.

3.

4.

5.

8-9

(b) Since 64 = 26 hence we will have total 6 stages in the tournament with last 6th stage is the nal match. (a) Total number of matches is 32+16+8+4+2+1 = 63 Or else since total number of players is 64 hence number of matches must be 64-1 = 63 (c) Seed 9 played with seed 56 in stage 1, with seed 24 in stage 2, But seed 11 can reach the nal if he beats seeds 6, 3 and 2 in stage 3 4 and 5 respectively. (a) If all the matches in stage 1 are an upset except the last match where seed 32 won, in stage 2 seed 32 is the highest seeded player who can win the tournament without causing an upset. (d) From the solution of previous question we have seen that seed 32 can win the tournament without causing an upset by him. So seed 15 can also win the tournament without causing an upset by him.

B

C

D

E

F

G

H

No of win

W

W

L

W

L

L

L

3

B

L

XXXX

L

W

W

W

L

L

3

C

L

W

XXXX

W

L

L

L

L

2

D

W

L

L

XXXX

W

L

W

L

3

E

L

L

W

L

XXXX

L

W

L

2

F

W

L

W

W

W

XXXX

L

L

4

G

W

W

W

L

L

W

XXXX

L

4

H

W

W

W

W

W

W

W

XXXX

7

6. 7. 8. 9. 10.

(c) (b) (b) (d) (a)

From the table F won 4 matches. From the table G lost 3 matches. From the table E won against C and G C and E won 2 matches. 2 teams won 2 matches each Concept Builder (CB)

Solution (11 to 15) Let us represent 8 players as R1, R2 etc then result would be as follows-

R1 R1 XXXX

R2

R3

R4

R5

R6

R7

R8

W

W

W

W

W

L

L

R2

L

XXXX

W

W

W

W

W

L

R3

L

L

XXXX

W

W

W

W

W

R4

L

L

L

XXXX

L

L

R5

L

L

L

L

L

R6

L

L

L

XXXX

L

L

R7

W

L

L

W

W

W

XXXX

W

R8

W

W

L

W

W

W

L

XXXX

XXXX

11. (b) For minimum number of matches let’s take example of R4 he won 0 matches in stage 1 and 5 matches in stage 2; so required minimum number is 5.

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182 Koncepts of Logical Reasoning 12. (c) From table option (C) is correct 13. (d) From the table any option can satisfy the given condition. 14. (a) No player can win 3 matches 15. (c) Minimum is 5 and maximum is 10 represented by option (C) Solution (16 to 20) 16. (a) Number of matches in stage 1 is 2(8C2) = 2(7x8/2) = 56 Stage 2 will be a knockout tournament with 8 teams so in this stage number of matches will be 7. Total number of matches is 56 + 7 = 63 17. (b) In the 1st round in one group number of matches is (7x8/2) = 28 Let’s consider bottom 3 they will have 3 matches between them, so remaining 28-3 = 25 matches have involvement of top 5 teams,

if they won equal number of matches, i.e. 5 each then decision will be taken based on tie breaker rule, hence a team may be eliminated even after winning 5 matches. 18. (c) From the solution of previous question even after winning 5 matches a team can get eliminated so to be sure a team must win 6 matches. 19. (a) Consider top three team if they won maximum number of matches then points with them is 7+6+5 = 18, remaining 28-18 = 10 points can be distributed to bottom 5 teams, so even after getting 2 points or 2 wins a team can advanced to next stage by tie breaker rule. 20. (c) From the solution of previous question a team even with two wins can advanced to next stage where it has to play 3 matches so total number of matches is 5.

Concept Cracker (CC) with XAT 21. (d) From the given conditions we can form the table as followsKarnataka

Punjab

Jharkhand

Gujarat

Ini al points

Karnataka

X

1

2

1

19

19

27

Punjab

1

X

2

1

29

29

37

Jharkhand

2

2

X

2

32

32

44

Gujarat

1

1

2

X

18

18

26

From the table we can conclude that Karnataka and Punjab have 1 match to play against each other. It is also given that a win can fetch 2 points, and a loss, 0 point. The maximum points that Gujarat (with lowest point) = 18 + 4 x 2 = 26 and it may not be eliminated, Similarly there are certain conditions exist where every team has a chance to survive. 22. (d) Let total number of teams participated in tournament is n+10 There are 10 teams in the bottom group then n teams in the top group. Since the bottom group gets 45 points since we have 1 point per match therefore 45 matches playing amongst themselves. Therefore they should get 45 points from their matches against the top group i.e., 45 out of the 10n points. The top group get nC2 points from the matches among themselves. They also get 10n – 45

Minimum points Maximum points

points against the bottom group, which is half their total points. Hence nC2 = 10n – 45 or n(n + 1) = 20n – 90 or n2 – 21n + 90 = 0 hence n = 6 or 15 If n = 6, the top group would get nC2 + 10n – 45 = nC2 + 10(6) – 45 = 30 points, or an average of 5 points per team, while the bottom group would get (45 + 45)/10 or an average of 9. This is not possible. Hence, n = 15. Then total number of teams is 10 + 15 = 25. Solution (23 to 24) 23. (c) Each match has 4 points and each team plays 4 matches. Hence total number of points is 40 Paraguay wins all matches. So it scores 3 × 4 = 12 points. Turkey loses all matches. So it scores 1 × 4 = 4 points. 16 points are scored by Turkey and Paraguay The other teams score 40 – 16 = 24 points

Games and Tournaments 183 As all other teams score equal points, each team scores 24/3 = 8 points 24. (d) The total score is 40 points. If each team scores equally, each team scores 40/5 = 8 points Solution (25 to 29) First Stage: There is two groups of 8 teams each. In each group, each team plays with every other team and hence total number of matches are 8c2 = 8x7/2 =28 matches So, in both the groups the total number of matches played at the rst stage are 28. And hence 56 matches are played in 1st stage Second Stage- In this stage there are 8 teams playing in such a way that in one round 4 teams play with 4 other teams. 4 teams win and go to the next round.That is called knock out tournament. In the 1st round no of matches 8/2= 4, in the 2nd round no of matches = 4/2=2, in the third or the last round number of match =2/2=1, so total no of matches in 2nd stage is 4+2+1= 7 Hence total match in the tournament = 56 +7=63 25. (c) 26. (b) From the given information we can complete the following table it is clear that if a team wins 5 games, then also there is no guarantee of its advancement to the next stage, since only 4 teams can go to the next stage.

Team 1 2 3 4 5 1 × W L L W 2 L × W W L 3 W L × W L 4 W L L × W 5 L W W L × 6 W W W W W 7 W W W W W 8 W W W W W Note: In the table W → Wins, L → Loose, × → No match (as example there can not be match between 1& 1, 2& 2 and so on) The above is one of such combination. Since after winning 5 matches too, there is no guarantee to advancement, so the answer must be 6, because no two teams can get 7 points each. 27. (a) The team which gets 5 point at 1st stage would be eliminated because the combination may be 5 points for each of 1st team and 1 point each for remaining. There are some more cases that supports the idea. 28. (c) Since there are 8 teams and we know that 8= 23 hence there are 3 rounds in 2nd stage. 29. (c) From the discussion above we can say that It is possible that the winner will have the same number of wins in the entire tournament as a team eliminated at the end of the rst stage.

Concept Deviator (CD) Solution (30 to 33) Lets see the table and try to understand the facts.

Teams Germany Argen na Spain Pakistan New Zealand South Africa

Games Played 2 2 2 2 2 2

Won 2 2 1 1 0 0

Drawn 0 0 0 0 0 0

Here Goals for will denoted as G.F. and goals against as G.A. From the goals for and goals against we can conclude that Germany has played a total of two games and has lost none, i.e. we can conclude that its two wins can be in one of the two combinations which can be given as Won 1-0/2-1 or Won 2-1/1-0 against two teams which are Spain and either Pakistan (PAK) or New Zealand (NZ) or South Africa (SA). If Germany wins by 2-1 then Spain wins its other match

Lost 0 0 1 1 2 2

Goals For 3 2 5 2 1 1

Goals Against 1 0 2 1 6 4

Points 6 6 3 3 0 0

by 4-0 or if 1-0 then Spain wins by 5-1 according to G.F./ G.A. From the goals against only New Zealand and South Africa have greater than or equal to 4 goals. Since South Africa has concealed 4 goals against itself hence it lost in both rounds. So we can conclude that that Spain played its second round against New Zealand and if this deduction is true then no other team can play NZ in round two. Hence we can draw the following possibilities from the above deductions:

EBD_7743

184 Koncepts of Logical Reasoning Pakistan

New Zealand

XXXX

D R3

W R2 5-1 or 4-0

Pakistan

D R3

XXXX

New Zealand

L R2 1-5 or 0-4

Germany

Germany

Argentina

XXXX

Draw R3

Argentina

Spain

South Africa

W R1 2-1 or 1-0

XXXX L R1 1-2 or 0-1

Spain

South Africa In the above table W R1 represents Won in Round 1 and then goal difference is given. Now not decided is Germany: Round 2: vs. PAK/SA → won 2-1 or 1-0. NZ: Round 1: vs. ARG/PAK → Lost 1-5/0-4. But from goal for and goal against we can draw following conclusionPAK won round 1 by 2-0 and lost second by 0-1 (G.F./ G.A. → 2/1 ). But NZ played Round 1 against PAK/ARG it could not have lost 1-2 (because if PAK had won 2-1 against NZ in Round 1, its second round would be a draw and Arg has conceded two goals so it cannot win against NZ by 2-1). Hence NZ won round 1 against ARG by 0-1 and lost 2nd round by 1-5 against Spain. Or we can make another table as-

Round 1

Germ- Spain (1-0)

Arg- NZ (1-0)

Pak- SA 92-0)

Round 2

Spain – NZ (5-1)

Arg-Pak (1-0)

Germ- SA (2-1)

Round 3

Germ- Arg Draw

Spain- Pak Draw

Nez – SA Draw

XXXX

D R3

D R3

XXXX

According to the information for the fourth and fth round matches following deductions can be made: Germany – PAK, Loss (0-1) and NZ won (3-0) Argentina – Spain, Loss by x goals and SA won by y goals. Spain - Argentina, won by x goals and SA won by y goals. PAK-Germany won (1-0) and NZ won (1-0) NZ – Germany lost (0-3) and PAK lost (0-1) SA – Argentina lost (0-3) and Spain lost by y goals. Goal differences for the teams: Germany +1 +1+0-1+3=+4 Argentina+1+1+0-x+3=5-x=Max. 4 or less. Spain-1+4+0+x+y=3+x+y=Min. 5 or more. Pakistan+2-1+0+1+1=+3 New Zealand-1-4+0-3-y=-6-y So nal situation is

Germany

Argen na

Spain

Pakistan

Germany

XXXX

Draw R3

W R1 1-0

L (0-1)

W (3-0)

W R2 2 -1

Argen na

Draw R3

XXXX

WR2 1-0

W R1 1-0

W (3-0)

W R2 5-1 W (1-0)

W R1 2-0

Spain Pakistan New Zealand South Africa

L R1 0-1 W (1-0) L (0-3)

L R2 0-1 L R1 0-1

L R2 1-2

L (0-3)

New Zealand

South Africa

XXXX

D R3

D R3

XXXX

L R2 1-5

L (0-1)

XXXX

D R3

L R1 0-2

D R3

XXXX

Now from the table30. (d) 31. (b) 32. (d) Solution (34 to 37) 34. (d) 1st let us understand the concept of the tournament.

ROUND 1 Match Seed





ROUND 2 Sum

Match Seed

  Games and Tournaments 33. (c)

ROUND 3 (QF) Sum

Match Seed

ROUND 4 (SF)

Sum

Match Seed

Sum

1

1+32

33

1

1+16

17

1

1+8

9

1

1+4

5

2

2+31

33

2

2+15

17

2

2+7

9

2

2+3

5

3

3+30

33

3

3+14

17

3

3+6

9

4

4+5

9

15

15+18

33

7

7+10

17

16

16+17

33

8

8+9

17

Try to understand the table, in the 1st round the sum of seed is always 33, the same for round 2 is 17 and so on considering there is no upset. Lindsay Davenport (Seed 2) was scheduled to meet seed 7, i.e., Justine Henin in the quarter finals. But in the second round since match 7 - involving seed 7 (Justine Henin) and seed 10 (Venus Williams) - resulted in an upset and Venus Williams won the game, hence Davenport will meet Venus Williams in the quarter finals. 35. (c) Maria Sharapova, seed 1, would meet the winner of the match between seed 8 and seed 9, but it is given that Serena Williams, seed 8 lost against seed 9 Nadia Petrovain the second round. Hence Nadia Petrova (seed 9). 36. (a) All odd numbered seeds up to 15 reach the 2nd round while instead of seeds 2 to 16 the players who reach the second round are seeds 31, 29, 27, 25, 23, 21, 19 and 17. The second round matches are between seeds 1-17, 31-15, 3-19, 29-13, 5-21, 27-11, 7-23 and 25-9. Since there are no upsets in that round the winners are seeds 1, 15, 3, 13, 5, 11, 7 and 9. The quarter final line up would be 1–9, 15–7, 3–11 and 13–5. If seed 13 wins her match in the quarters she will next meet seed 1, i.e. Marie Sharapova in semifinals. 37. (c) If the top eight seeds make it to quarterfinals, then from the table we can find that

185 

Sharapova being top seed will meet seed 8 in the quarterfinals, and in the semifinals she will meet the winner of the match between seeds 4 and 5. So she will not meet seeds 4, 5 or 8 in the finals. Solution (38 to 40) We can sum up the given information as bellow (i) No team plays against the same team more than once. (iv) No ties permitted. As per the information given for stage – I, we can conclude and draw the table that (a) As B lost at least one match, hence A won all the 3 matches. (b) The two teams who lost all the matches cannot be A (as it won all 3 matches), cannot be B (since it is given that E lost to B), and cannot be D (since D won against C & F). Hence, the two teams must be C and F. (c) F did not play against the top team (i.e. A). We can tabulate all the information as



A

B

C

D

A

X

W

W

W

B

L

X

C

L

X

L

D

L

W

X

E

L

F

L

W L

E

F

W

W

L W X

W

L

X

In the above table a blank space means there was no match in this stage (i.e. there was no match between A and E in this stage), Result of a match for a team is shown in its row,

EBD_7743

186 Koncepts of Logical Reasoning As per the information given for Stage-II, we can conclude (d)

A lost both its matches against E and F.

(e)

F won against A, hence is the bottom team (out of C & F) which won both the matches F won against C as well. This also means that C lost both its matches against B and F.

(f)

Apart from A and C, one more team lost both the matches in Stage-II. That team can neither be E (since A lost to E), nor B (since C lost to B), nor F (as F won both its matches). Hence, the team must be D. Now we can complete the table for Stage II. A A

B

C

D

X

B

X

W

C

L

X

D

L

E

W

F

W

E

F

L

L

43. (c)

44. (d)

W L X

L

W

X

W

45. (c)

X

Hence we can answer the questions now38. (e) 39. (b) 40. (d) 41. (e) Solution (42 to 46) In stage I, in a group in total 6 matches and that will have 12 points in total. Total number of matches in this stage is 4x6 =24 In stage II, total 8 players and they play each other except a player who comes from same group but point of their match in stage I is carried forward hence in stage II total points is 28x2 = 56. Number of matches in stage II is 7x8/2 – 4 = 24 Concept Eliminator (CE) with CAT Guru Gajen 42. (a) In this case we have to minimize the points scored by 4th ranker (Who advanced to Semi Final). In order to minimize the points scored by Ricky Singh top three players should get maximum points, Let points achieved by top three player at the end of stage II is 14, 12

46. (b)

and 10 summing up to 14+12+10 = 36 and remaining points 56-36 = 20 got distributed to bottom 5 players. To minimize the points scored by Ricky Singh, all of the bottom ve got same point (20/5 =4 points) and with complex tie breaker Ricky is advanced to Seminal. In stage 2 total points are 56. Consider bottom 3 players, they must play three matches among themselves and hence they have minimum 6 points with them, Remaining points 56-6 =50 if equally divided between top 5 then one would get 10 points, so to guarantee a seat in seminal Ricky Singh has to score 12 points. Total number of matches is as followsStage I:- 24 Stage II:- 24 And required ratio is 1:1 Total number of matches is as followsStage I:- 24 Stage II:- 24 Seminal:- 2 For 3rd place (Looser of semi nal):- 1 Final:- 1 Total number of matches is 52 and hence total number of points is 52 × 2 = 104 In stage 1 a player has to win minimum 1 match to advance to the next stage. Now consider the different groups, South Kolkata:- 1, 8, 9, 16 [seed 9 can advance after beating seed 16 without an upset] North Kolkata:- 2, 7, 10, 15 [seed 10 can advance after beating seed 15 without an upset] Central Kolkata:- 3, 6, 11, 14 [seed 11 can advance after beating seed 14 without an upset] Salt lake:- 4, 5, 12,13 [seed 12 can advance after beating seed 113 without an upset] So seed 12 is the lowest seed who can advance to stage II without causing Upset in stage I, and can win the tournament [Please note here that in stage II or seminal there may be an Upset]

Games and Tournaments 187 Solution (47 to 51) From the given information we can draw following table. Amit Amit

Bhanu

Dripto

X

Bhanu

X

Gandhi

Hemant

W L

W

Eshan

X

L

2

L

0

L X

Gandhi

W

Hemant

W

X W

Firoz

Firoz

7

L X

L

Eshan

W

Chandra Dripto

Chandra

L

X W

X

Since each player got different points hence points that they got must be 7, 6, 5, 4, 3, 2, 1, and 0. That means 1 student won all the matches while 1 student lost all the matches, the student who won all the matches must be Amit as rest all lost at least 1 match, while Firoz lost all the matches, Since Eshan got either 5, 6 or 7 points but he already lost against Amit and Gandhi so he won 5 points. Similarly Gandhi got 6 points Amit

Amit

Bhanu

Chandra

Dripto

Eshan

Firoz

Gandhi

Hemant

X

W

W

W

W

W

W

W

7

Bhanu

L

X

w

L

L

W

L

W

3

Chandra

L

L

X

L

L

W

L

W

2

Dripto

L

W

w

X

L

W

L

W

4

Eshan

L

W

W

W

X

W

L

W

5

Firoz

L

L

L

L

Lg

X

L

L

0

Gandhi

L

W

W

W

W

W

X

W

6

Hemant

L

L

L

L

L

W

L

X

1

47. (a) Amit own maximum number of matches. 48. (c) Firozown least number of matches. 49. (c) 50. (b) 51. (b) Solution (52 to 56) Total number of matches in stage 1 is 2(7 × 8/2) = 56 52. (b) If a player eliminated in 1st stage even after scoring maximum possible point then it is possible when top 5 has same point and Rajesh got eliminated with tie breaker rule. In this case bottom three got points because of matches between them. Out of 56 matches there are 6 matches played among bottom three hence total points in remaining 56-6 = 50 matches is 50x3 = 150 that is equally divided among top 5 players equally i.e 30 points each, So Rajesh can not get advanced even after getting 30 points.

53. (c) In this case top 3 players should get maximum possible points and remaining 5 should get equal points and Rajesh got advanced with tie breaker rule. Bottom 5 players have in total 4x5 = 20 matches and each match will fetch minimum possible points when its result is tie. So minimum point in 20 matches is 20x2 = 40 when distributed equally to bottom 5 each of them will get 40/5 = 8 points. 54. (a) Since total number of matches in the stage 1 is 56 and for minimum value let us assume that all the matches end with a tie, so total points is 56x2 = 112 and it is distributed equally among 8 players i.e each player would get 112/8 = 14 points. Let minimum number of matches that gave result is k then since each indecisive match give us 2 points but decisive match will give us 3 points hence total k decisive matches will increase the total number of points to 56+k

EBD_7743

188 Koncepts of Logical Reasoning Here we have assumed that all the matches are draw and now if we change the result of one match then that will increase the total point of winner by 2 and decrease the total point of looser by 1. 1 win and 1 loss will increase total point of a player by 1. 2 win will increase total point of a player by 4 2 win and 1 loss will increase total points of a player by 3 3 win will increase total point of a player by 6 3 win and 1 loss will increase total points of a player by 5 For minimum points scored by the highest scorer, their points should be as followsPlayer 1:- 14 + 5 = 19 (3 win and 1 loss) Player 2:- 14+4 =18 (2 win) Player 3:- 14+3 = 17 (2 win and 1 loss) Player 4:- 14+2 = 16 (1 win) Player 5:- 14 (no change) Player 6:- 14 – 1 = 13 (1 loss) Player 7:-14 – 2 = 12 (2 loss) Player 8:- 14 – 3 = 11 (1 loss) So highest point of highest scorer is 19 55. (d) From solution of previous question maximum points scored by lowest scoreris 11 56. (b) From solution of previous question number of decisive matches is 8 then indecisive matches is 56-8 = 48 ands required ratio is 6;1 Solution (57 to 61) Lets sum up the information, The teams who played seminal round is B, F, G and H Given that F lost both the matches to G i.e F and G is in the same group, then B and H is in the same group. Since number of matches won by A and G is same hence they must belong to two different groups, and C won only I match against D hence they must be in the same group. So we can divide the group nowGroup 1: F, G, C, D Group 2: B, H, A, E In one group number of matches is 2(3x4/2) = 12 since each team won a different number of matches and total number of matches is 12 so we have following possibilities for number of matches won by 4 teamsCase (i) 6, 4, 2, 0 Case (ii) 6, 3, 2, 1 Case (iii) 5, 4, 2, 1

Case (iv) 5, 4, 3, 0 Since number of matches won by A and G is the same but then also A eliminated so only possibility is case (ii) and (iv) and they won 3 matches. Lets sum up the conclusion till now Group 1: F, G, C, D and their points F = 6, G = 3, C = 1 and D=2 Group 2: B, H, A, E and their points B and H 5 and 4 in any order and A = 3, E =0 Now lets draw the table for Group 1 F

G

C

D

F

X

WW

WW

WW

6

G

LL

X

WW

WL

3

C

LL

LL

X

WL

1

D

LL

LW

LW

X

2

Now lets draw the table for Group 2 B/H

H/B

A

E

B/H

X

WW

LW

WW

5

H/B

LL

X

WW

WW

4

A

WL

LL

X

WW

3

E

LL

LL

LL

X

0

57. (a) 58. (d) 2nd Highest is either B or H so cant determine. 59. (c) E lost the maximum number of matches. 60. (d) A lost both the matches to B or H not B and H hence option C is incorrect. 61. (b) Solution (62 to 66) 62. (b) Number of matches in stage 1 is 2(7x8/2) = 56, at seminal stage we have 3 matches (2 seminal and 1 match for 3rd place) and 1 nal match, hence total number of matches is 56+3+1 = 60 63. (a) Seed 9 will play with seed, 1, 3, 5, 7, 11, 13, and 15 without an upset seed 9 can win with seed 11, 13, and 15, for minimum number of upset let seed 1 won all the matches and seed 9 won against seed 3 and 5, in that case number of wins of seed 3 and 9 is 5 but with tie breaker rule seed 9 will advance to the next stage. 64. (d) Total number of matches is 60 and out of those more than 45 matches are upset. But seed 1 need only 9 matches to win the tournament hence seed 1 may win the tournament.

Games and Tournaments 189 65. (b) Total number of matches in the 1st stage is 4x7 = 28, lets consider group 1 here if seed 1 won all the matches then remaining 21 matches or points can be equally distributed to 7 player (3 points each) and the lowest possible player would advance to next stage with tie breaker rule. In this stage seed 13 can get 3 points after 2 upsets caused by him. So from this group seed 1 and 13 would advance to the next stage. Similarly from 2nd group seed 2 and 14 would advance to the next stage. Now as per the rule seed 1 will play with seed 14 and seed 2 will play with seed 13, If seed 13 and 4 meet in the tournament then seed 13 will win with 3 upset. 66. (a) Here only problem is stage 1, so a player has to qualify for seminal without any upset. In previous question we have seen that with 3 points/win a team can advanced to next stage. Consider group 1:- Here players are seeded 1, 3, 5, 15. The player who can win 3 matches without an upset is seed 9. Consider group2:- same as previous case seed 10 can advanced to next stage without causing an upset. So seed 10 is the answer. Consider 1 such case- seminal round Group 1:- seed 13 and 15 Group 2: seed 2 and 10 Final match:- seed 10 & 15 Solution (67 to 71) Since total number of goals scored and goals against is same hence we can nd goals against Marconi which is (11+9+5+1+7+4) – (5+9+7+4+5) = 37 – 30 = 7

Since Bose sadan scored 11 goals while goals against is 5 it is possible when all the matches are won with (3-1) (4-2) and (4 -2) Since Marconi got 4 points which is possible with 1 win (3 points) 1 draw (1 point) and 1 loss (0 point) Similarly Rankin got 3 points and also Goals for is 4 and against is 5 so all three matches can not end with draw (It is when goals for is equal to goals against) so 3 points is possible with 1 win and 2 loss. Faraday scored only 1 goal, goal against is 4 and won 1 match it is possible only when he won by (1-0) and lost two matches with (0-2) each[ as with the combination (0-1) and (0-3) is not possible since (0-3) goal difference is 3] Since Edison won on Wednesday with (2-0) and total goals for is 5 while against is 7, so in other two game it lost with (2-4) and (1-3) Draw is casued only with Diesel and Marconi hence they must have played with each other and the match was tie. On Monday Diesel played with Bose, on Tuesday Diesel played with Edison while on Wednesday Diesel played with Marconi. On Wednesday Bose must have played with Marconi Faraday played against Edison on Wednesday and lost with (0-2) Marconi played against Faraday on Monday while against Bose on Tuesday. Since Rankin scored 4 goals and conceded 5 goals result of Bose and Rankin (3-1) similarly result of Rankin and Edison (3 – 1) Now we can conclude that Bose sadan won against Diesel and Marconi with (4-2) and (4-2) Diesel won against Edison with (4-2) So matches on Monday:-,

So nal result is

Monday Tuesday Wednesday

Bose & Diesel (4-2) Bose & Marconi (4-2) Bose & Rankin (3-1)

Number of points are as follows:- Bose (3+3+3 =9), Diesel (0 + 3 +1 =4) , Edison (0+0+3=3), Faraday (0+3+0=3), Marconi (3+0+1=4) , and Rankin (3+0+0=3) 67. (b) Total number of matches is 9, out these 9 matches, the matches that end with goal difference of 2 is Monday:- all the three matches, Tuesday 2 matches and Wednesday 2 matches so required percentage is 7x100/9 = 77.77%

Marconi & Faraday (2-0) Diesel & Edison (4-2) Diesel & Marconi (3-3)

Edison & Rankin (1-3) Faraday & Rankin (1-0) Edison & Faraday (2-0)

68. (c) Out of 9 matches only 1 match end up with tie so total number of points is 8x3 +2 = 26 Option (C) 69. (a) From the table Diesel team won against Edison. 70. (d) Since total number of points is 26 so number of points more than 2.6 but less than 5.2 is (i.e points 3. 4, 5) 5.

EBD_7743

190 Koncepts of Logical Reasoning 71. (b) The new condition would be-

Monday Tuesday Wednesday

Bose & Diesel (2-4) Bose & Marconi (2-4) Bose & Rankin (1-3)

Marconi & Faraday (0-2) Diesel & Edison (2-4) Diesel & Marconi (3-3)

Edison & Rankin (3-1) Faraday & Rankin (0-1) Edison & Faraday (0-2)

Initially total number of goals made by the teams are Bose (11), Diesel (9) , Edison (=5), Faraday (=1), Marconi (=7) , and Rankin (= 4) Now total number of goals made by the teams are Bose (5), Diesel (9) , Edison (7), Faraday (4), Marconi (7) , and Rankin (5) So only Diesel and Marconi have the same number of goals. Solution (72 to 76) Total number of matches is 21 when divided in 3 stages we will get 7 matches in one stage. Ahaskar

Bhanu

Chandan

Dripto

Eshan

Faisal

Gaurav

Ahaskar

XXXXX

L2

W2

L2 /W2

W3

W1

L1

3 or 4

Bhanu

W2

XXXXX

W3

W3

W3

W1

W1

6

Chandan

L2

L3

XXXXX

L1

W3

W2

L1

2

Dripto

W2/L2

L3

W1

XXXXX

W1

W2

L3

4 or 3

Eshan

L3

L3

L3

L1

XXXXX

W2

L3

1

Faisal

L1

L1

L2

L2

L2

XXXXX

L2

0

Gaurav

W1

L1

W1

W3

W3

W2

XXXXX

5

72. 73. 74. 75.

(b) (a) (d) it can be 3 or 4 . (b) The new result will be as follows Ahaskar

Bhanu

Chandan

Dripto

Eshan

Faisal

Gaurav

Ahaskar

XXXXX

W2

L2

L2 /W2

W3

W1

L1

3 or 4

Bhanu

L2

XXXXX

W3

W3

W3

W1

W1

5

Chandan

W2

L3

XXXXX

L1

W3

L2

L1

2

Dripto

W2/L2

L3

W1

XXXXX

W1

L2

L3

2 or 3

Eshan

L3

L3

L3

L1

XXXXX

L2

L3

0

Faisal

L1

L1

W2

W2

W2

XXXXX

W2

4

Gaurav

W1

L1

W1

W3

W3

L2

XXXXX

4

76. (a) From the table of previous question Ahaskar, Faisal and Gauravwon 4 matches then Driptowon only 2matches. Option (A)

Cubes

Cubes Exam

Importance

Exam

191

10 Importance

CAT

Important

IBPS/Bank PO

Very Important

XAT

Important

BANK Clerkt

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt Exams

Very Important

Other Aptitude Test

Very Important

To solved Input / Output question we have to observe the pattern looking at output and input. There is no general rule of pattern but in general we have following patterns.

INTRODUCTION The questions based on cubes test the ability of the students to visualize and deal with the multidimensional problems/situations. If you consider all the exams together, CAT/XAT/IIFT/FMS/JMET/CET/SNAP/MAT you will get at-least one question from this topic. •

In a cubeNumber of vertices V = 8 Number of Faces F = 6 Number of Edges E= 12 We know the Euler’s formula F + V = E + 2

•

If we make a cut in one plane then we divide the cube in two parts (as shown in g 1), if we make 2 cuts in same plane then we will get 2 + 1 = 3 pieces (as shown in g 2) similarly if we make ‘p’ cuts in one plane then we will get ‘p+1’ pieces as shown in Fig

•

If we make a cut in 2nd plane then we divide the cube in 2(p + 1) parts (as shown in g 4), if we make 2 cuts in same plane then we will get (2 + 1) (p + 1) pieces (as shown in g 5) similarly if we make ‘q’ cuts in 2nd plane then we will get ‘(p + 1)(q + 1)’ pieces as shown in Fig 6

•

If we make a cut in 3rd plane then we divide the cube in 2(p + 1)(q + 1) parts (as shown in g 7), if we make 2 cuts in same plane then we will get (2 + 1) (p + 1)(q + 1) pieces (as shown in g 8) similarly if we make ‘q’ cuts in 3rd plane then we will get ‘(p + 1)(q + 1)(r + 1)’ pieces as shown in Fig 9

•

To summarize :-’a’ Number of cuts in one plane gives (a + 1) pieces, while ‘b’ cuts in another plane gives (b + 1) pieces, and ‘c’ cuts in 3rd plane gives (c + 1) pieces. Therefore a, b and c cuts in three dimensions will give (a + 1)(b + 1)(c + 1) pieces. If number of cuts is given then maximum number of pieces can be obtain when a=b=c or these three are as close as possible while for minimum number of pieces can be obtain when all the cuts are made is in one plane.

EXAMPLE 1 If total number of cuts is 10 then nd the minimum and maximum number of pieces that can be obtained.

Solution When all the cuts are in one plane then total number of pieces = 11 For maximum number of pieces a = 4, b = 3 and c = 3 then total number of pieces= 5 × 4 x 4 = 80

EBD_7743

192 Koncepts of Logical Reasoning

EXAMPLE 2

Cubes 193

If total number of pieces are 45 then nd the possible number of cuts.

Solution Since 45 = 1 × 1 × 45 = 1 × 3 × 15 = 1 × 9 × 5 = 3 × 3 × 5, and hence corresponding value of (a, b, c) = (0,0,44), (0,2,14), (0, 8, 4) & (2,2,4) and hence total number of cuts = 44, 16, 12 or 8 •

If all 6 faces of cube are painted with 6 different colours and then it was cut in 125 smaller cubes thenNumber of smaller cubes that has 3 faces painted = 8 (as there are 8 vertices and these will give us 8 cubes with 3 faces painted. Number of smaller cubes that has 2 faces painted = 12 × 3 = 36 ( From each edge we will get 3 cubes that has 2 faces painted and we have total 12 edges) Number of cubes that has only one face painted- 6 × 32 = 6 × 9 = 54 (From each face we will get 3 2 or 9 cubes that has only one face painted and total number of faces = 6) Number of cubes that has no face painted- (5-2)3 = 33 = 27 if a big cube is painted and cut in to n^3 number of smaller cubes then:Number of cubes with 3 face painted is = 8 (Same as number of vertices) Number of cubes with 2 face painted is = 12(n – 2) (since 12 is the number of edges) Number of cubes with 1 face painted is = 6(n – 2)2 (since 12 is the number of edges) Number of cubes with 2 face painted is = (n – 2)3 From algebraic formula we will get n3 = (n – 2)3 + 6(n – 2) + 12(n – 2) + 8 Removal of a corner cube:- If a corner cube is removed and then all the exposed surface painted then changes due to the removal of a corner cube is as follows- (Lets take an example of 4x4x4 cube ) Total surface area will remain unchanged. Total number of cubes N3 – 1 or 43 – 1 = 63 cubes Number of cubes whose three face is painted will increase by 2 so number of such cubes is 10 Number of cubes whose two face is painted will decrease by 3 so number of such cubes is 21 Number of cubes whose 1 face is painted will remain unchanged Number of cubes whose no face is painted will remain unchanged

EBD_7743

194 Koncepts of Logical Reasoning

1.

2.

3.

4.

5.

6.

1

If total number of cuts is 10 then nd the minimum number of pieces that can be obtained. (a)

10

(b)

11

(c)

25

(d)

None of these

If total number of cuts is 10 then nd the maximum number of pieces that can be obtained. (a)

80

(b)

48

(c)

120

(d)

None of these

(a)

208

(b)

144

(c)

210

(d)

None of these

10. How many of the cubes have at most 2 faces painted? (a)

104

(b)

144

(c)

120

(d)

None of these

Directions for Question Nos. 11 to 15 343 cubes of similar size are arranged in the form of a bigger cube (7 cubes on each side, i.e., 7 × 7 × 7) and kept at the corner of a room, all the exposed surfaces

(a)

448: 63

(b)

448:21

are painted then-

(c)

122:21

(d)

None of these

If total number of pieces (Smaller cubes/ cuboids) is 45 then nd the possible number of cuts. (a)

18 or 16

(b)

8 or 12

(c)

12 or 18

(d)

None of these

Find the maximum number of cuts required to get 50 pieces. (a)

50

(b)

51

(c)

49

(d)

None of these

Find the minimum number of cuts required to get 50 pieces. (a)

10

(b)

11

(c)

9

(d)

None of these

216 cubes of similar size are arranged in the form of a bigger cube (6 cubes on each side, i.e., 6 × 6 × 6) all the exposed surfaces are painted.

8.

How many of the cubes have at most 2 faces painted?

If total number of cuts is 20 then nd the ratio of maximum and minimum number of pieces that can be obtained.

Directions for Question Nos. 7 to 10

7.

9.

How many of the cubes have 0 faces painted? (a)

64

(b)

125

(c)

27

(d)

None of these

How many of the cubes have 2 faces painted? (a)

144

(b)

25

(c)

96

(d)

None of these

11. How many of the cubes have 0 faces painted? (a)

64

(b)

125

(c)

240

(d)

None of these

12. How many of the cubes have 2 faces painted? (a)

14

(b)

18

(c)

16

(d)

None of these

13. How many of the cubes have at most 2 faces painted? (a)

208

(b)

244

(c)

342

(d)

None of these

14. How many of the cubes have at least 2 faces painted? (a)

19

(b)

144

(c)

120

(d)

None of these

15. How many of the cubes have 3 painted? (a)

0

(b)

3

(c)

5

(d)

None of these

faces

Directions for Question Nos. 16 to 20 343 cubes of similar size are arranged in the form of a bigger cube (7 cubes on each side, i.e., 7 × 7 × 7) and kept alongside an edge (or side) of a room, all the exposed surfaces (in this case there are 4) are painted.

Cubes 195 16. How many of the cubes have 0 faces painted? (a)

64

(b)

125

(c)

240

(d)

None of these

17. How many of the cubes have 2 faces painted? (a)

23

(b)

29

(c)

31

(d)

None of these

341

(b)

244

(c)

342

(d)

None of these

31

(b)

44

(c)

12

(d)

None of these

20. How many of the cubes have 3 painted? (a)

0

(b)

3

(c)

5

(d)

None of these

faces

21. How many of the cubes have 0 faces painted? (b)

150

(c)

240

(d)

None of these

22. How many of the cubes have 2 faces painted? (a)

23

(b)

29

(c)

44

(d)

None of these

23. How many of the cubes have at most 2 faces painted? (a)

339

(b)

244

(c)

342

(d)

None of these

24. How many of the cubes have at least 2 faces painted? (a)

48

(b)

44

(c)

12

(d)

None of these

25. How many of the cubes have 3 painted? (a)

0

(b)

3

(c)

5

(d)

None of these

64

(b)

125

(c)

27

(d)

None of these

(a)

48

(b)

44

(c)

45

(d)

None of these

(a)

205

(b)

144

(c)

210

(d)

None of these

29. How many of the cubes have at least 2 faces painted?

343 cubes of similar size are arranged in the form of a bigger cube (7 cubes on each side, i.e., 7 × 7 × 7) and kept on the surface of a room, all the exposed surfaces (in this case there are 5) are painted.

64

(a)

28. How many of the cubes have at most 2 faces painted?

Directions for Question Nos. 21 to 25

(a)

exposed surfaces are painted.

27. How many of the cubes have 2 faces painted?

19. How many of the cubes have at least 2 faces painted? (a)

216 cubes of similar size are arranged in the form of a bigger cube (6 cubes on each side, i.e., 6 × 6 × 6) one cube from a corner is removed and then all the

26. How many of the cubes have 0 faces painted?

18. How many of the cubes have at most 2 faces painted? (a)

Directions for Question Nos. 26 to 30

faces

(a)

55

(b)

44

(c)

52

(d)

None of these

30. How many of the cubes have exactly 4 face painted? (a)

55

(b)

44

(c)

52

(d)

None of these

Directions for Question Nos. 31 to 35 Four colours namely Blue, Green, Red and White are used to paint a cube such that each face is painted in exactly one colour and each colour is painted on at least one face. The cube is now cut into 120 identical pieces by making least number of cuts. 31. What is the number of cubes with no face painted? (a)

24

(b)

36

(c)

48

(d)

None of these

32. What is the least possible number of piece which have at most one colour on them? (a)

76

(b)

44

(c)

52

(d)

None of these

33. If K is the number of cuboids which have more than one face painted in the same colour then nd the maximum value of k. (a)

24

(b)

36

(c)

13

(d)

None of these

34. What is the ratio of maximum and minimum number of cuboids with red colour on them? (a) 18:7 (b) 18 :5 (c) 16:5 (d) None of these 35. What is the maximum number of cuboids with only two face painted one face is painted red and other green? (a) 18 (b) 16 (c) 14 (d) None of these Directions for Question Nos. 36 to 40: 125 Cubes of similar size are arranged in the form of a bigger cube (5 cubes on each side, i.e., 5 × 5 × 5). From one corner of the top layer of this cube, four smaller cubes (2 × 2 × 1) are removed. From the column on the opposite side, two cubes (1 × 1 × 2) are removed, and from the third corner, three cubes (1 × 1 x 3) are removed and from the fourth column four cubes (1 × 1 × 4) are removed. All exposed faces of the block thus formed are coloured red. 36. How many small cubes are left in the block? (a) 109 (b) 114 (c) 112 (d) 110 37. How many cubes do not have any coloured face? (a) 38 (b) 44 (c) 25 (d) 35 38. How many cubes have only two coloured faces? (a) 33 (b) 36 (c) 18 (d) 29 39. How many cubes in the top layer have three red faces each? (a) 6 (b) 8 (c) 3 (d) 4 40. How many cubes with 3 face coloured? (a) 10 (b) 11 (c) 14 (d) 12 Directions for Question Nos. (41 to 45) 343 Small unpainted cubes are arranged to form a large cube. All the six faces of the large cube are painted white. Now, a 3*3 cube, comprising 27 small cubes, is removed out from one of the corners of the large cube. The 3*3 cube is now painted blue on all six faces, while all the three surfaces (each of which a is a 3*3 square) of the large cube exposed due to the removal of the 3*3 cube are painted black. Then, the 3*3 cube is put back in its original position in the large cube and the large cube is nally painted yellow on all the six faces.

41. What is the number of small cubes which have exactly three faces painted? (a)

8

(b)

16

(c)

18

(d)

19

42. What is the number of small cubes with exactly one face painted ? (a)

136

(b)

144

(c)

142

(d)

45

43. What is the number of small cube with no face painted (a)

106

(b)

118

(c)

120

(d)

114

44. How many cubes are painted with yellow and blue? (a)

12

(b)

18

(c)

15

(d)

14

45. How many cubes are painted two faces only one with yellow and one with blue? (a)

12

(b)

11

(c)

5

(d)

6

Directions for Question Nos. 46 to 50 125 cubes of similar size are arranged in the form of a bigger cube (5 cubes on each side, i.e., 5x5x5). All the small cubes lying on the edge of the top layer of the bigger cube are removed and also cubes lyingat the four corners of the bottom face are removed. All exposed faces of the block thus left are coloured red. 46.

How many small cubes are left in the block? (a)

20

(b)

93

(c)

96

(d)

105

47. How many cubes have three red faces each? (a)

20

(c) 8 48. How many painted?

cubes

(b)

12

(d)

16

are

with

(a)

20

(b)

24

(c)

18

(d)

16

two

faces

49. How many cubes are with one face painted? (a)

34

(b)

24

(c)

28

(d)

16

50. How many cubes are with no face painted? (a)

34

(b)

24

(c)

28

(d)

27

EBD_7743

196 Koncepts of Logical Reasoning

Cubes Directions for Question Nos. 51 to 55 343 cubes of similar size are arranged in the form of a bigger cube (7 cubes on each side, i.e., 7 × 7 × 7).Its all the 6 faces are painted with Green, Red and blue colour such that each colour is used on exactly 2 faces. 51. If ‘K’ represents the number of cubes painted with red and blue but not green then which of the following set represents the value of ‘K’ most appropriately? (a)

{10, 17, 20, 22}

(b)

{ 10, 20, 22}

(c)

{ 10, 18, 22}

(d)

None of these

52. If ‘K’ represents the number of cubes painted with 3 different colours then what could be the value of ‘K’? (a)

8 or 5

(b)

4 or 5

(c)

2 or 8

(d)

None of these

53. If ‘K’ represents the number of cubes painted with red and blue then which of the following set represents the value of ‘K’ most appropriately? (a)

{112, 13, 24, 26}

(b)

{ 18, 14, 26, 28}

(c)

{ 12, 14, 24, 28}

(d)

None of these

54. If ‘K’ represents the number of cubes painted with only red colour then what could be the value of ‘K’? (a)

50 or55

(b)

45 or 50

(c)

45 or 55

(d)

None of these

55. How many cubes are with only one colour on it? (a)

150 or155

(b)

150 or 175

(c)

150 or 165

(d)

None of these

Directions for Question Nos. 56 to 60 216 cubes of similar size are arranged in the form of a bigger cube (6 cubes on each side, i.e.,6 × 6 × 6).Its all the 6 faces are painted with Green, Red, blue, black, white, orange colours. 56. How many cubes are painted with red or blue? (a)

60 or 72

(b)

66 or 72

(c)

62 or 72

(d)

None of these

57. How many cubes are painted with red or blue or green?

197

58. How many cubes are painted with red or blue but not green? (a)

55or 65

(b)

72 or 66

(c)

55 or 60

(d)

None of these

59. Which of the following statement is correct (i) At least 1 cube is painted with red, green and blue. (ii) At most 1 cube is painted with red, green and blue. (iii) At most 6 cubes are painted with red and green. (iv) At least 6 cubes are painted with red and green. (a)

Only (i) & (iii)

(b)

Only (ii) & (iii)

(c)

Only (ii) & (iv)

(d)

None of these

60. How many cubes are painted with red, blue, green and black? (a)

8

(b)

2

(c)

1

(d)

None of these

Directions for Question Nos. 61 to 65 N3 number of cubes of similar size are arranged in the form of a bigger cube (N cubes on each side, i.e., N × N× N) and kept at the corner of a room, all the exposed surfaces are painted with colour 1, then all the coloured smaller cubes are removed and all the exposed surfaces are painted with colour 2, then all the coloured smaller cubes are removed and all the exposed surfaces are painted with colour 3, this process is repeated ‘K’ number of times.

61. If number of cubes painted with colour 3 is 217 then how many cubes are painted with colour5 (a)

125

(b)

127

(c)

64

(d)

None of these

62. If after 7th step number of cubes painted in exactly 2 faces with colour 7 is 21, then what is the number of cubes removed in 3rd step. (a)

469

(b)

455

(c)

433

(d)

None of these

63. Which of the following can be the number of cubes removed from the original N3 number of cubes. (i) 37, (ii) 61 (iii) 98

(a)

91 or 96

(b)

90 or 84

(a)

only (i) & (ii)

(b)

Only (ii) & (iii)

(c)

96 or 108

(d)

None of these

(c)

all three

(d)

None of these

64. If number of cubes painted on exactly one face with colour 1 and colour 2 is 150, then how many cubes are painted with only colour 4? (a)

45

(b)

(c)

18

(D) 75

67. If number of cuboids with exactly 3 faces painted is 4 then how many cubes are painted on exactly 1 face.

72

65. Of all the removed cubes which one of the following could be the number of cubes with exactly 2face painted after 3 steps?

(c)

33 or 65

(d)

None of these

68. If number of cuboids with exactly 2 faces painted is 36 then how many cubes are painted on exactly 1 face.

114

(a)

46 or 65

(b)

57 or 66

(c)

116

(d)

118

(c)

33 or 65

(d)

None of these

(a)

51

(b)

57

(c)

48

(d)

None of these

69. If number of cuboids with exactly 1 face painted is 57 then how many cubes are painted on exactly 2faces. (a)

46

(b)

57

(c)

44

(d)

None of these

70. If number of cuboids with exactly 1 face painted is 44 then how many cubes are painted on exactly 2 faces. (a)

46

(b)

57

(c)

44

(d)

None of these

If 1 × 3 × 15 then we require 2 cuts in one plane and 14 cuts in other plane so total number of cuts is 2 + 14 = 16.

Concept Applicator (CA) (b) If total number of cut is 10 then minimum number of pieces is 11 when cut is made in one plane only.

If 1 × 5 × 9 then we require 4 cuts in one plane and 8 cuts in other plane so total number of cuts is 4 + 8 = 12

(a) If total number of cut is 10 then for maximum number of pieces these cuts have to be well distributed in three planes. For 10 cuts, 3, 3 and 4 is the distribution of cuts.

If 3 × 3 × 5 then we require 2 cuts in one plane, 2 cuts in 2nd plane and 4 cuts in 3rd plane so total number of cuts is 2 + 2 + 4 = 8.

Hence total number of pieces is (3+1)(3+1) (4+1) = 4x4x5 = 80

5.

(c) For maximum number of cuts it has to be in one cut only, so number of cuts is 49

(b) For maximum number of pieces cuts has to be 6, 7 and 7 and maximum number of pieces is (6+1)(7+1)(7+1) = 7x8x8 = 448.

6.

(c) For minimum number of cuts we will get 50 from 2x5x5 and cuts is 1 + 4 + 4 = 9

Solution (7 to 10)

Minimum number of pieces is 20+1 = 21.

Since total number of cubes is hence in the formula we will substitute n = 6

Hence required ratio is 448 : 21 4.

57 or 66

(b)

66. If number of cuboids with exactly 2 faces painted is 44 then how many cubes are painted on exactly 1 face.

3.

(b)

112

A big cube is painted with three colours red, green and blue such that each of the colour is on two faces. Now this cube is cut into 105 identical cuboids.

2.

51 or 65

(a)

Directions for Question Nos. 66 to 70

1.

(a)

(b) If 45 = 1x1x45 then we require only 44 cuts in one plane.

7.

(a) Number of the cubes with 0 faces painted is (6 – 2)3 = 43 = 64

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198 Koncepts of Logical Reasoning

Cubes 199 (d) Number of the cubes with 2 faces painted is 12(6 – 2) = 12 × 4 = 48 (a) At most 2 faces painted means number of cubes with 0 face painted + number of cubes with 1 face painted + number of cubes with 2 face painted = 64 + 48 + 96 = 208

15. (d) From the above explanation number of the cubes with 3 faces painted is 1.

10. (b) At least 2 faces painted means number of cubes with 2 face painted + number of cubes with 3 face painted = 96 + 8 = 104.

Number of vertices with three faces exposed (Painted) is 2

8. 9.

Solution (11 to 15) Out of 6 faces of 3 faces are exposed and those were painted. Number of vertices with three faces exposed (Painted) is 1 Number of vertices with 2 faces exposed (Painted) is 3 Number of vertices with 1 faces exposed (Painted) is 3 Number of vertices with 0 faces exposed (Painted) is 1 Number of sides with 2 sides exposed (Painted) is 3 Number of sides with 1 sides exposed (Painted) is 6 Number of sides with no sides exposed (Painted) is 3 From the above observation ;Number of cubes with 3 faces Painted is 1 Number of cubes with 2 faces Painted is given by sides which is exposed from two sides and there are 3 such sides and from one side we will get 6 such cubes hence required number of cubes is 6 x 3 = 18 Number of cubes with 1 face Painted is given by faces which is exposed from one sides and there are 3 such faces hence required number of cubes is 36 x 3 = 108 Number of cubes with 0 face Painted is given by difference between total number of cubes – number of cubes with at least 1 face painted = 343 – 1 – 18 – 108 = 216 In other words number of cubes with 0 painted is (7-1)3 = 216. 11. (d) From the above explanation number of the cubes with 0 faces painted is 216. 12. (b) From the above explanation number of the cubes with 2 faces painted is 18. 13. (c) From the above explanation number of the cubes with at most 2 faces painted is 216 + 108 + 18 = 342. Or else 343 -1 = 342 14. (a) From the above explanation number of the cubes with at least 2 faces painted is 18 + 1 = 19.

Solution (16 to 20) Out of 6 faces of 4 faces are exposed and those were painted.

Number of vertices with 2 faces exposed (Painted) is 4 Number of vertices with 1 faces exposed (Painted) is 2 Number of vertices with 0 faces exposed (Painted) is 0 Number of sides with 2 sides exposed (Painted) is 5 Number of sides with 1 sides exposed (Painted) is 6 Number of sides with no sides exposed (Painted) is 1 From the above observation ;Number of cubes with 3 faces Painted is 2 Number of cubes with 2 faces Painted is given by sides which is exposed from two sides and required number of cubes is 6 x 4 + 1x5 = 29 since there are 4 edges will give us 6 cubes from 1 edge and 1 edge (between two vertices which are painted or exposed from 3 sides) will give us only 5 cubes. Number of cubes with 1 face Painted is given by faces which is exposed from one sides and required number of cubes is 36x2 + 30x2 = 132 Number of cubes with 0 face Painted is given by difference between total number of cubes – number of cubes with at least 1 face painted = 343 – 2 – 29 – 132= 180 In other words number of cubes with 0 painted is 6x6x5 = 180 16. (d) From the above explanation number of the cubes with 0 faces painted is 180. 17. (b) From the above explanation number of the cubes with 2 faces painted is 29. 18. (a) From the above explanation number of the cubes with at most 2 faces painted is 180 + 132 + 29 = 341. Or else 343 -2 = 341

19. (a) From the above explanation number of the cubes with at least 2 faces painted is 29 + 2 = 31. 20. (d) From the above explanation number of the cubes with 3 faces painted is 2. Solution (21 to 25) Out of 6 faces of 5 faces are exposed and those were painted. Number of vertices with three faces exposed (Painted) is 4

23. (a) From the above explanation number of the cubes with at most 2 faces painted is 150 + 145 + 44 = 339. Or else 343 – 4 = 339 24. (a) From the above explanation number of the cubes with at least 2 faces painted is 44 +4 = 48. 25. (d) From the above explanation number of the cubes with 3 faces painted is 4. Solution (26 to 30)

Number of vertices with 2 faces exposed (Painted) is 4

Let us see the changes due to removal of 1 cube from corner-

Number of vertices with 1 faces exposed (Painted) is 0

Number of vertices with three faces exposed (Painted) is 7 + 3 = 10

Number of vertices with 0 faces exposed (Painted) is 0

Number of Cubes with 2 sides exposed (Painted) :- In general one edge give us 4 (n-2 in general case) cubes with two face painted but in this case out of 12 edges only 9 edges will give us 4 cubes in one edge and remaining 3 edges will give us 3 cubes from one edge, hence total number of edge is 9x4 + 3x3 = 45

Number of sides with 2 sides exposed (Painted) is 8 Number of sides with 1 sides exposed (Painted) is 4 Number of sides with no sides exposed (Painted) is 0 From the above observation ;Number of cubes with 3 faces Painted is 4 Number of cubes with 2 faces Painted is given by sides which is exposed from two sides, out of 8 such edges 4 vertical edges will give us 6 cubes per edge and 4 edges from top surface will give us 5 such cubes from each edge and required number of cubes is 6 x 4 + 4x5 = 44. Number of cubes with 1 face Painted is given by faces which is exposed from one sides four vertical faces will give us 6x5 = 30 cubes per face and top face will give us 5x5 = 25 and required number of cubes is 30x4 + 25x1 = 145 Number of cubes with 0 face Painted is given by difference between total number of cubes – number of cubes with at least 1 face painted = 343 – 4– 44– 145 = 150 In other words number of cubes with 0 painted is 6x5x5 = 150 21. (b) From the above explanation number of the cubes with 0 faces painted is 150. 22. (c) From the above explanation number of the cubes with 2 faces painted is 44.

Number of Cubes with 1 side exposed (Painted):- it will remain same as normal case i.e 6(42) = 96 Number of Cubes with no sides exposed (Painted) is 43 =64 From the above observation ;26. (a) From the above explanation number of the cubes with 0 faces painted is 64. 27. (c) From the above explanation number of the cubes with 2 faces painted is 45. 28. (a) From the above explanation number of the cubes with at most 2 faces painted is 64 + 96 + 45 = 205. Or else 215 – 10 = 205 29. (a) From the above explanation number of the cubes with at least 2 faces painted is 45 + 10 = 55. 30. (d) No cubes are with 4 face painted. Solution (31 to 35) For least number of cuts 120 = 4x5x6 i.e number of cuts must be 3 , 4 and 5 in three planes in this case number of cubes on a face is either 6x5= 30 or 6x4 = 24 or 4x5 = 20 cubes. And number of cuboids on an edge is 4 or 5 or 6

EBD_7743

200 Koncepts of Logical Reasoning

Cubes 201 31. (a) Number of cuboids with no face painted is (4-2) (5 – 2)(6 – 2) = 2x3x4 =24

38. (a) Only two faces are cloured is when cubes are at the edges (baring the corner cubes)

32. (a) To satisfy this case all the cuboids on the edges and corners must have more than one colour on them. And in that case opposite faces must have painted in the same colour.

If no cubes have been removed then on each edges we will get 3 cubes that has exactly 2 faces coloured, hence total number of such cubes = 12x3 =36, because we have 12 edges.

In that case number of cuboids with 3 colours on them = 8

Out of these 3 cubes are removed hence required number of cubes = 36-3 =33

In that case number of cuboids with 2 colours on them = 4 x(2 +3+4) = 36

39. (b) Each has Red faces on top layer = all edges cube = 2 + 2 + 2 + 2 = 8

Hence number of cuboids with at least 1 colour on them is 120 – 36 – 8 = 76 33. (c) In this case when k is maximum, one particular colour is used on three faces such that any two faces are adjacent to each other. Required number of cuboids will come from edges but not from vertex = 3 + 4 + 5 + 1 = 13 34. (b) Maximum number of cuboid with red colour is possible when cube is painted with red colour in 3 sides with minimum number of common edges (which is equal to 2)

40. (c) Number of cubes with 3 face coloured = 3 (Bottom cubes) + 8 top cubes +3 (coloumn cubes) = 14 Solution (41 to 45) Initial total number of cubes = 343, Number of cubes removed = 27 Smaller 27 cubes painted blue Exposed faces of original big cube (3 faces with 9 cube on each face i.e total 27 cubes) painted with black

Hence required maximum value is 6 (5 + 5 + 4 – 2) = 72

41. (c) Since 7 corner (Vertices) of bigger cube is untouched hence they are painted with three faces.

For minimum number of such cuboid Red colour is used only once and minimum number of cubes in that case is 20

Now consider the corner from where we have removed 3x3x3 cubes,

Hence required ratio is 72 : 20 = 18 : 5 35. (b) In this case we have to use red and green twice and same colour should be on opposite faces then required cube is given by 4 edges (but not corner ), maximum number of cubes from one edge is 6 – 2 =4 so required number of cubes is 4 × 4 = 16 Solution (36 to 40) 36. (c) Total no. of cubes=53=125, Some cubes from different corners are removed and the number removed cubes are 2, 3, 4 and 4. Remaining number of small cubes:= 125 – 2 – 3 – 4 – 4 =125 – 13 =112 37. (c) In any plane, leave 4 sides cube and select (3x3x3) inter section .But the cubes 2 × 2 ×1 give 2 less cube because that part we have already removed. No. of cubes = (3 x 3 x 3) – 2 = 25.

After removal 3 new corners of the bigger cube will be generated that will be painted with 3 faces and 8 corners from smaller cube of 3x3x3 painted with 3 faces. So the such total number of such cubes is 7 + 3+8 =18. 42. (b) In original big cube number of faces with one colour is 3(7 – 2)2 = 75 (here we have considered only 3 untouched faces of big cube) And 3 × 17 = 51 cubes from other 3 faces. But here we have removed a cubes of the form of 3x3x3 and again put it back so out of three new exposed faces of big cube we will have 4 cubes in each face that is painted with one colour hence number of cubes from these three surfaces is 3x4 = 12 Now consider out of 3x3x3 cubes we will have 6 cubes (one in each face) which are painted only one face. Hence total number of cubes = 75 +51 + 12 + 6 = 144

43. (a) Without any changes number of cubes with no face colour is given by (7 – 2)3 = 125 Now because of removal of 3x3x3 cubes from one of the corner from each face that were not painted earlier got exposed and will get painted, so from 3 × 3 × 3 cubes, 7 cubes got painted, and a 4 × 3 = 12 cubes from 3 exposed faces of big cube got painted. Total number of cubes with no face painted is 125 – 7 – 12 = 106 44. (a) Out of 27 small cubes from 3 × 3 × 3, outer 27 cubes are 1st painted with blue and then it is kept back with original cube and painted with yellow so out of 26 cubes only 6 edges will give us cubes with both the colours and number of such cubes are 12 45. (d) out of 12 cubes in previous question there are 6 cubes with 2 faces yellow so number of cubes painted two faces only one with yellow and one with blue is 12 – 6 = 6 Solution (46 to 50) 46. (d) Number of cubes removed from top face = 16 Number of cubes removed from bottom face = 4 Number of cubes left = 125 – (16 + 4) = 105 47. (a) Number of cubes with three coloured face on the top side = 4 Number of cubes with three coloured face on the 2nd from top side = 4 Number of cubes with three coloured face on the bottom side = 12 Total number of such cubes = 12 + 8 = 20 48. (b) Number of cubes with two face painted from the top side (Which is a square of 3x3 = 9 cubes) is 4. Number of cubes with two face painted from the 2nd from top side (Which has four edges and each edge has 3 such cubes) is 4x3 = 12. Number of such cubes from vertical edges is 4x1 = 4 Number of such cubes from bottom face is 4x1 =4 Hence total such cubes is 4 +12 + 4 +4 = 24 49. (a) From top face (out of 3x3 square face) only one cube is with one face painted. From 4 vertical faces each face will give us 6 cubes hence total number of cubes from vertical faces is 6 × 4 = 24.

From bottom face we will get 3 × 3 = 9 cubes So total number of cubes with one face painted is 1 + 24 + 9 = 34 50. (d) Number of cubes with no face painted is 105 – 34 – 24 – 20 = 27 Or else all the 3x3x3 inner cubes will remain coloured . Solution (51 to 55) Here we have following casesCase (i) :- when same colour is on opposite face. Case (ii) :- When two red colours are on opposite face and blue & green on adjacent faces Case (iii):- When two green colours are on opposite face and blue &red on adjacent faces. Case (iv) :- When two blue colours are on opposite face and red & green on adjacent faces. Case (v) :- When same colours are on adjacent faces. 51. (a) We will evaluate the value of ‘K’ in each and every caseCase (i) :- In this case number of cubes is given by 4 common edges except all 8 corner ones so number of cubes is 5x4 = 20 Case (ii) :- :- In this case number of cubes is given by 4 common edges (From one edge we will get 5 cubes with 2 face painted) except 6 corner ones (2 corner cubes are painted with only red and blue) so number of cubes is 5x4 +2= 22 Case (iii) :- :- In this case number of cubes is given by 2 common edges (From one edge we will get 5 cubes with 2 face painted) so number of cubes is 5x2= 10. Case (iv) :- :- In this case number of cubes is given by 4 common edges (From one edge we will get 5 cubes with 2 face painted) except 6 corner ones (2 corner cubes are painted with only red and blue) so number of cubes is 5 × 4 + 2 = 22 Case (v) :- :- In this case number of cubes is given by 3 common edges (From one edge we will get 5 cubes with 2 face painted) except 6 corner ones (2 corner cubes are painted with only red and blue) so number of cubes is 5x3+2= 17 Hence option (a) gives the all possible value of ‘K’

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202 Koncepts of Logical Reasoning

Cubes 203 52. (c) We will evaluate the value of ‘K’ in each and every caseCase (i) :- In this case number of cubes is all 8 corner ones so number of cubes is 8 Case (ii) :- :- In this case number of cubes is 4 corner ones so number of cubes is 4 Case (iii) :- :- In this case number of cubes is 4 corner ones so number of cubes is 4 Case (iv) :- :- :- In this case number of cubes is 4 corner ones so number of cubes is 4 Case (v) :- :- :- In this case number of cubes is 2 corner ones so number of cubes is 2 Hence option (c) gives the all possible value of ‘K’ 53. (b) We will evaluate the value of ‘K’ in each and every caseCase (i) :- In this case number of cubes is given by 4 common edges so number of cubes is 7x4 = 28 Case (ii) :- :- In this case number of cubes is given by 4 common edges, out of these 4 edges there are 2 corner cubes common with these 4 edges so number of cubes is 7x 4 – 2 = 26 Case (iii) :- :- In this case number of cubes is given by 2 common edges so number of cubes is 7x 2 = 14. Case (iv) :- In this case number of cubes is given by 4 common edges, out of these 4 edges there are 2 corner cubes common with these 4 edges so number of cubes is 7x4 -2= 26 Case (v) :- :- In this case number of cubes is given by 3 common edges, out of these 3 edges there are 2 corner cubes common with these 4 edges so number of cubes is 7x3 -3= 18 Hence option (b) gives the all possible value of ‘K’ 54. (a) We will evaluate the value of ‘K’ in each and every caseCase (i) :- In this case number of cubes is given by 25 cubes from 1 face so number of cubes is 25 × 2 = 50 Case (ii) :- In this case number of cubes is given by 25 cubes from 1 face so number of cubes is 25 × 2 = 50 Case (iii):-In this case number of cubes is given by 25 cubes from 1 face and 5 cubes from common edge so number of cubes is 25 × 2 + 5 = 55

Case (iv) :- In this case number of cubes is given by 25 cubes from 1 face and 5 cubes from common edge so number of cubes is 25 × 2 + 5 = 55 Case (v) :- In this case number of cubes is given by 25 cubes from 1 face and 5 cubes from common edge so number of cubes is 25 × 2 + 5 = 55 Hence option (a) gives the all possible value of ‘K’ 55. (c) From the solution of previous question required number of cubes is 3x50 = 150 or 3x55 = 165 Solution (56 to 60) Here on each face 6x6 = 36 cubes that are painted with one colour. 56. (b) Case (i) :- when red and blue are adjacent to each other then from one face we will get 6x6 =36 cubes but out of them 6 cubes from common edge is common so number of cubes are 2x6x6 -6 = 66 Case (ii) ;- When red and blue are opposite to each other then required number of cubes is 2x6x6 = 72 57. (a) Case (i) :- when these three colour are adjacent to each other then from one face we will get 6x6 =36 cubes but out of them 6x3 = 18 cubes from common edge is common so number of cubes are 3x6x6 -6x3 = 90 2 one corner cube 50 total 90 + 1 = 91 cube Case (ii) ;- When red and blue are opposite to each other (or any two of the given three) then required number of cubes is 3x6x6 – 2x6 = 96 58. (c) Case (i) :- when red and blue are opposite to each other then from one face we will get 6x6 =36 cubes but out of them 2x6 cubes from common edge with green painted face is common so number of cubes are 2x6x6 -2 x6 = 60 Case (ii) ;- When red and blue are adjacent to each other then green is either adjacent to these or opposite to any one of red or blue, in 1st condition number of cubes is 2 x 6x6 – 2x6 – 11 = 55 cubes or in 2nd condition 2x6x6 – 6 – 6 = 60, required number of cubes is 55 or 60 59. (b) From solution of previous questions statements (ii) and (iii) are correct. 60. (d) None of the cubes can be painted in four faces.

Solution (61 to 65) Consider the 1st step, .initial number of cubes N3 after removal of 1st set of coloured cubes number of cubes left out is (N – 1)3 hence number of cubes removed in 1st step (i.e with colour 1) is N3 – (N – 1)3 = 3N2 – 3N + 1 Similarly number of cubes removed in 2nd step (i.e with colour 2) is 3(N – 1)2 – 3(N – 1) + 1

63. (c) The required number of cubes must be equal to difference between two positive integer Since 64 – 27 = 37 125 – 27 = 98 125 – 64 = 61 64. (c) Number of cubes with only face is painted with colour 1 is Number of cubes with only face is painted with colour 2 is

Similarly number of cubes removed in 3rd step

From the given condition (3N2 – 9N + 6) + (3N2 – 15N + 18 from this we will get N =7.

is (i.e with colour 3) and so on. Number of cubes remaining after 1st step is (N – 1)3

Number of cubes left after step 3 is 4x4x4 = 64

Number of cubes remaining after 2ndstep is (N – 2)3 and so on.

When all the exposed faces are painted with colour 4 then number of cubes with only one face painted is 3x2x3 =18

61. (b) from the given condition

From the observation for 1st step we have seen that number of cubes is 3(N-2)(N-1) or in other words 3 times the product of 2 consecutive integer that is satised only by 18 which is 3 times of 2x3.

Number of cubes removed in 3rd step (i.e with colour 3) is = 217 hence N = 11 So number of cubes with colour 5 is = 127 62. (a) Total number of Cubes left after 7th step in (N-7)3 in the form of (N-7) x (N-7) x (N-7) cubes. And out of these number of cubes whose two sides are painted is given by three edges with each edge has (N-8) so total number of cubes is 3x(N-8)

65. (b) After step 1 number of cubes with exactly 2 face painted is 4(N – 1) + (N – 2) = 5N -6 Similarly after 2nd step number of cubes with exactly 2 face painted is 5(N – 1) -6 = 5N -11 And after 3rd step number of cubes with exactly 2 face painted is 5(N – 2) -6 = 5N -16

From the given information 3(N-8)= 21 or N = 15

So total number of such cubes is 15N –33 out of the given options only option B satisfy the given condition.

Number of cubes removed in 3rd step (i.e with colour 3) is = 469 Solution (66 to 70)

If C3 means number of cuboids with 3 face painted, C2 means number of cubes with 2 face painted and so on Case

P= Q=

R=

C3

C2

C1

C0

(i)

1 × 1 × 105

0

0

104

2

102

0

0

(ii)

1 × 3 × 35

0

2

34

4

68

33

0

(iii)

1 × 5 × 21

0

4

20

4

44

57

0

(iv)

1 × 7 × 15

0

6

14

4

36

65

0

(v)

3×5×7

2

4

6

8

36

46

15

66. (b) From the table condition is same as that of case (iii) then number of cubes painted with 1 face is 57 67. (c) From the table condition is same as that of case (ii), (iii) or (iv) then number of cubes painted with 1 face is either 33 or 57 or 65 68. (a) From the table condition is same as that of case (iv) or (v) then number of cubes painted with 1 face is either 46 or 65 69. (c) From the table condition is same as that of case (iii) then number of cubes painted with 1 face is either 46 or 65 70. (d) From the table we can say that no such exists so data is inconsistence.

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204 Koncepts of Logical Reasoning

Calendar 205

Calendar Exam

Importance

11

Exam

Importance

CAT

Average

IBPS/Bank PO

Very Important

XAT

Average

SSC

Very Important

IIFT

Important

CSAT

Very Important

SNAP

Important

Other Govt Exams

Very Important

NMAT

Important

Other Govt Exams

Very Important

HISTORY It takes the earth about 8765.8128 Hrs ( 365.2422 days ) to go around the sun, but in a normal calendar year we have only 365 days. That means we are leaving 0.2422 days (5.8128 Hrs) every year uncounted. The extra fraction of a day 0.2422 days (5.8128 Hrs) adds up for four calendar years would be 9688 days (23.2512 hrs) is almost a whole day, so every four years we add an extra day to our calendar, February 29. We call that year leap year. To make things easier, leap years are always divisible by four: 1992, 1996 and 2008 are leap years. For hundreds of years, people used a calendar called the Julian calendar that followed this rule, adding a leap year every four years. However, because .9688 isn't exactly a whole day, the Julian calendar slowly began to disagree with the real seasons. In 1582, Pope Gregory xed this problem by ordering everyone to use a new set of rules. These rules are named the Gregorian calendar, after him. They work like this: Rule

Examples

If numerical value of year is divisible by 4 then 2004, 2008, and 2012 are leap years. it is a leap year. If numerical value of century is divisible by 400 1900 and 2100 are not leap years. But 2000, then it is a leap year 1600 are leap years

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206 Koncepts of Logical Reasoning How do I nd the day of the week for any date? Method 1:- Zeller's Rule The following formula is named Zeller's Rule after a Reverend Zeller. [x] means the greatest integer that is smaller than or equal to x. e.g [2.1] =2, [3.7] =3, [4.99] =4 and so on.

é13* m - 1 ù é D ù éC ù f =k + ê + D + ê ú + ê ú - 2* C . ú 5 ë û ë4û ë4û Ø

k is the day of the month. Let's take 25th June 2013 as an example. For this date, k = 25.

Ø

m is the month number. Months have to be counted specially for Zeller's Rule: March is 1, April is 2, and so on to February, which is 12. (This makes the formula simpler, because on leap years February 29 is counted as the last day of the year) Because of this rule, January and February are always counted as the 11th and 12th months of the previous year. Means for January 2011 we will use month m = 11 but year we will take as 2010, as new year will begin from March 2011. In our example, m = 4. D is the last two digits of the year, D = 13. C stands for century: it's the rst two digits of the year. In our case, C = 20. Now let's substitute our example numbers into the formula.

é13* m - 1 ù é D ù éC ù f =+ k ê D + + úû êë 4 úû + êë 4 úû - 2* C. 5 ë é13* 4 - 1 ù é13 ù é 20 ù f= 25 + ê + 1 3 + úû êë 4 úû + êë 4 úû - 2*20. ë 5 é 51ù f= 25 + ê ú + 13 + [3.25]+ [5]- 40. ë5û F = 25 + 10+ 13 +3+5-40= 16 Now divide this value of f= 16 by 7 remainder is 2. A remainder of 0 corresponds to Sunday, 1 means Monday, etc Here remainder 2 means Tuesday. In this formula we may get value of f as negative if this is the case then we will add multiple of 7 to make it positive. e.g if f = -17 then f = -17 + 3x7 = 4 so f = -17 is same as that of 4. Method 2:- The Key Value Method In this method we have to memorize few codes but this method is quick and fast in calculation. We'll take an example of December 20, 1984 as an example. Ø

Take the last 2 digits of the year. In this case it is 84.

Ø

Divide it by 4, and nd quotient, since [84/4] = 21

Ø

Add the day of the month. In our example, 20 + 21 = 41

Calendar 207 Ø

Add the month's key value, from the following table. Jan

Feb

Mar

Apr

May

June

July

Aug

Sept

Oct

Nov

Dec

1

4

4

0

2

5

0

3

6

1

4

6

Ø

The month for our example is December, with a key value of 6. Hence 41+ 6 = 47.

Ø

If month is January or February of a leap year, subtract 1. We're using December, so it is not applicable here.

Ø

Add the century code from the following table. 1700s

1800s

1900s

2000s

4

2

0

6

Ø

Our example year is 1984, and the and get the code 0. Now we add this to our running total: 47 + 0 = 47.

Ø

Add the last two digits of the year. 47 + 84 = 131.

Ø

Divide by 7 and take the remainder. This time, 1 means Sunday, 2 means Monday, and so on. A remainder of 0 means Saturday. 131 / 7 = 18, remainder 5, so December 20 of 1984 will be on the fourth day of the week-Wednesday.

Some more interesting points:A month has either 28 (Non leap year February), 29 (Leap year February), 30 or 31 days. If a month start with Monday, Tuesday, or Wednesday the month will have 4 Saturdays and 4 Sundays i.e total 8 weekends. If a month start with Thursday the month will have 5 Saturdays and 4 Sundays i.e total 9 weekends. If month has 31 days. If a month start with Friday the month will have 5 Saturdays and 4 Sundays i.e total 9 weekends if month has 28, 29 or 30 days and will have 5 Saturdays and 5 Sundays i.e total 10 weekends if month has 31 days. If a month start with Saturday the month will have 5 Saturdays and 4 Sundays i.e total 9 weekends if month has 28 or 29 days and will have 5 Saturdays and 5 Sundays i.e total 10 weekends if month has 30 or 31 days. If a month start with Sunday the month will have 4 Saturdays and 5 Sundays i.e total 9 weekends. 1.

Counting of Odd Days: 1.

1 ordinary year = 365 days = (52 weeks + 1 day.) \

2.

1 leap year = 366 days = (52 weeks + 2 days) \

3.

ordinary year has 1 odd day. leap year has 2 odd days.

100 years = 76 ordinary years + 24 leap years = (76 × 1 + 24 x 2) odd days = 124 odd days. = (17 weeks + days) 5 odd days. Number of odd days in 100 years = 5. Number of odd days in 200 years = (5 × 2) 3 odd days. Number of odd days in 300 years = (5 × 3) 1 odd day. Number of odd days in 400 years = (5 x 4 + 1) 0 odd day. Similarly, each one of 800 years, 1200 years, 1600 years, 2000 years etc. has 0 odd days.

EBD_7743

208 Koncepts of Logical Reasoning

1.

1

If 14th October 2005 is Friday then which day is 14th October 2006? (a) Tuesday (c) Thursday

2.

3.

(b) Wednesday (d) Saturday

If 1st Jan 2001 is Monday then which day was on 1st Jan 2013. (a) Tuesday (c) Thursday

2

7.

8.

(b) Wednesday (d) Saturday

11. Today is Sunday what day of the week was 79 days back. (a)

Tuesday

(b)

Friday

(c)

Thursday

(d) Saturday

12. If 2nd June 2013 is Sunday then which day was on 2nd June 2010? (a)

Wednesday

(b)

Friday

(c)

Thursday

(d) Saturday

13. Which of the following day of the week was 15th august 1947?

Which of the following is not a leap year? (a) 2100

(b)

1200

(c)

(d)

2000

1600

Which of the following is a leap year? (a) 1146

(b)

1254

(c)

(d)

1128

2100,

If Jan 1, 2006 was Sunday. What was the day of the week Jan 1, 2010

(b) Wednesday (d) Friday

If 2nd June 2012 is Saturday then which day is 7th July 2012? (a) Tuesday (c) Thursday

5.

(b) Wednesday (d) Friday

If 14th October 2005 is Friday then which day is 14th October 2009? (a) Tuesday (c) Thursday

4.

(b) Saturday (d) Friday

If today is Monday then which day of the week is after 59 days. (a) Tuesday (c) Thursday

6.

9.

(a) Tuesday

(b)

(c)

(d) None of these

Thursday

Wednesday

Today is Sunday. After 69 days, it will be (a) Tuesday

(b)

(c)

(d) None of these

Thursday

Saturday

10. If 6th March, 2005 is Sunday, what was the day of the week on 6th March, 2002? (a) Tuesday

(b)

Saturday

(c)

Thursday

(d) None of these

(a)

Wednesday

(b)

(c)

Thursday

(d) Saturday

Friday

14. Which of the followings day could be the 18th October 2050? (a) Tuesday (b) Friday (c) Thursday (d) Saturday 15. If 5th march of a particular year is Friday then which day of the week will be on 5th November. (a) Tuesday (b) Sunday (c) Thursday (d) None of these

Calendar 209 16. How many weekends are there in March 2009? (a) 9 (b) 10 (c) 8 (d) None of these. 17. Which of the following two months in a particular year will have same calendar. (a) (b) (c) (d)

January and August January and October March & November None of these

(a)

Tuesday

(b)

Sunday

(c)

Thursday

(d) None of these

19. How many days could be in K weeks and K days. (a)

96

(b)

92

(c)

90

(d) None of these

20. How many days are between Kth day of the Kth week to 2Kth day of the 2Kth week

18. On 14th Feb the valentine day , 2009 it was Saturday. What was the day of the week on 14th Feb, 2008?

3

21. On what dates of April, 2001 did Friday fall?

(a) (c)

96 90

(b) 92 (d) None of these

(a) 1981

(b)

(c)

(d) None of these

1990

1985

26. Which of the following year has 53 Sundays?

(a)

5th , 12th, 19th, 26th April

(b)

4th , 11th, 18th, 25th April

(c)

3rd , 10th, 17th, 24th April

(a) 2001

(b)

(d)

None of these

(c)

(d) None of these

22. How many Sundays will be in a period of 100 years. (a)

5217

(b)

5219

(c)

5217 or 5218

(d)

5218 or 5219

23. If year 194X starts with Sunday then which one of the following could be the value of x.

(a)

9

(b)

6 or 9

(c)

1 or 5

(d) None of these

24. Calendar of which one of the following 2 years is similar? (a) 1940, 1946

(b)

(c)

(d) None of these

1912, 1616

1977, 1982

25. Calendar of 2013 is related to calendar of 2015 same as calendar of 1977 to which one of the following year?

2009

2006

27. If in a particular year ‘X’ there are 53 Sundays then how many Sundays will be there in a period of four years X to X+3 year. (a) 208

(b)

(c)

(d) none of these

208 or 209

209

28. If in a particular month there are 5 Thursday then how many Fridays are there in next month. (a) 4

(b)

(c)

(d) None of these

4 or 5

5

29. Monday falls on 20th March, 1995. What was the day on 3rd November, 1994? (a) Tuesday

(b)

(c)

(d) None of these

Thursday

Sunday

30. If 94 days back it was Monday then after how many days we will get a Sunday? (a) 5

(b)

(c)

(d) None of these

27

24

EBD_7743

210 Koncepts of Logical Reasoning

4

(a) 2014

(b)

2018

(c)

(d)

2022

31. The last day of century can be which of the following day? (a) Tuesday

(b)

(c)

(d) Saturday

Thursday

Friday

32. The calendar for the year 2011 will be the same for the year 2021

33. If in a particular year June month has 5 Mondays then that year started with which

36. If in a particular year there are 53 weekends (i.e Saturday and Sunday) then how many weekend January of that year has.

(a) 8

(b)

(c)

(d) None of these

10

37. If 1st day of a century is Thursday then what could be the 1st day of next century.

(a) Monday or Tuesday

one of the following days?

(b) Sunday or Monday

(i)

(c)

Thursday or Friday

(ii) Friday or Wednesday

Tuesday or Wednesday

(d) None of these

(iii) Thursday or Saturday

38. If 1st January of a particular year (say x) is Monday then what will be the 1st day of the year 100 years after (i.e year x+100).

(a) Only (i) (b) Only (i) and (ii) (c)

9

only (i) and (iii)

(d) None of these 34. If in a particular year November has 5 Mondays then How many weekends (Saturday and Sunday) does April of that month has?

(a) 8

(b)

(c)

(d) 8 or 9

10

9

35. If in a year December has 9 weekends (Saturday & Sunday) then in that year January had how many weekends?

(a) 8

(b)

(c)

(d) None of these

10

9

(a) Friday (b) Friday or Saturday (c)

Saturday or Sunday

(d) None of these 39. 1st day of century can not start with which of the following day? (a) Wednesday, Friday, & Sunday (b) Wednesday, Friday, & Saturday (c)

Wednesday, Thursday, & Sunday

(d) None of these 40. Calendar of month July is same as that of which month of previous year. (a) May

(b)

September

(c)

(d)

None of these

April

  Calendar

211 

Solutions Concept Applicator (CA) 1. (b) Since in 1 year we have one odd day while in one leap year we have two odd days.

Number of odd days are 1 (for 2006)



1 days more than Friday means Saturday .

2. (c) Since a day of the week repeat after every 7 days, divide 59 by 7, remainder is 3 so number of odd days is 3, and 3 days after Monday is Thursday. 3. (b) Since in 1 year we have one odd day while in one leap year we have two odd days.



Number of odd days are 1 (for 2006)+1 (for 2007) + 2 (For leap year 2008) + 1 (for year 2009) so total number of odd days are 1 + 1 + 2 + 1 = 5 days,

8. (d) On 31st December, 2005 it was Saturday.

Number of odd days from the year 2006 to the year 2009 = (1 + 1 + 2 + 1) = 5 days.



On 31st December 2009, it was Thursday.



Thus, on 1st Jan, 2010 it is Friday.

9. (b) When 69 is divided by 7 we will get remainder 6 so the day will be 6 days more than Sunday i.e Saturday. 10. (d) Number of odd days between 6th march 2002 to 6th march 2003 is 1

Number of odd days between 6th march 2003 to 6th march 2004 is 2



Number of odd days between 6th march 2004 to 6th march 2005 is 1



Hence total Number of odd days between 6th march 2002 to 6th march 2005 is 1+2+1 =4 odd days so 6th march 2002 is 4 days before Sunday i.e Wednesday.

5 days more than Friday means Wednesday.

4. (d) Number of days in June is 30 – 2 = 28, and number of days in July is 7, total number of days is 28 + 7 = 35, when we divide 35 by 7 remainder is 0, or number of odd days is 0 hence 7th july must be the same day as that of 2nd June i.e Saturday in this case. 5. (a) Number of years between 2001 to 2013 is 12 years (2013–2001 = 12 years)

Number of leap years between 2001 to 2013 is 3, since a year has 1 odd days and a leap year has 2 odd days, so total number of odd days is 12 + 3 = 15 odd days.



When we divide 15 by 7 remainder is 1 or in total we have 1 odd day so 1st Jan 2013 is Tuesday.

6. (a) The century divisible by 400 is a leap year. 2100 is not divisible by 400 hence is not a leap year. 7. (d) The century divisible by 400 is a leap year. 2100 is not divisible by 400 hence is not a leap year so 2100 is not a leap year, out of remaining 1128 is divisible by 4, so 1128 is a leap year.

Concept Builder (CB) 11. (b) When we divide 79 by 7 we will get remainder 2 so we have 2 odd days, so required day must be 2 days back from today (i.e Sunday) and that day should be Friday. 12. (a) Consider from 2nd June 2010 to 2nd June 2013 we have total 2 non leap year and one leap year so number of odd days are 1+1+2 =4 so 2nd June 2010 must be 4 days back from Sunday and that day is Wednesday.

From Zeller’s Formula:–

13* m − 1  D  C  f =k +  + D +   +   − 2* C.  5   4 4

In this case k = 2 (since 2nd June)



Month m = 4 (As march =1, April = 2, May = 3, June = 4)



D is the last two digit of year here D = 10



(As year is 2010)



C is 1st two digit of century here C = 20



(As year is 2010)

13* 4 − 1 10   20  f =2 +  + 10 +   +   − 2* 20.  5   4 4

 51 f =2 +   + 10 + [ 2.5] + [ 5] − 40. 5

f = 2 + 10 + 10 + 2 + 5 – 40 = –11



This – ve value of f can be made positive by adding multiple of 7



So f = – 11 + 14 = 3



When divided by 7 we will get remainder 3, hence number of odd days is 3,



So 2nd june 2010 is 3 days more than Monday, i.e Wednesday.

13. (b) From Zeller’s Formula:–

13* m − 1  D  C  + D +   +   − 2* C. f =k +   5   4 4



D is the last two digit of year here D = 50 (As year is 2050)



C is 1st two digit of century here C = 20 (As year is 2050)

13*8 − 1  50   20  f =18 +  + 50 +   +   − 2* 20.   5  4 4

103  f =18 +  + 50 + [12.5] + [ 5] − 40.  5 

f = 18 + [103/5] + 50 + [12.5] + [5] – 40.



f = 18 + 20 + 50 + 12 + 5 – 40 = 65 When divided by 7 we will get remainder 2, hence number of odd days is 2,



So 18th October 2050 is 2 days more than Sunday, i.e Tuesday.

15. (d) Here we have to find the number of odd days between, 5th march and 5th November,

Number of days in March is 26 or 5 odd days (Here we have not included 5th march)



In this case k = 15 (since 15th August)



Number of days in April is 30 or 2 odd days



Month m = 6 (As march = 1, April = 2, May = 3, August = 6)



Number of days in May is 31 or 3 odd days



Number of days in June is 30 or 2 odd days



D is the last two digit of year here D = 47 (As year is 1947)



Number of days in July is 31 or 3 odd days



C is 1st two digit of century here C = 19(As year is 1947)



Number of days in August is 31 or 3 odd days



Number of days in September is 30 or 2 odd days



Number of days in October is 31 or 3 odd days



Number of days in November is 5 or 5 odd days (Here 5th November is included)



So total number of odd days = 5 + 2 + 3 + 2 + 3 + 3 + 2 + 3 + 5 = 28 when divided by 7 gives remainder 0 hence 5th November will be same as that of 5th march.

13*6 − 1  47  19  f =15 +  + 47 +   +   − 2*19.  5   4 4

142  f =+ 1   + 39 + [ 9.75] + [ 4.75] − 38. 5 

f = 15 + 15+ 47 +11+4–38= 54 When divided by 7 we will get remainder 5, hence number of odd days is 5



A remainder of 0 corresponds to Sunday, 1 means Monday,



So 15th august 1947 is 5 days more than Sunday, i.e Friday.

14. (a) From Zeller’s Formula:–

13* m − 1  D  C  f =k +  + D +   +   − 2* C.  5   4 4

In this case k = 18 (since 18th October)



Month m = 8 (As march =1, April =2, May =3, October =8)

16. (b) From Zeller’s Formula we can find that 1st March 2009 is Sunday so we will have 4 Saturdays and 5 Sundays in total 9 weekends. 17. (c) January and July or October depends on leap year or non leap year. 18. (c) The year 2008 is a leap year. It has 2 odd days.

The day on 14th Feb, 2008 is 2 days before the day on 14th Feb, 2009.



Hence, this day is Thursday.

EBD_7743

212  Koncepts of Logical Reasoning 

19. (a) Number of days in K weeks is 7K hence total number of days is 7K + K = 8k or number of days must be a multiple of 8. 20. (a) Number of days in K weeks is 7K hence total number of days is 7K + K = 8k

  Calendar

213 



So number of odd days is 0,



So 1st April 2001 is Sunday ,



So 1st Friday is on 6th April, so next Fridays is 13th, 20th, 27th April.



Similarly number of days in 2Kth day of the 2Kth week is 2k x 7 +2k = 16k



Required number of days is 16K – 8K = 8k

22. (c) In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap year, as 1900 was not a leap year)



or number of days must be a multiple of 8.



Number of days in a period of 100 years is 365x100 + 23 or 365 × 100 + 24.

Concept Cracker (CC) with XAT



If century year is not a leap year then number of days = 365 × 100 + 23, and number of weeks is 5217 and 4 odd days and for leap year it will be 5217 weeks and 5 odd days, hence number of Sundays is either 5217 or 5218.

21. (d) Lets find Formula:–

out

1st

April

from

Zeller’s

13* m − 1   D  C  f =k +  + D +   +   − 2* C.  5   4 4

In this case k = 1 (since 1st April)



Month m = 2 (As march =1, April = 2,)



D is the last two digit of year here D = 01



(As year is 2001)



C is 1st two digit of century here C = 20



(As year is 2001)

13* m − 1   D  C  + D +   +   − 2* C. f =k +   5   4 4

13* 2 − 1  01  20  1  f =+ + 01 +   +   − 2* 20.  5   4 4

103  f =18 +   + 50 + [12.5] + [ 5] − 40. 5  f = 1 + 5 + 01 + 0 + 5 – 40 = – 28



This –ve value of f can be made positive by adding multiple of 7



So f = –28 +28 = 0



Then from next year we can make table. Odd Day

1940 2

Day of 1st Jan Monday

1941

1942



In this case k = 1 (since 1st January)



Month m = 11 (As march =1, April =2, and hence January = 11)



D is the last two digit of year here D = 39 (As year 1940 will start from March)



C is 1st two digit of century here C = 19 (As year is 1940)

142  f =+ 1   + 39 + [ 9.75] + [ 4.75] − 38. 5  



Year

23. (d) From Zeller’s formula lets find the 1st day of 1940



Or f = 1+ 28 + 39 + 9 + 4 – 38 = 43 when divided by 7 gives remainder 1 hence it has one odd day or 1st January 1940 was Monday .

1943

1944

1945

1946

1947

1948

1949

1

1

1

2

1

1

1

2

1

Wed

Thu

Fri

Sat

Mon

Tue

Wed

Thu

Sat

So in the range from 1940 t0 1949 not a single year started with Sunday.

EBD_7743

214 Koncepts of Logical Reasoning Case (ii) When year start with Saturday and then we have 53 Sundays that means year is a leap year then next 4 years will always have 52 Sundays hence total number of Sundays are 53+ 3x52 = 209 Sundays.

24. (d) Calendar of 2 years is similar if number of odd days between these two years is zero. Consider options one by one – Option (a) Between 1940 to 1946 we have two leap years 1940 and 1944 so number of odd days is 2 + 1 + 1 + 1 + 2 + 1 = 8 or 1 hence calendar of these two years is not similar.

28. (c) If month ends with Thursday then next month will start with Friday and it may have 5 Friday otherwise it may have 4 Fridays.

Option (b) Between 1977 to 1982 we have one leap years 1980 so number of odd days is 1 + 1 + 1 + 2 + 1 = 6 calendar of these two years is not similar.

29. (c) Counting the number of days after 3rd November, 1994 we have: Number of Days in November = 27 or 6 odd days. Number of Days in December = 31 or 3 odd days.

Option (c) Between 1912 to 1916 we have one leap years 1912 so number of odd days is 2 + 1 + 1 + 1 = 5 hence calendar of these two years is not similar.

Number of Days in January = 31 or 3 odd days. Number of Days in February = 28 or 0 odd days. Number of Days in March = 20 or 6 odd days.

25. (c) Number of odd days between 2013 and 2015 is 2, so we have to nd the year which will have 2 odd days between 1977 and required year.

So total number of odd days = 6 + 3 + 3 + 0 + 6 = 18 when divided by 7 gives remainder 4. Number of odd days = 4.

Consider options one by one–

The day on 3rd November, 1994 is (4) days beyond the day on 20th March, 1995.

Option (a) Number of odd days between 1977 and 1981 is 1 + 1 + 1 + 2 = 5 odd days

So, the required day is Thursday.

Option (b) Number of odd days between 1977 and 1985 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 10 odd days or 3 odd days

30. (b) 94 when divided by 7 gives remainder 3 hence today it must be Thursday, after 3 more days we will get a Sunday, next Sunday will be after 3 + 7 = 10 days and so on so we will get Sunday after a day 7K +3 days.

Option (c) Number of odd days between 1977 and 1990 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 = 16 odd days or 2 odd days, hence calendar of 1990 is the answer.

Concept Deviator (CD) 31. (b) 100 years contain 5 odd days.

26. (b) In a year we have 53 Sundays only when year start with Sunday (For non leap year) and either with Saturday or Sunday (For leap year).

\ Last day of 1st century is Friday. 200 years contain (5 × 2) = 3 odd days. \ Last day of 2nd century is Wednesday.

From zeller’s formula 1st January 2001 was Monday

300 years contain (5 × 3) = 15 = 1 odd day. \ Last day of 3rd century is Monday.

From number of odd days 1st January 2006 will start with Sunday so it had 53 Sunday

400 years contain 0 odd day. \ Last day of 4th century is Sunday.

Similarly 1st January 2009 will start with Thursday.

This cycle is repeated that means last day of century is Friday, Wednesday, Monday, or Sunday and it repeats the cycle.

27. (b) Let us take two cases– Case (i) When year start with Sunday then next 4 years will always have 52 Sundays hence total number of Sundays are 53+ 3x52 = 209 Sundays Year

2012 2013

2014

2015

1

2

1

1

1

2

1

1

Total odd day 1

3

4

5

6

8 (=1)

2

3

Odd day

2011

32. (d) Count the number of odd days from the year 2011 onwards. Calendar will repeat when we will have number of odd days = 0. 2016

2017

2018

2019

2020

2021

1

2

1

4

6

7

So after ending of 2021 the new year 2022 will start and that will have same calendar as that of 2011.

Calendar 215 33. (c) If June month has 5 Mondays then dates must be 1st , 8th , 15th , 22nd , and 29th or second group of dates must be 2nd, 9th, 16th, 23rd, and 30th

Number of days in July :– 31 or number of odd days :– 3

Now lets nd out the number of days or odd days between 1st January till 1st june,

Number of days in September :– 30 or number of odd days :– 2

Number of days in january :– 30 or number of odd days are 2 (here we have not included 1st jan so number of days is 31–1 = 30)

Number of days in October :– 31 or number of odd days :– 3

Number of days in February :– 28 or number of odd days :– 0 (Assuming non leap Year) Number of days in March :– 31 or number of odd days :– 3 Number of days in April :– 30 or number of odd days :– 2 Number of days in May :– 31 or number of odd days :– 3 Number of days in June :– 1 or number of odd days :– 1 So total number of odd days are 2 + 0 + 3 + 2 + 3 + 1 = 11 when divided by 7 gives remainder 4, Hence 1st January must be 4 days back from Monday i.e Thursday. And if 2nd January is Monday then 1st January must be Friday. Now consider a leap year then number of odd days between 1st January and 1st June is 5 days hence in that case 1st January must be either Friday or Saturday 34. (d) If November month has 5 Mondays then dates must be 1st , 8th , 15th , 22nd , and 29th or second group of dates must be 2nd, 9th, 16th, 23rd, and 30th Now lets nd out the number of days or odd days between 1st April till 1st November, Number of days in April :– 29 or number of odd days are 1 (here we have not included 1st April so number of days is 30 –1 = 29) Number of days in May :– 31 or number of odd days :– 3 Number of days in June :– 30 or number of odd days :– 2

Number of days in August :– 31 or number of odd days :– 3

Number of days in November :– 1 or number of odd days :– 1 So total number of odd days are 1 + 3 + 2 + 3 + 3 + 2 + 3 + 1 = 18, when divided by 7 gives remainder 4, Hence 1st April must be 4 days back from Monday i.e Thursday. Then weekend will be 3rd, 4th, 10th 11th, 17th,18th,24th, 25th, And if 2nd November is Monday then 1st April must be Friday. Then weekend will be 2nd, 3rd, 9th,10th, 16th, 17th, 23rd, 24th, and 30th total 9 weekends Hence number of weekends may be 8 or 9 35. (d) Here we have two cases– Case (i) If a month start with Thursday the month will have 5 Saturdays and 4 Sundays i.e total 9 weekends. In this case 31st December will be Sunday hence 1st January will be either Sunday (For non leap year) or Saturday (For leap year) then January will have 9 or 10 weekends. If a month start with Sunday the month will have 4 Saturdays and 5 Sundays i.e total 9 weekends. In this case 31st December will be Wednesday hence 1st January will be either Wednesday (For non leap year) or Tuesday (For leap year) then January will have 8 or 9 weekends. Hence number of weekends in January will be 8 or 9 or 10. 36. (c) If there are 53 Saturday and Sunday that means year is a leap year and last day of year is Sunday so 1st day of January must be Saturday then January will have 10 weekends.

EBD_7743

216 Koncepts of Logical Reasoning 37. (a) In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap year, as 1900 was not a leap year) Number of odd days in a period of 100 years is 100 +23 or 100 + 24 or 123 / 124 odd days, Number of odd days in 100 years is when 123 / 124 divided by 7, remainder is 4 or 5. So next century should start with Monday or Tuesday. 38. (b) In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap year, as 1900 was not a leap year)

Number of odd days in a period of 100 years is 100 + 23 or 100 + 24 or 123 / 124 odd days, Number of odd days in 100 years is when 123/124 divided by 7, remainder is 4 or 5. So next century should start with Friday or Saturday. 39. (a) Last day of century can be Friday, Wednesday, Monday, or Sunday and it repeats the cycle. So 1st day of next century must be either Saturday, Thursday, Tuesday, or Monday and so it must not be Wednesday, Friday, & Sunday . 40. (a) If present year is leap year then calendar of May is similar to July .

Path and Route 217

12

Path and Route

Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Important

XAT

Important

BANK Clerk

Important

IIFT

Very Important

SSC

Important

SNAP

Very Important

CSAT

Important

NMAT

Very Important

Other Govt. Exams

Important

Other Aptitude Test

Very Important

Questions based on network and path is frequently asked in CAT and other aptitude test examination. In general there are three types of questions on this topic:(i)

Number of paths from one point of network to other part.

(ii) Minimum/maximum distance from one point to other. (iii) Flow of liquid in pipe line with its maximum/minimum capacity. For number of paths from one point on network to other point has two casesCase (i): When paths are perpendicular to each other. Let there are h number of horizontal steps and v number of vertical steps then number of ways to reach from one corner to other corner is h+vCh = h+vCv

EBD_7743

218 Koncepts of Logical Reasoning Here in the below gure number of horizontal steps are h =7 and number of vertical steps v =5. Then number of ways to reach one corner from the other is given by 7+5C7 = 7+5C5

The reason behind this is that if we have total h + v steps, out of these steps if we x all horizontal steps then we have only one way to select the vertical step, hence total number of way is given by selecting h steps out of h = v steps OR by selecting v steps out of h = v steps or we can write h+vCh = h+vCv Case (ii): In general case when paths are irregular then we have to count the number of paths.

Path and Route 219

1

3.

Directions for Question Nos. 1 to 3

Natural gas once extracted from a source is puried for commercial use at natural gas plants. From gas plants it is pumped to various destinations through pipelines. There are pumping stations, at intermediate places to maintain recommended pressure in the pipelines. The pumping stations do not produce or process any natural gas. They pump out exactly the quantity they receive from plants or other pumping stations. The following gure depicts a network gas pipelines. The circles denote the locations of gas plants, pumping stations or cities with big demand for natural gas. One location can be only one of these three. The numbers on the arrows are the capacities (in appropriate units) of the pipeline that carry gas in the direction of the arrow. Currently the demand supply situation is such that the capacity utilization of the pipelines is very close to 100%.

11 units

(b)

7 units

(c)

9 units

(d) 6 units

Directions for Question Nos. 4 to 5 In the following gure that represents a network of roads, more than one road meets at intersections identied by numbers written on them. The trafc department is considering installation of surveillance cameras at these intersections that are capable of identifying all trafc violations on all the roads converging there. The cost of installing a surveillance camera at an intersection is equal to the intersection number (in lakhs of rupees). The Trafc Department wants all roads to be monitored at the minimum cost. 3

6

5 1

Q 3

4 2

9

1 2

P 1

2

5

(a)

6

N

M

What is the maximum quantity of natural gas than can be transported from M to R?

O

5

7

2 S

4.

6 R

4

Statement 1: It is more cost-effective to have surveillance cameras on intersections ‘2 and 4’ than on intersections ‘7 and 3’. Statement 2: Intersection 5 must have a surveillance camera. Which of the following is true?

1.

2.

What is the maximum quantity of natural gas S can receive? (a) 13 units

(b)

(c)

(d) 17 units

16 units

15 units

For which two cities it can be safely concluded that they have natural gas plants?

5.

(a)

Statement 1 is true but statement 2 is false.

(b)

Statement 1 is false but statement 2 is true.

(c)

Statement 1 is true and so is statement 2.

(d)

Statements 1 and 2 are false.

Which nodes should be selected installation of surveillance cameras?

(a) M and P

(b)

M and O

(a)

3, 6 and 7

(b)

(c)

(d) M and N

(c)

2, 5 and 4

(d) None of these

P and N

3, 1 and 4

for

EBD_7743

220 Koncepts of Logical Reasoning

2

9.

Directions for Question Nos. 6 to 8

In a city there are three bus routes 1, 2 and 3 between A and F. Route-1 has intermediate stops and at B and D. Route-2 has stops at C and D. The shortest route-3 with a length of 10 km, stops at C only, which is exactly at the middle of this route. The longest route has 3 km more length than the shortest one. The distances between C and D, B and D and F and D are 4, 3 and 2 kilometres respectively. [SNAP 2007] 6.

What is the distance between A and B? (a) 5 km (b) 6 km (c) 7 km (d) 8 km

7.

What is the length of route-2? (a) 11 km (b) 12 km (c) 13 km (d) Insufcient information

8.

10. How many different starting cities are possible such that the above restriction is satised? (a) One (b) Zero (c) Three (d) Two 11. What is the total number of ways to reach A to B in the network given? [CAT 1991]

What is the number of routes from P to Q?

B

A

4

1 P

Q

2 5

3 (a) 5 (c) 9

For a route that satises the above restrictions, which of the following statements is true? (a) There is no route that satises the above restriction. (b) A route can either start at C or end at C, but not both. (c) D can be only an intermediate city in the route. (d) The route has to necessarily end at E.

(b) 6 (d) 12

Directions for Question Nos. 9 to 10

(a) 12 (c) 20

(b) 16 (d) 22

12. Four cities are connected by a road network as shown in the gure. In how many ways can you start from any city and come back to it without travelling on the same road more than once? [CAT 1993]

There are 5 cities, A, B, C, D and E connected by 7 roads as shown in the gure below: A

A E

C

D

Design a route such that you start from any city of your choice and walk on each of the 7 roads once and only once, not necessarily returning to the city from which you started. [CAT 1990]

(a) 8 (c) 16

(b) 12 (d) 20

13. Eight cities A, B, C, D, E, F, G and H are connected with one-way roads R1, R2, R3, R4, R5 and R6 in the following manner: [XAT 2007]

Path and Route 221 Vaishali

R1 leads from A to C via B;

Jyotishmati

Panchal

R2 leads from C to D and then via B to F; R3 leads from D to A and then via E to H; R4 leads from F to B via G;

Avanti

R5 leads from G to D; and R6 leads from F to H.

The minimum number of road segments that have to be blocked in order to make all trafc form B to D impossible is (a) 5 (b) 4 (c) 3 (d) 2 (e) 1 Directions for Question Nos. 14 to 16 The following sketch shows the pipelines carrying material from one location to another. Each location has a demand for material. The demand at Vaishali is 400, at Jyotishmati is 400, at Panchal is 700, and at Vidisha is 200. Each arrow indicates the direction of material ow through the pipeline. The ow from Vaishali to Jyotishmati is 300. The quantity of material ow is such that the demands at all these locations are exactly met. The capacity of each pipeline is 1,000. [CAT 2001]

3

Directions for Question Nos. 17 to 21

A signicant amount of trafc ows from point S to point T in the one-way street network shown below. Points A, B, C, and D are junctions in the network, and the arrows mark the direction of trafc ow. The fuel cost in rupees for travelling along a street is indicated by the number adjacent to the arrow representing the street. A 9 S

2

5

2 3

B

C 1

7 D

6

2

T

Vidisha 14. The quantity moved from Avanti to Vidisha is (a) 200 (b) 800 (c) 700 (d) 1,000 15. The free capacity available at the AvantiVaishali pipeline is (a) 0 (b) 100 (c) 200 (d) 300 16. What is the free capacity available in the Avanti-Vidisha pipeline? (a) 300 (b) 200 (c) 100 (d) 0

Motorists travelling from point S to point T would obviously take the route for which the total cost of travelling is the minimum. If two or more routes have the same least travel cost, then motorists are indifferent between them. Hence, the trafc gets evenly distributed among all the least cost routes The government can control the ow of trafc only by levying appropriate toll at each junction. For example, if a motorist takes the route S-A-T (using junction A alone), then the total cost of travel would be Rs 14 (i.e. Rs 9 + Rs 5) plus the toll charged at junction A. [CAT 2006] 17. If the government wants to ensure that all motorists travelling from S to T pay the same amount (fuel costs and toll combined) regardless of the route they choose and the street from B to C is under repairs (and hence unusable), then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is:

EBD_7743

222 Koncepts of Logical Reasoning (a) 2, 5, 3, 2

(b)

0, 5, 3, 1

(c)

1, 5, 3, 2

(d)

2, 3, 5, 1

(e)

1, 3, 5, 1

18. If the government wants to ensure that no trafc ows on the street from D to T, while equal amount of trafc ows through junctions A and C, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is (a) 1, 5, 3, 3

(b)

1, 4, 4, 3

(c)

1, 5, 4, 2

(d)

0, 5, 2, 3

(e)

0, 5, 2, 2

19. If the government wants to ensure that all routes from S to T get the same amount of trafc, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is: (a) 0, 5, 2, 2

(b)

0, 5, 4, 1

(c)

1, 5, 3, 3

(d)

1, 5, 3, 2

(e)

1, 5, 4, 2

20. If the government wants to ensure that the trafc at S gets evenly distributed along streets from S to A, from S to B, and from S to D, then a feasible set of toll charged (in rupees) at junctions A, B, C, and D respectively to achieve this goal is: (a) 0, 5, 4, 1

(b)

0, 5, 2, 2

(c)

1, 5, 3, 3

(d)

1, 5, 3, 2

(e)

0, 4, 3, 2

21. The government wants to devise a toll policy such that the total cost to the commuters per trip is minimized. The policy should also ensure that not more than 70 per cent of the total trafc passes through junction B. The cost incurred by the commuter travelling from point S to point T under this policy will be: (a) Rs 7

(b)

(c)

Rs 10

(d) Rs 13

(e)

Rs 14

Directions for Question Nos. 22 and 23 The gure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town [CAT 2008] A

C

D P

B

22. Neelam rides her bicycle from her house at A to her ofce at B, taking the shortest path. Then the number of possible shortest paths that she can choose is (a) 60 (b) 75 (c) 45 (d) 90 (e) 72 23. Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is (a) 1170 (b) 630 (c) 792 (d) 1200 (e) 936 24. In the adjoining gure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A? [CAT 2004] B

North West

Rs 9 (a) 15 (c) 120

(b) 56 (d) 336

A

4

Path and Route 223

Directions for Question Nos. 25 to 30

The gure (not drawn on scale) given below shows the aerial distance (in km) between 8 cities, e.g. distance between H and F is 600 km etc. 700

A 600

1300

600

200 400

C

500 D

B

400

500

E

400 500

300

F

400

600

G

30. Find the maximum distance between A and H if condition is that traveling a city more than once is not allowed. (a) 3800 (b) 3700 (c) 3200 (d) None of these Directions for Questions 31 to 35: A city has hexagonal ring road as shown in the gure below. Mr. Ricky stays at X near city and his ofce is located at Y. The arrow in the gure shows ow of trafc at ofce hours. The numerical value near each path shows the cost of travel along that road/stretch of the road. There is a x toll tax at each of toll booths T1, T2… T8. The total cost of travel includes cost incurred along the path and at the toll booth.

400

H

25. Find the ratio of maximum to minimum distance between A and H if condition is that traveling a city more than once is not allowed. (a) 37:14 (b) 37:23 (c) 39:14 (d) None of these 26. Find the total number of paths from city H to city A if condition is that traveling a city more than once is not allowed. (a) 44 (b) 55 (c) 59 (d) None of these 27. Find the minimum distance between H to A without visiting city E. (a) 1200 (b) 1300 (c) 1400 (d) None of these 28. What is the minimum distance between H to A through D? (a) 1800 (b) 2000 (c) 2200 (d) None of these 29. If path HEC is taken and no city is visited twice then nd the total number of such paths. (a) 3 (b) 4 (c) 5 (d) None of these

Y

100

T5

80

40

T6

40

20

T7

30 T8

T2

T3 60

60 T1

30

30

30 40

T4

X 80

31. What is the difference between maximum amount and minimum amount that Mr. Ricky can spend on a day if he wants to travel from his home (X) to ofce (Y) if toll tax at each toll booth is Rs 50? (a) 180 (b) 210 (c) 240 (d) None of these 32. If toll tax is devised in such a way that total cost of travel remains same irrespective of the path then what is the minimum cost of travel from X to Y? (a) 240 (c) 310

(b) 260 (d) None of these

EBD_7743

224 Koncepts of Logical Reasoning 33. Which one of the following could be the value of toll taxes at T1, T2… T8 such that total cost of travel is same irrespective of route? (a) 50, 10, 0, 10, 60, 20, 0, 10 (b) 50, 20, 0, 10, 60, 20, 0, 10 (c) 50, 20, 0, 10, 40, 20, 0, 10 (d) None of these 34. If toll tax at all of the toll booths is same and is devised such that the ratio of maximum cost to minimum cost of travel from X to Y is

28:19 then which one of the following is the cost of travel from X to Y through toll Both T3? (a) 320 or 510 (b) 330 or 560 (c) 560 or 510 (d) None of these 35. If a new path is constructed from T 7 to Y then nd the total number of paths from X to Y? (a) 7 (b) 6 (c) 8 (d) None of these

  Path and Route

225 

Solutions Concept Applicator (CA) 1. (b) From the figure the maximum quantity of natural gas that S can receive = 6 + 2 + 9 = 17 units. Option (4) 2. (a) We know that cities M and P are gas plants. Option (1) 3. (d) From the figure and the conditions R can receive only 6 units (5 + 1) of natural gas if utilization is 100%. Option (4) 4. (c) As per the given conditions, we require surveillance cameras that would cover all roads converging at an intersection. The cost for covering all the roads has to be minimum.

If we put a camera on ‘1’, roads from ‘7’ and ‘6’ to 1 will be covered.



If we put a camera on intersection ‘2’, roads converging from ‘3’ to ‘2’ will be covered.



If we put a camera on ‘3’, roads from ‘3’ to ‘5’, ‘6’ to ‘5’ and ‘7’ to ‘5’ will be covered.



Now we are left with roads from 7 to 4 and from 6 to 4 both of them converge from 4 hence these will be covered by putting a camera on ‘4’. And then all the roads will be covered and the minimum cost is 2 + 5 + 1 + 4 = 12 lakhs

5. (c) Similar to the above answer cameras are put on 2, 5, 1 and 4 to minimize the cost.

Concept Builder (CB) 6. (d) From the above result Route 1 = 5 + AB = 13 or AB = 8 km 7. (a) From the above result length of route 2 = 11 km 8. (b) 9. (b) Here we will eliminate the options one by one. Option (A) cannot be true as there are many routes that satisfy the given condition. Option (C) is also not true as we can have

a route starting from D (e.g. DEBDCBAC). The route need not necessarily end at E, which is apparent. Option B satisfies all the conditions. 10. (d) City A is connected by 2 roads, B by 4 roads, C by 3 roads, D by 3 roads and E by 2 roads. For a city to be starting city for such a route, it has to be connected by odd number of roads. Hence the required answer is 2, i.e. C and D. 11. (b) There are four ways to go from A to the first level of nodes. Each of these 4 nodes in turn leads into two more ways to go to the second level nodes. Each of the second level nodes leads into two more ways to go to the third level nodes. And from here we have only one way each to go to B. Hence by fundamental principal of counting, total number of ways = 4 x 2 x 2 x 1 = 16 ways. 12. (b) It can be seen that every city is connected to all the other cities (i.e. 3 other cities). A

D B

C

Step 1: Let starting point is A, there are 3 ways in which we could proceed, viz. AB, AD or AC. Step 2: Once we are at any of these cities (B, D or C), each one of them is connected to 3 other cities. But since we cannot go back to A the originating city, there are only 2 ways in which we could proceed from here. Step 3: let us assume that we are at B, we can only go to D or C by taking BD or BC respectively. From this point we a choice of either directly going back to A (thus skipping 4th city or go to 4th city and come back to A. Step 4:- Now if we are at D, we can either take DA or DCA. So there are 2 more ways to go from here. So total number of ways = 3 x 2 x 2 = 12 ways.

EBD_7743

226 Koncepts of Logical Reasoning 13. (d) We can block B to D if A to B means R1 and B to F means R2 is blocked.Therefore minimum 2 ways needed to be block. Option (D) 14. (d) Since ow from Vaishali to Jyotishmati is 300 where as demand is 400 so the decient 100 would be met by ow from Vidisha. Again the demand of 700 in Panchal is again to be met by ow from Jyotishmati which can get it from Vidisha. Thus, the quantity moved from Avanti to Vidisha is 200 + 100 + 700 = 1000 15. (d) Free capacity at Avanti-Vaishali pipeline is 300, since capacity of each pipeline is 1000 and demand at Vidisha is 400 and 300 ows to Jyotishmati. Thus, free capacity = {1000 – (400 + 300)} = 300 16. (d) Free capacity in Avanti-Vidisha is zero. Explanation is similar as in previous answer.

Solutions (17 to 21) Let the toll charged at junctions A, B, C and D be a, b, c and d respectively. Then 1st we will list down all the routes and corresponding cost of travel. A

S

2

3

B

C 1

7

Option (2)/(3) Inconsistent options 18. (e) Since the cost of travel including toll on routes S-A-T, S-B-C-T, S-B-A-T and S-D-C-T is the same. And D-T has no trafc due to high toll charge at D. From the last solutions we will get b = 5, 14 + a = 7 + b + c = 12 + c, or a + 2 = c 7 + b + c = 10 + c + d = 12 + c or d = 2, hence the result is B = 5, d = 2 and c-a = 2 that is satised by option (E). 19. (d) Here in this case all 5 routes have the same toll charge hence-

20. (a) If all the ve routes have the same cost, then there will be an equal ow in all the ve routes, i.e. 20% in each route. But then the percentage of trafc in S-A = 20% (Only one route involving S-A) S-B = 40% (As there are two routes involving S-B)

5

2

Hence Options 4 and 5 ruled out, now if we check option rest 3 options we will nd out that option 2 and 3 both are correct.

14 + a = 7 + b + c = 13 + d = 9 + a + b = 10 + c + d After solving we will get a = 1, b = 5, c = 3 and d = 2 Option (D)

Concept Cracker (CC) with XAT

9

Similarly 10 + c + d = 13 + d or c = 3

2

T

6

D Route

Cost

S-A-T

9 + 5 + a = 14 + a

S-B-A-T

2+2+5+a+b=9+a+b

S-B-C-T

2+3+2+b+c=7+b+c

S-D-C-T

7 + 1 + 2 + c + d = 10 + c + d

S-D-T

7 + 6 = 13 + d

17. (b/c) Since Route BC is under repair hence route S-B-C-T is not in use. Rest all four have the same toll charges hence 14 + a = 9 + a + b or b=5

S-D = 40% (As there are two routes involving S-D) But here the given condition that trafc in S-A is equal to that in S-B, which in turn is equal to S-D is not satised. Of the routes, that can be used the number of routes involving S-A must be the same as S-B, which in turn is same as that as S-D. That is possible only when we block the junction C and that can be done by taking higher toll charge at C to achieve this goal c > 3. 21. (c) There must be one other route other than those involving B. We must take S-D-C-T as the other route. S-B-C-T, if toll at B = 3, total cost = 10 S-D-C-T, if toll at D and C is 0, total cost is 10. Hence 10 is the least cost.

Path and Route 227 Starting from HGC: HGCA, HGCBA, HGCDA, HGCEDA, HGCEFDA (Total 5 paths)

22. (d) Map of the town is given as below: A

C

Starting from HGE: HGECA, HGECBA, HGECDA, HGEDCA, HGEDCBA, HGEDA, HGEFDCBA, HGEFDCA, HGEFDA (Total 9 paths)

D

E P

Starting from HEG: HEGBA, HEGCBA, HEGCA, HEGCDA (Total 4 paths)

F

Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths)

B

Starting from HED: HEDCBGA, HEDCBA, HEDCA, HEDA (Total 4 paths)

It is given that Neelam took the shortest path hence she has to pass through path EF. Number of ways to reach E from A is 4C = 6. 2

2+2C 2

Starting from HEF: HEFDCGBA, HEFDCBA, HEFDCBA, HEFDCA, HEFDA (Total 5 paths)

=

Number of ways to reach B from F is 4+2C2 = 6C = 15. 2

Starting from HFE: HFEGBA, HFECGBA, HFECBA, HFECA, HFEDCGBA, HFEDCBA, HFEDCA, HFEDA, (Total 8 paths)

Hence total number of ways to reach B from A is 6 x 15 = 90.

Starting from HFD: HFDEGBA, HFDEGCA, HFDEGBCA, HFDECGBA, HFDECBA, HFDECA, HFDCEBGA, HFDCGBA, HFDCBA, HFDCA, HFDA (Total 11 paths)

23. (a) From the answer of previous question total number of ways from A to B is 90. Now we need to calculate total number of ways from B to C that is 13 (on the same logic) Hence total number of paths from A to C is 90 x 13 = 1170. 24. (b) Here number of vertical steps v = 3 Number of horizontal steps h = 5 Then in this case total number of ways is given by h+vCh = h+vCv = 8C3 = 6 x 7 x 8/6 = 56.

Concept Deviator (CD) 25. (a) Maximum distance is when path taken is- A – D – C – B – G – E – F – H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 km

Total number of paths is: 5 + 5 + 9 + 4 + 4 + 4 + 5 + 8 + 11 = 55 27. (c) For minimum distance path is HGCA = 400 + 400 + 600 = 1400 28. (d) The minimum distance between A and H through D is HEDCA and distance is 500 + 500 + 500 + 600 = 2100. 29. (b) From solution of question number 26 Starting from HEC: HECGBA, HECBA, HECA, HECDA (Total 4 paths) 30. (b)

From the solution of question number 25 maximum distance is when path taken is: A – D – C – B – G – E – F – H = 1300 + 500 + 200 + 400 + 400 + 300 + 600 = 3700 km.

Solutions (31 to 35)

Minimum distance is when path taken is- A – C – G – H = 600 + 400 + 400 = 1400 km

The paths available and cost of travel associated with it are:

Required ratio is 37:14.

Path 1: X – T1 – T8 – T6 – Y and cost associated is 80 + 60 + 20 + 80 + t1 + t6 + t8 = 240 + t1 + t6 + t8

26. (b) Number of paths is as follows: Starting from HGB: HGBA, HGBCA, HGBCDA, HGBCEDA, HGBCEFDA (Total 5 paths)

Path 2: X – T2 – T4 – T7 – T8 – T6 – Y and cost associated is 60 + 30 + 40 + 30 + 20 + 80 + t2 + t4 + t6 + t7 + t8 = 260 + t2 + t4 + t6 + t7 + t8

EBD_7743

228 Koncepts of Logical Reasoning Path 3: X – T2 – T4 – T3 – T7 – T8 – T6 – Y and cost associated is: 60 + 30 + 30 + 10 + 30 + 20 + 80 + t2 + t3 + t4 + t6 + t7 + t8 = 260 + t2 + t3 + t4 + t6 + t7 + t8

33. (b) From the solution of previous question we have following results: t3 = t7 = 0, t1 = 20 + t2 + t4, t5 = 30 + t6 + t8.

Path 4: X – T2 – T4 – T3 – T8 – T6 – Y and cost associated is 60 + 30 + 30 + 40 + 20 + 80 + t2 + t3 + t4 + t6 + t8 = 260 + t2 + t3 + t4 + t6 + t8

From the given options option B satises the condition.

Path 5: X – T2 – T4 – T5 – Y and cost associated is 60 + 30 + 40 + 100 + t2 + t4 + t5 = 230 + t2 + t4 + t5 31. (a) Maximum expenditure is in path 3 and that is 260 + 50 x 6 = 260 + 300 = 560 Minimum expenditure is in path 5 and that is 230 + 50 x 3 = 230 + 150 = 380 Required difference is 560 – 380 = 180 32. (b) From the given condition cost incurred in path 2 is equal to that in path 3 then t3 = 0 Similarly equating cost of path 3 and path 4 t7 = 0 Equating cost of path 4 and path 5 we will get 260 + t6 + t8 = 230 + t5 or t5 = t6 + t8 + 30 Equating cost of path 1 and path 2 we will get 240 + t1 = 260 + t2 + t4 or t1 = t2 + t4 + 20 For minimum total cost t1 = 20 and t5 = 30 then cost incurred is Rs260.

34. (c)

If toll tax at all of the toll booths is same and let it be k then expense in each path is as follows: Path 1: 240 + 3k Path 2: 260 + 5k Path 3: 260 + 6k Path 4: 260 + 5k Path 5: 230 + 3k So from the given condition On solving we will get k = 50 Tool booth 3 will come in Path 3 and Path 4. So cost incurred is: Path 3: 260 + 6k = 260 + 300 = 560 Path 4: 260 + 5k = 260 + 250 = 510

35. (a) Total number of paths from X to Y is 7.

Clocks 229

Clocks

Exam

Importance

13

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Important

IBPS/Bank PO

Very Important

XAT

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Very Important

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Other Govt. Exams

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Other Aptitude Test

Important

SALIENT FEATURES OF A CLOCK (i)

A clock has 12 divisions (equally spaced) hence each division is 360/12 = 30o.

(ii) A clock has two hands. Smaller hand is called short hand or hour hand, while larger hand is called long hand or minute hand. (iii) Angle traced by minute hand in 60 minutes is 360o that means speed of minute hand is 6o/min. (iv) Angle traced by hour hand in 12 hours is 360° that means speed of minute hand is ½ o/min. (v)

Relative speed between minute hand and hour hand is (6 – ½) = 11/2 o/min

(vi) Both the hands of clock coincide in one hour. (vii) In each 60 minutes, the minute hand gains 55 minutes on the hour hand

EBD_7743

230 Koncepts of Logical Reasoning (viii) When both the hands are at right angle, they are 15 minutes space apart (ix) Both hands are in straight line when they are opposite to each other or coincident to each other. (x)

Clock hands coincide 11 times in every 12 hours, as they coincide only one time between 11 and 1, at 12 o’clock. Hence, both hands coincide 22 times in 24 hours or in a day.

(xi) In 12 hours both hands are in straight line (either coincides or in opposite direction) 22 times. Hence in 24 hours both hands are in straight line 44 times. (xii) In 12 hours both hands are at right angled 22 times and 44 times in a day (as in the case of coincide). (xiii) In 12 hours both hands are in opposite direction 11 times, between 5 to 7, they are opposite at 6 o’clock only. Hence, in a day they are 22 times in opposite directions. Since relative speed between minute hand and hour hand is 11/2o per min, or in other words, minute hand will travel 11/2o more than hour hand. Let’s these two hands are together, they will again be together when minute hand will travel 360o more than hour hand i.e. after (360)/(11/2) = 720/11 or after 65 5/11 minute. So we can conclude that after every 65 5/11 minute these two hands will be together. In a day, i.e. 12 hours or 12 x 60 minute (12 x 60)/(720/11) = 11 times these two hands will be together.

ANGLE BETWEEN TWO HANDS: Let us have to nd angle between minute hand and hour hand at ‘P hours and Q minutes’. Consider from 12:00 when these two hands are together. Total time passed since 12:00 to P hour Q minute is (60P + Q) minute. In this time angle moved by hour hand is (60P + Q)/2 degrees. And angle moved by minute is 6Q (from the 12:00 position hence angle between them is 6Q – (60P + Q)/2 = 11/2Q – 30P. Angle between two hands at P hours Q minute is given by 11 Q - 30P . 2 Example: Find the angle between minute hand and hour hand at 4:40 PM. Solution: From the formula angle between these two hands is 11 x 40/2 – 30 x 4 = 220 - 120 = 100o Example: Find the angle between minute hand and hour hand at 4:15PM Solution: From the formula angle between these two hands is 11 x 15/2 – 30 x 4 = 82.5 – 120 = -37.5 o ignoring the negative sign the required angle is 37.5o.

1.

2.

Clocks 231

1

Find the angle between minute hand and hour hand at 6:30 AM. (a) 25o (b) 22.5o (c) 15o (d) None of these

Find the angle between minute hand and hour hand at 11:50 AM. (a) 55o (b) 22.5o o (c) 15 (d) None of these

3.

Find the angle between minute hand and hour hand at 4:20 AM. (a) 10o (b) 12.5o (c) 15o (d) None of these

4.

Find the angle between minute hand and hour hand at 1:40 AM. (a) 180o (b) 190o (c) 175o (d) None of these

5.

Find the angle between minute hand and hour hand at 10:10 AM. (a) 245o (b) 195o o (c) 175 (d) None of these

6.

At what time between 3 PM and 4 PM minute hand and hour hand will coincide? 4 4 (a) 3:16 (b) 3:15 11 11 (c) 3:17 4 (d) None of these 11

2

11. At what time between 2 PM and 3 PM the angle between minute hand and hour hand is 100o?

(a) 2:29

1 11

(b)

(c)

6 11

(d) None of these

2:18

2:12

6 11

7.

At what time between 3PM and 4PM minute hand and hour hand will be opposite to each other? 1 1 (b) 3: 48 11 11 1 (c) 3:49 (d) None of these 11 At what time between 4 PM to 5 PM minute hand and hour hand will be at the right angle

(a) 3:47

8.

9.

to each other? 5 5 (a) 4:7 (b) 4: 4 11 11 5 (c) 4:9 (d) None of these 11 At what time between 9 PM to 10 PM minute hand and hour hand will coincide?

1 1 (b) 9: 48 11 11 1 (c) 9:44 (d) None of these 11 10. At what time between 9 PM and 10 PM minute hand and hour hand will be opposite to each (a) 9:49

other? (a) 9:15

4 11

(b)

(c)

4 11

(d) None of these

9:12

9: 16

4 11

12. Find the angle between minute hand and hour hand at 2:18 AM. (a) 37o (b) 33o (c) 39o (d) None of these 13. Find the percentage change in angle between minute hand and hour hand from 2 PM to 6 PM. (a) 100 (b) 200 (c) 250 (d) None of these

EBD_7743

232 Koncepts of Logical Reasoning 14. Find the angle between minute hand and hour hand at 2:45. (a) 180o

(b)

190o

175o

(d)

None of these

(c)

15. At what time between 9 and 10 o’clock will the hands of a clock be in the same straight

20. How many times in a day (24 Hrs) are the hands of a clock coincide? (a) 20 (b) 22 (c) 24 (d) 48 21. How many times in a day (24 Hrs) are the hands of a clock are in a straight line?

line but not together? (a) 9:16

4 11

(b)

9: 15

4 11

4 (d) None of these 11 16. At what time between 8 and 9 o’clock will the (c)

9:17

hands of a clock be at right angle? (a) 8:16

3 11

(b)

8: 27

3 11

8:17 3 (d) None of these 11 17. By how many degrees does the minute hand move in the same time, in which the hour (c)

hand moves by 28 ° ?

18.

(a) 168

(b)

(c)

(d) 376

196

336

At what time between 1’o clock and 2’o clock the hands of the clock are opposite to each other? (b) 38 ( /11 ) past 1’oclock 56 (8/11) past 1’oclock

(d) 64 (9/11) past 1’oclock 19. How many times in a day (24 hours) are the hands of a clock in straight line but opposite in direction? (a) 20

(b)

(c)

(d) 48

24

22

(b) 22 (d) 48

22. How many times in a day (24 Hrs) are the hands of a clock are in a perpendicular to each other? (a) 20 (b) 22 (c) 44 (d) 48 23. A watch, which gains uniformly, is 2 min, slow at noon on Sunday, and is 4 min 48 seconds fast at 2 PM on the following Sunday when was it correct? (a) 2:00 PM on Tuesday (b) 12 noon on Monday (c) 2:00 AM on Tuesday (d) None of these 24. A clock is set at 10 a.m. The clock loses 16 minutes in 24 hours. What will be the true time when the clock indicates 3 a.m. on 4th day? (a) 9 p.m. (c) 11 p.m.

(a) 34 (6/11) past 1’oclock (c)

(a) 20 (c) 44

(b) (d)

10 p.m. 12 p.m.

25. A watch which gains uniformly is 2 minutes slow at noon on Sunday and is 4 min. 48 sec fast at 2 p.m. on the following Sunday. When it has shown the correct time? (a) 2 p.m. on Monday (b) 2 p.m. on Tuesday (c) 3 p.m. on Wednesday (d) 1 p.m. on Thursday

Clocks 233

They will be opposite to each other when the minute hand will travel (90 + 180 =) 270o more than the hour hand. And time taken for this is (270)/(11/2) = 540/11 or 49 1/11 minute.

Concept Applicator (CA) 1.

(c) At 6:30AM from the formula here P = 6 and Q = 30 so required angle is 11 x 30/2 – 30 x 6 = 165 -180 = -15 or ignoring the negative sign the required angle is 15o.

2.

(a) At 11:50AM from the formula here P = 11 and Q = 50 so required angle is 11 x 50/2 – 30 x 11 = 275 – 330 = -55 or ignoring the negative sign the required angle is 55o.

3.

(a) At 4:20AM from the formula here P = 4 and Q = 20 so required angle is 11 x 20/2 – 30 x 4 = 110 – 120 = -10 or ignoring the negative sign the required angle is 10o.

4.

(b) At 1:40 from the formula here P = 1 and Q = 40 so required angle is 11 x 40/2 – 30 x 1 = 220 – 30 = 190 the required angle is 190o.

5.

(a) At 10:10 from the formula here P = 10 and Q = 10 so required angle is 11 x 10/2 – 30 x 10 = 55 – 300 = -245 the required angle is 245o.

6.

(a) Consider exactly at 3:00, minute hand is at position ‘12’ and hour hand is at ‘3’ hence angle between these two hands at 3:00 is 900.

7.

Hence they will be opposite to each other at 3: 49 minute:0.09 second. 8.

(d) Consider exactly at 4:00, minute hand is at position ‘12’ and hour hand is at ‘4’ hence angle between these two hands at 4:00 is 1200. Since relative speed between them is 11/2o per minute or every minute the minute hand will travel 11/2o more than the hour hand. They will be perpendicular to each other when the minute hand will travel (120 – 90 =) 30o more than the hour hand. And time taken for this is (30)/(11/2) = 60/11 or 5 5/11 minute.

9.

Hence they will be perpendicular to each 5 other at 4 : 5 11 (c) Consider exactly at 9:00, minute hand is at position ‘12’ and hour hand is at ‘9’ hence angle between these two hands at 9:00 is 270o.

Since relative speed between them is 11/2o per minute or every minute the minute hand will travel 11/2o more than the hour hand.

Since relative speed between them is 11/2o per minute or every minute the minute hand will travel 11/2o more than the hour hand.

They will be together when the minute hand will travel 90o more than the hour hand. And time taken for this is (90)/(11/2) = 180/11 or 16 4/11 minute.

They will be together when the minute hand will travel 270o more than the hour hand. And time taken for this is (270)/(11/2) = 540/11 or 49 1/11 minute.

Hence they will coincide at 3:16 minute:21.81 second.

Hence they will be opposite to each other at 9: 49 minute:0.09 second.

(c) Consider exactly at 3:00, minute hand is at position ‘12’ and hour hand is at ‘3’ hence angle between these two hands at 3:00 is 900. 11/2o

Since relative speed between them is per minute or every minute the minute hand will travel 11/2o more than the hour hand.

Concept Builder (CB) 10. (c) Consider exactly at 9:00, minute hand is at position ‘12’ and hour hand is at ‘9’ hence angle between these two hands at 9:00 is 270o.

EBD_7743

234 Koncepts of Logical Reasoning

Since relative speed between them is 11/2o per minute or every minute the minute hand will travel 11/2o more than the hour hand.

They will be together when the minute hand will travel (270-180 =) 90o more than the hour hand. And time taken for this is (90)/(11/2) = 180/11 or 16 4/11 minute. Hence they will be opposite to each other at 9: 49 minute: 21.81 second. 11. (a) Same as previous question; option A is correct. 11 12. (c) 18 × – 2 × 30 = 99 – 60 = 30 2 13. (b) At 2PM angle between minute hand and hour hand is 60o and that at 6PM is 180o hence percentage change is 200%. 14. (d) From the formula the required angle is 187.5o. 90 180 4 = = 16 15. (a) 11 / 2 11 11 16. (b) Consider exactly at 8:00, minute hand is at position ‘12’ and hour hand is at ‘8’ hence angle between these two hands at 8:00 is 240o. Case (i) when minute hand is behind hour hand then required time is (240 – 90)/(11/2) = 150/(11/2) = 300/11 = 27 3/11 minute. Case (ii) when minute hand is ahead of hour hand then required time is (240 + 90)/(11/2) = 330/(11/2) = 330 × 2/11 = 60 minute, i.e. at 9:00 PM. 17. (b) 28 * 2 * 6 = 336 degree 18. (b) The minutes hand to coincide with the hour hand. It should trace at rst 5 minute spaces and then the hands of the clocks to be opposite to each other minute hand should trace 30 minute spaces, i.e. totally it should gain 5 + 30 = 35 minute spaces to be opposite to that of hour hand. We know that minute hand gains 55 minute spaces over hour hand in 1 hour. Therefore, minute hand gains 40 minute spaces over hour hand in 35 × (60/55) = 38(2/11). Hence the hand of the clock will minutes be opposite to each at 38 (2/11) past 1 o’clock. Therefore, correct option is B.

19. (b) The hands of a clock point in opposite directions (in the same straight line) 11 times in every 12 hours. (Because between 5 and 7 they point in opposite directions at 6 o’clock only). So, in a day, the hands point in the opposite directions 22 times. 20. (b) 21. (c) In 12 hours both hands are in straight line (either coincides or in opposite direction) 22 times. Hence in 24 hours both hands are in straight line 44 times. 22. (c) 23. (a) From Sunday noon to the following Sunday at 2 PM there are 7 days 2 hours or 170 hours. The watch gains 2 + 44/5 minute in 170 hours. Therefore, the watch gains 2 minute in 2 *170 5 × hours, i.e. 50 hours. 34 Now 50 hours = 2 days 2 hours. Therefore, 2 days 2 hours from Sunday noon = 2 PM on Tuesday. 24. (c) Time from 10 AM on a day to 3 AM on 4th day = 24 x 3 + 17 = 89 hours. Now 23 hours 44 minute of this clock = 24 hours of correct clock. Or 356/15 hours of faulty clock = 24 hours of correct clock. 89 hours of faulty clock = (24 x 15/356 x 89) hours = 90 hours. So, the correct time is 11PM. 25. (b) Time from 12 PM on Sunday to 2 PM on the following Sunday = 7 days 2 hours. = 24 x 7 + 2 = 170 hours. The watch gains = (2 + 4 x 4/5) minute = 34/5 minute in 170 hours. Since, 34/5 minutes are gained in 170 hours. 2 min are gained in (170 x 5/34 x 2) hours = 50 hours, i.e. 2 days 2 hours after 12 PM on Sunday, i.e. it will be correct at 2 PM on Tuesday.

Blood Relation and Family Tree 235

14

Blood Relation and Family Tree Exam

Importance

Exam

Importance

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Very Important

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Very Important

XAT

Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt. Exams

Very Important

Other Aptitude Test

Very Important

TIPS ON SOLVING THIS QUESTION TYPE This question type is based on our understanding of the family tree. To solve questions related to family tree or blood relation we should follow some pictorial representation of the given information. Here we have some tips to handle these questions. 1.

To solve these types of questions put all the information in a family tree diagram.

2.

Identify the gender and mark the same with a symbol. Example: male can be identied by putting a box around that name or relation while a female can be identied by circling the same.

EBD_7743

236 Koncepts of Logical Reasoning Example: If A is a male then we can represent it as A If it is known that B is a female then represent as B 3.

Care should be exercised while interpreting gender neutral relations such as siblings, parent, spouse etc.

4.

Two siblings (Brothers, sisters) should be joined with a straight line while a couple should be joined with two parallel lines. Example: A is brother of B. Here it is known that A is a male but we don’t know about B hence we can represent this information asA

B

If it is given that C is husband of D then C 5.

D

Distinguish generations by depicting them on different levels. For Example grandfather, grandmothers have to be placed on one level. Mother, father, uncle and aunts are to be placed on level two, and sister, brother; niece and nephew have to be placed on level three. Example: If it is given that C is father of A and A is son of D Here C must be husband of D hence we can represent this information as-

6.

Last but not at all the least do not assume genders based on the name only.

  Blood Relation and Family Tree

1

Concept Applicator

Directions for Question Nos. 1 to 5

Directions for Question Nos. 10 to 15

In a family of 5, A is father of B who is brother of C

Ashok Mehta has three children, Usha, Ramchander and Sunil. Sunil married Rita, the eldest daughter of Mr. and Mrs. Mathur. The Mathurs married their youngest daughter to the eldest son of Mr. and Mrs. Saxena, and they had two children named Sanjay and Sunita. The Saxena have two more children, Rakesh and Bindu, both younger to Sanjay. Sonu and Surinder are sons of Sunil and Rita. Lata is the daughter of Sanjay and Rima. Now answer the

who is daughter of D. E is son in law of A. 1.

How B is related to E? (a) Brother (b) Brother in law (c) Son in law (d) None of these

2.

How many husband wife pairs in family? (a) 1 (b) 2 (c) 1 or 2 (d) None of these

3.

How many sister/s are in the family? (a) 1 (b) 2 (c) 1 or 2 (d) None of these

4.

How A is related to E? (a) Father (b) Father in law (c) Son in law (d) None of these

5.

How D is related to B? (a) Mother (b) Mother in law (c) Son in law (d) None of these Directions for Question Nos. 6 to 9

6 persons are present in a family gathering it is known that E is B’s daughter. C is D’s sister. C’s daughter is E and son is F. G is E’s maternal aunt. D is brother of C. 6.

How many husband wife pairs in family? (a) 1 (b) 2 (c) 1 or 2 (d) None of these

7.

How B is related to G? (a) Brother (b) Brother in law (c) Son in law (d) None of these

8. How many gathering? (a) 2 (c) 2 or 3

237 

sisters

are

in

the

family

(b) 3 (d) None of these

9. How many brothers are in the family gathering? (a) 2 (b) 3 (c) 2 or 3 (d) None of these

following questions: 10. What is surname of Lata?

(a) Mehta



(b) Mathur



(c) Saxena



(d) Can’t be uniquely determined

11. How is Rakesh related to lata?

(a) Uncle

(b) Maternal Uncle



(c) Cousin

(d) None of these

12. How Mrs. Mathur is related to Sanjay?

(a) Mother

(b) Mother in law



(c) Sister in law

(d) None of these

13. As per the given information how many husband wife pairs exist from the given names?

(a) 2

(b) 3



(c) 4

(d) 5

14. Which surname is associated with maximum number of persons (As per the given information)?

(a) Mehta



(b) Mathur



(c) Saxena



(d) Can’t be uniquely determined

15. How is Sunil related to Lata?

(a) Uncle

(b) Maternal uncle



(c) Cousin

(d) None of these

Directions for Question Nos. 16 to 20 Mr. and Mrs. Sharma have two children Asha and Shashi. Shashi married Radha, daughter of Mrs. Mahajan. Suresh, son of Mrs. Mahajan marries Rita. Sonu and Rocky are sons to Suresh and Rita. Uma and Sudha are the daughters of Shashi and Radha.

16. How many grand children in the family?

(a) 2

(b) 3



(c) 4

(d) None of these

17. How many husband wife pairs are there in the family?

2



(a) 1 (c) 1 or 2

(b) 2 (d) None of these

18. How is Shashi related to Sonu? (a) Uncle (b) Maternal uncle (c) Cousin (d) None of these 19.

How is Mrs. Mahazan related to Uma? (a) Grandmother (b) Maternal grandmother (c) Mother (d) None of these

20. How many brothers are there in the family? (a) 2 (b) 3 (c) 4 (d) None of these

Concept Builder

Directions for Question Nos. 21 to 24

Directions for Question Nos. 25 to 27

Radha and Minnilal have two children—Simmi and Divya. Divya is married to Anuj who is the son of Madhu and Jabbar. Resham is the daughter of Anuj. Kiran who is Anuj’s sister is married to Subodh and has two sons Tarun and Aman. Tarun is grandson of

A, B and D meet their relatives C, E, F and G while visiting the trade fair. A is the brother of B and D is the father of A. F is the only son of C and E. E, who is the brother-in-law of G, is the father-in-law of B.

Madhu and Jabbar. 21. What is the relationship between Aman and Resham?

(a) Cousins

(b) None of these



(c) Uncle - Niece

(d) Father-daughter

22. How is Subodh related to Jabbar?

(a) Father-in-law

(b) None of these



(c) Son-in-law

(d) Son

23. How is Kiran related to Divya?

(a) Sister

(b) None of these



(c) Grandmother

(d) Sister-in-law

24. Which of the following statements is definitely true?

(a) Resham is the cousin of Kiran



(b) All the three are true



(c) Aman is the son of Simmi



(d) Madhu is the mother-in-law of Subodh

25. How many female members are there? (a) 2 (b) 3 (c) 4 (d) Can’t be determined 26. How is G related to A? (a) Uncle (b) Father-in-law (c) Mother-in-law (d) None of these 27. Who is the spouse of F? (a) B (b) E (c) C (d) G 28. Sanjay says, “I have as many sisters as brothers.” Sarita says, “Each of us sisters has only half as many sisters asbrothers.” Assuming that Sanjay and Sarita are brother and sister, how many brothers and sisters

are there inthe family? (a) 6 brothers and 4 sisters (b) 4 brothers and 6 sisters (c) 3 brothers and 4 sisters (d) 4 brothers and 3 sisters



EBD_7743

238  Koncepts of Logical Reasoning 

29. Pointing to a photograph, Devika says, “This man’s son’s son’s son is my grandson”. What is Devika’s husband’s relation with the man in the photograph?

(a) Son

(b) Nephew



(c) Father

(d) Grandson

Directions for Question Nos. 30 to 32 Six directors of a private limited company namely A, B, C, D, E and F are playing Golf. A and E are brothers. F is the sister of E. C is the only son of A’s uncle. B and D are the daughters of the brother of C’s father.



(a) A – B + C # D* E (b) A*B # C *D–E



(c) A # B *C +D-E

(d) A + B – C*D # E

36. At a family reunion were the following people: one grandfather, one grandmother, two fathers, two mothers, four children, three grandchildren, one brother, two sisters, two sons, two daughters, one father-in-law, one mother-in-law, and one daughter-in-law. But not as many people attended as it sounds. How many persons were there? [SNAP]

30. How is C related to F?

(a) Cousin

(b) Brother



(c) Son

(d) Uncle

31. How many male executives are there in the group?

(a) Two

(b) Three



(c) Four

(d) One

32. How many of these executives are real brothers?

(a) Two

(b) Three



(c) Cannot be said

(d) None of these

33. A,B,C,D,E,F and G are the members of a family consisting of 4 adults and 3 children, two of whom, F and G are girls. A and D are brothers and A is doctor. E is an engineer married to one of the brothers and has two children. B is married to D and G is their child. Who is C? [SNAP 2006]

(a) G’s father

(b) F’s father



(c) E’s daughter

(d) A’s son

Directions for Question Nos. 34 and 35 P # Q means P is the father of Q P + Q means P is the mother of Q P – Q means P is the brother of Q P * Q means P is the sister of Q

  Blood Relation and Family Tree 239  35. Which of the following shows that A is the Aunt of E?



(a) 5

(b) 10



(c) 15

(d) 7

37. Each child in a family has at least 4 brothers and 3 sisters. What is the smallest number of children the family might have? [SNAP 2008]

(a) 7

(b) 8



(c) 9

(d) 10

38. Consider

the

following

statements

answer the question.

and

[SNAP 2008]



M, N, O and P are all different individuals.



M is the daughter of N



N is the son of O



O is the father of P



Which among the following statements is contradictory to the above premises?



(a) P is the father of M



(b) O has three children



(c) M has one brother



(d) M is the granddaughter of O Directions for Question Nos. 39 to 41

[SNAP 2006]

34. If A + B # C – D, then A is D’s ____________.

(a) Sister

(b) Grandfather



(c) Grandmother

(d) Father

There are only four members of a family viz., A, B, C and D and there is only one couple among them. When asked about their relationships, following were their replies: A. A: B is my son. D is my mother B. B: C is my wife. D is my father C. C: D is my mother-in-law. A is my daughter. D. D: A is my grand-daughter. B is my daughterin-law.

[SNAP 2011]

39. Who always speaks truth?

(a) A

(b) B



(c) C

(d) D

41. Which of the following statements must be true?

40. How is B related to C?

(a) Father

(b) Mother



(c) Wife

(d) Husband

3

EBD_7743

240  Koncepts of Logical Reasoning 



(a) A’s grandmother alternates between the truth and lie.



(b) C’s wife always speaks the truth.



(c) A’s grandfather always speaks the truth.



(d) B’s daughter always tells lies.

Concept Cracker

Mr. Bedi’s family members went on a picnic. There were two grandfathers and four fathers and two grandmothers and four mothers in the group. There was at least one grandson or a granddaughter present in this group. There were two husband-wife pairs in this group. The single grandfather (whose wife was not present) had two grandsons and a son present in the picnic. The single grandmother (whose husband was not present) had two granddaughters present. A grandfather or a grandmother present with their spouses did not have any grandson or granddaughter present.  42. What was the minimum number of people present in this picnic group? [IIFT 2007]

(a) 14

(b) 10 



(c) 12

(d) 16



(a) Paternal uncle

(b) Maternal uncle



(c) Grand Father

(d) None of the above

45. Mr. Raju took the members of his family for a picnic. His father’s mother and mother’s father including their two children were in one car. His father’s son and sister’s husband, brother’s wife were in second car. He along with his wife, wife’s sister, wife’s brother and son’s wife with a kid was in the third car. How many members of Mr. Raju’s family were there in the picnic along with Mr. Raju and how many were left behind (assuming all members of the third generation are married)?

43. Pointing to Priya, father of Pritu says, she is the daughter of the daughter of the wife of the only son of the grandfather of my sister. How is Sushma related to Priya if Sushma is the sister of Pritu?

[IIFT 2008]

(a) Mother

(b) Aunt



(c) Niece

(d) None of the above

44. Pointing to a photograph Yuvraj says, he is the only brother of the only daughter of my sister‘s maternal grandmother. Pointing to another photograph Sourav says, he is the only brother of the only daughter of my sister‘s maternal grandmother. If among the two photographs, one was either of Sourav or Yuvraj, and the photograph, towards which Yuvraj was pointing, was not of Sourav, then how is Yuvraj related to Sourav?

[IIFT 2009]



(a) 13 and 4

(b) 14 and 5



(c) 12 and 5

(d) 13 and 6

Directions for Question Nos. 46 to 48 A, B, C, D, E and F are a group of friends. There are two housewives, one professor, one engineer, one accountant and one lawyer in the group. There are only two married couples in the group. The lawyer is married to D, who is a housewife. No woman in the group is either an engineer or an accountant. C, the accountant, is married to F, who is a professor. A is married to a housewife. E is not a housewife.

[IIFT 2008]





[CAT 1999] 46. Which of the following is one of the married couples?

(a) A and B

(b) B and E



(c) D and E                   (d) A and D

 

Blood Relation and Family Tree 241 47. What is E’s profession? (a) Engineer

(b)

(c)

(d) Accountant

Professor

Lawyer

48. How many members of the group are males? (a) 2 (b) 3 (c)

4

(d) Cannot be determined 49. In a family gathering there are two males who are grandfathers and four males who are fathers. In the same gathering there are two females who are grandmothers and four females who are mothers. There is at least one grandson or a granddaughter present in

4

Directions for Question Nos. 50 to 53

A, B, C, D, E, F and G are brothers. Two brothers had an argument and A said to B, “You are as old as C was when I was twice as old as D, and will be as old as E was when he was as old as C is now”. B said to A, “You may be older than F but G is as old as I was when you were as old as G is, and D will be as old as F was when F will be as old as G is”. 50. Who is the eldest brother? (a) A (b) E (c)

C

(d) Cannot be determined 51. Who is the youngest brother? (a) B (b) D (c)

F

(d) Cannot be determined 52. Which two are probably twins?

this gathering. There are two husband-wife pairs in this group. These can either be a grandfather and a grandmother, or a father and a mother. The single grandfather (whose wife is not present) has two grandsons and a son present. The single grandmother (whose husband is not present) has two granddaughters and a daughter present. A grandfather or a grandmother present with their spouses does not have any grandson or granddaughter present. What is the minimum number of people present in this gathering? (CAT 2001) (a) 10

(b)

(c)

(d) 16

14

12

53. Which of the following is false? (a) G has 4 elder brothers. (b) A is older than G but younger than E. (c)

B has three elder brothers.

(d) There is a pair of twins among the brothers Directions for Question Nos. 54 to 58 Chandra Mohan and his wife Kamini have a family of three generations comprising thirteen members of whom six are female members. Some of Chandra Mohan’s children are married and none of his grand children are married. Kamini has a daughter-in-law named Fullara and two sons-in-law, one being Eeshwaran Harihar's brother is Devesh, who has two nephews and two nieces- one being Leela. Bandana Devesh’s sister has two children. Fullara, who is sister-in-law to Devesh has four nephews and nieces. Manohar who is married to Geeta in the family has a daughter Indira and a son. Joy has a sister and two cousins, Akash and Indira.

(a) D and G (b) E and C (c)

A and B

(d) Cannot be determined

54. Akash is the son of: (a) Geeta

(b)

(c)

(d) Bandana

Devesh

Harihar

EBD_7743

242 Koncepts of Logical Reasoning 55. Leela is the niece of: (a) Bandana

(b)

(c)

(d) Kamini

Harihar

Eeshwaran

56. Which of the following pairs is a brother and sisters? (a) Manohar and Fullara (b) Devesh and Geeta (c)

Joy and Indira

(d) Eeshwaran and Bandana 57. Amongst the following, which one is false? (a) Bandana is Kamini’s child (b) Joy is Geeta’s child (c)

58. Statement: (a) Fullara is Harihar’s wife (b) Harihar’s son’s name is Akash Bandana is Leela’s mother

Based on the above, which one of the following is true? (a) (a) and (b) only but not (c) (b) (b) and (c) only but not (a) (c)

(a) and (c) only but not (b)

(d) All

5

Directions for Question Nos. 64 to 68

Mohan is Rakesh’s paternal uncle. Ritu is Rakesh’s daughter. Madhav is Ritu’s younger brother. Mukesh is Ritu’s husband. John is Sunita’s son. Tinu is Madhav’s sister. Kulwant is Rakesh’s father. Pushpa is Kulwant’s sister. Kaushalaya is Mohan’s wife. Gagan is Mohan’s son. Sunita is Kulwant’s daughter. 64. How is John related to Kulwant? (a) Nephew

(b)

(c)

(d) Great Grandson

Grandson

In a family, Ishita is the granddaughter of Ashima. Deepika is the mother of Hannah. Chetan is the son of Anil. Rohini is the mother of Ishita. Deepika is the sister of Vivek and Chetan. Nilesh has two children, Gauri and Hannah. Elesh is the only grandson in the family. Chetan is not married. Rohini is the daughter-in-law of Anil. 59. Who is married to Rohini? (a) Anil

(b)

(c)

(d) Vivek

Chetan

Nilesh

60. Who is the daughter of Anil?

Akash is Harihar’s nephew

(d) Geeta has less than two nephews

(c)

Directions for Question Nos. 59 to 63

Grandnephew

(a) Gauri

(b)

(c)

(d) Deepika

Ishita

Hannah

61. Who is the son –in-law of Ashima? (a) Elesh

(b)

Vivek

(c)

(d)

Anil

Nilesh

62. How many Children does Deepika have? (a) One

(b)

(c)

(d) None

Three

Two

63. Who is the father of Ishita? (a) Nilesh

(b)

Vivek

(c)

(d)

Anil

Elesh

65. Tinu is John’s ____________. (a) Aunt (b) Cousin Sister (c) Sister (d) Niece 66. Mukesh is Gagan’s ____________. (a) Nephew (b) Brother (c) Brother-in-law (d) Nephew-in-law 67. From this data, which of the following statements can be said to be correct? (a) Mohan is younger than Kulwant (b) Mukesh and Gagan do not share the same surname (c) Madhav is younger than Rakesh (d) Pushpa is younger than Mohan

Blood Relation and Family Tree 243 68. Manish is Chitra’s son, Chitra and Simran are sisters, Dibleen is Simran’s mother. Sagar is the son of Dibleen. Which of the following statements is true? (a) Sagar and Manish are cousins.

the lawyer. The engineer has 2 daughters. Radha is a businessman. Sumit is a non earning member of the family. Rima and Seema are the housewives and the name of one more member of the family is Ajit.

(b) Sagar is the paternal uncle of Manish. (c)

Simran is Manish’s grandmother.

(d) Chitra and Sagar are siblings. Directions for Question Nos. 69 to 72 Nine people are present in a family gathering. Members from 3 generation were present in the gathering. There are 3 married couples. No one is married in the 3rd generation. In the family there are 2 housewives, 1 doctor, 1 engineer, 1 architect and 1 lawyer, 1 businessman, 1 policeman and 1 student. Sunita is an engineer is the sister in law of the housewife which has only 1 child Rajesh. The businessman is the niece of Amit who is a doctor. The student has the only one cousin and is in police. One of the housewives has 2 children and 3 grandchildren. Rakesh is a Lawyer and is grandfather of the policeman. The architect is the son in law of

69. How many brothers in laws in the family gathering? (a) 1

(b)

(c)

(d) None of these

3

2

70. How is Architect related to Rima? (a) Brother in law

(b)

(c)

(d) Can’t be determined

Brother

Son in law

71. How is Ajit related to policemen? (a) Maternal uncle

(b)

(c)

(d) None of these

uncle

Cousin

72. How is Lawyer related to Radha? (a) Husband

(b)

(c)

(d) None of these

Grand father

Uncle

EBD_7743

244 Koncepts of Logical Reasoning

Concept Applicator (CA) Solutions (1 to 5) The family tree is as follows: A

B

D

C

Solutions (6 to 9) The family tree is as shown below. Since maternal aunt means mother’s sister (In Hindi it is termed as Mausi) or mother’s sister in law (In Hindi it is termed as Mami), hence we are not sure about G’s relationship, she can be wife of D or sister of C and D. B

C

F

E

G

E

1.

(b) From the family tree B is brother in law of E.

2.

(b) As per the family tree given, husband wife pairs are (A and D) and (C and E).

3.

(a) Only C is the sister.

4.

(b) From the family tree B is father in law of E.

5.

(b) From the family tree B is father in law of E.

6.

(c) (B and C) is a married couple but (D and G) can be cannot be hence number of married couple is 1 or 2.

7.

(b) From the family tree B is brother in law of G.

8.

(c) From family tree there may be 2 or 3 sisters.

9.

(b) There are two brothers in the family.

Solutions (10 to 15)

10. (c) From the above family tree Surname of Lata is Saxena. 11. (a) From the above family tree

Rakesh is Lata’s

uncle.

12. (b) From the above family tree

Mrs. Mathur is

Mother in law of Sanjay.

13. (c) From the given information there exist 4 husband wife pairs. 14. (c) From the given information Mehta Family: 7, Mathur family: 2, Saxena Family: 8

15. (b)

D

Solutions (16 to 20)

Blood Relation and Family Tree 245 16. (c) From the given family tree in total 4 grand children. 17. (d) There are three husband wife pairs in family. 18. (b) 19. (b) 20. (c) Total four brothers in the family. Concept Builder (CB) Solutions (21 to 24) From question we get the following family tree-

29. (a) From the given information we can conclude that, Devika’s grandson is the man’s son’s grandson. Thus, Devika’s husband is the man’s son. Solutions (30 to 32) 30. (a) Six directors A, B, C, D, E and F are playing Golf and out of A and E are brothers and F is their sister. It is also given that C is the only son of A’s uncle. Hence C can be their maternal or paternal cousin. B and D are daughters of the brother of C’s father. So B and D can be the sisters of A, E and F, but this cannot be denitely said as we don’t know the number of brother’s of C’s father. C is F’s cousin. Hence family tree would be:

21. (a) Relationship between Aman And Reshma— Cousins [from 3rd level] 22. (c) Subodh related to Jabbar -Son-in law [from 2nd and 1st level] 23. (d) Kiran related to Divya- Sister-in law [from 2nd level] 24. (d) From above tree rst 3 options are not true. Solutions (25 to 27) G

C

E

F

D

A

B

25. (d) From the family tree we don’t know the gender of G hence can’t be determined. 26. (d) From the above family tree the correct answer is option D 27. (a) From the above family tree the correct answer is option A 28. (d) Sarita B (Sanjay) S

B

B

B

S

Including Sanjay and Sarita there has to be 4 brothers and 3 sisters. B and S stand for brother and sister respectively.

D

B

A

E

F

C

31. (b) There are 3 male executives in the group, A, E and C. 32. (a) From the explanation given in the rst question of the set, we get that A and E are real brothers. 33. (d) From the given information we can conclude that A and E is husband wife pair with A (husband and doctor), E (engineer and A’s wife), and D and B is husband wife pair with D (husband), B (wife of husband). G being the child of B-D and being a girl would be B and D’s daughter. Similarly, F (girl) and C (boy) would be A’s and E children. Thus, C is A’s son. 34. (c) A + B # C – D converts to A is the mother of B, B is the father of C, C is the brother of D. Thus, A is D’s granddaughter. 35. (b)

36. (d)

37. (c) If a child is a boy, he needs to have at least 4 brothers, means that there are at least 5 boys in the family. By the same reasoning, if the child is a girl, she needs to have at least 3 sisters. This means that there must be at least 4 girls in the family. The minimum number of children the family might have is 5 boys + 4 girls = 9 children.

EBD_7743

246 Koncepts of Logical Reasoning 38. (a) The family tree would be like: Generation

People

Dened relationships

First generation

0 (male)

O is the father of N and P, M is the granddaughter of O.

Second generation

P (sex N is the father of M unknown) and son of O. Also, N is N (make) brother of P.

Third generation

M (female)

45. (a) From the given information we can draw the following conclusion: In the rst car: Maternal grandfather + Paternal grandmother + Mother + Father + So total members in Car 1 = 4. In Car 2: Brother, Sister-in-law (brother’s wife) and Brother-in-law (sister’s husband). Total members in Car 2 = 3. In Car 3: Raju, his wife, sister-in-law (wife’s sister), brother-in-law (wife’s brother), daughterin-law (son’s wife) and grandson.

M is granddaughter of O and daughter of N.

39. (d) D always speaks truth.

Total members in Car 3 = 6

40. (c) B is the wife of C. 41. (d) Once we realize that D is always true, we also realize that both of A’s statements must be false. Also, if D’s statements are true, A is B’s daughter. Concept Cracker (CC) with XAT 42. (c) From the given information we can draw the following diagram. A (M) P (F) | B (M) Q (F) | C (M) D (F) R (F) S (M) | E (M), F (M) T (F), U (F) There will be minimum 12 people in the gathering and they will satisfy all the constraints of the problem. 43. (d)

Total members who went to picnic = 4 + 6 + 3 = 13. The member, missing are – Raju’s sister, wife’s sister’s husband, wife’s brother’s wife, Raju’s son = 4 members. Solutions (46 to 48) Number of HF-2, Law-1, Prof-1, Engg-1, Acc-1. There are 2 married couples. We will solve this question by matrix method. D is HF, so, we put tick her box of HF. C is accountant and F is professor so, those box also closed. Again C and F is the couple. Here number of HF is 2, so, in HF column there must be another name. A is male, married with D. So, A must be lawyer. Engineer can’t be female. E can’t be HF, so he must Engineer. B is HF.

A

HF

Acc

X

X

B (female) C

Engg

X

X

X

X

X

X

X

X

X

X

X

X

X

X

D (female)

44. (b) The photograph towards which Yuvraj was pointing was not of Sourav. This means that the photograph towards which Sourav was pointing was of Yuvraj. So, Yuvraj is the maternal uncle of Sourav.

Prof

X

Law

E

X

X

X

F

X

X

X

X

46. (d) Married couples are C and F and A and D. 47. (a) E is engineer. 48. (b) Number of males among 6 members is 3.

Blood Relation and Family Tree 247 49. (a) From the given information we can draw the following diagram. Grandfather (single) Son

Wife

Son

Grand Son

Grand Son

Grandfather Daughter Grand daughter

Husband Grand daughter

There will be minimum 10 people in the gathering and they will satisfy all the constraints of the problem. Concept Deviator (CD) Solutions (50 to 53) From the 1st statement: B is now as old as C was in the past. Hence B is younger to C or B < C. Also sometime in the past A was twice as old as D. So A is elder to D or A > D. C will be as old as E in future, hence C < E. The second statement suggests: A > F. A was as old as G in the past. Hence A > G. D will be as old as F in future. Hence F > D. F will be as old as G now in future. Hence G > F. G was as old as B, when A was as old as G. Hence A = B. Combining both the results, we get: and E > C > B = A > G > F > D (Note by A=B, it is meant that they are of similar age group, not necessarily the same).

50. 51. 52. 53.

(b) (b) (c) (c)

E is the eldest brother D is the youngest brother. Only A and B could probably be twins. Only statement (c) is false as B has only 2 elder brothers and not 3.

Solutions (54 to 58) Given Chandra Mohan and Kamini have a family comprising thirteen members. From the rst point it can be inferred that there are three married couples in the second generation, i.e. a total of 6 persons. Form the second point it can be inferred that the third generation consists of 4 persons. The remaining one person must be of the second generation. From second and sixth points, Leela, Joy, Akash and Indira belong third generation. From second and fourth points, Harihar and Devesh are the sons of Chandra Mohan, Fullara is the wife of Harihar. From the fth and sixth points, Geeta is the daughter of Chandra Mohan and is married to Manohar, whose children are Indira and Akash. From the fourth point it can be inferred that the remaining two children, i.e. Joy and Leela are not the Children of Fullara and Harihar. They must be the Children to the remaining couple Eeshwaran and Bandana.

The following charts illustrate their family tree.

54. (a)

55. (c)

56. (b)

57.

(b)

60. (d)

61. (c)

62. (b)

58.

(c)

Solutions (59 to 63)

59. (d)

63. (b)

EBD_7743

248 Koncepts of Logical Reasoning Solutions (64 to 68) As per the given information we can draw following conclusion the family tree is drawn here with taking some conventional notation.

From the above table we can get the answer of the followings-

Solutions (69 to 72) Housewife

Concept Eliminator (CE)

Lawyer (Rakesh)

64. (c) 65. (b)

Housewife

66. (d)

Doctor (Amit)

Engineer Sunita

Architect (Ajit)

Student (Sumit)

Businessman Radha

67. (c) 68. (d) From the given information we can draw the following:

Policeman (Rajesh)

Dibleen (f) 69. (b) Amit and Ajit are the brother-in-laws. Chitra (f)

Simran (f)

Sagar (m)

Manish (m) From this, option (D) is correct.

70. (d) We cannot uniquely determine as we don’t know the name of housewives. 71. (a) From the above family tree, Ajit is maternal uncle of policeman. 72. (c) From the family tree, option C is correct.

Coding and Series 249

15

Coding and Series

Exam

Importance

Exam

Importance

CAT

Important

IBPS/Bank PO

Very Important

XAT

Important

BANK Clerk

Very Important

IIFT

Important

SSC

Very Important

SNAP

Important

CSAT

Very Important

NMAT

Important

Other Govt. Exams

Very Important

Other Aptitude Test

Important

Coding and series is one of the most important chapters for aptitude tests, but it has shown very less presence in CAT exam and has shown a signicant presence in SNAP, IIFT exams. There are numerous ways of coding and hence there is no specic rule for the same. But as far as aptitude test is concern questions come from some specic types of coding only. Let’s see some general rule for the coding decoding series. Unless stated in the question assume letters of the English alphabets according to their position, as A = 1, B =2, C =3, D = 4, and so on X = 24, Y = 25, Z = 26.

EBD_7743

250 Koncepts of Logical Reasoning The following is the complete table A=1

N = 14

B=2

O = 15

C=3

P = 16

D=4

Q = 17

E=5

R = 18

F= 6

S = 19

G=7

T = 20

H=8

U = 21

I=9

V = 22

J = 10

W = 23

K = 11

X = 24

L = 12

Y = 25

M = 13

Z = 26

Sometimes, coding has an extension as well, means after Y = 25, Z = 26, again A = 27, B = 28 and Z = 52 and so on. Following different types of coding we may encounter – 1.

Coding according to numerical valueK

O

L

K

A

T

A

11

15

12

11

1

20

1

So here we can say that KOLKATA is coded as 111512111201. 2.

Coding according to square of numerical valueK

O

L

K

A

T

A

11

15

12

11

1

20

1

121

225

144

121

1

400

1

So here we can say that KOLKATA is coded as 12122514412114001. We may have some different relations based on the numerical value of alphabets. 3.

Coding according to 1 or 2 place advance or back or mixture of thatK

O

L

K

A

T

A

11

15

12

11

1

20

1

+1

+1

+1

+1

+1

+1

+1

12

16

13

12

2

21

2

L

P

M

L

B

U

B

Coding and Series 251 So here we can say that KOLKATA is coded as LPMLBUB. K

O

L

K

A

T

A

11

15

12

11

1

20

1

+1

-1

+1

-1

+1

-1

+1

12

14

13

10

2

19

2

L

N

M

J

B

S

B

So here we can say that KOLKATA is coded as LNMJBSB. 4.

Coding is done after interchanging the position of letters on a specic patterns – 1st place

2nd place

3rd place

4th place

5th place

6th place

7th place

K

O

L

K

A

T

A

Interchange of letters happens diagonally (Or letters are arranged in reverse order) A

T

A

K

L

O

K

So here we can say that KOLKATA is coded as ATAKLOK And there are numerous ways of coding and series. We will see more types in the questions below.

EBD_7743

252 Koncepts of Logical Reasoning

1

Read the following carefully and answer the question: 1.

2.

3.

4.

5.

AZ, GT, MN, ?, YB which one of the following can replace the question mark? (a) KF (b) RX (c) SH (d) TS J2Z, K4X, I7V, ?, H16R, M22P which one of the following can replace the question mark? (a) I11T (b) L11S (c) L12T (d) L11T

Fill in the blanks gfe _ ig _ eii_ fei _ gf _ ii (a) eig (b) ifgie (c) gie (d) ige If 18514 stands for AHEAD, what does 31385 stand for? (a) CATCH (b) CASSET (c) CONQUER (d) CACHE In a particular code, the digits from 0 to 9 are each represented by a different letter of the alphabet, the letter always representing the same digit. In case the following sum holds true when it is expressed in digits, which of the following cannot be correct? B C D E + I (a) (b) (c) (d)

6.

A D E D H G G must be even C + A must be greater than 9 D must be greater than 3 B must be smaller than I

Complete the sequence of numbers below: 1, 11, 21, 1211, 111221,……… (a) 312211 (b) 311211 (c) 11133212 (d) 1223123 Directions for Question Nos. 7 to 9

The letters of the English alphabet are numbered from 26 to 1 such that 26 stands for A, 25 stands for B, and so on. The assigned numbers are used to write the letters of the original alphabet.

7.

8.

9.

What will be the sum total of all vowels in the sequence? (a) 78

(b)

(c)

(d) None of these

76

84

Which of the following sequences denotes a valid word? (a) 6-12-17-23

(b)

5-11-18-22

(c)

(d)

5-12-17-2.3

5-12-18-23

If each of the alphabets stands for the number which denotes it, what will be the next term in the following sequence? Z, W, R, K, -(a) A

(b)

(c)

(d) D

C

B

Directions for Question Nos. 10 to 12 In the English alphabet, letters from A to M denote numeric values from-13 to 1 (such that A is -13, B is -12, .........) and letters from N to Z denote values from 1 to 13 (such that N means 1, O means 2, .........). 10. Which of the following words will have a negative numeric value as a product of the numeric codes? (a) HIS

(b)

(c)

(d) None of these

ROD

HELL

11. Which one of the following words will have an absolute numeric value equal to that of BELL as a product of the numeric codes? (a) YELL

(b)

(c)

(d) None of these

RODE

HELL

12. Assuming that the salaries and perks are basically coded with the help of designations using the code given above, who will be drawing the highest salary amongst the following? (a) PEON

(b)

(c)

(d) None of these

HEAD

CLERK

Coding and Series 253 13. Out of the given four groups of letters, three groups follow a certain code. Which one of these is the odd group that does not follow the code? (a) DSFU

(b)

(c)

(d) BUDW

INKR

PGRI

Directions for Question Nos. 14 and 15 A + B means A is the mother-in-law of B; A - B means A is the daughter-in-law of B; and A*B means A is the sister-in- law of B.

2

Directions for Question Nos. 16 to 18

Director of a drama group has to assign different roles to two artists - Paramjeet and Kamaljeet to play in a drama depending on four different symbols - @ for father, $ for wife, # for brother and * for daughter. There were four combinations decided by the director showing the following result. (IIFT 2007) Answer the following questions on the basis of results I, II, III, IV Ι. Paramjeet @ Kamaljeet stands for Paramjeet is father of Kamaljeet ΙΙ. Paramjeet $ Kamaljeet implies Paramjeet is the wife of Kamaljeet ΙΙΙ. Paramjeet # Kamaljeet stands for Paramjeet is brother of Kamaljeet ΙV. Paramjeet * Kamaljeet stands for Paramjeet is daughter of Kamaljeet 16. If Daljeet # Chiranjeet $ Baljeet which of the following statement is true? (a) Daljeet is the brother of Baljeet (b) Daljeet is the father-in-law of Baljeet (c) Daljeet is the father of Baljeet (d) Daljeet is the brother-in-law of Baljeet 17. If Manjeet * Chiranjeet @ Daljeet @ Baljeet, which of the following is not true? (a) Manjeet is the mother of Baljeet (b) Chiranjeet is the grandfather of Baljeet (c) Manjeet and Daljeet are siblings (d) Manjeet is the aunt of Baljeet

14. P*Q+R means: (a) P is the mother-in-law of Q (b) P is the mother-in-law of Rand Q (c) R is the daughter-in-law (d) None of these 15. Given X - Y + Z, which of the following is certainly true? (a) Y is the mother-in-law (b) X is the mother-in-law (c) Z is the mother-in-law (d) None of these

18. If Abhijeet # Chiranjeet * Baljeet, which of the following is not true? (a) Baljeet is the parent of Abhijeet (b) Abhijeet and Chiranjeet are siblings (c) Abhijeet is the son of Baljeet (d) Baljeet is the mother-in-law of Chiranjeet Directions for Question Nos. 19 to 22 All the vowels in the English Alphabet are retained in their original places and the remaining letters are written in the reverse order. The new sequence looks like: A, Z, Y, X, E, W, ... … … … . Now each of the letters in the new sequence represents the letter in the original sequence. 19. Which one of the following is a valid word in the new sequence? (a) PUNYT (b) VTFHC (c) PREHG (d) None of the above 20. Which one of the following represents the name of a country? (a) ITXNA (b) YTOHV (c) YTINA (d) None of the above 21. Which one of the following is the name of a popular brand? (a) FILLS (b) NIPKA (c) AKIEQ (d) None of the above 22. Which one of the following is not a valid word in the new sequence? (a) PUJYQE (b) WQOOX (c) JHINH (d) None of the above

EBD_7743

254 Koncepts of Logical Reasoning 23. In a certain code, the word CARTOON is written as BBQUNPM. Using this code, OSHNZSX will stand for the word : (a) CURSORY

(b)

(c)

(d) None of these

NETWARE

CORRECT

Directions for Question Nos. 24 to 26 In the English alphabet, letters from A to M denote numeric values from 1 to 13 (such that A is 1, B is 2, … … …) and letters from N to Z denote numeric values from -13 to –1 (such that N is -13, 0 is –12, … … …).

3

27. If in a certain code BEAUTIFUL is coded as 573041208, BUTTER as 504479, how is FUTURE coded in that Code? (SNAP 2009) (a) 201497 (b) 204097 (c) 704092 (d) 204079 28. Replace the question mark with the right option. (SNAP 2008) 4, 32, 288, ?, 31680 (a) 25600 (c) 7420

24. The numeric value of which of the following equations will be a whole number? (a) KISS/RAPP (b) HIS/HELL (c) HISS/YOUR (d) None of these 25. Assuming that the salaries are basically coded with the help of employee names using the code given above, who among the following will be drawing the highest salary? (a) PREM (b) SHAN (c) RAMU (d) None of the above 26. Following the above mentioned code, which of the following will be true? (a) GS - TSZ = 0 (b) PRO = DLW (c) ROD = YET (d) None of the above

30. ‘Developers are following laws’ would be correctly written as: (a) ‘bop cop uop eop’ (b) ‘lop bop eop uop’ (c)

‘oup cop lop aop’

(d) None of the above 31. ‘qop gop cop eop’ would correctly mean: (a) Protable laws were stopped

(b) (d)

2880 10000

29. In each of the following two sets I and II, nd the word or pair of words different from the other three words or pair of words. (SNAP 2008) I. J. Lake K. Brook L. Stream M. River II. J. Weighty – Heavy K. Broad - Wide L. Big – Large M. Tiny - Small (a) I-J, II-J (b) I-K, II-M (c) I-K, II-J (d) I-J, II-K Directions for Question Nos. 30 to 31 In a certain code language the following lines written as -‘lop eop aop fop’ means ‘Traders are above laws’ ‘fop cop bop gop’ means ‘Developers were above protable’‘aop bop uop qop’ means ‘Developers stopped following traders’‘cop jop eop uop’ means ‘Following maps were laws’ (IIFT 2012)

(b) Developers stopped following laws (c)

Traders were above protable

(d) None of the above Directions for Question Nos. 32 to 33 In each of the following letter series, some of the letters are missing, which are given below it. Choose the correct alternative. [IIFT 2011] 32. D_F_DEE_D_EF_DE_F (a) EFFDED (b) EFFDDF (c)

EFFDFE

(d) None of the above 33. _OPO_QOPQ_RQPO_POR_O (a) APRQO

(b)

(c)

(d) None of the above

QPROO

QPORO

Coding and Series 255 Directions for Question Nos. 34 to 35 In each of the following questions, nd the relationship that can denitely be deduced on the basis of the relations given. The symbols used to dene the relationship are as follows: @ means ‘greater than’ # means ‘less than’ $ means ‘not equal to’ % means ‘equal to’ (IIFT 2011) 34. If it is given that, 3 M % 2 N and N % 3 O,

(a) R-3 and 5; C-8 and 9 (b) R-9 and 10; C-4 and 5 (c)

R- 4 and 5; C-7 and 8

(d) 4 and 5; C-7 and 8 39. In the following series, what numbers should replace the question marks? -1, 0, 1, 0, 2, 4, 1, 6, 9, 2, 12, 16, ??? (a) 11, 18, 27

then:

(b) -1, 0, 3

(a) O @ M

(c)

(b) M # O

(d) Cannot be ascertained

(c)

2O%M

40. Here are some words translated from an articial language.

(d) None of the above 35. If it is given that, N @ P, P # O, O @ M and N %

Di onot means Oak tree Bly crin means maple leaf

(a) O @ N

(b)

(c)

(d) None of the above

O#N

36. If the word ‘EXAMINATION’ is coded as

Which word could mean “maple syrup”? (a) Bly muth

56149512965, then the word ‘GOVERNMENT’

(b) Hupp onot

is coded as:

(c)

(IIFT 2012)

Parri crin

(d) Crin weel

(a) 7645954552 (b) 7654694562 (c)

(IIFT 2012)

Bly onot means Oak leaf

M, then: O$N

3, 20, 25

41. If IQS : LNV, then JRM : ?

7645965426

(IIFT 2012)

(d) 7654964526 37. In a certain code language ‘HORSE’ is written as 71417184, then the word ‘MONKEY’ is coded as:

(IIFT 2012)

(a)

11141216425

(b)

12141310424

(c)

12151411325

(d)

12151210424

(a) OKS

(b)

(c)

(d) MOQ

NIP

MOP

42. P≠Q implies that Q is standing 2 km to the right of P

(IIFT 2012)

P*Q implies that Q is standing 2 km to the left of P

38. In the word HEIRARCHICAL, if the rst and

P @ Q implies that Q is standing 2 km below P

second, third and forth, fourth and fth, fth

P $ Q implies that Q is standing 2 km above P

and sixth words are interchanges up to the

If F ≠ S $ B * V, in which direction is F with respect

last letter, which are the two positions from

to V?

the left on which R would appear and on

(a) North

(b)

(c)

(d) West

which positions would C appear twice?

East

South

EBD_7743

256 Koncepts of Logical Reasoning

4

43. In a coding language, the letters of the English alphabet are arranged in such a manner that all the vowels are put in the end and the remaining letters are arranged from the rst letter onwards. The rearranged alphabets are used to denote the position occupied by letters in the original alphabets.

What is the code of “META”? (a) LWPV

(b)

(c)

(d) TEAM

PWLV

QGYB

44. If CLOSE is coded as DNRWJ, then APART will be(a) BRDVY

(b)

(c)

(d) BTDYV

BSKYV

BRBVY

45. Four pairs of numbers are given below of which the last number is missing. In all pairs, the number on the left has a certain relationship with the number on the right. Select the option that has the same relationship with the given number. 4:8; 6:10; 12:14; 18: ? (a) 20

(b)

(c)

(d) 60

42

30

46. A pair of grouping of symbols is given below. Choose a pair of symbols that best expresses the relationship closest to the original pair LLI : UQR. (a) DEF : ABC

(b)

AXE : TIX

(c)

(d)

LMN : AEF

AEF : BGO

47. If TSEREVE and NOITACUDE stands for EVEREST and EDUCATION, how will you code REDFORT (a)

ARECFORT

(b)

BREDFNRT

(c)

CREDFORT

(d)

DRECFNRT

48. If PAINT is coded as 74128 and EXCEL IS CODED AS 93596 then how would you encode ACCEPT? (a)

455978

(b)

547978

(c)

554978

(d)

735961

49. In a certain code, 15789 is written as EGKPT and 2346 is written ALUR. How is 23549 written in that code? (a) ALEUT (b) ALGTU (c)

ALGUT

(d) ALGRT 50. If white is called blue, blue is called

red, red is called yellow, yellow is called green, green is called black, black is called violet and violet is called orange, what would be the colour of human blood? (a) Red

(b)

Green

(c)

(d)

Violet

Yellow

Coding and Series 257

Concept Applicator (CA) 1.

2.

3.

4.

5.

(c) AZ, GT, MN, ___ , YB In this series letter contain AZ, second letter GT. From this we can see that A to G increasing order but Z to T decreasing order. 3rd letter also maintains same pattern respect to 2nd letter. Now, we are considering position of each letter according to the alphabetical order. A + 6 = G, G + 6 = M = > M + 6 = S Z – 6 = T, T – 6 = N = >N – 6 = H So, fourth term will be = SH (d) J2Z, K4X, I7V, ___, H16R, M22P Here rst letter contain 2 letters and one numeric value. 1st term of each letter: J K I __ H M Here, j to k increasing order but k to I decreasing order, again h to k also increasing order. Each step +1, -2, +3, -4 …… J + 1 = K, K – 2 = I, I + 3 = L, L – 4 = H, H +5 = M, 2nd term also increases by +2, +3, +4, +5 order. 2 + 2 = 4, 4 + 3 = 7, 7 + 4 = 11, 11 + 5 = 16, 16 + 6 = 22 3rd term decreases by 2 xed orders. Z – 2 = X, X – 2 = V, V – 2 = T 4th letter will be L11T. (b) gfe ___ ig ___ eii ___ fei ___ gf ___ ii Above series cannot be solved directly because some alphabets are repeated. So, take each option individually and put it. Among all 4 given options only option 2 satised the series: gfeii/gfeii/…. (d) 18514 denotes AHEAD which means 1-®A, 8--®H according to the alphabetical position. Same as 31385, 3àA, 1àC, 3àC, 8àH, 5àE 31385 Þ CACHE. (c) Since E + E = G and maximum carry over is 1 then 1+D + D =H

6.

From here we cannot conclude that D must be greater than 3. (a) Given sequence is 1, 11, 21, 1211, 111221, ……… Now analyse the terms1st term = 1 2nd term- It describes the 1st term that it has “one” “one” or 11 3rd term It describes the 2nd term that it has “two” “one” or 21 4th term It describe the 3rd term that it has “one””2” and “one” “1” or 1211 5th term describes the 4th term that it has “one” “1”, “one” “2” and “two” “1” or 111221 Hence the 6th term would be- “Three” “1”, “two” “2” and “one” “1” or 312211

Solutions (7 to 9)

7.

A

26

J

17

S

8

B

25

K

16

T

7

C

24

L

15

U

6

D

23

M

14

V

5

E

22

N

13

W

4

F

21

O

12

X

3

G

20

P

11

Y

2

H

19

Q

10

Z

1

I

18

R

9

(b) Sum of total vowels in the sequence = A (26) + E (22) + I (18) + O (12) + U (6) = 26 + 22 + 18 + 12 + 6 = 84;

8.

(c) 5-12-18-23 Þ V-O-I-D Þ VOID, is a valid word.

9.

(b) Z, W, R, K, -- Þ 1, 4, 9, 16, -- Þ 12, 22, 32, 42, -Observing the given sequence, we conclude that 52 = 25 = B should be the next valid alphabet

EBD_7743

258 Koncepts of Logical Reasoning Solutions (10 to 12) As per the given conditions-

10.

11.

12.

13. 14.

Letter Number Letter Number A –13 N 1 B –12 O 2 C –11 P 3 D –10 Q 4 E –9 R 5 F –8 S 6 G –7 T 7 H –6 U 8 I –5 V 9 J –4 W 10 K –3 X 11 L –2 Y 12 M –1 Z 13 (c) Evaluate the given words. HIS ≡ (−6) × (−5) × 6 = 180 HELL ≡ (−6) × (−9) × (−2) × (−2) = 216 ROD ≡ 5 × 2 × (−10) = −100 (a) Product of the numeric value of BELL will be = (-12) x (-9) x (-2) x (-2) = 432. Product of the numeric value of YELL will be 12 x (-9) x (-2) x (-2) = -432. Hence, the answer is option (1). (c) Multiplying the numeric codes which are assigned to those alphabets for the designations PEON ≡ 3 × (−9) × 2 × 1 = −54 CLERK ≡ (−11) × (−2) × (−9) ×5 × (−3) = 2970 HEAD ≡ (−6) × (−9) × (−13) × (−10) = 7020 Thus, the highest salary is for the designation HEAD. (c) INKR is the answer as all other three are having the same sum, which is equal to –4. (d) From the given information P * Q + R means: P is the sister-in-law of Q and Q is the motherin-law of R.

Hence, P is not the mother-in-law of Q and R. ∴ Option (a) and (b) can be eliminated. Q is the mother-in-law of R, but we don’t know the gender of R i.e. whether R is the daughter-in-law or son-in-law of Q. ∴ Option (c) is also incorrect. Option (d) satises the given condition. 15. (a) From the given information X – Y + Z means: Y is the mother-in-law of Z and X is the daughter-in-law of Y. ∴ Y is denitely the mother-in-law. Concept Builder (CB)

16. (d) From the given relation Daljeet # Chiranjeet $ Baljeet means Daljeet is brother of Chiranjeet who is wife of Baljeet. Hence Daljeet must be brother-in-law of Baljeet. Hence option (d) is correct. 17. (a) Manjeet is daughter of Chiranjeet who is father of Daljeet who is father of Baljeet. Hence, Manjeet cannot be mother of Baljeet. 18. (d) Abhijeet is brother of Chiranjeet who is daughter of Baljeet. Hence Baljeet cannot be mother-in-law of Chiranjeet. 19. (a) The valid word will be formed by PUNYT that will become MUNCH. 20. (c) Here from the given options YTINA becomes CHINA which is a name of country. 21. (c) From the given options AKIEQ represent ‘ARIEL’ which is a famous brand. 22. (d) In the rearrangement we can get the word MUSCIE, FLOOD, STINT (which are meaningful) from options (1), (2), (3), respectively. 23. (d) Original word: C A R T O O N Pattern: –1 +1 –1 +1 –1 +1 –1 Coded word: B B Q U N P M Similarly OSHNZSX stands for PRIMARY.

Solutions (24 to 26) The letters and their numeric values are as follows. Letter A B C D E F G H

I

Value

9 10 11 12 13 –13 –12 –11 –10 –9 –8 -7 –6 –5 –4 –3 –2 –1

1

2

3

4

5

6

7

8

J

K L M N

24. (d) From the given options only option 4 gives us a whole number. 25. (a) Option (1) PREM has the highest salary of 6435.

O

P

Q

R S

T U V W X Y

Z

26. (a) Evaluating the options we will get option (1): GS – TSZ = (7) (–8) – (–7) (–8) (–1) = 0 is correct

Coding and Series 259 Concept Cracker (CC) with XAT 27. (b) looking at BEAUTIFUL = 573041208 and BUTTER = 504479, one realizes that the coding in this question is the digit code in the code for the word is in the exact same place of the letter it is replacing. This can be seen if we look at B = 5 and T = 4 in the two words. So, f = 2, U = 0, T = 4, R = 9 and E = 7, hence code FUTURE = 204097, option (b) 28. (b) 4 x 8 = 32 32 x 9 = 288 288 x 10= 2880 2880 x 11 = 31680, option (b) 29. (a) Brook is a small stream of river. Brook, stream and river are owing water bodies but lake is not. In II, ‘weighty-heavy’ is different because others are adjectives used for length but (J) is for weight. So Option (a) is the answer. 30. (b) From (A) and (D), laws is coded as eop. From (B) and (C), developer is coded as bop From (A) and (C), traders is coded as aop. From (A) and (B), above is coded as fop. From (B) and (D), maps is coded as jop. From (C) and (D), following is coded as uop. Now, from (A), are is coded is lop. From (C), stopped is coded as uop. From (D), were is coded as cop. From (B), protable is coded as gop. The option (b) is the correct code. 31. (a) 32. (c) If we split the alphabets in a group of 3 we get following arrangement DEF/FDE/EFD/DEF/FDE/EF Here, DEF are rotating in cyclic order. 33. (d) Analyze the options one by oneOption (A): A O P O P Q O P Q R R Q P OQPOROO Option (B): Q O P O P Q O P Q O R Q P ORPOROO Option (C): Q O P O P Q O P Q R R Q P OOPOROO None of (A), (B) and (C) follows any logical pattern. 34. (c) 3M%2N Þ 3M = 2N. N%3O Þ N = 3O From above two information 3M = 2(3O) Þ M = 2O Þ M%2O.

35. (a) N@P Þ N > P P#O Þ P < O O@M Þ O > M .... (1) N%M Þ N = M .... (2) N > P, P < O, O > M and N = M. So, clearly O > N and O is not equal to M O>NÞO@N 36. (a) The code of each letter is the sum of its place values (if the letter has two digits as its place value) \ The code for GOVERNMENT is 7, (1 + 5) (2 + 2), 5, (1 + 8), (1 + 4), (1 + 3), 5 (1 + 4) and (2 + O) i.e. 7645954552. 37. (b) The code of each letter is one less than its place value. \ The code for MONKEY is (13 - 1), (15 - 1), (14 -1), (11 - 1), (5 - 1), (25 - 1) i.e. 12141310 424. 38. (a) The word is HEIRARCHICAL, according to the given condition the 1st and 2nd, 3rd and 4th, 4th and 5th, 5th and 6th letters and so on are interchanged, here it should have been 1st and 2nd, 3rd and 4th, 5th and 6th, 7th and 8th and so on then the newly formed term is EHRIRAHCCILA. Then the positions of R and C are 3, 5, 8 and 9, i.e. given in 39. (d) The given series –1, 0, 1, 0, 2, 4, 1, 6, 9, 2, 12, 16, ___, ___, ___ is an example of alternate series. The rst number in each group –1, 0, 1, 2, 3 follows one series. The second number in each group 0+2, 2+4, 6+6, 12+8, 20 follows one series. The third number in each group 1+3, 4+5, 9+7, 16+9, 25 follows one series. The third number in each group 1+3, 4+5, 9+7, 16+11, 27 follows one series. 40. (c) It is given that (1) di onot oak tree (2) bly onot oak leaf (3) bly crin maple leaf From statements (2) and (3), we can conclude that the code for leaf is bly. But the order is given in the reverse Hence, from statement (3), the code for maple is crin. The code for maple syrup should have crin as 2nd part hence option C is correct.

EBD_7743

260 Koncepts of Logical Reasoning 41. (b) The given analogy is I Q S I R M +3 –3 +3 Similarly, +3 –3 +3 L N V M O P 42. (b) The given expression is F # S $ B * V The given information can be represented as follows. 2 kms V B

44. (b) 45. (a) 2*2 : 2*4, 2*3 : 2* 5, 2*6 : 2* 7, 2* 9 : 2*10 46. (b) There is same number of alphabets (8) between (L, U) and (I, R). Only option (b) follows the same pattern. The same number of alphabets (18) between rst and third pair of symbols, i.e. (A, T) and (E, X). Hence 47. (c) We can observe that the code is the reverse of the given word. The answer is obviously TROFDER which is the reverse of REDFORT. 48. (a) Clearly, in the given code, the alphabets are coded as follows:

F

2 kms

S

F is towards South with respect to V. Concept Deviator (CD) 43. (b) The English alphabet and their rearranged order are given below

P A I N T E X C E L 7 4 12 8 9 3 5 9 6 Like A C C E P T 4 5 5 9 7 8 49. (c) In the given codes, the numbers are coded as shown:

Rearranged order B C D F G H J K

1 5 7 E G K i.e., 2 as A,

Original

3 as L,

Order I J K L M N O P

5 as G,

Rearranged order L M N P Q R S T

4 as U and

Original

9 as T.

Order Q R S T U V W X Y Z

So, 23549 is coded as ALGUT.

Original Order A B C D E F G H

Rearranged order V W X Y Z A E I O U The code r META can be obtained from the above table as QGYB.

8 P

9 T

2 A

3 L

4 U

6 R

50. (c) The colour of the human blood is ‘red’ and as given, ‘red’ is called ‘yellow’. So, the colour of human blood is ‘yellow’.

Statement Conclusion and Binary Logic 261

16

Statement Conclusion and Binary Logic Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt. Exams

Very Important

Other Aptitude Test

Very Important

EBD_7743

262 Koncepts of Logical Reasoning

1.

1

In a cricket team, three batsmen Ricky, Sachin and Brian are the top three runscorers in any order. Each of them gives two replies to any question, one of which is true and the other is false, again, in any order. When asked about who the top scorer was, following were the replies they gave:

― In the present period of economic hardships, education and small family norm may lead the nation to progress and prosperity. Assumptions: A. Education and small family norms are directly related to nation’s progress. B. Big families nd it difcult to bear the cost of education.

[SNAP 2010]

(a)

Only A is implicit.

(b)

Only B is implicit.

Brian: I got the top score. Sachin was second.

(c)

Both A and B are implicit

Ricky: I got the top score. Sachin was third.

(d) Neither A nor B is implicit.

Sachin: I got the top score. Ricky was second.

Which of the following is the correct order of batsmen who got the top score, second best and third best score respectively? (a)

4.

Brian, Ricky, Sachin

were:

(b) Brian, Sachin, Ricky (c)

Asit: None of us is interested.

Ricky, Sachin, Brian

(d) Sachin, Brian, Ricky 2.

Danny: Three of us are interested. Eshita: Four of us are interested. From his experience, Mr. Basu knows that those who are interested only tell the truth and other lie. How many of them are interested in the new project? [XAT 2007]

[SNAP 2008]

A.

Only conclusion I follows.

B.

Only conclusion II follows.

C.

Both conclusions I and II follow.

D.

Neither conclusion I nor conclusion II follows.

Statements: Some doctors are fools. Joshi is a doctor. Conclusions: I. Joshi is a fool. II. Some fools are doctors.

3.

Barun: One of us is interested. Chandra: Two of us are interested.

In the following question two statements are followed by two conclusions numbered I and II. Assume the two statements are true even if they are at variance with commonly known facts. Then pick the correct answer from the choices given below.

(a) A

(b) B

(c) C

(d) D

Mr. Basu, the managing director of XYZ Company asked ve persons – Asit, Barun, Chandra, Danny and Eshita – about their interest in new project. The replies he got

Consider the statement and decide which of the assumptions are implicit:-

5.

(a) 4

(b)

3

(c)

2

(d) 1

(e)

0

Five persons – A, B, C, D and E – are either guards or thieves. The guards always tell the truth, whereas thieves always lie. A claims that B is a guard. B claims that C is a thief. C claims that D is a thief. E claims that A is a guard. D claims that B and E are different kinds. The number of thieves is:

[XAT 2007]

(a) 1

(b)

2

(c)

3

(d) 4

(e)

5

Statement Conclusion and Binary Logic 263 Directions for Question Nos. 6 to 8 Kya–Kya is an obscure island which is inhabited by two types of people: the ‘Yes’ type and the ‘No’ type. Native of type ‘Yes’ ask only questions the right answer to which is ‘Yes’ while those of type ‘No’ ask only questions the right answer to which is ‘No’. For example, the ‘Yes’ type will ask questions like “Is 2 plus 2 equal to 4?” while the ‘No’ type will ask questions like “Is 2 plus 2 equal to ve?” The following questions are based on your visit to the island of Kya–Kya. [CAT 1992] 6.

7.

8.

If an islander asks, “Do I belong to the ‘No’ type”, which of the following is correct? (a) He is a ‘No’. (b) He is a ‘Yes’. (c) It is impossible for him to have asked such a question. (d) His type cannot be identied. Ram and Laxman are brothers from the Island. Laxman asks you, “Is at least one of us brothers of type ‘No”? You can conclude that: (a) Ram is ‘NO’, Laxman is ‘Yes’. (b) Both are ‘Yes’. (c) Ram is ‘Yes’, Laxman is ‘No’. (d) Both are ‘No’. You are approached by one of the islanders and asked, “Am I of type ‘Yes”? You can infer that: (a) He was a ‘No’. (b) He was a ‘Yes’. (c) Such a situation is not possible. (d) No conclusion is possible. Directions for Question Nos. 9 to 12

“Kya–Kya” is an island in the South Pacic. The inhabitants of “Kya–Kya” always answer any question with two sentences, one of which is always true and the other always false. [CAT 1993] 9.

You nd that your boat is stolen. You question three inhabitants of the island and they reply as follows: John says, “I didn’t do it. Mathew didn’t do it.” Mathew says. “I didn’t do it. Krishna didn’t do it.” Krishna says. “I didn’t do it. I don’t know who did it.”

Who stole your boat? (a) John

(b)

(c)

(d) None of them

Krishna

Mathew

10. There is only one pilot on the island. You interview three men, Koik, Lony and Mirna. You also notice that Koik is wearing a cap. Mirna says, “Lony’s father is the pilot. Lony is not the priest’s son.” Koik says, “I am the priest. On this island, only priests can wear caps.” Lony says, I am the priest’s son. Koik is not the priest.” Which of the following is true? (a) Lony is not Koik’s son. (b) Koik is the pilot. (c)

Mirna is the pilot.

(d) Lony is the priest. 11. You are walking on the road and come to a fork. You ask the inhabitants Ram, Laxman and Lila. ”which road will take me to the village?” Ram says, “I never speak to strangers, I am new to these parts.” Laxman says, “I am married to Lila. Take the left road.” Lila says, “I am married to Ram, He is not new to this place.” Which of the following is true? (a) Left road takes you to the village. (b) Right road takes you to the village. (c)

Lila is married to Laxman.

(d) None of these. 12. You want to speak to the chief of the village. You question three inhabitants Amar, Bobby and Charles. Only Bobby is wearing a red shirt.” Amar says. “I am not Bobby’s son. The chief wears a red shirt.” Bobby says, “I am Amar’s father, Charles is the chief.” Charles says, “The chief is one among us. I am the chief.” Who is the chief? (a) Amar

(b)

(c)

(d) None of them

Charles

Bobby

EBD_7743

264 Koncepts of Logical Reasoning Directions for Question Nos. 13 to 16 The table below presents the revenue (in million rupees) of four in three states. These rms, Honest Ltd., Aggressive Ltd., Truthful Ltd. and Protable Ltd. are disguised in the table as A, B, C and D, in no particular order. [CAT 2005] States

(a) Both statements could be true. (b) At least one of the statements must be true. (c)

At most one of the statements is true.

(d) None of the above. 16. If Protable Ltd.’s lowest revenue is from UP, then which of the following is true?

Firm A Firm B Firm C Firm D

UP

49

82

80

55

Bihar

69

72

70

65

(a) Truthful Ltd.’s lowest revenues are from MP.

MP

72

63

72

65

(b) Truthful Ltd.’s lowest revenues are from Bihar.

Further, it is know that:

(c)

In the state of MP, Truthful Ltd. has the highest market share.

(d) No denite conclusion is possible.

Aggressive Ltd.’s aggregate revenue differs from Honest Ltd.’s by Rs.5 million. 13. What can be said regarding the following two statements? Statement 1: Protable Ltd. has the lowest share in MP market. Statement 2: Honest Ltd.’s total revenue is more than Protable Ltd. (a) If statement 1 is true then statement 2 is necessarily true. (b) If statement 1 is true then statement 2 is necessarily false. (c) Both statement 1 and statement 2 are true. (d) Neither statement 1 nor statement Solution: 14. What can be said regarding the following two statements? Statement 1: Aggressive Ltd.’s lowest revenues are from MP. Statement 2: Honest Ltd.’s lowest revenues are from Bihar. (a) If statement 2 is true then statement 1 is necessarily false. (b) If statement 1 is false then statement 2 is necessarily true. (c) If statement 1 is true then statement 2 is necessarily true. (d) None of the above. 15. What can be said regarding the following two statements?

Truthful Ltd.’s lowest revenues are from UP.

Directions for Question Nos. 17 and 18 In the following questions, a statement is given followed by three assumption is something to be taken for granted. Considering the statement and the assumptions are implicit in the statement, decide which one of the answers is correct answer?

17. Statements: Prakash decided to get his railway reservation done in the month of May for the journey he wants to undertake in July to Mumbai. Assumptions: I.

The railways open reservation two months in advance.

II.

There is more than one train to Mumbai.

III. There will be a vacancy in the desired class. (a) II and III

(b)

I

(c)

(d)

III

I and II

18. Statement: In a recently held all-India conference, the session on Brand Management in India surprisingly attracted a large number of participants and also received excellent media coverage in the leading newspapers. Assumptions: I.

Nobody expected such an encouraging response to Brand Management.

II.

Brands are not managed properly in India.

Statement 1: Honest Ltd. has the highest share in the UP market.

III. The Media is always very positive towards Brands.

Statement 2: Aggressive Ltd. has the highest share in the Bihar market.

(a) I

(b)

I and II

(c)

(d)

I, II and III

I and III

  Statement Conclusion and Binary Logic Directions for Question Nos. 19 to 21 A statement followed by three assumptions numbered. I, II and III are given. An assumption is something supposed or taken for granted. You have to consider the statement and the following assumptions and decide which of the assumptions is implicit in the statement, then decide which of the answers (A), (B), (C), or (D) is correct. 19. Statement: As our business is expanding we need to appoint more managers — says Mr. Adani, owner of Crayon Limited. Assumptions: I. The present managers are not competent. II. More staff will further expand the business. III. Suitable persons to be taken as managers will be available. (a) Only I is implicit (b) Only II is implicit (c) Only III is implicit (d) None is implicit 20. Statement: The Government of Rajasthan, in a report, revealed that due to social disturbances and bomb blasts in Jaipur, the number of foreign tourists has decreased considerably, resulting in a loss of Rs. 200 crores.

Actions:

265 



I.

Government should provide financial support to the tourism sector.



II. Foreign tourists should be advised to visit the country at their risk.



(a) Only I follows.



(b) Only II follows.



(c) Either I or II follows.



(d) Neither I nor II follows.

21. Statement: The Indian textile industry venturing into the Western European market faces tough competition from China. Actions:

I.

India should search for other international markets for its textile products.



II. India should improve the quality and reduce costs to compete with China in capturing the market.



(a) Only I follows.



(b) Only II follows.



(c) Both I and II follow.



(d) Neither I nor II follows.

Solutions Concept Applicator (CA) 1. (a) From the given information we have 2 cases for Sachin: Case (i) First statement is true and the second one is false.           ⇒ Sachin – Top, Ricky – Third, and Brian – Second. Going by the above, both statements of Brian become false. Hence, this case is not possible and hence ruled out. . Case (ii) First statement is false and the second one is true. ⇒ Sachin – Third, Ricky – Second, Brian – Top Going by this, Brian’s first statement is true and the second one is false. Ricky’s first statement is false and the second one is true.           ∴ The order is Brian, Ricky and Sachin.

2. (b) 3. (a) 4. (b) All statements are contrary to each other hence if one is correct rest all have to be wrong so only one of them must be true and only one is interested in the project. Let us assume that Asit is interested in the project then he will speak truth, but as per his statement no one is interested hence it is contradictory to his own statement and he cannot be the one who speaks truth. Then let us assume Barun speaks truth then as per his statement only one is interested and that person is Barun himself, hence this condition is possible. Then assume that Chandra is interested and he speaks truth then as per him two of them are interested that means two of the statements are correct but we have seen that

EBD_7743

266 Koncepts of Logical Reasoning

5.

(d)

6.

(c)

7.

(a)

8.

(d)

it is not possible, hence this condition is ruled out. On the similar logic rest all are eliminated and we can conclude that only Barun speaks truth and he is the only one who is interested in the project. Here A and D is guard, C is thief. If A is thief then B is not the guard. Therefore E must be thief he is telling lie. So there is one guard and four thieves. This type of question should be solved by assuming one of the statement true and the prove this assumption true or false. The answer to this question can be neither ‘No’ nor ‘Yes’ as both would contradict the given conditions; hence option (c) is correct. The answer to the question has to be ‘yes’ implying that Laxman is ‘Yes’ hence Ram has to be ‘No’ hence option (a) is correct The answer to this question can be ‘No’ as well as ‘Yes’; hence option (d) is correct.

Solutions (9 to 12) The best way to solve these kinds of question is to assume that one of the statements is either true or false and hence gure out whether there is consistence in what everyone is saying. 9. (b) 10. (b) 11. (a) 12. (b) Solutions (13 to 16) Given that, In the state of MP, Truthful Ltd. has the highest market share. Hence, Truthful Ltd must be either Firm A or Firm C as these have the highest market share in MP. So we have two case to consider Case (i) Truthful Ltd is Firm A Revenue Truthful

Firm A

190

Protable

Firm D

185

Honest

Firm B/ Firm C

217/222

Aggressive

Firm C/ Firm D

222/217

13. (b) Now consider the given statements. Statement 1: Protable Ltd has the lowest share in M.P. market i.e., protable Ltd is Firm B, with total revenue of 217 million rupees. And Honest Ltd. is either rm A or rm D with total revenue of 185 million rupees or 190 million rupees and In any case, the total revenue of Honest Ltd cannot be more than the revenue of Protable Ltd. Hence, if statement 1 is true then statement 2 is false. 14. (c) Statement 1: If statement 1 is true, i.e. Aggressive Ltd.’s lowest revenues are from M.P., Aggressive Ltd. must be Firm B. In that case Honest Ltd must be Firm C and for rm C i.e., the lowest revenues are from Bihar i.e., statement 2 is true.Hence if statement 1 is true, statement 2 must be true. From the given options, only option (3) is satised. 15. (c) Statement 1: Honest Ltd has the highest share in UP market then, Honest Ltd must be Firm B and Case (i) from the above two cases is applicable and Honest Ltd has a total revenue of 217 million rupees. Statement 2: Aggressive Ltd has the highest share in Bihar market then Aggressive Ltd is Firm B and has a total revenue of 217 million rupees. Now considering both the statements, at most one of the statements can be true as only one among Aggressive Ltd and Honest Ltd can have total revenue of 217 million rupees (see table). 16. (c) Protable Ltd has its lowest revenue from UP. Then Protable Ltd must be Firm D. Then from table Truthful is rm A and its lowest revenue is from U.P. 17. (b) The statement is referring one particular region and the competition faced there. I is based on the assumption that India has not ventured in to other international markets.

Case (ii) Truthful Ltd is Firm C

I does not follow. Reduction in price and increase in quality would make the product easily marketable. Hence, II follows. Only, II follows.

Revenue Truthful

Firm C

222

Protable

Firm B

217

Honest

Firm A/ Firm D

190/185

Aggressive

Firm D/ Firm A

185/190

18. (a) 20. (d)

19. (c) 21. (c)

Direction 267

17

Direction

Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt. Exams

Very Important

Other Aptitude Test

Very Important N

INTRODUCTION W

There are four main directions - East, West, North and South, as per the convention we generally take upward direction of a paper as north direction and remaining directions as shown below: Other than 4 main directions, there are four cardinal directions - North-East (N-E the direction exactly between North and East), North-West (N-W the direction exactly between North and West), South-East (S-E the direction exactly between South and East), and South-West (S-W the direction exactly between South and West) as shown below:

E S

N-W

N

W S-W

N-E E

S

S-E

EBD_7743

268 Koncepts of Logical Reasoning The main formula for Direction questions is Pythagoras Theorem. As per the Pythagoras theorem a2+ b2= c2 a

SOME PYTHAGOREAN TRIPLETS ARE

b

3, 4, 5 and its multiples, c

5, 12, 13 and its multiples, 7, 24, 25, and its multiples, 8, 15, 17 and its multiples, 9, 40, 41 and its multiples, 11, 60, 61 and its multiples, 12, 35, 37 and its multiples.

Example 1: Rajesh, at the time of morning walk, goes 5 km in the East, after that he turns to his left and goes 4 km. Finally he turns to his left and goes 5 km. Now how far is he from his house and in what direction?

Solution: the picture below represents each actionHere

I is: Moving 5 km east. II is: After travelling in east direction turning left means in north direction (4 km). III is: After travelling in north direction turning left means in west direction (5 km). 5 km

4 km

House

5 km I

5 km II

N 4 km

5 km III

From third position, it is clear he is 4 km from his house and is in north direction.

W

E S

1.

2.

1

Direction 269

Rajesh, starting from his house, goes 4 km in the East. Then he turns to his right and goes 3 km. What is his nal distance from his house? (a) 4 km

(b)

(c)

(d) None of these

5.5 km

5 km

Ricky after travelling for 5 km took right turn and travelled 6 km before taking left turn and then travelled for 3 km. nd his 7.

4.

(a) 25 km

(b)

(c)

(d) 40 km

4km

2 km

In a game “Pass the ball” position of some players are as follows-

starting point?

'A' is 20 meters to the north of 'B' who is 18 meters to the east of 'C' who is 12 meters to the west of 'A'. If the ball was initially with B and is passed to C, in which direction is it from its starting point?

(a) 16 km

(b)

(c)

(d) 4 km

A country has the following types of trafc signals.

(a) North - East

(b)

3 red lights = stop;

(c)

(d) North

(a) 10 km

(b)

(c)

(d) None of these

9 km

South - East

12 km

North - West

Rimpy on the way to her school starts walking from her home towards south. After walking 15 meters she turns towards north. After walking 20 meters, she turns towards east and walks 10 meters. She then turns towards south and walks 5 meters. How far is she from his

2 km

12 km

Directions for Question Nos. 8 to 11

2 red lights = turn left; 3 green lights = go at 100 kmph speed; 1 red light = turn right; 1 green light = go at 20 kmph speed 2 green lights = go at 40 kmph speed

(b) 10 meters, South - East

A motorist starts at a point on a road and follows all trafc signals literally. His car is heading towards the north. He encounters the following signals (the time mentioned in each case below is applicable after crossing the previous signal).

(c)

Starting point – 1 green light;

original position and in which direction? (a) 10 meters, East 10 meters, West

(d) 10 meters, North - East 5.

Rimpy, on her morning walk, starting from home she walks to the North for 250 m, then she turns to her right and travels 20 m and then she again turns to the right and drives straight another 250 m. How much distance has she now to cover to go back to her home?

Amar on his new car 1st drives towards North 4 km and turns right and drives 5 km. Then he turns towards South and drives 2 km. Then he takes a right turn and walks 6 km. What is the distance of Amar from his

nal distance from home.

3.

6.

An ant moves 10 cm towards east and turns to the right hand moves 3 cm. Then it moves to its right and moved 3 cm. It then turns to his left and moves 2 cm. Finally, it turns to his right and travel's 7 cm. How far and in which direction is it now from the starting point? (a) 10 cm, East

(b)

(c)

(d) 5 cm, West

8 cm, West

9 cm, North

After half an hour, 1st signal – 2 red & 2 green lights; After 15 minutes, 2nd signal – 1 red light; After half an hour, 3rd signal – 1 red and 3 green lights; After 24 minutes, 4th signal – 2 red and 2 green lights; After 15 minutes, 5th signal – 3 red lights [CAT 2002]

EBD_7743

270  Koncepts of Logical Reasoning 

8. The total distance traveled by the motorist from the starting point till the last signal is:

(a) 90 km.

(b) 100 km.



(c) 120 km.

(d) None of these.

9.

What is the position (radial distance) of the motorist when he reaches the last signal?



(a) 45 km directly north of starting point.



(b) 30 km directly to the east of the starting point.



(c) 50 km. away to the northeast of the starting point.



(d) 45 km away to the northwest of the starting point.

10. After the starting point if the 1st signal were 1 red light and 2 green lights. What would be the final position of the motorist?

(a) 30 km to the west and 20 km to the south.



(b) 30 km to the west and 40 km to the north.



(c) 50 km to the east and 40 km to the north.



(d) 30 km directly to the east.

11. If at the starting point, the car was heading towards south, what would be the final position motorist?

(a) 30 km to the east and 40 km to the south.



(b) 50 km to the east and 40 km to the south.



(c) 30 km to the west and 40 km to the south.



(d) 50 km to the west and 20 km to the north. Directions for Question Nos. 12 to 14 In a motor race competition certain rules are given for the participants to follow. To control direction and speed of the motorists, guards are placed at different signal points with caps of different colour. Guard with red cap indicates the direction of participant‘s movement and guards with green cap indicates the speed of the participant‘s movement. At any signal point presence of three guards, two guards and one guard with red cap means the participant must stop, turn left and turn right respectively. Signal points with three guards, two guards and one guard with green cap means the participants must move at 10, 4 and 2 km/hour respectively. Kartikay, one of the participants, starts at a point where his car was heading towards north and he encountered signals as follows: at start point one

guard with green cap; after half an hour two guards with red cap and two guards with green cap at first signal; after fifteen minutes one guard with red cap at second signal; after half an hour one guard with red cap and three guards with green caps at third signal; after 24 minutes two guard with red cap and two guards with green cap at fourth signal; after 15 minutes three guard with red cap at fifth signal. (Time mentioned in each case is applicable after crossing the previous signal).

[IIFT 2008]

12. Total distance travelled by Kartikay from starting point till last signal is:

(a) 9 km

(b) 10 km



(c) 8 km

(d) 12 km

13. What would be the final position of Kartikay if one guard with red cap and two guards with green caps were placed at the first signal point after the starting point?

(a) 3.0 km to the west and 2.0 km to the south



(b) 3.0 km to the west and 4.0 km to the north



(c) 5.0 km to the east and 4.0 km to the north



(d) 2.0 km to the west and 4.0 km to the south

14. If at the starting point Kartikay was heading towards south what would be his final position?

(a) 3.0 km to the east and 4.0 km to the south



(b) 5.0 km to the east and 4.0 km to the south



(c) 3.0 km to the west and 4.0 km to the south



(d) 5.0 km to the west and 2.0 km to the north

15. Mr. Raghav went in his car to meet his friends John. He Drove 30 km towards north and then 40 km towards west. He then turned to south and covered 8 km. Further he turned to east and moved 26 km. Finally he turned right and drove 10 km and then turned left to travel 19 km. How far and in which direction is he from the starting point? [IIFT 2009]

(a) East of starting point, 5 km



(b) East of starting point, 13 km



(c) North East of starting point, 13 km



(d) North East of starting point, 5 km

16. A Retail chain has seven branches in a city namely R1, R2, R3, R4, R5, R6, R7 and a central distribution center (DC). The nearest branch to the DC is R6 which is in the south of DC and is 9km away from DC. R2 is 17 km away from DC in the west. The branch R1 is 11 km away from R2 further in the west. The branch R3 is 11km in the north east of R1. R4 is 13 km away from R3 in the east. R5 is 11 km in the north east of the distribution center. In the north east of R6 is R7 and distance between them is 15 km. The distance between R1 and R6, R2 and R6, R6 and R5 is 23km, 19km, 13km respectively. R3 is 14 km away from the DC in the northwest direction, while R2 is also 14 Km away from R4 in the north east

  Direction 271  direction of R2. A truck carrying some goods starts from the distribution center and has to cover at least four stores in a single trip. There is an essential good that has to be delivered in the store R7 but the delivery at R7 has to be done in the end, so what is the shortest distance the truck would travel? [IIFT 2010]

(a) 55

(b) 56



(c) 63

(d) 66

17. In a code, south-east becomes west, northeast becomes south and so on. What will west become?

(a) South-east

(b) North-east



(c) East

(d) North

EBD_7743

272 Koncepts of Logical Reasoning

Concept Applicator (CA) 1.

82 + 62 = 10km. 5

(b) Condition is as shown in picture- from Pythagoras theorem nal distance is 5 km. 4

6 3

5

2.

(a) Condition is as shown in picture- from Pythagoras theorem nal distance is

3.

(b)

8.

(a)

4.

(a)

3 5.

(d)

6.

(b)

7.

(b)

FINISH

IIIrd Signal

N W

E S

20 km

IInd Signal

@100 kmph t = 24 mins s = 40 km

10 km

40 km

@40 kmph t = 30 mins \ s = 20 km

Vth Signal @40 kmph t = 15 minutes \ s = 10 km IVth Signal

@40 kmph t = 15 mins \ s = 10 km 10 km 10 km

Ist Signal Moves @20 kmph t = ½ hr = 30 mins \ s = 20 × 30/60 = 10 km

START Note: s = Distance covered; v = Velocity (km/hr); t = Time taken; s = v × t. Total distance travelled by the motorist from the starting point till last signal is 10 + 10 + 20 + 40 + 10 = 90 km.

9. (c) 30 km

T

F

40 km 20 km

TF = 50 km; ST = 40 km.



Considering the figure given above, option (c) is correct, i.e. 50 km to the east and 40 km to the north.

11. (c) From the above we can conclude that option (c) is correct.

IV

40 km

S

I III

10 km

10 km

273 



10 km

II

  Direction

START

10 km 40 km

I



S

II

10 km

20 km 30 km

Note: According to Pythagoras’ Theorem, for a right-angled triangle. C

BC = AB2 + AC2



SF = √(ST2 + TF2) = √(402 + 302)



III

F FINISH

13. (a) As Kartikay started heading north, he will be 3.0 km to the west and 2.0 km to the south if one guard with red cap and two guards with green caps were placed at the first signal point. Option (a)

BC2 = AB2 + AC2

40 km

12. (a) Total distance travelled by Kartikay from starting point till last signal = (1 + 1 + 2 + 4 + 1) km = 9 km. Option (a)

B

A

IV

= √2500 = 50 km

10. (a) For the case where 1st signals were 1 red light and 2 green light, the surface diagram will be as follows.

14. (c) As Kartikay started heading south, he will be 3.0 km to the west and 4.0 km to the south. Option (c) 15. (c) From the given information we can draw the diagram: 40 kms

50 km

T

F

8 kms 26 kms

10 km

III 40 km 4 km

10 kms

IV

5 kms 19 kms

20 km

12 kms

I 10 km II 10 km S

30 kms



13 kms

From the diagram we can conclude that finally he is 13 km in the north-east direction from the starting point. Option (c)

EBD_7743

274 Koncepts of Logical Reasoning 16. (c) From the given information we can draw the following diagram:

Thus, the shortest path is DC-R3 -R1 –R6 –R7. The distance travelled by truck = 14 + 11 + 23 + 15 = 63 km. NB: As per the given condition when the distance from R1 to DC is 28 km, the distance from R1 and R6 must be greater than 28 km, but it is given as 23 km. Similarly, distance between R5and R6 must be greater than 17 km, but it is given as 13 km. The answer is obtained by ignoring this fact. Option (c) 17. (b)

Data Sufciency 275

18

Data Sufficiency

Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt Exams

Very Important

Other Aptitude Test

Very Important

Data sufciency is one of the most important as well as most difcult to handle. Questions based on data sufciency require deeper knowledge in the subject area. Data sufciency questions can be from any topic viz. numbers, arithmetic, algebra, geometry, and puzzles. These questions are not difcult but confusing. Students generally get confuse with the given information. Some general tips to solve the data sufciency questions. Tip 1: 1st and foremost is understanding of options. In GMAT options are xed but in other aptitude test exams these options vary. Tip 2: Identify the parent question, means the main question that is asked. Tip 3: Look at the statement (i) and information given in that, see whether it is sufcient enough to answer the parent question. Tip 4: Look at the statement (ii) and information given. In that, most important is that when you see statement (ii), just forget the information given in statement (i). Then see whether it is sufcient enough to answer the parent question. Tip 5: When the given statements are not sufcient then use information given in both the statements.

EBD_7743

276 Koncepts of Logical Reasoning

1

Directions for Question Nos. 1 to 7

(a)

If statement I alone is sufcient to answer the question. (b) If statement II alone is sufcient to answer the question. (c) If both statements I and II together are necessary to answer the question. (d) If both statements I and II together are not sufcient to answer the question. [SNAP 2010] 1.

Is ‘b’ positive? (I) a + b is positive. (II) a – b is positive.

2.

In a general body election, 3 candidates, p, q and r were contesting for a membership of the board. How many votes did each receive? (I) p received 17 votes more than q and 103 votes more than r. (II) Total votes cast were 1703.

3.

If C1 and C2 are the circumferences of the outer and inner circles respectively. What is C1:C2? (I) The two circles are concentric. (II) The area of the ring is 2/3 the area of greater circle.

4.

What is the middle number of 7 consecutive whole numbers? (I) Product of number is 702800. (II) Sum of the number is 105.

5.

Total marks obtained by P, Q, R and S in Mathematics is 360. How many marks did P secure in Mathematics? (I) P secured one-third marks of the total of Q, R and S. (II) Average marks obtained by Q and R are 20 more than that secured by S.

6.

How many ice cubes can be accommodated in a container? (I) The length and breadth of the container is 20 cm and 15 cm respectively. (II) The edge of the ice cube is 2 cm.

7.

Ram got Rs 1500 as dividend from a company. What is the rate of interest given by the company? (I) The dividend paid last year was 10%. (II) Ram has 350 shares of Rs 10 denomination. Directions for Question Nos. 8 to 11

Answer (a) if data in statement I alone are sufcient to answer the question but the data in statement II alone are not sufcient to answer the question. Answer (b) if data in statement II alone are sufcient to answer the question but the data in statement I alone are not sufcient to answer the question. Answer (c) if data in statement I and II together are necessary to answer the question. Answer (d) if data in statement I and II together are not sufcient to answer the question. [SNAP 2010] 8.

∆ABC and ∆PQR are congruent. (I) Area of ∆ABC and ∆PQR are same. (II) ∆ABC and ∆PQR are right angle triangles.

9.

Salary of A and B is in ratio 3:4 and expenditure is in ratio 4:5. What is the ratio of their saving? (I) B’s saving is 25% of his salary. (II) B’s salary is Rs 2500.

10. What is the average height of the class? (I) Average height of the class decreases by 1 cm if we exclude the tallest person of the class whose height is 56 cm. (II) Average height of the class increases by 1 cm if we exclude the shortest person of the class whose height is 42 cm. 11. Ram is taller than Shyam and Jay is shorter than Vikram. Who is the shortest among them? (I) Ram is the tallest. (II) Shyam is taller than Vikram.

2

  Data Sufficiency

277 

Concept Builder

Directions for Question Nos. 12 to 13

Are followed by two statements labelled as I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below:   (a) Statement I alone is sufficient to answer the question. (b) Statement II alone is sufficient to answer the question. (c) Statement I and Statement II together are sufficient, but neither of the two alone is sufficient to answer the question. (d) Either Statement I or Statement I alone is sufficient to answer the question. (e) Neither Statement I nor Statement 11 is necessary to answer the question.  12. Let PQRS be a quadrilateral. Two circles 01 and 02 are inscribed in triangles PQR and PSR respectively. Circle 01 touches PR at M and circle 02 touches PR at N. Find the length of MN. I. A circle is inscribed in the quadrilateral PQRS. II. The radii of the circles 01 and 02 are 5 and 6 units respectively. [5] 13. Given below is an equation where the letters represent digits. (PQ). (RQ) = XXX. Determine the sum of P + Q + R + X. I X = 9. II. The digits are unique. Directions for Question Nos. 14 to 15 Followed by two statements labelled as I and II. You have to decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from options given below: (a) If statement I alone is sufficient to answer the question. (b) If statement II alone is sufficient to answer the question. (c) If statement I and statement II together are sufficient but neither of the two alone is sufficient to answer the question.

(d) If either statement I or statement II alone is sufficient to answer the question. (e) Both statements I and II are insufficient to [XAT 2008] answer the question. 14. The base of a triangle is 60 cm, and one of the base angles is 60°. What is length of the shortest side of the triangle? (I) The sum of lengths of other two sides is 80 cm. (II) The other base angle is 45°. 15. A, B, C, D, E and F are six integers such that E < F, B > A, A < D < B. C is the greatest integer. Is A the smallest integer? (I) E + B < A + D (II) D < F Directions for Question Nos. 16 to 19 Are followed by two statements labelled as (1) and (2). You have to decide if thesestatements are sufficient to conclusively answer the question. Choose (a) If statement (1) alone is sufficient to answer the question. (b) If statement (2) alone is sufficient to answer the question. (c) If statement (1) and Statement (2) together are sufficient but neither of the two alone is sufficient to answer the question. (d) If either statement (1) or Statement (2) alone is sufficient to answer the question. (e) Both statement (1) and statement (2) are insufficient to answer the question. [XAT 2007] 16. What is the maximum value of a/b? (I) a, a + b and a + 2b are three sides of a triangle. (II) a and b are both positive. 17. ABC is a triangle with ∠B = 90°. What is the length of the side AC? (I) D is the midpoint of BC and E is the midpoint of AB. (II) AD = 7 and CE = 5

18. Five integers A, B, C, D and E are arranged in such a way that there are two integers between B and C and B is not the greatest. There exists one integer between D and E and D is smaller than E. A is not the smallest integer. Which one is the smallest? (I) E is the greatest. (II) There exists no integer between B and E.

3

Concept Cracker

Directions for Question Nos. 20 to 29

Choose (a) if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone. Choose (b) if the question can be answered by using either statement alone. Choose (c) if the question can be answered by using both statements together, but cannot be answered using either statement alone. Choose (d) if the question cannot be answered even by using both statements together.

19. Le f: N → N (N is the set of all natural numbers). How many solutions are there to the equation f(x) = 1485? (I) For a, b ∈N f(10a + b) = f(a) + 12b (II) The maximum value of b is 9.

[CAT 1999]  

20. The average weight of students in class is 50 kg. What is the number of students in the class? A. The heaviest and the highest members of the class weigh 60 kg and 40 kg respectively. B. Exclusion of the heaviest and the lightest members from the class does not change the average weight of the students.    21. A small storage tank is spherical tank in shape. What is the storage volume of the tank? A. The wall thickness of the tank is 1cm. B. When the empty spherical tank is immersed in a large tank filled with   water, 20 litres of water overflow from the large tank. 22. Mr. X starts walking northwards along the boundary of a field, from point A on the boundary, and after walking for 150 m reaches B, and then walks westwards, again along the boundary, for another 100 m when he reaches C. What is the maximum distance between any pair of points on the boundary of the field?



A. The field is rectangular in shape.   B. The field is a polygon, with C as one of its vertices and A the mid-point of a side.   

23. A line graph on a graph sheet shows the revenue for each year from 1990 through 1998 by points and joins the successive points by straight line segments. The point for revenue of 1990 is labelled A, that for 1991 as B, and that for 1992 as C. What is the ratio of growth in revenue between 91-92 and 90-91? A. The angle between AB and X-axis when measured with a protractor is 40 degrees, and the angle between CB and X-axis is 80 degrees. B. The scale of Y-axis is 1 cm = 1000 Rs. 24. There is a circle with centre C at the origin and radius r cm. Two tangents are drawn from an external point D at a distance d cm from the centre. What are the angles between each tangent and the X axis? A. The coordinates of D are given B. The X-axis bisects one of the tangents. 25. Find a pair of real numbers x and y that satisfy the following two equations simultaneously. It is known that the values of a, b, c, d, e and f are non-zero. ax + by = c dx + ey = f         A. a = kd and b = ke, c = kf, k ≠ 0 B. a = b = 1, d = e = 2, f ≠ 2c 26. Three professors A, B and C are separately given three sets of numbers to add. They were expected to find the answers to 1+1, 1+1+2, and 1+1 respectively. Their respective

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278  Koncepts of Logical Reasoning 

  Data Sufficiency 279  are watching TV

answers were 3, 3, and (2) How many of the professors are mathematicians? A. A mathematician can never add two numbers correctly, but can always add three numbers correctly. B. When a mathematician makes a mistake in a sum, the error is +1 or -1.

32. How many people programme P? A. Number of people watching TV programme Q is 1000 and number of people watching both the programmes, P and Q, is 100. B. Number of people watching either P or Q or both is 1500.

27. How many among the four students A, B, C and D have passed the exam? A. The  following is a true statement: A and B passed the exam. B. The following is a false statement. At least one among C and D has passed the exam. 

33. Triangle PQR has angle PRQ equal to 90°, what is the value of PR + RQ? A. Diameter of the inscribed circle of the triangle PQR is equal to 10 cm. B. Diameter of the circumscribed circle of the triangle PQR is equal to 18 cm.

28. What is the distance x between two cities A and B in integral number of km? A. x satisfies the equation log2x = √ x  B. x ≤ 10 km 

34. Harshad bought shares of a company on a certain day, and sold them the next day. While buying and selling he had to pay to the broker one percent of the transaction value of the shares as brokerage. What was the profit earned by him per rupee spent on buying the shares? A. The sales price per share was 1.05 times that of its purchase price. B. The number of shares purchased was 100.





29. Mr. Mendel grew one hundred flowering plants from black seeds and white seeds, each seed giving rise to one plant. A plant gives flowers of only one colour. From a black seed comes a plant giving red or blue flowers. From a white seed comes a plant giving red or white flowers. How many black seeds were used by Mr. Mendel?   A. The number of plants with white flowers was 10. B. The number of plants with red flowers was 70. Directions for Question Nos. 30 to 39 Choose (a)   if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone. Choose (b) if the question can be answered by using either statement alone. Choose (c) if the question can be answered by using both statements together, but cannot be answered using either statement alone. Choose (d) if the question cannot be answered even by using both statements together.

[CAT 2000]

30. Consider three real numbers, X, Y and Z. Is Z the smallest of these numbers? A. X is greater than at least one of Y and Z. B. Y is greater than at least one of X and Z. 31. Let X be a real number. Is the modulus of X necessarily less than 3? A. X (X + 3) < 0 B. X (X – 3) > 0

35. For any two real numbers: a ⊕ b = 1 if both ‘a’ and ‘b’ are positive or both a and b are negative. = –1 if one of the two numbers ‘a’ and ‘b’, is positive and the other negative. What is (2 ⊕ 0) ⊕ (–5 ⊕ –6)? A. a  ⊕ b is zero if a is zero B. a  ⊕ b = b ⊕ a 36. There are two straight lines in the x-y plane with equations: ax + by = c dx + ey = f Do the two straight lines intersect? A. a, b, c, d, e and f are distinct real numbers. B. c and f are non-zero 37. O is the centre of 2 concentric circles. ae is a chord of the outer circle and it intersects the inner circle at points b and d. C is a point on the chord in between b and d. What is the value of ac/ce? A. bc/cd = 1 B. A third circle intersects the inner circle at b and d and the point c is on the line joining the centres of the third circle and the inner circle. 

38. Ghosh Babu has decided to take a non-stop flight from Mumbai to No-man’s-land in South America. He is scheduled to leave Mumbai at 5 am, Indian Standard Time on December 10, 2000. What is the local time at No-man’s-land when he reaches there? A. The average speed of the plane is 700 kilometres per hour. B. The flight distance is 10,500 kilometres. 39. What are the ages of two individuals, X and Y? A. The age difference between them is 6 years. B. The product of their ages is divisible by 6. Directions for Question Nos. 40 to 46 Choose (a) if the question can be answered by one of the statements alone but cannot be answered using the other statement alone. Choose (b) if the question can be answered by using either statement alone. Choose (c) if the question can be answered by using both the statements together, but cannot be answered by using either statement alone. Choose (d) if the question cannot be answered even by using both the statements together. [CAT 2001] 40. What are the value of m and n? A. n is an even integer, m is an odd and m is greater than n. B. Product of m and n is 30. 41. Is Country X’s GDP higher than country Y’s GDP? A. GDP’s of the countries X and Y have grown over the past 5 years at compounded annual rate of 5% and 6% respectively. B. Five years ago, GDP of country X was higher than that of country Y. 42. What is the value of X? A. X and Y are unequal even integers less than 10 and X/Y is an odd integer. B. X and Y are even integers each less than 10 and product of X and Y is 12. 43. On a given day a boat ferried 1500 passengers across the river in 12 hrs. How many round trips did it make? A. The boat can carry 200 passengers at any time. B. It takes 40 minutes each way and 20 minutes for waiting time at each terminal. 44. What will be the time for downloading software?



A. Transfer rate is 6 kilobytes per second. B. The size of the software is 4.5 megabytes.

45. A square is inscribed in a circle. What is the difference between the area of the circle and that of the square?

A. The diameter of the circle is 25 2 cm. B. The side of the square is 25 cm.

46. Two friends, Ram and Gopal, bought apples from a wholesale dealer. How many apples did they buy? A. Ram bought one-half the number of apples that Gopal bought. B. The wholesale dealer had a stock of 500 apples. Directions for Question Nos. 47 to 54 Choose a if the question can be answered by one of the statements alone but not by the other. Choose b if the question can be answered by using either statement alone. Choose c if the question can be answered by using both the statements together, but cannot be answered by us either statement alone. Choose d if the question cannot be answered by either [CAT 2002] of the statements. 47. In a hockey match, the Indian team was behind by 2 goals with 5 minutes remaining. Did they win the match? A. Deepak Thakur, the Indian striker, scored 3 goals in the last five minutes of the match. B. Korea scored a total of 3 goals in the match. 48. Four students are added to a dance class. Will the teacher be able to divide her students evenly into a dance team (or teams) of 8? A. If the 12 students are added, the teacher can put everyone in teams of 8 without any leftovers. B. The number of students in the class is currently not divisible by 8. 49. Is x = y? A. (x + y)( 1 / x + 1 / y) = 4 B. (x – 50)2 = (y – 50)2 50. A dress was initially listed at a price that would have given the store a profit of 20 percent of the wholesale cost. What was the wholesale cost of the dress? A. After reducing the listed price by 10 percent, the dress sold for a net profit of 10 dollars. B. The dress sold for 50 dollars.

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280  Koncepts of Logical Reasoning 

51. Is 500 the average (arithmetic mean) score on the GMAT? A. Half of the people who take the GMAT score above 500 and half of the people score below 500. B. The highest GMAT score is 800 and the lowest score is 200. 52. Is |x - 2| < 1? A. |x| > 1 B. |x - 1| < 2 53. People in a club either speak French or Russian or both. Find the number of people in a club who speak only French. A. There are 300 people in the club and the number of people who speak both French and Russian is 196. B. The number of people who speak only Russian is 58.



  Data Sufficiency 281  after any of the tosses. Ram has incurred a loss of Rs.50 by playing this game. How many times did he toss the coin? A. The game ended normally. B. The total number of tails obtained in the game was 138.

57. Each packet of SOAP costs Rs.10. Inside each packet is a gift coupon labelled with one of the letters S, O, A and P. If a customer submits four such coupons that make up the word SOAP, the customer gets a free SOAP packet. Ms. X kept buying packet after packet of SOAP till she could get one set of coupons that formed the word SOAP. How many coupons with label P did she get in the above process? A. The last label obtained by her was S and the total amount spent was Rs.210 B. The total number of vowels obtained was 18.

54. A sum of Rs. 38,500 was divided among Jagdish, Punit and Girish. Who received the minimum amount? A. Jagdish received 2/9 of what Punit and Girish together received. B. Punit received 3/11 of what Jagdish and Girish together received.

58. If A and B run a race, then A wins by 60 seconds. If B and C run the same race, then B wins by 30 seconds. Assuming that C maintains a uniform speed what is the time taken by C to finish the race? A. A and C run the same race and A wins by 375 m. B. The length of the race is 1 km.

Directions for Question Nos. 55 to 58

Directions for Questions 59 to 64

Choice (a) if the question can be answered by one of the statements alone but not by the other. Choice (b) if the question can be answered by using either statement alone. Choice (c) if the question can be answered by using both the statements together but cannot be answered using either statement alone. Choice (d) if the question cannot be answered even by using both the statements A and B. [CAT 2003] 55. F and M are father and mother of S, respectively. S has four uncles and three aunts. F has two siblings. The siblings of F and M are unmarried. How many brothers does M have? A. F has two brothers. B. M has five siblings. 56. A game consists of tossing a coin successively. There is an entry fee of Rs.10 and an additional fee of Re.1 for each toss of the coin. The game is considered to have ended normally when the coin turns heads on two consecutive throws. In this case the player is paid Rs.100. Alternatively, the player can choose to terminate the game prematurely

Choose 1 if the question can be answered by using one of the statements alone but not by using the other statement alone. Choose 2 if the question can be answered by using either of the statements alone. Choose 3 if the question can be answered by using both statements together but not by either statement alone. Choose 4 if the question cannot be answered on the basis of the two statements. [CAT 2004] 59. Four candidates for an award obtain distinct scores in a test. Each of the four casts a vote to choose the winner of the award. The candidate who gets the largest number of votes wins the award. In case of a tie in the voting process, the candidate with the highest score wins the award. Who wins the award? A. The candidates with the three scores each vote for the top scorer amongst the other three. B. The candidate with the lowest score votes for the player with the second highest score.

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282 Koncepts of Logical Reasoning 60. Zakib spends 30% of his income on his children’s education, 20% on recreation and 10% on healthcare. The corresponding percentages for Supriyo are 40%, 25%, and 13%. Who spends more on children’s education? A. Zakib spends more on recreation than Supriyo. B.

Supriyo spends more on healthcare than Zakib.

61. Tarak is standing 2 steps to the left of a red mark and 3 steps to the right of a blue mark. He tosses a coin. If it comes up heads, he moves one step to the right; otherwise he moves one step to the left. He keeps doing this until he reaches one of the two marks, and then he stops. At which mark does he stop? A. He stops after 21 coins tosses. B.

He obtains three more tails than heads.

62. In a class of 30 students, Rashmi secured the third rank among the girls, while her brother Kumar studying in the same class secured the sixth rank in the whole class. Between the two, who had a better overall rank? A. Kumar was among the top 25% of the boys merit list in the class in which 60% were boys. B.

There were three boys among the top ve rank holders, and three girls among the top ten rank holders.

63. Nandini paid for an article using currency notes of denominations Re. 1, Rs. 2, Rs. 5 and Rs. 10 using at least one note of each denomination. The total number of ve and ten rupee notes used was one more than the total number of one and two rupee notes used. What was the price of the article? A. Nandini used a total of 13 currency notes. B.

The price of the article was a multiple of Rs. 10.

64. Ravi spent less than Rs. 75 to buy one kilogram each of potato, onion, and gourd. Which one of the three vegetables bought was the costliest? A. 2 kg potato and 1 kg gourd cost less than 1 kg potato and 2 kg gourd. B. 1 kg potato and 2 kg onion together cost the same as 1 kg onion and 2 kg gourd.

Directions for Question Nos. 65 to 68 Mark (a) if the question can be answered by using the statement A alone but not by using the statement B alone. Mark (b) if the question can be answered by using the statement B alone but not by using the statement A alone. Mark (c) if the question can be answered by using either of the statements alone. Mark (d) if the question can be answered by using both the statements together but not by either of the statements alone. Mark (e) if the question cannot be answered on the basis of the two statements. [CAT 2007] 65. Five students Atul, Bala, Chetan, Dev and Ernesto were the only ones who participated in a quiz contest. They were ranked based on their scores in the contest. Dev got a highest rank as compared to Ernesto, while Bala got a higher rank as compared to Chetan. Chetan’s rank was lower than the median. Who among the ve got the highest rank? A. Atul was the last rank holder. B. Bala was not among the top two rank holders. 66. Thirty per cent of the employees of a call centre are males. Ten per cent of the female employees have an engineering background. What is the percentage of male employees with engineering background? A. Twenty ve per cent of the employees have engineering background. B. Number of male employees having an engineering background is 20% more than the number of female employees having an engineering background. 67. In a football match, at the half-time, Mahindra and Mahindra Club was trailing by three goals. Did it win the match? A. In the second-half Mahindra and Mahindra Club scored four goals. B. The opponent scored four goals in the match. 68. In a particular school, sixty students were athletes. Ten among them were also among the top academic performers. How many top academic performers were in the school? A. Sixty per cent of the top academic performers were not athletes. B. All the top academic performers were not necessarily athletes.

Data Sufciency 283 Directions for Question Nos. 69 to 70 Mark (a) if question can be answered from A alone but not from B alone. Mark (b) if question can be answered from B alone but not from A alone. Mark (c) if question can be answered from A alone as well as from B alone. Mark (d) if question can be answered from A and B together but not from any one of them alone. Mark (e) if question cannot be answered even from A and B together. [CAT 2008] In a single elimination tournament, any player is eliminated with a single loss. The tournament is played in multiple rounds subject to the following rules: (a) If the number of players, say n, in any round is even, then the players are grouped in to pairs.

(b)

The players in each pair play a match against each other and the winner moves on to the next round. If the number of players, say n, in any round is odd, then one of them is given a bye, that is, he automatically moves on to the next round. The

remaining (n − 1) players are grouped into pairs. The players in each pair play a match against each other and the winner moves on to the next round. No player gets more than one bye in the entire tournament. Thus, if n is even, then n/2 players move on to the next round while if n is odd, then (n + 1)/2 players move on to the next round. The process is continued till the nal round, which obviously is played between two players. The winner in the nal round is the champion of the tournament 69. What is the number of matches played by the champion? A. The entry list for the tournament consists of 83 players. B. The champion received one bye. 70. If the number of players, say n, in the rst round was between 65 and 128, then what is the exact value of n? A. Exactly one player received a bye in the entire tournament. B. One player received a bye while moving on to the fourth round from third round

Now, by considering Statement II alone, p + i + r = 170 Hence, we cannot nd how many each received. So, this statement is not sufcient enough Using I and II together, we get p + (p – 17) + (p – 103) = 170. Solving the above equation we get the value of p and the values of q and r.

Concept Applicator (CA) 1.

2.

(d) If we look at Statement I then we will get If a = 3 and b = 2, a + b > 0. Here b > 0 If a = 3 and b = –2, a + b > 0. Here b < 0 Hence I alone is not sufcient. Now if we look at Statement II only then we will get If a = 3 and b = 2, a – b > 0. Here b > 0 If a = 3 and b = –2, a – b > 0. Here b < 0 Hence II alone is not sufcient. Now by using statements I and II together If a = 3 and b = 2, a – b > 0 and a + b > 0. Here b>0 If a = 3 and b = –2, a – b > 0, and a + b >0. Here b < 0. Hence I and II together are also insufcient. (c) If we look at Statement I i = p – 17 and r = p – 103 Hence, we cannot nd how many each received so this statement is not sufcient enough.

3.

(b) If we look at Statement I It is given that the circles are concentric. But nothing is given about their dimensions. Hence I alone is not sufcient. In statement II ratio of area is given hence we can nd the required ratio.

4.

(b) Let the 7 consecutive whole numbers be (n ± 3), (n ± 2) (n ± 1), n. Now, if we consider Statement I alone Product of these 7 integers = 702800 Since 702800 = 24 52 (251)(7), it cannot be the product of 7 consecutive whole numbers.

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284 Koncepts of Logical Reasoning

5.

(a)

6.

(d)

7.

(b)

8.

(d)

Hence I alone is insufcient. Now if we consider Statement II alone Given that their sum = 105 = 7n or n = 15 and then 7 consecutive integers are 12, 13, 14, 15, 16, 17, 18 So, II alone is sufcient. Since sum is 360 hence P + Q + R + S = 360 From statement I alone we will get P = (Q + R + S)/3 from this we can nd the value of P hence statement I alone is sufcient enough. From statement II alone we cannot nd the value of P. Statement I is not sufcient as the size of the ice cube and height of the container is not known hence statement I is not sufcient alone. Statement II is also not sufcient as the dimension of the container is not known. We cannot answer the question even by combining both the statements as the height of the container is not known. It is given that Ram got a dividend of Rs. 1500. Statement I Knowing the dividend paid last year, we cannot nd the dividend paid this year. Statement II Given that Ram bought 350 shares of face value `10, and so, their total face value is `3500. So here we know the investment and the return hence we can nd out the rate of interest. Consider Statement I alone. Given that Area (∆ABC) = Area (∆PQR) since nothing about the sides or angles is mentioned, we cannot say if they are congruent. Hence, I alone is not sufcient. Consider Statement II alone ∆ABC and ∆PQR are right triangles. Nothing about the sides is given; hence, II alone is not sufcient. Now using both I and II

9.

Now, we have two right angled triangles with same area we may have different combination as only product of base and height is same. Hence even by using both the statement we cannot nd the answer. (a) Given that their salaries are in the ratio of 3:4and expenditure is in the ratio of 4:5; hence we can assume that salary of A and B are 3x and 4x and their expenditures are 4y and 5y. Now we need to nd the ratio of (3x-4y)/ (4x-5y)

Consider statement I alone: Saving of B is 25% of his salary hence his expenditure must be 75% so ¾(4x) = 5y or 3x = 5y from this we can nd the required ratio hence this statement is sufcient. Consider statement II alone: Given that 4x = 2000 or x = 500 but from this we cannot nd the value of y and hence we cannot nd the ratio of their savings. 10. (c) Let x be the average height of the class and n be the number of students in the class. Consider statement I alone:xn – 56 = (x – 1)(n – 1) Þ x + n = 57 ---- (1) Hence, the value of x cannot be found. So, I alone is not sufcient. Consider statement II alone:xn – 42 = (x + 1)(n – 1) Þ x – n = 41 ---- (2) Hence, the value of x cannot be found. So, II alone is not sufcient. Both the statements together are sufcient as the value of x can be found by solving (1) and (2) 11. (b) Given that Ram > Shyam, Vikram > Jay. Hence from this we can conclude that neither Ram nor Vikram is the shortest. And we have to nd the shortest among them. Considering statement I alone: We know that Ram is not the shortest, either Shyam or Jay is the shortest. Hence (I) alone is not sufcient. Consider statement II alone, Shyam > Vikram. From the given information and the information in (II), we get Ram > Shyam > Vikram > Jay. Hence, (II) alone is sufcient. Concept Builder (CB) 12. (a) Statement I alone is sufcient. Statement II alone is not sufcient, for we can have more than one value of MN possible. 13. (e) Given relationship is (PQ)(RQ) = XXX Since X can take 9 values from 1 to 9 hence we have 9 possibilities 111 = 3 ´ 37 222 = 6 ´ 37 333 = 9 ´ 37

444 = 12 ´ 37 555 = 15 ´ 37 666 = 18 ´ 37

777 = 21 ´ 37 888 = 24 ´ 37 999 = 27 ´ 37

But out of these 9 cases only in 999, we get the unit’s digit of the two numbers the

same. Since it is a unique value, hence we need neither statement Ι nor statement ΙΙ to answer the question. C 14. (d)

a cms

b cms

60° A

60 cms

B

Let a cm and b cm be the two unknown sides, as shown in the figure. From statement 1, a + b = 80 cm, hence b = (80 – a) cm Now, using cosine rule, Cos 60° = (AB2 + AC2 - CB2)/2 AB ∴ ½ = [602 + b2 - (80 - b)2]/120 By solving this we get, b = 28 cm. Hence, statement 1 is sufficient to answer. From statement 2, Since ∟B = 45°hence ∟C = 75° According to sine rule: we know that a/ = b / = c/ a/ sin45° = b/ sin60° = 60/ sin75° From this we can find the value of the sides. Hence statement 2 is sufficient to answer the question. 15. (a) From statement I, E + B < A + D, we easily say that E is less than A, because B > D and as the statement suggest E + B < A + D ∴ E < A. ∴ A is not the smallest integer. Statement I is sufficient to answer. From statement II, D < F This statement is not sufficient to find the relation between A and E. 16. (a) Since ‘b’ is the common difference of three sides of a triangle, a/b can take any values in the given range of real; numbers. Suppose a = 1000 b=1 Then there sides will be 1000, 1001, 1002 Here a/b = 1000/1 = 1000 Hence the values of ‘a’ and ‘b’ can be varied to any values making the ratio a/b undeterminable. 17. (c) From statement 1: Let AB = 2x and BC=2y, then AC2 = 4(x2+y2), since we don’t know the values of x and y hence we cannot calculate the AC. Therefore, this statement is not sufficient enough. From statement 2: AD = 7 and BD = 5; we don’t get answer from this.

  Data Sufficiency 285  Now combining both the information we can write (2x + y)2 = 49 and (2y + x)2 = 25, hence we can calculate the value of x and y and hence we can calculate the value of AC. 18. (e) We have three possible cases: Case (i) DBEAC Case (ii) BADCE Case (iii) CADBE From statement (1), either case (ii) or case (iii) could be valid hence we cannot get a unique value from this statement alone. From statement (2), either case (i) or case (ii) is possible hence we cannot get a unique value from this statement alone. If we combine the two statements, only case (iii) is possible; hence (c) is correct option. 19. (e) From statement 1: F(x = 10a + b) = f(a) + 12b = 1485. From this equation we will have different values of a, and b and hence many values of x. From statement 2: Maximum value of b = 9, since we don’t know the exact value of b so we cannot determine. From both equations, we cannot solve due to unknown value of f(a). Concept Cracker (CC) with XAT 20. (d) Statement A alone – it gives the highest and lowest weight of members of the class. This does not give any other information so we cannot determine the number of students. Hence this statement is not sufficient enough. Statement B alone – It gives that by eliminating the highest and lowest members, the average weight does not change. This is also not sufficient to answer anything. Both the statements together cannot determine the number of students as the average weight of students (after eliminating highest and lowest) cannot determine the total number. 21. (c) To find the volume, we need radius of sphere. Statement A alone – does not give any useful information. Statement B alone – it says 20 litres of water are displaced when the tank is fully immersed. Hence we will get the outer volume of the spherical tank. Now with this the outer radius of the tank can be found. But this is not sufficient to get an answer.



On combining we can find the inner radius because the wall is 1cm thick. And hence the storage volume can be determined.

22. (c) Statement A alone – It is given here that the shape of the field is rectangular but exact location of point A, B and C is not given hence we cannot find the from this information. Statement B alone – It is given that the shape is polygon but we don’t know whether it is polygon of n = 4, 5, 6 or anything else, hence this is not sufficient to get an answer. By combining statement A and B we know that shape is rectangle and C and B are its vertices and A is the mid-point of other side, hence together both the statement is sufficient to get answer. 23. (a) Here we need to find the ratio of growth and that will be equal to the ratio of the slope of the line. Statement A: From this statement we know the slope and hence we can find out the ratio of slopes. So this is sufficient to get answer. Statement B: From the scale only we cannot calculate the ratio. 24. (b) statement A alone: With the coordinates of point D given it helps to find out the equation of the straight line passing from the two points where the tangents meet the circle, i.e., (r , 0) and (0, r). And if we know the equation of a straight line then we can calculate its slope and angle between tangent and x- axis and hence this statement is sufficient enough. Statement B does not give any answer. So answer is (a). 25. (a) Statement A gives (kd)x + (ke)y = kf, k ≠ 0 ⇒ dx + ey = f ⇒ These represent one and  the same equations, hence infinite pairs can satisfy; hence this is not sufficient to get an answer. Statement B gives x + y = c, 2x + 2y = f ⇒ f = 2c. But it is given that f ≠ 2c ⇒ Inconsistent. Combining the two statements A and B also we cannot determine the answer. 26. (d) Statement A alone: it says a mathematician can never add two numbers correctly, but can always add three numbers correctly. Let there are three professor A, B and C, A gives an answer 1 + 1 = 3, hence he is a mathematician. B gives an answer 1 + 1 + 2 = 3, hence he is not a mathematician. C gives an answer 1 + 1 = 3, hence he is a mathematician. So this statement is sufficient enough. Statement B says when a mathematician makes a mistake in a sum, the error is +1 or –1. Here two professors madea mistake of +1

and –1, the third got it correctly. Hence this statement is sufficient enough. Hence, each statement individually is able to answer the question. 27. (c) Statement A states that A and B passed. Statement B actually means neither C nor D has passed. Hence from both the statements we can determine that the total number of students who passed was 2 and these are A and B. 28. (c) Statement A alone – it gives the following equation Log2  X = √X; X = 2√ X. This is not sufficient to answer the question. Because there are many values of X may satisfy the equation, e.g., 4, 16 etc. Statement B alone – it says X ≤ 10. This is not sufficient by itself to answer the question. But on combining both the statement A and B we get only the integer 4 that shall satisfy the equation. 29. (d) Since Black Seed gives Blue flower or Red flower. White seed gives White flower or red flower. Statement A alone – it says plants with white flowers are 10. Hence this statement is not sufficient enough to answer the question. Statement B alone – it says number of plants with red flowers is 70. Again nothing can be concluded. On combining we get 10 plants with white flowers and 70 with Red flowers. So plants with Blue flowersshould be 20 (As total number of plants is 100). Now blue flowers shall come from Black seeds only. So there are at least 20 Black seeds used. 10 white flowers can come only from white seeds. So, at least 10 white seeds have been used. Now the remaining 70 red could have come from either black or white seeds. So data of this statement is insufficient. Even by using both the statement we cannot determine the answer. 30. (c) Statement A  alone:  X may be greater than Z. But it doesn’t say anything about Z being smallest. Statement B alone: Y may be greater than Z. But it says nothing about Z being smallest. Now, we will use both statements A and B. If X > Y, then Y > Z and   X > Y > Z. But if X > Z, then Y > X and Y > X > Z. In either of these two cases, Z is the least and hence the question can be answered.     31. (a) From statement A alone: x(x + 3) < 0; hence 0 > x > – 3

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286  Koncepts of Logical Reasoning 

Data Sufciency 287 Hence, |x| is necessarily less than 3 in this range. From statement B alone: x (x – 3) hence x > 3 or x < 0 In this condition, any given |x| value need not necessarily be less than 3. So B is not sufcient enough. 32. (c) Consider statement A alone: from the formula n(P U Q) = n(P) + n(Q) – n (P ∩ Q) We get n (P U Q) = n (P) + 1000 – 100. From A alone, we get 2 unknowns and 1 equation. Hence, A alone is not insufcient enough. Consider B alone: n(P U Q) = 1500 given, but from Balone, n(P) cannot be determined. Consider both together: 1500 = n(P) + 1000 – 100 and n(P) = 1500 – 900 = 600 Hence both the statement required to solve the question. Alternately, from statement A alone: Q = 1000

a

b

34. (a)

35. (c)

c

P Q = 1000 It is given that b + c = 1000 and b = 100 hence c = 900 but we cannot calculate a, so this statement is not sufcient enough. From statement B alone, a + b + c = 1500, and this statement is also not sufcient enough to nd the value of a + b. If we use both the statements then we can calculate the value of a + b.

36. (d)

33. (c) 37. (b)

Statement A: Gives us the radius of inscribed circle = 5, but with this information we cannot calculate PR and QR.

Statement B: Gives hypotenuse PQ = 18 cm (diameter of circumcircle). Now, ∟ P + ∟ Q = 180 – 900 = 900 So ∟P/2 + ∟Q/2 = 450 but this is not sufcient enough. Now, If IS = 5cm (in radius = 5), then Using both the information we can calculate Consider A alone: Given S.P/C.P = 1.05 and brokerage of 1% Let CP = 100 x brokerage = x so total CP = 101x, and SP = 105x A is sufcient. Consider B alone: No. of shares = 100, since no price is given, prot cannot be calculated hence this statement is not sufcient enough. From the main information a @ b is not dene when a or b is zero. Consider A alone: from this information we see that a @ b = 0 only if a is zero. However, in the expression (2 + 0): a ≠ 0. A is not sufcient. B alone is not sufcient enough as it does not provide important information. Consider A and B together a @ b = b @ a hence 2 @ 0 = 0 @ 2 So (2 @ 0) = 0 and (– 5 @ – 6) =1 Hence A and B together is sufcient enough Two straight line intersect if they are not parallel hence if a/d ≠ b/e Now consider statement A alone, it does not say specically that a/d ≠ b/e From statement B alone, if c and f both are zero then also we cannot say that a/d ≠ b/e Even with both the statements together we cannot determine whether the two straight lines intersect or not. From statement A

We see that BC = CD or C is at the center of B and D









So C will also be the centre of A and E. Hence (A) is sufficient to answer. From statement (B)

A





B

O C I

D

E

Let T be the center of the third circle. Then since C isthe line joining O and T implies C is midpoint of B and D or bisects B and D.            Similar to last case now we can say that this statement is also sufficient enough. 38. (d) Consider statement A  alone:  Avg. speed of plane says nothing about local time in noman’s land. Statement B alone: Distance also says nothing about the local time. Now if we use both the statement together:  Time required for flight can be obtained but still we cannot find the local time. 39. (d) From statement A alone it is given that x-y = 6 or y-x = 6, from this we cannot find out a unique value of x or y, hence this statement is not sufficient enough. From statement B xy = 6k, again we cannot find out a unique value of x or y, hence this statement is also not sufficient enough. Now if we use both the statement together then, let x-y = 6 and xy = 6k, or x = 6k/y or 6k/y-y 6 or y2 + 6y = 6k, and again we will not get a unique value of x and/or y. 40. (c) Statement (A): it is not sufficient to give the answer as we will get the pairs (15,2), (10,3), (6,5). There are infinitely many values. But using the statement (B) we can say the that value of (m, n) may be, (1,30), (2, 15), (3, 10), (5, 6) hence this is not sufficient enough. By combining A and B we will get unique answer m = 15, n = 2. 41. (d) Statement A: As we don’t know that what were their GDPs 5 years ago. Hence we cannot compare its present value. Statement B- we know that GDP of country X is greater than that of Y but we do not know by how much. Hence even using both the statement we cannot determine.

42. (a) Considering statement (A) alone the possible pairs are (5,1), (7,1), (9,1), (2, 6), (9, 3). So there is only one pair with both even numbers (2,6). Now considering statement (B) the possible pairs are (2, 6) and (6, 2). So, there is no unique solution by using (B) alone. We can not get answer from statement A & B. 43. (a) From statement A: We cannot make out the number of rounds, as maximum capacity is given 200, but actual value is not given. Statement (B) However, we can find out the trip time using the given values and can thus calculate the total number of trips since total time is fixed 12 hours. 44. (c) Statement (A) alone - it is not sufficient enough because it gives only the data transfer rate but does not give the size of the software to be downloaded. Statement (B) alone - it is also not sufficient because it gives only the size of software but does not give the rate of data transfer. Combining both the statements, we can get the answer. 45. (c) We know that if diameter is given, then side of square can be found. Similarly if side of the square is given, the radius of circle can be found. Hence, both the statements individually can provide the difference in areas. 46. (d) Ratio of number of apples bought by Ram and Gopal is given and total quantity at dealer is given, but what part of the total quantity they bought is not given hence even both statements are not sufficient to get the number of apples bought by Ram and Gopal. 47. (d) We do not know whether Korea scored a goal in the last 5 minutes or it may be possible that India at beginning has 0 goal and Korea has 2, so at the end they may draw , even from both statement we cannot conclude. 48. (a) From statement A - it gives us that by adding 4, the number (4, 12, 20…) would be divisible by 8. Hence it is sufficient enough 49. (a) From statement A - we get (x + y) (x + y/xy) = 2 4; or (x + y) = 4xy ; 2 Or (x – y) = 0, hence x = y. From statement B- (x-50)2 = (y-50)2, or x-50 = +/-(y-50) Hence we will get x = y or x + y = 100; hence, we cannot get the answer from the second statement.

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288  Koncepts of Logical Reasoning 

50. (a) Statement A – After reducing the listed price by 10 percent, the dress sold for a net profit of 10 dollars and given that net profit is 20% hence we can get the CP. But statement (2) just gives the SP but we do not know the discount. 51. (d) We cannot arrive at the even with the use of both the statement. Average since we do not know individual scores or number of students. 52. (c) Both the statements combine together are sufficient to answer the question. Hence option (c) is the right answer. 53. (c) If we draw a Venn diagram we will find that, we need both statements. Hence by using both the statement we can solve it. 54. (c) From first statement we get only J’s share. Statement B also alone cannot solve it But, by combining the statements we get the values of each student. 55. (a) From the given condition, Father and Mother of S is F and M respectively. Total no. of uncle and aunts are 4 and 3 respectively. A tells us F has 2 brother then they must be unmarried. Rest two uncle must be brother of M and married .Rest aunt of s should be M’s sibling. We get answer from A. From B is not correct to get answer. 56. (a) Let the total no. of time coin tosses be n. Total amount spent by ram = (10 + n x 1) = 10 + n First he paid Rs. 100 but he suffered Rs. 50 losses. Then amount left with him, (10 + n) – 100 = 50 ⇒ n = 140. We get result using A. For statement B we have 2 cases: Case (1) Game ended normally – this case is similar to statement A Case (2) Game ended prematurely – this case is not possible with 138 tails. Hence both the statements are sufficient enough 57. (c) Price of each soap = Rs. 10 From statement 1: She spent total Rs 210. So, she bought 210/10 = 21 soaps. And get last letter S. We cannot say anything from it. From statement 2: Total vowels = 18 means he got no. of A and O = 18

  Data Sufficiency 289  From it we cannot say anything But using both, no. of P she got = 21-(18 + 1) = 2 58. (c) Race of A and C, A wins by 375 m and A wins by (60 + 30) = 90s Speed of c = (375/90) m/s. Time cannot be calculated because we don’t know the length of track. From statement B, length of track = 1 Km. from it nothing is concluded. But using A and B we get speed of C = 1000/ (375/90). 59. (a) Assume A, B, C, D get score 10, 8, 6, 4 ABCD 10 8 6 4 Statement A: With the conditions A will give vote to B With the conditions B will give vote to A With the conditions C will give vote to A Even if D gives to A/B/C — 2 situations arises. Either A will win or there will a tie when D gives vote to B. Even then A will win. So we are getting the answer. Statement B: Cannot conclude anything. 60. (a) We have to find out, which one greater: 30% of Z or 40% of S. Statement A: 20% of Z > 25% of S. So, we cannot conclude anything about 30% of z : 40% of s; Statement B: 13 % of S > 10% of Z => 39 % of S > 30 % of Z Which further implies, 40% of S > 30 % of Z 61. (b) Statement A: We can find, there are 12 Tails and 9 Heads. After tosses he will reach at blue point. So statement A is sufficient. Statement B: 3 more Tails greater than Heads. So he will reach at blue point after tosses. So statement B is also sufficient. 62. (a) Statement A: Cannot say anything. Statement B: Because amongst the Top 5 → 3 are boys, 2 are girls. And Rashmi is third among the girls and Kumar is sixth. So, B is sufficient. 63. (d) Statement A: 13 currency notes will give different possible Values and hence it is not sufficient. Statement B: Multiple of 10 and will give many values. Even if you combine the statement, we can have various values.

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290 Koncepts of Logical Reasoning 64. (c) Statement A: 2 kg potato cost + 1 kg gourd cost < 1 kg potato cost + 1 kg gourd cost or 1 kg potato cost < 1 kg gourd cost. Hence statement A is not sufcient. Statement B: 1 kg potato cost + 2 kg onion cost = 1 kg onion cost + 2 kg gourd cost Hence 1 kg potato cost + 1 kg onion cost = 2 kg gourd cost. So statement B is also not sufcient as we cannot conclude. Combining both statements we get 1 kg potato cost < 1 kg gourd cost (From statement A) …(i) 1 kg potato cost + 1 kg onion cost = 2 kg gourd cost (From statement B) …(ii) So we can conclude that the onion is costliest. 65. (d) Let the abbreviation of their name as A, B, C, D and E, From main information D > E, B > C, and the rank of C is either 4th or 5th in descending order. And we need to nd the highest ranker. Now From Statement 1- it is given that A ranked 5 and any one of the B or D can be the top ranker, this statement is not sufcient enough as any one of B or D can be top ranker. Statement 2- B’s rank is either 3, 4 or 5, again we don’t have clear information that who is the topper. Combining the statement 1 & 2 we can conclude that D is the top ranker as from statement 2 B cannot be the topper. 66. (c) Let the number of employee be 100x Then the number of female employee is 70x, and those with engineering background 7x Statement I- the number of employee with engineering background is 25x The % of male employees with engineering background out of the males = (25x-7x)/30x X 100 Similarly the % of employee out of total employees can be calculated. Hence this statement is sufcient enough.

67. (e)

68. (a)

69. (d)

70. (d)

Statement II- the number of male employees having engineering background is 84x Similar to above calculation this statement is also sufcient enough. Statement I- this statement doesn’t give information about the number of goals scored by the opponent team, hence this statement is not sufcient enough. Statement II- This statement doesn’t give information about the number of goals scored by Mahindra & Mahindra club, hence this statement is not sufcient enough. Combining both the statement even it is not given the number of goals scored by Mahindra & Mahindra in 1st half, hence even by using both the information we cannot nd the result. From statement I- 40% of total is given 10 hence total is 10/0.4 = 25. This statement is sufcient enough. Statement II- It does not give any signicant information. Statement A alone: Since we do not know the number of byes got by the champion. Hence, statement A alone is not sufcient to answer the question. Statement B alone: Since we do not know the exact number of players in the tournament. Hence, statement B alone is not sufcient to answer the question. Combining both the statements together - If there are 83 players, there will be 6 rounds in the tournament. We know that the champion received only one bye, therefore the total number of matches played by the tournament will be 6 – 1 = 5. Given in option (d) Statement A alone: When n = 127, exactly one bye is given in round 1. When = 96, exactly one bye is given in round 6. As no unique value of n can be determined hence, statement A alone is not sufcient. Statement B alone: Since we do not know exactly how many bye’s are given, so, we cannot determine the value of n, uniquely. Combining statement A and B: using both the statement. There is a unique value of n = 120, for which exactly 1 bye is given from the third round to the fourth round.

Puzzles 291

Puzzles

Exam

Importance

Exam

19 Importance

CAT

Very Important

IBPS/Bank PO

Very Important

XAT

Very Important

BANK Clerk

Very Important

IIFT

Very Important

SSC

Very Important

SNAP

Very Important

CSAT

Very Important

NMAT

Very Important

Other Govt. Exams

Very Important

Other Aptitude Test

Very Important

Puzzle is an integral part of any aptitude test exam. These are basically combination of one or more concepts of logical reasoning. There are more than 1000 types of puzzles with 1000 different concepts; so, let’s concentrate on previous year actual questions that was asked in different aptitude test exam.

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292  Koncepts of Logical Reasoning 

1.

1

Concept Applicator

A group of students is organized in two rows, crossing each other. If a student is tenth from all ends, how many students are there in the

Directions for Question Nos. 6 to 7 Three men and three women are travelling in two cars (red and blue). Each car has exactly three persons. The cars cannot have all women or all men passengers. Mala, Ajit and Suman know how to drive a car. Mala and Sapna are women. Manjit and Sarat are not in the same car. Ajit and Sarat are men.

group?



(a) 27

(b) 47

(c) 37

(d) 39

2. A grocer sells half of the eggs that he has and another half an egg to Anurag. Then he sells half of the balance eggs and another half an egg to Deepak. Then he sells half of the balance eggs and another half an egg to Shivani. In the end he is left with just 7 eggs and he claims that he never broke an egg. How many eggs did he start with?

(a) 66

(b) 63



(c) 33

(d) 68

3.

A hunter has just returned after a day’s bird shooting. He was asked how many birds he had in his bag. He answered, “They are all sparrows but six,” all pigeons but six, and all doves but six. How many birds did he have in his bag?



(a) 18

(b) 9



(c) 27

(d) None of these

4. Arun started walking positioning his back towards the sun. After some time, he turned left, then turned right and then towards the left again. In which direction is he going now? (i) North

(ii) West

(iii) South

(iv) East



(a) (i) or (ii)

(b) (i) or (iii)



(c) (ii) or (iii)

(d) (iii) or (iv)

5. In a row at a bus stop, A is 9th from the right and B is 7th from the left. They both interchange their positions. If there are 20 people in the row, what will be the new position of B from the left’?

(a) 11th

(b) 12th



(c) 13th

(d) 10th

6.

7.

[2010] If the red car can have either Manjit or Sarat but not both while Ajit drives it and Mala is driving the blue car, which of the following can be true? (a) Suman is a man (b) Manjit is a woman             (c) Both (a) and (b) are correct (d) (a) is correct but (b) is not If Suman and Manjit are the two passengers in the red car, who can be passengers of the blue car? (a) Sapna and Sarat (b) Mala and Ajit (c) Sapna and Ajit (d) Mala and Sarat

8. In a game show, participants are asked to build two towers of different designs using plastic bricks. These plastic bricks are of two types: Red and Blue. Participants are given a fixed number of bricks for building these towers. The height of each tower is measured and 2 points are awarded for every 10 centimetres of first building and 1 point is awarded for every 10 centimetres of second building. One participant has 4 red and 3 blue bricks left with him. If he wants to increase the height of his first tower, he would need 1 red and 2 blue bricks for every 10 cm. For the second tower, to raise the height he would need 2 red and 1 blue bricks, for every 10 cm. What is the best option for him now?

[2009] (a) Raise the height of first tower by 20 cm (b) Raise the height of second tower by 20 cm. (c) Raise the heights of first and second towers by 10 cm and 20 cm respectively (d) None of these

9.

2

Puzzles 293

What number should replace the question mark? [SNAP 2011]

6

(a) 1 (c) 12

4

3

2

5

3

5

1

1

1

2

8

3

3

7

2

8

4

3

9

?

3

1

Directions for Question Nos. 13

(b) 4 (d) 6

10. There are two cups, one containing orange juice and one containing an equal amount of lemonade. One teaspoon of the orange juice is taken and mixed with the lemonade. Then a teaspoon of this mixture is mixed back into the orange juice. Is there more lemonade in the orange juice or more orange juice in the lemonade? [SNAP 2008] (a) More orange juice in the lemonade (b) More lemonade in the orange juice (c) Equal amount of each juice between the two cups (d) None of these 11. ABCD and E play a game of cards. A tells B that if B gives him ve cards, A will have as many cards as E has. However, if A gives ve cards to B then B will have as many cards as D. A and B together have 20 cards more than what D and E together have. B has four cards more than what C has and total number of cards are 201. How many cards does B have?

[IIFT 2009] (a) 185 (c) 175

order to avoid report to the manager she had to pay each security guard half of the money she had in her purse and 2 rupees more besides. She found only one rupee with her at the end. How much money Ms. Rani had before entering the ofce on the annual day? [IIFT 2008] (a) Rs. 40 (b) Rs. 36 (c) Rs. 25 (d) Rs. 42

(b) 37 (d) 27

12. All employees have to pass through three consecutive entrance doors to enter into the ofce and one security guard is deployed at each door. These security guards report to the manager about those who come to ofce after 10 AM. Ms. Rani is an employee of this ofce and came late on the annual day. In

A school in Bhopal decided to stage a historical drama, involving a battle between two ancient kingdoms. As a part of the battle, four students, namely - Aslam, Bimal, Chris and Dilip dressed as soldiers, marched through the stage at one point. When the make-up man dressed these four students, he put a helmet on each one’s head, without any one realizing the colour of their respective helmets. The make-up man selected the helmets for these four students from 3 gold-plated helmets, 2 silver-plated helmets and one copper-plated helmet at the makeup room. Now, when Aslam, Bimal, Chris and Dilip marched in that order, Aslam being the rst person in the queue could not see the helmet on the heads of the other three. Bimal saw the colour of helmet on Aslam’s head; Chris saw the same on Aslam’s head and Bimal’s head and Dilip saw the helmets on the heads of all others. After the drama, a classmate of them asked whether they were aware of the colour of the helmet on their own head, starting from Dilip. No one could answer the question. [IIFT 2006] 13. Mark all the incorrect statements. (a) Dilip did not observe the helmets on the heads of the other three actors, two of which were silver-plated and one copper-plated. (b) Bimal did not observe Aslam wearing either silver-plated or copper-plated helmet. (c)

Chris did not observe the helmets worn by Aslam and Bimal, one of which could be silver-plated and the other copper-plated or both could be silver-plated helmets.

(d) None of the above.

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294 Koncepts of Logical Reasoning 14. On the 1st March, Timon arrived in a new city and was looking for a place to stay. He met a landlady who offered to rent her apartment at a reasonable price but wanted him to pay the rent on a daily basis. Timon had a silver bar of 31 inches, and an inch of the silver bar was exactly equal to a day’s rent. He agreed to pay an inch of the silver bar towards the daily rent. Timon wanted to make the minimum number of pieces of the silver bar but did not want to pay any advance rent. How many pieces did he make? [XAT 2011] (a) 5 (c) 16 (e) 31

(b) 8 (d) 20

would he 437. What is the right position for the whole group of 545 as per the modied rule? (a) 3 (b) 194 (c) 249 (d) 437 (e) 543 17. Little Pika who is ve and half years old has just learnt addition. However, he does not know how to carry. For example, he can add 14 and 5, but he does not know how to add 14 and 7. How many pairs of consecutive integers between 1000 and 2000 (both 1000 and 2000 included) can Little Pika add? [XAT 2011]

Directions for Question Nos. 15 to 16 From a group of 545 contenders, a party has to select a leader. Even after holding a series of meetings, the politicians and the general body failed to reach a consensus. It was then proposed that all 545 contenders be given a number from 1 to 545. Then they will be asked to stand on a podium in a circular arrangement, and counting would start from the contender numbered as 1. The counting would be done in clockwise fashion. The rule is that every alternate contender would be asked to step down as the counting continued, with the circle getting smaller and smaller, till only one person remains standing. Therefore, the rst person to be eliminated would be the contender numbered as 2. [XAT 2011] 15. Which position should a contender choose if he has to be the leader? (a) 3 (b) 67 (c) 195 (d) 323 (e) 451 16. One of the contending politicians, Mr. Chanaya, was quite procient in calculations and could correctly gure out the exact position. He was the last person remaining in the circle. Sensing foul play the politicians decided to repeat the game. However, this time, instead of removing every alternate person, they agreed on removing every 300th person from the circle. All other rules were kept intact. Mr. Chanaya did some quick calculations and found that for a group of 542 people the right position to become a leader

(a) 150 (b) 155 (c) 156 (d) 258 (e) None of the above 18. Prof Bee noticed something peculiar while entering the quiz marks of his ve students into a spreadsheet. The spreadsheet was programmed to calculate the average after each score was entered. Prof. Bee entered the marks in a random order and noticed that after each mark was entered, the average was always an integer. In ascending order, the marks of the students were 71, 76, 80, 82 and 91. What were the fourth and fth marks that Prof Bee entered? (a) 71 and 82 (c) 71 and 80 (e) 91 and 80

(b) (d)

[XAT 2011] 71 and 76 76 and 80

19. In the gure, number in any cell is obtained by adding two numbers in the cells directly below it. For example, 9 in the second row is obtained by adding the two numbers 4 and 5 directly below it. The value of X – Y is [XAT 2008]

(a) 2 (c) 4 (e) 6

(b) 3 (d) 5

Puzzles 295 Directions for Question Nos. 20 to 22 Substitute different digits (0, 1, 2 ….9) for different letters in the problem below, so that the corresponding addition is correct and it results in the maximum possible value of MONEY.

P

M

A

Y

M

E

R

E

A

L

O

N

E

Y [XAT 2007]

3

Directions for Question Nos. 23 to 24

Five horses, Red, White, Grey, Black and Spotted participated in a race. As per the rules of the race, the persons betting on the winning horse gets four times the bet amount and those betting on the horse that came in second gets thrice the bet amount. Moreover, the bet amount is returned to those betting on the horse that came in third, and the rest lose the bet amount. Raju bets Rs. 3000, Rs. 2000 Rs. 1000 on Red, White and Black horses respectively and ends up with no prot and no loss. [CAT 2008]

20. The letter ‘Y’ should be (a) 0 (b) 2 (c) 3 (d) 7 (e) None of the above 21. There are nine letters and ten digits. The digit that remains unutilized is: (a) 4 (b) 3 (c) 2 (d) 1 (e) None of the above 22. The resulting value of ‘MONEY’ is (a) 10364 (b) 10563 (c) 10978 (d) 19627 (e) None of the above

(d) Black came in second (e)

There was one horse between Black and White

25. Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over personto-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

23. Which of the following cannot be true?

[CAT 2005]

(a) At least two horses nished before Spotted

(a) 5

(b)

(b) Red nished last

(c)

(d) 15

(c)

There were three horses between Black and Spotted

(d) There were three horses between White and Red (e)

Grey came in second

24. Suppose, in addition, it is known that Grey came in fourth. Then which of the following cannot be true?

9

10

26. Each family in a locality has at most two

adults, and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls, and more girls than families. Then the minimum possible number of families in the locality is [CAT 2004]

(a) Spotted came in rst (b) Red nished last

(a) 4

(b)

(c)

(c)

(d) 3

White came in second

2

5

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296 Koncepts of Logical Reasoning 27. On her walk through the park, Hamsa collected 50 coloured leaves, all either maple or oak. She sorted them by category when she got home, and found the following: [CAT 2001] The number of red oak leaves with spots is even and positive. The number of red oak leaves without any spot equals the number of red maple leaves without spots. All non-red oak leaves have spots, and there are ve times as many of them as there are red spotted oak leaves. There are no spotted maple leaves that are not red. There are exactly 6 red spotted maple leaves. There are exactly 22 maple leaves that are neither spotted nor red. How many oak leaves did she collect? (a) 22

(b)

(c)

(d) 18

25

17

Directions for Question Nos. 28 to 32 Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament is conducted in two stages. In the rst stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of the rst stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprises of several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated. The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup. The tournament rules are such that each match results in a winner and a loser with no possibility of a tie. In the rst stage a team earns one point for each win and no points for a loss. At the end of the rst stage teams in each group are ranked on the basis of total points to determine the qualiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams from each group advance to the next stage. [CAT 2000] 28. What is the total number of matches played in the tournament? (a) 28

(b)

(c)

(d) 35

63

55

29. The minimum number of wins needed for a team in the rst stage to guarantee is advancement to the next stage is:

(a) 5 (c) 7

(b) 6 (d) 4

30. What is the highest number of wins for a team in the rst stage in spite of which it would be eliminated at the end of rst stage? (a) 1 (b) 2 (c) 3 (d) 4 31. What is the number of rounds in the second stage of the tournament? (a) 1 (b) 2 (c) 3 (d) 4 32. Which of the following statements is true? (a) The winner will have more wins than any other team in the tournament. (b) At the end of the rst stage, no team eliminated from the tournament will have more wins than any of the teams qualifying for the second stage. (c) It is possible that the winner will have the same number of wins in the entire tournament as a team eliminated at the end of the rst stage. (d) The number of teams with exactly one win in the second stage of the tournament is 4. 33. My bag can carry no more than ten books. I must carry at least one book each of management, mathematics, physics and ction. Also, for every management book I carry I must carry two or more ction books, and for every mathematics book I carry I must carry two or more physics books. I earn 4, 3, 2 and 1 points for each management, mathematics, physics and ction book, respectively, I carry in my bag. I want to maximise the points I can earn by carrying the most appropriate combination of books in my bag. The maximum points that I can earn are: [CAT 2000] (a) 20 (b) 21 (c) 22 (d) 23 34. Five persons with names P, M, U, T and X live separately in any one of the following: a palace, a hut, a fort, a house or a hotel. Each one likes two different colours from among the following: blue, black, red, yellow and green. U likes red and blue. T likes black. The person living in a palace does not like black or blue. P likes blue and red. M likes yellow. X lives in a hotel. M lives in a: [CAT 2000] (a) hut (b) palace (c) fort (d) house

Puzzles 297 35. There are ten animals–two each of lions, panthers, bison, bears, and deer–in a zoo. The enclosures in the zoo are named X, Y, Z, P and Q; and each enclosure is allotted to one of the following attendants: Jack, Mohan, Shalini, Suman and Rita Two animals of different species are housed in each enclosure. A lion and a deer cannot be together. A panther cannot be with either a deer of a bison. Suman attends to animals from among bison, deer, bear and panther only. Mohan attends to a lion and a panther. Jack does not attend to deer, lion or bison. X, Y, and Z are allotted to Mohan, Jack and Rita respectively. X and Q enclosures have one animal of the same species. Z and P have the same pair of animals. The animals attended by Shalini are: [CAT 2000] (a) bear and bison

(b)

(c)

(d) bear and panther

bear and lion

40. What was the amount with Uma at the end of the second round? (a) 36 (b) 72 (c) 16 (d) None of these Directions for Question Nos. 41 to 42 In a game played by two people there were initially N match sticks kept on the table. A move in the game consists of a player removing either one or two matchsticks from the table. The one who takes the last matchstick loses. Players make moves alternately. The player who will make the rst move is A. The other player is B. [CAT 1990]

bison and deer

36. Eighty kilograms (kg) of store material is to be transported to a location 10 km away. Any number of couriers can be used to transport the material. The material can be packed in any number of units of 10, 20, or 40 kg. Courier charges are Rs.10 per hour. Couriers travel at the speed of 10 km/hr if they are not carrying any load, at 5 km/hr if carrying 10 kg, at 2 km/hr if carrying 20 kg and at 1 km/hr if carrying 40 kg. A courier cannot carry more than 40 kg of load. The minimum cost at which 80 kg of store material can be transported to its destination will be: [CAT 2000] (a) Rs. 180

(b)

Rs. 160

(c)

(d)

Rs. 120

Rs. 140

39. Who started with the highest amount? (a) Suvarna (b) Tara (c) Uma (d) Vibha

Directions for Question Nos. 37 to 40 Four sisters—Suvarna, Tara, Uma and Vibha are playing a game such that the loser doubles the money of each of the other players from her share. They played four games and each sister lost one game in alphabetical order. At the end of fourth game, each sister had Rs. 32. [CAT 1995]

41. The smallest value of N (greater than 5) that ensures a win for B is (a) 7 (b) 6 (c) 10 (d) 8 42. The largest of N (less than 50) that ensures a win for B is (a) 46 (b) 47 (c) 48 (d) 49 43. Consider the following set of three statements: (i) There are three statements in this set. (ii) Two of them are not true. (iii) Your IQ score will go up by 20 if you learn to play game Z. For these three statements to be consistent: (a) Only statements (ii) and (iii) must be true. (b) Statement (iii) may be either true or false (c) Statements (ii) and (iii) must be false. (d) Statement (iii) must be false. 44. Three married couples on a journey come to a river where they nd a boat that cannot carry more than two persons at a time. An additional condition is that a lady cannot be left on either bank where other men are present, without her husband.

37. How many rupees did Suvarna start with? (a) Rs. 60 (b) Rs. 34 (c) Rs. 66 (d) Rs. 28

(a) These people will not be able to cross the river.

38. Who started with the lowest amount? (a) Suvarna (b) Tara (c) Uma (d) Vibha

(c)

(b) They will be able to cross the river in 9 steps. They will be able to cross the river in 10 steps.

(d) They will be able to cross the river in 13

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298 Koncepts of Logical Reasoning

Finally he is going to south direction. Now, assume that Arun’s back in the west direction.

Concept Applicator (CA) 1.

(c) Suppose, the student is standing on the centre as shown in the diagram.

E S

N

9

W (10)

9

So, nally he is going in the North direction. So, he can go either in north or the south direction.

9

9 Hence, the total number of students in the group = (9 + 9 + 9 + 9 + 1) = 37. 2.

(b) It is better to solve this type of question from the end. As per the given condition after selling half an egg more than half the number of eggs to Shivani, the grocer was left with 7 eggs. Therefore, before selling eggs to Shivani the grocer had (7 + 1/2)2, i.e. 15 eggs. So, after selling half an egg more than half the number of eggs to Deepak, the grocer was left with 15 eggs. Therefore, before selling eggs to Deepak, the grocer had (15 + 1/2)2, i.e. 31 eggs. Similarly, before selling eggs to Anurag (i.e. at the start) the grocer had (31 + 1/2) 2, i.e. 63 eggs.

3.

(b) Let the number of sparrows, pigeons and doves be a, b and c respectively. Then from the given condition a + b, b + c = 6, and c + a = 6 on solving we will get a = b = c = 3. The total number of birds is 9.

4.

(b) Arun’s back was positioned towards the sun. Thus, depending on the time of the day, let us say Arun’s back is the in east direction. W N

S East

E

5.

(b) After they interchange their positions, B is 9th from the right. Also, there are 20 people in the row. So, B’s position is 12th from the left. Option (b)

Solutions (6 and 7) As per the given conditions are – (i)

Mala, Ajit and Suman- know how to drive a car

(ii) Females are Mala and Sapna and Males are Ajit and Sarat. Condition 1: At least one person should be in each car who knows driving. Condition 2: Manjit and Sarat are not in the same car. Condition 3: Each car has three persons and the cars cannot have all women or all men passengers. Condition 4: Each car has exactly three persons. 6.

(c) If Manjit is in red car, Sarat will be in blue car, then Red car ⇒ Manjit, Ajit (M)

Blue car ⇒ Sarat (M), Mala (F)

As per the given condition since at least one man (Sarat) and one woman (Mala) go in blue car. So, one man and one woman will certainly go in red car. Therefore, ‘Manjit is a woman’ can be a true statement. Similarly, ‘Suman is a man’ can be a true statement. Option (c)

7. () It is given that Suman and Manjit are the two passengers in the red car. And Suman can drive the car. And Manjit is in red car, so Sarat must go in the blue car. Also, only one out of Mala and Ajit can go in the red car (as three persons in one car) or Sapna can go in the red car. It satisfies all the conditions.

Here all the four given options are correct.

8. (a) As per the given conditions, if we have 4 red and 3 blue bricks then the best possible option is to raise 10 cm for both the towers, so 3 points will be awarded. Now evaluate the options one by one

Option (a) 2 red and 4 blue bricks, but it is not a possible case hence ruled out. Option (b) gives as 2 points, so it is less then what can be achieved. Option (c) since as per the given condition we require 5 red bricks hence it is not possible.



  Puzzles 299  Option (b): Now in this condition since Bimal knows that Chris and Dilip cannot tell their helmets colour. If Bimil observed the helmet colour of Aslam, then he could tell his helmet colour, i.e. gold. But he did not give the right answer hence option B is also correct.



Option (c): Here since Chris knows that Dilip cannot tell his helmet colour. Now if Chris observed the other two actors’ helmets colour like one silver and one copper or both silver. Then he can identify his helmet colour, i.e. Gold. But he did not give right answer. So it is the right statement.



Option (d): None of the statement is wrong. So option D is correct.

14. (a) The parts should be made of lengths: 1, 2, 4, 8, and 16. On day 1, he will give first piece. On day 2, he will take first piece back and will give second piece of length 2. Similar pattern continues. Thus minimum parts = 5. 15. (b) Here we will try to find the pattern

Concept Builder (CB) 9. (d) The number at the center of each row is the sum of the rest of the numbers divided by 2. Hence in last column we should have (9 + 3)/2 = 6 10. (c) Equal amount of each juice between the two cups 11.

Data given is ambiguous

12. (b) At first, Ms. Rani had x rupees with her. Then, equation will be:

1st round - 2x (All the numbers in the form of 2x eliminated) 2nd round = 1 + 4x (All the numbers in the form of 1 + 4x eliminated) similarly 3rd round = 3 + 8x 4th round = 7 + 8x 5th round = 15 + 16x 6th round = 31 + 32x 7th round = 63 + 64x

(½ x + 2) + ½ (½ x -2) + 2 + ½ (x/4 – 3) + 2 + 1 =x

8th round = 127 + 128x

After solving, x = Rs 36



9th round = 255 + 256x  Now we will eliminate options one by one,

13. (d) We will evaluate each of the options one by one-



(a) The number 3 is in the form of 3 + 8x (when x = 0) hence eliminates in the 3rd round

Option (a): As per the given condition if Dilip observed the other three actors helmets colour like 2 silver and 1 copper. Then he could be sure that now only one type of helmet that is gold plated is left, and he could tell his helmet colour, i.e. Gold. But he did not give right answer. So option (a) is the correct statement.



(c) Eliminates in 3rd round



(d) Eliminates in 3rd round



(e) Eliminates in 9th round



Thus (b) will be the leader.



16. (c) Similar to above solution Option (c) is correct.

17 (c) He can add 30 numbers in each group 1000 – 1099, 1100 – 1199, 1200 – 1299, 1300 – 1399, 1400 – 1499, i.e. 150 numbers. He can also add 1099 + 1100, 1199 + 1200

20. (e) Summation of ‘E’ and ‘L’ must give zero in the end so that ‘y’ comes as it is in the fourth row. (E + L = 10) The correct assignment of digits is





725



16



9624



10365

1299 + 1300, 1399 + 1400 1499 + 1500 and 1999 + 2000, i.e. 6 more numbers. Total numbers = 156.

18. (c) We can solve this question by elimination. Here we need to find the last 2 numbers. As per the given condition after putting the 3 numbers average is an integer hence the sum of three numbers must be divisible by 3.

Option (A)- in this case the 1st 3 numbers are 76, 80 and 91 and their sum 241 and it is not divisible by 3 hence it is ruled out. Similarly option B, D and E are eliminated. Option (c) is correct. Option (c)

23. (d) The maximum amount that he has invested is Rs 3000, on red horse, so let us start eliminating from Red horse.



No matter what rank White and Black take he cannot end up with No-Profit-No-Loss



Hence this condition ruled out.



If Red got the 3rd rank then his return = 3000





Continue filling the boxes from the bottom, we get the above one.

Now he can end up with No-Profit-No-Loss if lose his amount on other two horses, i.e. white or black come either 4th or 5th rank. This is a possible condition.



If Red got 4th or 5th rank in that case he will lose Rs 3000

68



So his gain from other two should be Rs 3000



That is possible in two cases

Y+29

9 Y

4

7

X+2

5

2

X

Y+29 16 9 Y

4

Case (i) White got 2nd rank then return = 3 x 2000 = 6000

X+9 7

5

X+2 2

X



Proceeding upward we get the above one. Then, Y+29 = 16+ X+ 9 / X – Y = 4



We can sum up this result in tabular form



Concept Cracker (CC) with XAT

If Red got 2nd rank then his return = 3 x 3000 = 9000

68



22. (e) From the solution of last question.



19. (c) We have two variables here and if we know the numbers at the bottommost row then we can calculate the numbers for any row.



21. (e) From the solution of last question.

And Black got either 4th or 5th rank.

Case (ii) White got 3rd rank then return = Rs 2000

And Black got 1st rank then return = 4000

Amount Invested

Rank in case (i)

Rank in case (ii)

Rank in case (iii)

Red

3000

3

4/5

4/5

White

2000

4/5

2

3

Black

1000

2

4/5

1

Now from this table we can conclude that option 4 is correct.

EBD_7743

300  Koncepts of Logical Reasoning 

24. (c) If Grey came 4th then case (ii) from the table is not possible, and white cannot come 2nd.

Puzzles 301 26. (d) As per the given conditions:

25. (c) 2

1

E1

E2

E3 7

6

Family

Possible no. of adults

Possible no. of children

1

0, 1, 2

3, 4, 5, ……….

II

0, 1, 2

3, 4, 5, ……….

III

0, 1, 2

3, 4, 5, ……….

Also given: Adults (A) > Boys (B) > Girls (G) > Families (F)

8

Going by the options, when we take the minimum number of families as 2, then the maximum possible number of adults can be 4.

9

F1 3

F 4

Let us assume that the three English men are E1, E2, E3 and the Frenchmen are F1, F2, F3, respectively. Let us assume that E2 is the only English man who knows French. Since our aim is to use the minimum number of calls, and there is only one possible communicator (i.e. E2) between the two groups, all the information to be exchanged between the two groups can be done with a single call from F2 to E2. Before this E2 should get the secrets from E1 and E3 and F2 should get the secrets from F1 and F3. Hence four calls are used till now, two in each group. (Shown in the diagram, from 1-4)

Since in any family, there should be at least 2 boys and 1 girl, which means 4 boys in 2 families. It violates our condition of adults > boys. When we take minimum number of families as 3, then all the conditions are satised. 27. (b) From the given information we can conclude for oak leaves: Non Red Spotted Red a

b

c

d

Similarly for maple leaves: Non Red e

Spotted f

Red g

h

After this, E2 calls up F2 and both exchange the secrets of their groups.

Here c must be even and positive, and d = h, a = 0, b = 5c f = 0, g = 6, e = 22

Now E2 and F2 know all the secrets. So now E2 uses two calls (to E1 and E3 calls 6 and 7 shown in the diagram) to convey the French secrets. Same is the case with F2.

Given that total = 50 hence: a + b + c + d + e + f + g + h = 50

Total minimum number of phone calls = (2 + 2) + 1 + (2 + 2) = 9 (as shown in the diagram).

From the above equation we can conclude that d + 3c = 11 But we know that c is even positive so c must be 2 If c = 2 and d = 5 Hence total oak leaves = a + b + c + d = 17.

EBD_7743

302 Koncepts of Logical Reasoning 28. (c) First Stage- There is two groups of 8 teams each. In each group, each team plays with every other team and hence total number of matches are 8c2 = 8 x 7/2 = 28 matches So, in both the groups the total number of matches played at the rst stage is 28. And hence 56 matches are played in the 1st stage. Second Stage- In this stage there are 8 teams playing in such a way that in one round 4 teams play with 4 other teams. 4 teams win and go to the next round. That is called knock out tournament. In the 1st round no of matches 8/2= 4, in the 2nd round no of matches = 4/2 = 2, in the third or the last round number of match =2/2=1, so total no of matches in 2nd stage is 4 + 2 + 1 = 7 Hence total match in the tournament = 56 + 7 = 63. 29. (b) From the given information we can complete the following table it is clear that if a team wins 5 games, then also there is no guarantee of its advancement to the next stage, since only 4 teams can go to the next stage.

31. (c) Since there are 8 teams and we know that 8 = 23 hence there are 3 rounds in the 2nd stage. Option (c) 32. (c) From the discussion above we can say that it is possible that the winner will have the same number of wins in the entire tournament as a team eliminated at the end of the rst stage. 33. (c) For every Mathematics book he has to carry 2 more physics books and points earned with the combination of these three books are 4 + 2 x 1 = 6 points (for 3 books). For every management book he has to carry 2 more ction books and points earned with this combination he will earn 3 + 2 x 2 = 7 points (for 3 books). So to maximize the points he will carry 2 sets of management book and one set of mathematics book and total points earned is 7 x 2 + 6 = 20 (for 9 books), now as mathematics and management book he cannot select as these book come in a set of 3 books so the 10th book he will select physics book and total points earned = 22 + 2 = 22. 34. (b) Palace Hut Fort House Hotel

Team

1

2

3

4

5

1

X

W

L

L

W

P (Blue and Red)

2

L

X

W

W

L

M (Yellow)

3

W

L

X

W

L

4

W

L

L

X

W

5

L

W

W

L

X

6

W

W

W

W

W

7

W

W

W

W

W

8

W

W

W

W

W

Note: In the table W → Wins, L → Loose, × → No match (as example there cannot be match between 1 and 1, 2 and 2 and so on). The above is one of such combinations. Since after winning 5 matches too, there is no guarantee to advancement, so the answer must be 6, because no two teams can get 7 points each. 30. (a) The team which gets 1 point at 1st stage would be eliminated because the combination may be 6 points for the team and 2 times each for remaining. There are some more cases that support the idea. Option (a)

×

U (Red or Blue)

×

T (Black)

×

X

×

×

×

×

√

From the table we have eliminated all four except M for Palace, hence M will stay in Palace 35. (c) Write down the information given L→ -D (Lion and Dear cannot be together) P→ -D and -Bs (A Panthor cannot be with a Dear or a Bison) Suman → -L Mohan → L + P Jack → P or B Hence we can short out following table

Keeper

Animal

Enclosure

Mohan

L + P (given)

X (given)

Jack

B+P

Y (given)

Shalini

B+L

Q

Suman

D+B

P

Rita

D+B

Z (given)

Since Z and P have the same pair of animals, the other 3 enclosures viz., X, Y and Q must contain the other 3 species within themselves. So apart from Lion and Panther, Bear must be third species in X, Y, Q.

Jack gets Bear and Panther as he doesn’t handle lion (given) that leaves. Q with Lion and Bear; which goes to Shalini as Suman doesn’t handle lion (given).

36. (b) When it carries 10 kg, then speed is 5 kmph so time taken 2hr, and cost for 2 hr is Rs. 20, so if we send 8 couriers together then it will transfer 80 kg and costs 20 x 8 = 160

When it carries 20 kg, then speed is 2 kmph so time taken 5hr, and cost for 5 hr is Rs 50,

  Puzzles 303  so if we send 4 couriers together then it will transfer 80 kg and costs 20 x 4 = 200

When it carries 40 kg, then speed is 1 kmph so time taken is 10 hours, and cost for 10 hours is Rs. 100, so if we send 2 couriers together then it will transfer 80 kg and costs 100 x 2 = 200



There are some more combinations but all of them has to be more than Rs. 160, hence Rs. 160 is the minimum.

Solutions (37 and 40) Students, please note that the best way to solve this question is by working backwards, e.g. after the 4th round, each one of them has Rs. 32. Since it is Vibha who lost in this round, all the remaining three must have doubled their share. In other words, they would have had Rs. 16 each after the 3rd round. Since the increase in Rs. 16 in ones share, i.e. Rs. 48 overall, comes from Vibha’s share, her share before the 4th round will be (32 + 48) = Rs. 80, after the 3rd round. Working backwards in this manner, we can get the following table.

Share of Each Suvarna

Tara

UMA

Vibha

4. Vibha

32

32

32

32

3. UMA

16

16

16

(32+48) = 80

2. Tara

8

8

(16+40+8+8) = 72

40

4 (4+34+18+10)=66

(8+4+36+20) = 68 34

36 18

20 10

1. Suvarna Initial

37. (c) Suvarna started with Rs. 66. 38. (d) It was Vibha who started with the lowest amount, viz. Rs. 10. 39. (a) It was Suvarna who started with the highest amount, viz. Rs. 66. 40. (a) At the end of the second round, Uma had Rs. 72. Solutions (41 and 42) Students please note that the best way to answer this question is by finding generally what would ensure a win for B. If B has to win, A has to pick up the last matchstick. This can be forced upon A if there are 2 or 3 matchsticks left on the table when it is B’s turn.



As then, B could pick up 1 or 2 matchsticks and force upon A to pick up the last one. For this to happen there should always be odd number of matchsticks initially, e.g. If there are 7 match sticks initially any of the following combinations will leave either 2 or 3 matchsticks on the table when it is B’s turn. A B B A 1 2

3

4

A B B A A 5

6

7

A A B A 1 2

3

4

1 2 3 4 5 6 7 A B B A A

5

6

7

1 2

3

4

5 6 7

Hence the smallest value of N (greater than 5) to ensure a win for B is 7. Also the largest value of N (less than 50) to ensure a win for B is 49.

EBD_7743

304 Koncepts of Logical Reasoning 41. (a)

42. (d)

43. (b) Given that,

44. (b) Let the three couples be represented as follows. (M1, F1), (M2, F2), (M3, F3)

X came from P to R. Y came from Q to R. Both of them arrived at R at the same time. We have to nd out, who among X and Y walked faster. From (1) alone as there is no information about the distance no relation can be made about their speeds. From (2) alone, the distance between Q and R is less than the distance between P and R. As there is no information about the total time taken, no relation between speeds can be made. Combining (1) and (2), there is no data given about the difference in distance between P and R and Q and R. Hence, the question cannot be answered. Option (b) is correct.

The following table represents the steps. People Set People on the People travelling People on the up side travelled from other other side side 1 M1 M2 M3 F3

F1 F2

-

-

2 M1 M2 M3 F3

-

F2

F1

3

M1 M3 F3

M2 F2

-

F1

4

M1 M3 F3

-

F1

M2 F2

5

M1 F1

M3 F3

-

M2 F2

6

M1 F1

-

F3

M2 F2 M3

7

F3

M1 F1

-

M2 F2 M3

8

F3

-

M3

M1 F1 M2 F2

9

-

M3 F3

-

M1 F1 M2 F2

So, 9 steps are required. Option (b) is correct

  Mathematical Reasoning

305 

20 Mathematical Reasoning

Exam

Importance

Exam

Importance

CAT

Very Important

IBPS/Bank PO

Important

XAT

Very Important

BANK Clerk

Important

IIFT

Very Important

SSC

Important

SNAP

Very Important

CSAT

Important

NMAT

Very Important

Other Govt. Exams

Important

Other Aptitude Test

Very Important

Application Of Percentage

Profit & Loss

If the original value P increased by x% followed by an increase of another y%, then final value Q is given by x  y   Q = P 1 + 1+  100   100 

• If cost price of two article is same .One of them is sold at x% profit while other at y% profit then overall profit is given by (x+y)/2. In case of loss just put –ve sign instead of +ve sign.

In this case overall percentage increase can be xy calculated as x + y + 100 Instead of 2 successive increases if there are 3 successive increases by x%, followed by y%, followed by z% then final value is given by x  y  z   Q = P 1 + 1+ 1+  100   100   100 

• Two successive discount of x% and y % is equivalent xy % to a single discount of x + y − 100 • If S.P of 2 articles are same but one of them was sold at P% profit while another one at P% loss then on overall transaction there will be a loss and Loss % = P2/100.

In each case value increased by Q-P Instead of two successive increase, one value decreases followed by an increase; then in that case just put (-) in place of (+), so in a case where x% increase followed by y% decrease is given by(just replace +y to –y) x  y   Q = P 1 + 1−  100   100 

And overall percentage change is given by

x (− y) xy = x− y− 100 100 x • If A is x% more than B then B is X 100% less 100 + x than A.

x + (− y) +

x • If A is x % less than B then B is X 100% more 100 − x than A. • If the price of a commodity increases reduction in consumption so that x remains constant is X 100%. 100 + x • If the price of a commodity decreases the increase in consumption so that x remains constant is X 100% 100 − x

by x%, then expenditure

by x%, then expenditure

If product of two quantities a and b is P that means P = ab

• One of them increases by x% and another one increases by y% then percentage change in product xy P = x+ y+ 100 • If decrease is given, then instead of increase in that case just put (–x) instead of x and (–y) instead of y. • If all the sides of a triangle, rectangle, any polygon (or any 2 dimensional figure) or radius of circle increased by x% then percentage change in area = 2 x +

x2 100

• If a trader uses a false weight of X gm instead of 1 kg and he keeps SP same as CP then his profit 1000 − X ×100 percentage is given by X

Average

• Average of the n numbers of a set is in between that smallest number of the set and the largest number of the set. • If a number k is added to each number of the set then their average will also increases by the same value k. • If a number k is subtracted to each number of the set then their average will also decreases by the same value k. • If a number k is multiplied to each number of the set then their average will also multiplied by the same value k. • If a number k( not equal to 0) is divided to each number of the set then their average will also divided by the same value k.

GEOMETRIC MEAN (GM) Geometric mean generally gives us the rate of growth. For n numbers x1, x2, ….xn geometric mean is the nth

root of the product of the numbers. That means GM = (x1. x2. ….xn)1/n

Geometric Mean (G.M) of two number a and b is (ab)1/2

HARMONIC MEAN (HM) Harmonic mean of a set of n numbers x1, x2, …..xn is calculated as n H.M = 1 1 1 + …. + x1 x2 xn

The harmonic mean H.M of two numbers a and b is 2ab/ (a+b)

EBD_7743

306  Koncepts of Logical Reasoning 

  Mathematical Reasoning • Relation between A.M., G.M. and H.M. A.M. ≥ G.M.≥ H.M.

of concequent and this property is called Alternendo.

PROPERTIES OF RATIO If we multiply or divide a ratio by the same number then the ratio remains same. a a aXk k So b = bXk = b k a c • The multiplication of two ratios and yield a b d ac new ratio . bd a c e then each ratio will become (sum of all • If = = b d f a+c+e . numerators)/(sum of all denominators) = b+d + f • Above rule is also true if we multiply the numerator and denominator of the same fraction by same number a c e g So if = = = then each ratio is equal to b d f h pa + qc + re + … pb + qd + rf + .. Provided all the numbers p, q, r are not zero. • Extension of above rule is that if all the ratio is raised to a power of n then… 1

 a n + c n + en  n a c e If = = then each ratio =  b n + d n + f n  b d f  

a c = then by adding 1 to both b d a+b c+d = this property is the ratio we will get b d called Componendo.

• Componendo. If

a c = then by subtracting 1 from b d both the ratio we will get

• Dividendo. If

a −b c−d = this property is called Divedendo. b d a c then by • Componendo and Dividendo. If = b d dividing the results of componendo and dividend

we will get

 pa n + qc n + ren  n Or each ratio =  pb n + qd n + rf n  provided all p, q,   r are not equal to zero.

a a2 • The duplicate ratio of = 2 b b a a3 = b b3



The triplicate ratio of



 a The sub duplicate ratio of a/b =   b

1/2

The sub triplicate ratio of a/b=(a/b)1/3 • If three quantities a, b and c are such that a: b:: b:c a b or = then these three numbers are in Continued b c proportion and b2=ac. a c b d = then = as per the • Invertendo. If b d a c property of Invertendo the inverse ratio of two equal ratios are also equal. a c a b • Alternando. If = = then = = if two ratios b d c d

a+b c+d = a−b c−d

Time & Work •

As number of persons involved in the work increases work done increases if time is constant, ie work done is directly proportional to number of persons involved in job if time remains constant, so w α m if time is constant.

•

If number of persons involved in job is constant then work done is directly proportional to time taken by them, so w α d when number of persons working the job is constant.



Adding the above two equations we will get w α md, or w=kmd where k is constant



So md/w= constant or m1d1/w1= m2d2/w2= m3d3/w3



Above formula is also known as MANDAYS formula. According to this formula mandays or product of number of man and number of days (time) require to complete a job is constant.

•

Above formula can be extended to m1d1h1/w1= m2d2h2/w2= m3d3h3/w3here h1 ,h2, and h3 is the number of working hours per day.

•

If work done is in fragment like digging a well with some length, breadth and depth then in that case work is taken as volume of well so work w=lbd or w1=l1b1d1 m1d1h1/l1b1d1= m2d2h2/l2b2d2

•

If two worker A and B can finish a job in p days when working together, while when working alone A can finish the same job in p+a days and when working alone B can finish the same job in p+b days then p=√ab

1



307 

are equal then ratio of antecedents are equal to ratio

EBD_7743

308  Koncepts of Logical Reasoning 

1

Concept Applicator

Directions for Question Nos. 1 to 2

 

In Patel Nagar police station, the requirements of constables who need to be on duty on weekdays are as follows: Monday – 2, Tuesday – 2, Wednesday – 2, Thursday – 2, Friday – 4, Saturday – 4 and Sunday – 3. There is a pool of constables who can be deployed. But the duty of a constable can start either on Friday or on Monday, who then works consecutively for five days. The police headquarters want minimum deployment of its manpower but at the same time it never compromises with requirements.

What is the minimum number of constables who should start duty on Monday?



(a) 0

(b) 1



(c) 2

(d) 3



(b) Sunday and Monday



(c) Sunday and Tuesday



(d) Current arrangement is the best

Indian Language Re25 lakhs search Center

2

(Q) New Course on Human Rights

27 lakhs

3

Research on Global Financial 22 lakhs Crisis

5

(R)

(S) New Course on Design

4.

Directions for Question Nos. 3 to 6

[2010]

3

4

5.

projects are as follows:

75 lakhs

57 lakhs



Central Delhi University has a special budget of one crore and five lakh rupees that it wants to distribute among its departments for innovative projects. Seven proposals have been received by the University, out of which some would be selected based on the funding requested and money available from the budget. An expert committee has evaluated all the projects and have given them scores on a scale of 1 to 5 (higher score implies better). A proposal can either be selected for full funding or dropped. The fund requests and expert committee scores for the seven

6.



1

(P) Disaster Management Center

3.

2. Instead of Fridays and Mondays on which other two days should constables start their 5-day long duty so that the objective of the police department is satisfied? (a) Friday and Saturday

(O)

Fund Score requested

Equipments for Astrophysics 35 lakhs Lab

(N) Biotech Centre

[2010]

1.



(M)

Proposals

65 lakhs

3

If the University has already decided to fund the Disaster Management Center, which are the other two projects that can also be selected? (a) O and Q (b) O and S (c) M and R (d) O and R If the University decides to select projects with total score between 7 and 9, which of the following combinations would require minimum total funding? (a) Q and R (b) O and R (c) M and R (d) O and Q If three projects are to be selected, which of the following combinations gives the maximum total score? (a) P, Q and R (b) O, Q and R (c) O, P and R (d) Q, R and S Ajay and Vijay work in the same office. Ajay sends a leave application for the day through Vijay. As per company policy, Ajay would receive a call from office if Vijay does not submit the leave application. Which of the following is true? (a)   Ajay receives a call from office, because the leave application was submitted. (b)  If Ajay does not receive a call from office, the leave application was not submitted. (c) Ajay receives a call from office, only if Vijay doesn’t submit the leave application. (d) None of the above

  Mathematical Reasoning

I. A has the highest amount amongst all persons which is Rs 40 more than what B has. II. C has Rs 30 less than what A has.

Directions for Question Nos. 7 to 9 A, B, C, D and E are five persons holding certain amount of money each (all different). When B, C, D and E exchanged their amounts amongst themselves so that no one had their original amount, it is observed that: I.

B possesses the highest amount amongst all persons.

II. D possesses the lowest amount amongst all persons, which is Rs 20 less than what A has. III. E and C possess Rs 50 and Rs 70 respectively. When A, C, D and E exchanged their initial amounts amongst themselves so that no one had their original amount, it is observed that:

2

309 

III. E has Rs 10 less than what D has. 7.

The initial amount possessed by A was:



(a) Rs. 80

(b) Cannot be determined



(c) Rs. 40

(d) Rs. 60

8.

The initial amount possessed by B was:



(a) Rs. 60

(b) Cannot be determined



(c) Rs. 40

(d) Rs. 50

9.

The initial amount possessed by C was:



(a) Rs. 80

(b) Cannot be determined



(c) Rs. 50

(d) Rs. 70

Concept Builder

Directions for Question Nos. 10 and 11

To get admission in a management course at Dadhichi Institute of Management (DIM) following criteria are given. A candidate must: 1. be a graduate from a recognized university with minimum 54 percent marks. 2. not be more than 33 years of age as on 1.4.2008. 3. have secured 60 percent or more marks in the entrance test. 4. pay one-time deposit fee of Rs. 2,00,000 at time of admission. 5. pay tuition fee of Rs. 4,000 per month. Any candidate who fails to fulfill the condition (4) at above, he/she may be referred to the chairmanadmission. Any candidate who has scored 80 percent mark in the entrance test but does not fulfill the condition (1) at above, he/she may be referred to the director. Any candidate having work experience of at least 10 years in supervisory cadre and does not satisfy the condition (2) at above, he/she may be admitted under sponsored quota. Given the above information and condition in each of the following questions, you have to decide which of the following course of action should be taken. You

should not assume anything in case of any of the candidates. Mark answer I. If the candidate is admitted II. If the candidate is not admitted III. If the candidate is referred to the director IV. If the candidate is referred to the chairmanadmission V. If the candidate is admitted under sponsor quota

[IIFT 2008]

10. Kamaljeet secured 60 percent marks in graduation and was born on 15th April 1976. He scored 56 percent marks in the entrance test. He can pay one-time deposit of Rs. 2,00,000 and monthly tuition fee of Rs. 4,000. (a) I (b) II (c) III (d) IV 11. Gourav is a first-class science graduate who obtained 81 percent marks in the entrance test. He has 12 years of work experience in supervisory cadre. He can pay the stipulated one-time deposit and monthly tuition fees. His date of birth is 20th October, 1970. (a) I (b) IV (c) III (d) V

24th July 2007. The film has been approved by the censor board of India.

Directions for Question Nos. 12 to 14 For selection of films produced before December 2007 for the national film festival of India, following criteria are given. 1. The film must be submitted to the National Film Development Corporation (NFDC) by 31.10.2007. 2. The production cost of the film should not exceed rupees five crore. 3. The director of the film should have passed a three rear course either from the Film and Television Institute of India (FTII) or from Satyajit Ray Film & Television Institute. 4. The length of the film should not exceed 150 minutes. 5. The film must have been approved by the film censor board of India. 6. However, if the film fulfils all the above criteria except (a) criteria 2 above, it must be sent to the finance secretary (b) criteria 3 above, the director has done at least a one year course from FTII or Satyajit Ray Film & Television Institute, the film is kept as a stand-bye On the basis of above information and information provided below, decide the course of action in each case. No further information is available. You are not to assume anything. Mark answer: I. if the film is to be selected II. if the film is not to be selected III. if the film should be sent to the kept finance secretary IV. if the film should be as a stand-bye V. if the data given about the film are not adequate to make a decision.

[IIFT 2008]

12. Film Dainandini was produced at the cost of Rupees 2.5 crore. It was submitted to the NFDC on 29th September 2007. The director of the film Govind Chadha passed a 3-year course from FTII. Length of film was 120 minutes and has been approved by the censor board of India.

(a) I

(b) II



(c) IV

(d) V

13. Bhadrasalam is a 135-minute film directed by Katyani, who was a student of Satyajit Ray Film & Television Institute from 1996 to 1999. The cost of producing the film was Rupees 2.3 crore and it was submitted to NFDC on



(a) I (c) III

(b) V (d) IV

14. Rakesh Mohan, the director of film Ek Bar Achanak, has successfully completed a 2-years course at Satyajit Ray Film & Television Institute. The 150-minute film was produced at rupees 4.85 crore. It has approved by the censor board of India and submitted to NFDC on 30th Nov. 2007. (a) I (b) III (c) IV (d) II Directions for Question Nos. 15 to 18 Four persons (1) Mohit, (2) Monohar, (3) Prasant and (4) Dinesh each had some initial money with them. They all were playing bridge in a way that the loser doubles the money of each of the other three persons from his share. They played four rounds and each person lost one round in the order 1, 2, 3 and 4 as mentioned above. At the end of fourth round each  

person had Rs. 32000/-

[IIFT 2007]

15. What was the amount with Mohit to start with? (a) Rs. 60000 (b) Rs. 34000 (c) Rs. 66000 (d) Rs. 28000 16. What was the amount with Monohar at the end of first round? (a) Rs. 68000 (b) Rs. 72000 (c) Rs. 64000 (d) Rs. 80000  17. Who had the lowest amount at any round of play throughout the tournament? (a) Mohit (b) Monohar (c) Prasant (d) Dinesh 18. What was the amount with Prasant at the end of the second round? (a) Rs. 36000 (b) Rs. 72000 (c) Rs. 16000 (d) Rs. 68000 19. The VC (Vice-Chancellor) of a university has to select four professors, out of eight professors for a committee. The VC decided to select these four professors in such a manner that each selected professor has a habit common with at least one of the other three professors selected. The selected professors must also share at least one of the non-common habits of any of the other three professors selected.

EBD_7743

310  Koncepts of Logical Reasoning 

  Mathematical Reasoning

Professor Arora likes surfing and smoking but hates gambling.



Professor Bhalla likes smoking and drinking but hates surfing.



Professor Chadha likes gambling but hates smoking.



Professor Dhyani likes movie but hates drinking.



Professor Eswar likes drinking but hates smoking and movie.



Professor Fazil likes surfing but hates smoking and movie.



Professor Goyal likes gambling and movie, but hates surfing.



Professor Hooda likes smoking and gambling but hates movie.

 

Who are the four professors selected by the VC for the committee?

[IIFT 2007]



(a) Prof. Chadha, Prof. Dhyani, Prof. Eswar, Prof. Goyal



(b) Prof. Arora, Prof. Bhalla, Prof. Eswar, Prof. Fazil



(c) Prof. Bhalla, Prof. Chadha, Prof. Goyal, Prof. Hooda



(d) Prof. Dhyani, Prof. Eswar, Prof. Fazil, Prof. Hooda Directions for Question Nos. 20 to 23 While selecting candidates for positions of engineers, a software firm followed criteria as given below. A candidate must: i. be an engineering graduate with at least 60% marks at degree and 80% marks at higher secondary level. ii. have at least one year’s experience of working iii. be ready to sign a bond of three yeas iv. must not be more than 28 years of age on 1.2.2007 However, if a candidate fulfills all the criteria except– (a) at (i) above but has obtained 50% marks at degree and 70% at higher secondary respectively and has at least three years experience of working, the case may be referred to the director of the firm. (b) at (iii) above, but is willing to pay an amount of 1 lakh if required to leave, the case may be referred to the president of the firm. (c) at (ii) above but is a computer engineer, the case may be referred to DGM.

311 

Based on the above criteria and the information given in each of the following cases, you have to take the decision on employing a candidate. You are not to assume anything and in the absence of adequate information, your answer will not be selected. The case is given to you as on 1.2.2007. The options available for you are provided in A, B, C and D. [IIFT 2007] 20. Amar is a mechanical engineer with 65% marks at degree and 88% marks at HSC. He completed his engineering degree in 2003 at the age of 22 years and immediately started working in an engineering firm. He is keenly interested in going to USA and is not ready to sign a bond. However, he does not mind paying an amount of Rs. 1 lakh: (a) if the case is to be selected (b) if the case is not be selected (c) if the case is to be referred to Director (d) if the case is to be referred to President 21. Rajkishore, a computer engineer, has just completed graduation in July 2006, at the age of 23 years obtaining 72% marks. He had obtained 92% marks in HSC. He is willing to sign a bond with the company. He has joined a software company as trainee in August 2006 and working there till date: (a) if the case is to be selected (b) if the case is not to be selected (c) if the case is to be referred to Director (d) if the case is to be referred to President 22. Madhuri is an electrical engineer and working as an assistant engineer for past two years. She had secured 85% and 69% marks at HSC and degree respectively. She has just completed 25 years of age. (a) if the case is to be selected (b) if the case is not to be selected (c) if the case is to be referred to Director (d) if the case is to be referred to President 23. Kamla is an engineering graduate with 66% marks at degree and 90% at HSC. She has joined engineering firm 2 years ago at the age of 24 years. She is ready to sign the bond: (a) if the case is to be selected (b) if the case is not be selected (c) if the case is to be referred to Director (d) if the case is to be referred to President

EBD_7743

312  Koncepts of Logical Reasoning 

3

Concept Cracker

Directions for Question Nos. 24 to 26

Alex Company has its office at the third floor of a multi-storied building in Mumbai. There are 5 rooms to be allotted to 5 managers (designated Ml to M5), each of whom will occupy one room. Each room has its own advantages and disadvantages. Some have the sea view, while others are closer to either the lift or the dining room, while some are more spacious. Each of the five managers was asked to rank the room preferences amongst the rooms 301, 302, 303, 304 and 305. Their preferences were recorded as follows: Preference

M1

M2

M3

M4

M5

1st

302

302

303

302

301

2nd

303

304

301

305

302

3rd

304

305

304

304

305

301

305

303

4th 5th

302

Some managers did not indicate any preference for some rooms, as they did want to he there under any condition. The company decided to allot rooms to managers in such a way that so that the managers get rooms as per their best preference or close to that.

[XAT 2011]

24. How many managers would get their rooms as per their best preference? (a) 1 (b) 2 (c) 3 (d) 4 (e) 5 25. If manager X gets his/her 1st choice, then his /her preference ranking is I and so on. Management decided to allot rooms so that the sum of the preference rankings of all the managers is minimized. What is the total preference ranking for the rooms allotted to all the managers? (a) 5 (b) 6 (c) 7 (d) 8 (e) 9 26. Suppose that Manager M2 decides not to join the new zonal office and Manager M6 takes his place. Manager M6 has the following preference ranking in decreasing order: 301.



303, 302, 304 — in this case what would be the sum of the preference rankings allotted to all the five managers? (a) 5 (b) 6 (c) 7 (d) 8 (e) 9 Directions for Question Nos. 27 to 29 Mrs. Sharma has a house which she wants to convert to a hostel and rent it out to students of a nearby women’s college. The house is a two storey building and each floor has eight rooms.. When one looks from the outside, three rooms are found facing North, three found facing East, three found facing West and three found facing South. Expecting a certain number of students Mrs. Sharma wanted to follow certain rules while giving the six teen rooms on rent: All sixteen rooms must he occupied. No room can be occupied by more than three students. Six rooms facing north is called north wing. Similarly six rooms facing east, west and south are called as east wing, west wing and south wing. Each corner room would be in more than one wing. Each of the wings must have exactly I students. The first floor must have twice as many students as the ground floor. However, Mrs. Sharma found that three fewer students have come to rent the rooms. Still, Mrs. Sharma could manage to allocate the rooms according to the rules. [XAT 2011]

27. How many students turned up for renting the rooms? (a) 24 (b) 27 (c) 30 (d) 33 (e) None of the above 28. If Mrs. Sharma allocates the north-west corner room on the ground floor to 2 students, then the number of students in the corresponding room on the first floor, and the number of students in the middle room in the first floor of the east wing are: (a) 2 and 1 respectively (b) 3 and I respectively (c) 3 and 2 respectively (d) Both should have 3 students (e) Such an arrangement is not possible.

  Mathematical Reasoning 29. If all the students that Mrs. Sharma expected initially had come to rent the rooms, and if Mrs. Sharma had allocated the northwest corner room in the ground floor to 1 student, then the number of students in the corresponding room on the first floor, and the number of students in the middle room in the first floor of the east wing would have been: (a) 1 and 2 respectively (b) 2 and 3 respectively (c) 3 and 1 respectively (d) Both should have 2 students. (e) Such an arrangement is not possible Directions for Question Nos. 30 and 31 Every Saturday, the members of Raja Harish Chandra Club meet in the evening. All the members of the club are honest and never lie. Last Saturday, the following conversation was heard at one of the tables with five members sitting around it.

4

313 

Satya Sadhan: In this club not all members are friends with each other. Satyabrata: None of the pair of friends in this club has any common friend. Satyajit: Every pair of members who are not friends has exactly two common friends in this club. Satya Pramod: There are fewer than 22 people in this club.

[XAT 2011]

30. How many members are there in the club?

(a) 5

(b) 7



(c) 15

(d) 16



(e) 20

31. How many friends does Satya Sadhan have in the club?

(a) 2

(b) 3



(c) 5

(d) 7



(e) 8

Concept Deviator

Directions for Question Nos. 32 to 35

A health-drink company’s R & D department is trying to make various diet formulations, which can be used for certain specific purposes. It is considering a choice of 5 alternative ingredients (O, P, Q, R and S), which can be used in different proportions in the formulations. The table below gives the composition of these ingredients. The cost per unit of each of these ingredients is O:150, P: 50, Q: 200, R: 500, S:100. [CAT 2007] Composition Ingredient Carbohydrate Protein Fat Minerals (%) (%) (%) (%) O 50 30 10 10 P

80

20

0

0

Q R

10

30

50

10

5

50

40

5

S

45

50

0

5

32. Which among the following is the formulation having the lowest cost per unit for a diet having 10% fat and at least 30%protein? The diet has to be formed by mixing two ingredients. (a) P and Q (b) P and S (c) P and R (d) Q and S (e) R and S 33. In what proportion P, Q and S should be mixed to make a diet having at least 60% carbohydrate at the lowest per unit cost? (a) 2:1:3 (b) 4:1:2 (c) 2:1:4 (d) 3:1:2 (e) 4:1:1 34. The company is planning to launch a balanced diet required for growth needs of adolescent children. This diet must contain at least 30% each of carbohydrate and protein, no more than 25% fat and at least 5% minerals. Which

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314  Koncepts of Logical Reasoning 

one of the following combination of equally mixed ingredients is feasible? (a) O and P (b) R and S (c) P and S (d) Q and R (e) O and S

35. For a recuperating patient, the doctor recommended a diet containing 10% minerals and at least 30% protein. In how may different ways can we prepare this diet by mixing at least two ingredients? (a) One (b) Two (c) Three (d) Four (e) None

On the fifth day, E co-authored a paper with F which reduced the group‘s average Erdös number by 0.5. The Erdös numbers of the remaining six were unchanged with the writing of this paper.  No other paper was written during the conference. 

[CAT 2006]

36. The person having the largest Erdös number at the end of the conference must have had Erdös number (at that time):

Directions for Question Nos. 36 to 40



(a) 5

(b) 7

Mathematicians are assigned a number called Erdös number, (named after the famous mathematician, Paul Erdös). Only Paul Erdös himself has an Erdös number of zero. Any mathematician who has written a research paper with Erdös has an Erdös number of 1. For other mathematicians, the calculation of his/ her Erdös number is illustrated below: Suppose that a mathematician X has co-authored papers with several other mathematicians. From among them, mathematician Y has the smallest Erdös number. Let the Erdös number of Y be y. Then X has an Erdös number of y + 1. Hence any mathematician with no co-authorship chain connected to Erdös has an Erdös number of infinity.  In a seven day long mini-conference organized in memory of Paul Erdös, a close group of eight mathematicians, call them A, B, C, D, E, F, G and H, discussed some research problems. At the beginning of the conference, A was the only participant who had an infinite Erdös number. Nobody had an Erdös number less than that of F.  On the third day of the conference F co-authored a paper jointly with A and C. This reduced the average Erdös number of the group of eight mathematicians to 3. The Erdös numbers of B, D, E, G and H remained unchanged with the writing of this paper. Further, no other coauthorship among any three members would have reduced the average Erdös number of the group of eight to as low as 3.  At the end of the third day, five members of this group had identical Erdös numbers while the other three had Erdös numbers distinct from each other.



(c) 9

(d) 14



(e) 15

37. How many participants in the conference did not change their Erdös number during the conference?

(a) 2

(b) 3



(c) 4

(d) 5



(e) Cannot be determined

38. The Erdös number of C at the end of the conference was:

(a) 1

(b) 2



(c) 3

(d) 4



(e) 5

39. The Erdös number of E at the beginning of the conference was:

(a) 2

(b) 5



(c) 6

(d) 7



(e) 8

40. How many participants had the same Erdös number at the beginning of the conference?

(a) 2

(b) 3



(c) 4

(d) 5



(e) Cannot be determined

  Mathematical Reasoning Directions for Question Nos. 41 to 44 The year is 2089. Beijing, London, New York, and Paris are in contention to host the 2096 Olympics. The eventual winner is determined through several rounds of voting by members of the IOC with each member representing a different city. All the four cities in contention are also represented in IOC. In any round of voting, the city receiving the lowest number of votes in that round gets eliminated. The survivor after the last round of voting gets to host the event. A member is allowed to cast votes for at most two different cities in all rounds of voting combined. (Hence, a member becomes ineligible to cast a vote in a given round if both the cities (s)he voted for in the earlier rounds are out of contention in that round of voting.) A member is also ineligible to cast a vote in a round if the city (s)he represents is in contention in that round of voting. As long as the member is eligible, (s)he must vote and vote for only one candidate city in any round of voting. The following incomplete table shows the information on cities that received the maximum and minimum votes in different rounds, the number of votes cast in their favour, and the total votes that were cast in those rounds. [CAT 2005] Round

Total Votes cast

1 2

83

3

75

Maximum votes cast No. of City votes

43. What percentage of members from among those who voted for Beijing in round 2 and were eligible to vote in round 3, voted for London? (a) 33.33 (b) 38.10 (c) 50 (d) 66.67 44. Which of the following statements must be true? (a) IOC member from New York must have voted for Paris in round 2.

London

30

New York

Paris

32

Beijing

Coach John sat with the score cards of Indian players from the 3 games in a one-day cricket tournament where same set of players played for India and all the major batsmen got out. John summarized the batting performance through three diagrams, one for each game. In each diagram, the three outer triangles communicate the number of runs scored by the three top scorers from India, where K, R, S, V, and Y represent kaif, Rahul, Saurav, Virender and Yuvraj respectively. The middle triangle in each diagram denotes the percentage of total score that was scored by the top three Indian scorers in that game. No two players score the same number of runs in the same game. John also calculated two batting indices for each player based on his scores in the tournament; the R-index of a batsman is the difference between his highest and lowest scores in the 3 games while the M-index is the middle number, if his scores are arranged in a non-increasing order.

No of votes 12 21

It is also known that: All those who voted for London and Paris in round 1, continued to vote for the same cities in subsequent rounds as long as these cities were in contention. 75% of those who voted for Beijing in round 1, voted for Beijing in round 2 as well. Those who voted for New York in round 1, voted either for Beijing or Paris in round 2. The difference in votes cast for the two contending cities in the last round was 1. 50% of those who voted for Beijing in round 1, voted for Paris in round 3.

[CAT 2005]

41. What percentage of members from among those who voted for New York in round 1, voted for Beijing in round 2? (a) 33.33 (b) 50 (c) 66.67 (d) 75

(b) IOC member from Beijing voted for London in round 3. (a) Only (a) (b) Only (b) (c) Both (a) and (b) (d) Neither (a) nor (b) Directions for Question Nos. 45 to 48

Eliminated City

315 

42. What is the number of votes cast for Paris in round 1? (a) 16 (b) 18 (c) 22 (d) 24

[CAT 2004] Y(40)

K(51)

90% V(130)



R(55)

70% K(28)

Pakistan

S(75)

80% R(49)

South Africa

Y(87)

S(50)

Australia

45. How many players among those listed definitely scored less than Yuvraj in the tournament? (a) 0 (b) 1 (c) 2 (d) More than 2

46. Which of the players had the best M-index from the tournament? (a) Rahul (b) Saurav (c) Virender (d) Yuvraj 47. For how many Indian players is it possible to calculate the exact M-index? (a) 0 (b) 1 (c) 2 (d) More than 2

52. While Balbir had his back turned, a dog ran into his butcher shop, snatched a piece of meat off the counter and ran out. Balbir was mad when he realised what had happened. He asked three other shopkeepers, who had seen the dog, to describe it. The shopkeepers really didn’t want to help Balbir. So each of them made a statement which contained one truth and one lie.

48. Among the players mentioned, who can have the lowest R-index from the tournament? (a) Only Kaif, Rahul or Yuvraj (b) Only Kaif or Rahul (c) Only Kaif or Yuvraj (d) Only Kaif Directions for Question Nos. 49 to 51 The table below provides certain demographic details of 30 respondents who were part of a survey. The demographic characteristics are: gender, number of children, and age of respondent. The first number in each cell is the number of respondents in that group. The minimum and maximum age of respondents in each group is given in brackets. For example, there are five female respondents with no children and among these five, the youngest is 34 years old, while the oldest is 49.

[CAT 2004]

No. of Children

Male

Female

Total

0

1(38,38)

5(34, 49)

6

1

1(32, 32)

8(35, 57)

9

2

8(21. 65)

3(37, 63)

11

3

2(32, 33)

2(27, 40)

4

Total

12

18

30

49. The percentage of respondents aged less than 40 years is at least (a) 10% (b) 16.67% (c) 20.0% (d) 30% 50. Given the information above, the percentage of respondents older than 35 can be at most (a) 30% (b) 73.33% (c) 76.67% (d) 90% 51. The percentage of respondents that fall into the 35 to 40 years age group (both inclusive) is at least (a) 6.67% (b) 10% (c) 13.33% (d) 26.67%





Shopkeeper number 1 said: “The dog had black hair and a long tail.”



Shopkeeper number 2 said: “The dog had a short tail and wore a collar.”



Shopkeeper number 3 said: “The dog had white hair and no collar.”

Based on the above statements, which of the following could be a correct description? (a) The dog collar. (b) The dog collar. (c) The dog collar. (d) The dog collar.

[CAT 2001] had white hair, short tail and no had white hair, long tail and a had black hair, long tail and a had black hair, long tail and no

Directions for Question Nos. 53 to 55 A and B are two sets (e.g. A = mothers, B = women). The elements that could belong to both the sets (e.g., women who are mothers) is given by the set C = A.B. The elements which could belong to either A or B, or both, is indicated by the set D = A È B. A set that does not contain any elements is known as a null set, represented by j (for example, if none of the women in the set B is a mother, then C = A.B is a null set, or C = j).  Let ‘V’ signify the set of all vertebrates; ‘M’ the set of all mammals; ‘D’ dogs; ‘F’ fish; ‘A’ Alsatian and ‘P’, a dog named Pluto.

[CAT 2001]

53. Given that X = M.D is such that X = D, which of the following is true? (a) All dogs are mammals. (b) Some dogs are mammals. (c) X = j. (d) All mammals are dogs.

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316  Koncepts of Logical Reasoning 

  Mathematical Reasoning 54. If Y = F.(D.V), is not a null set, it implies that

(a) All fish are vertebrates



(b) All dogs are vertebrates.         



(c) Some fish are dogs.



(d) None of the above.

Directions for Question Nos. 59 to 61 Recently, Ghosh Babu spent his winter vacation on Kyakya Island. During the vacation, he visited the local casino where he came across a new card game. Two players, using a normal deck of 52 playing cards, play this game. One player is called the Dealer and the other is called the Player. First, the Player picks a card at random from the deck. This is called the base card. The amount in rupees equal to the face value of the base card is called the base amount. The face values of Ace, King, Queen and Jack are ten. For other cards, the face value is the number on the card. Once, the Player picks a card from the deck, the Dealer pays him the base amount. Then the dealer picks a card from the deck and this card is called the top card. If the top card is of the same suit as the base card, the Player pays twice the base amount to the Dealer. IF the top card is of the same colour as the base card (but not the same suit) then the Player pays the base amount to the Dealer. If the top card happens to be of a different colour than the base card, the Dealer pays the base amount to the Player. Ghosh Babu played the game 4 times. First time he picked eight of clubs and the Dealer picked queen of clubs. Second time, he picked ten of hearts and the dealer picked two of spades. Next time, Ghosh Babu picked six of diamonds and the dealer picked ace of hearts. Lastly, he picked eight of spades and the dealer picked jack of spades. Answer the following questions based on these four games.

55. If Z = (P.D) È M, then

(a) The elements of Z consist of Pluto the dog or any other mammal.



(b) Z implies any dog or mammal.



(c) Z implies Pluto or any dog that is a mammal.



(d) Z is a null set.

56. If P.A =  j  and P  È  A = D, then which of the following is true?

(a) Pluto and alsatians are dogs.



(b) Pluto is an alsatian.



(c) Pluto is not an alsatian.



(d) D is a null set. Directions for Question Nos. 57 and 58 For any activity, A, year X dominates year Y if IT business in activity A, in the year X, is greater than the IT business, in activity A, in the year Y. For any two IT business activities, A & B, year X dominates year Y if i. the IT business in activity A, in the year X, is greater than or equal to the IT business, in activity A in the year Y, ii. the IT business in activity B, in the year X, is greater than or equal to the IT business in activity B in the year Y, and iii. there should be strict inequality in the case of at least one activity.

[CAT 2000]

57. For the IT hardware business activity, which one of the following is not true?

(a) 1997-98 dominates 1996-97



(b) 1997-98 dominates 1995-96



(c) 1995-96 dominates 1998-99



(d) 1998-99 dominates 1996-97

58. For the two IT business activities, hardware and peripherals, which one of the following is true?

(a) 1996-97 dominates 1995-96



(b) 1998-99 dominates 1995-96             



(c) 1997-98 dominates 1998-99



(d) None of these

317 

[CAT 1999] 59. If Ghosh Babu stopped playing the game when his gain would be maximized, the gain in Rs. would have been

(a) 12

(b) 20



(c) 16

(d) 4 

60. The initial money Ghosh Babu had (before the beginning of the game sessions) was Rs.X. At no point did he have to borrow any money. What is the minimum possible value of X?

(a) 16

(b) 8



(c) 100

(d) 24

61. If the final amount of money that Ghosh Babu had with him Rs.100, what was the initial amount he had with him?

(a) 120

(b) 8



(c) 4

(d) 96

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318  Koncepts of Logical Reasoning 

Solutions Concept Applicator (CA) Solutions (1 and 2) From the given information- Number of required constables on weekdays are→ Monday – 2, Tuesday – 2, Wednesday – 2, Thursday – 2, Friday – 4, Saturday – 4 and Sunday – 3 It is also given that the duty of a constable can start only on Monday or Friday and each constable has to work for 5 consecutive days, two constables can start their duty on Friday. Hence on Monday, 0 or 1 constable cannot start their duty because there is a need of more number of constables on Wednesday and Thursday. So, two constables can start duty on Monday and four constables can start duty on Friday. Then we can complete the table as followsDay On which duty start Day on which duty end Friday

Tuesday

Saturday

Wednesday

Sunday

Thursday

Tuesday

Saturday

1. (c) From the above result answer is Option (c). From the above result and the given options 2. (d) only option d satisfy all the condition. Solutions (3 to 6) 3. (d) As per the given data the Disaster Management Center (P) which requests for a fund of 57 Lakhs is already selected, the combination of the other two projects should have a request for total fund less than or equal to (105 – 57) lakhs i.e. 48 lakhs. Hence project O and R would be selected. 4.

(b) As per the given information University selects projects with total score between 7 and 9 i.e. a total score of 8. Hence Hence project O and R would be selected.

5. (c) Project O, P and R atisfy all the condition.

The given condition is “If Vijay does not 6. (c) submit the leave application, then Ajay will get a call”. This is the proposition ‘If p, then q’. Then the logical conclusion we can draw is – p→ Q or – Q → -X i.e. If Ajay does not get a call, then Vijay has submitted the application. This is not mentioned in any of the answer options. Solutions (7 to 9) 7.

(c) Money distributed among A/B/C/D/E.



After two times exchange A/B/C/D/E. any one get Rs. 50 /Rs.70



As per condition highest = C/ D/ E, lowest =B/ C/ E



Their differences = 10, 20, 30, 40,



And difference between C and A is 30 and D and E Rs.10



So, A / B/ C/ D/ E = 40/70/50/30/20



But B =70 impossible, Only choice 3) is possible.

8.

(b) Initial amount of B cannot be determined from above solution.

9.

(c) Initial amount of C must be RS.50 because, RS.80 is impossible and C got 70 after 2nd exchange. Concept Builder (CB)

Solutions (10 and 11) 10. (b) Kamaljeet satisfied all the conditions except the third so he cannot be admitted. 11. (d) Gourav satisfies criteria 1, 3, 4 and 5. He does not satisfy the 2nd criterion, but has 12 years of work experience in supervisory cadre. So, as per the given condition he should be admitted under sponsored quota. Solutions (12 to 14) 12. (a) From the given conditions we can observe that Film “Dainandini” satisfies all the criteria given here. So, it should be selected.

  Mathematical Reasoning 13. (b) Film “Bhadrasalam” satisfies criteria 1, 2, 4 and 5. But have a look on criteria 3 it is not clearly given that the director of the film Katyani had passed the three year course (or) not. So, the film is not to be selected.



14. (d) It does not satisfy criterion 1, although it satisfy rest all criteria, So, the film is not to be selected. Solutions (15 to 18) This type of question should be solved by back calculation method Mohit

Manohar Prashant

Round – 4

32

32

32

Round – 3

16

16

16

Round – 2

8

8

Round – 1

4

Dinesh 32

Initially All values given in the table is in thousand(k) This is the primary table that we can make from the given information. Since Mohit had 32k at the end of 4th round and he won the 4th, 3rd and 2nd round hence his amount must be as shown in the table. Now consider for Dinesh he lost in 4th round so he has to pay 16+16+16 =48k to each of other players and then he left with 32k hence amount with him after 3rd round = 32+48k = 80k then his amount in the subsequent round must be 40k, 20k and finally 10k Manohar Prashant

Dinesh

Round – 4

32

32

32

32

Round – 3

16

16

16

80

Round – 2

8

8

72

40

Round – 1

4

68

36

20

Initially

66

34

18

10

15. (c) 16. (a) 17. (a)

Now consider condition (ii) it is given that in 2003, he had started working for an engineering firm but we have no information on the duration of his employment in that firm, hence we cannot conclude that he satisfies the condition of having at least one year’s experience which is required as per condition (ii). Due to lack of this information, and it is given in instruction not to assume anything, hence he should not be selected. Hence (B) is the correct answer.

21. (b) In this case Rajkishore satisfies the conditions in (i), (iii) and (iv) but he does not satisfy the condition in (ii) but as he is a computer engineer so his case should be referred to DGM. As none of the answer options mention this, none of them is correct. 22. (b) Here in ths case Madhuri satisfies the conditions in (ii) and (iv) but She does not satisfy the condition in (i) and as she has scored less than 70% marks in her HSC exam.

But look at the condition (iii) nothing is said about that hence she should not be selected.

23. (a) Here Kamla satisfies all the conditions hence she should be selected.

Similarly we can complete the table for others Mohit

Concept Cracker (CC) with XAT 24. (c) It is given that three managers have given the same room as their best preference hence only one among the three and the two remaining managers can be given rooms as per their best preference.  Hence option C is correct. 25. (c) For sum of the preference rankings of all the managers to be minimum, the allotment has to be done in the following way-

18. (b)

Room No

Manager

19. (b) From the given information we can conclude the right combination is Prof. Arora, Prof. Bhalla, Prof. Eswar, Prof. Fazil.

301

M5

302

M1

303

M3

Solutions (20 to 23)

304

M2

305

M4

20. (b) Here in this case Amar satisfies the conditions in (i) and (iv) but not (iii) insead he is willing to pay an amount of Rs.1 lakh, if required to

319 

leave the software firm, hence his case could have been referred to the “President” of the firm.



The sum of the preference rankings

= 1 + 2 +1 + 2 +1 = 7

26. (e) As per the given condition if M6 takes the place of M2, one of the way of allotting rooms, so that sum of the preference rankings is the minimum is as follows Room No

Manager

301

M6

302

M4

303

M3

304

M1

305

M5

= 3 + 1 + 1+ 1 + 3 = 9

27. (b) As per the given condition there are 8 rooms on 1st floor and 8 on the 2nd. If total number of students are 27 and number of students on the 1st floor was double that on ground floor, hence number of students on the ground floor is 9 and that on the 1st floor is 18. Ground floor – 9 students

W Wing-1 W/S Wing-1

N/E Wing-1 E Wing-1

S Wing-1

S/E Wing-1

1st floor- 18 students N/W Wing-3

N Wing-1

W Wing-1 W/S Wing-3

N/E Wing-3 E Wing-1

S Wing-2

N Wing-1

W Wing-1

N/W Wing-2



N Wing-1

N/W Wing-1

N/E Wing-2 E Wing-1

S Wing-1

S/E Wing-1

1st floor- 20 students

The sum of the preference rankings

N/W Wing-2

Ground floor – 10 students

W/S Wing-2







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320  Koncepts of Logical Reasoning 

S/E Wing-3

28. (b) It is given that the north-west corner room is allotted to 2 students then as in the diagram in the previous question, the number of students in the corresponding room on the first floor and the number of students in the middle room in the first floor of the east wing (as east wing consists of room facing east, it would be at the west) would be 3 + 1 (the circled values in the diagram for the previous question. 29. (b) Had all the 30 students come, the arrangement according to all the conditions would have been as fallows

N Wing-3

W Wing-3 W/S Wing-2

N/E Wing-2 E Wing-3

S Wing-3

S/E Wing-2

Hence option B is correct.

Solutions (30 and 31) (This question can be treated as toughest question of this chapter) We will represent a person by a point and the relation of being friends by joining those points by a straight line. Now we will correlate the given information in graphical format(1) None of the pair of friends in this club has any common friend or graphically we can say there are no tri-

angles in the graph.

(2) Every pair of members who are not friends has exactly two common friends or graphically we can say this

pattern is a quadrilateral.

Consider any point (a person) X and let it is connected with m points (i.e he has m friends) and these points are A1, A2,………. Am. As per the given condition none of these points can be connected with each other. Therefore, corresponding to every pair Ai, Aj there have to be exactly two points that are connected to both one of them is X and other is say Bij. This has to be different B for different values of (i,j) otherwise Bij and X would have more than two common friends as they have already 2 common friends Ai, and Aj. Hence all the points connected to X (an Ai) or a point connected to Ai(Bij). If X and P are not connected (i.e non friends) they have to have only two common friends but all the friends of X are included in A1, A2,…………Am hence P has to be connected with some of the A’s. Also, there can be no other point, (other than X and the Bij’s) connected with any Ai. If there is such a point B, it can’t be connected to any Aj (j ≠ i) for it would then be a Bij. Thus X and B would have only 1 common friend.

  Mathematical Reasoning

321 

Now we need to determine the number of Bij we can list down it as



Option 2- P & S, None of them contain Fat hence this option also eliminated.

B12, B13, …. B1m = total (m-1) points



Option 3- P & R, these two must be mixed in the ratio of 3:1 to get Fat =10%, and then protein content is 110/400 < 30 hence eliminated



Option 4- Q & S they must be mixed in the ratio of 1:4 to get 10% Fat, in that case proteincontent =230/500>30 hence this combination is allowed. And then cost per unit would be(200+4x100)/500=6/5



Option 5- R & S, they must be mixed in the ratio of 1:3 to get 10% fat, then protein content will be (150+50) /400 ½ >30% hence this combination is also allowed. The cost per unit in this case = 800/400 =2 >6/5, hence option 4 gives us the minimum cost.

B23, B24, B2m = total (m-2) points Similarly last point would be B(m-1)m = 1 point Hence total number of Bij = 1+2+…..+ (m-1)

= m(m-1)/2

The points in the graph X,A1,A2,……….Am  B12,B13,… ………..B(m–1)m. Total number of Ai = m points The total number of points is n = 1 + m + m(m-1)/2. Here we have taken X as an arbitrary point and it has m friends, hence other points should also have m friends. Let consider a point B12 it should also have m friends. Out of total Bij some of them cannot be friends of B12 these points are B12, B13, …. B1m = total (m-1) points B23, B24, B2m = total (m-2) points The reason of this elimination is let B25 a friend of B12 but A2 is a common friend that violate the condition that no two friend has a common friend. Rest all Bij can be friend of B12 Hence total number of friends of B12 = 1+ 2+ ….+ (m-3) + 2 = (m-3)(m-2)/2 + 2 (this two friends are A1 & A2) But it has to be equal to m hence (m-3)(m-2)/2 +2 = m or m2 -7m +10 =0 or m= 2 or 5 Then the value of n = 1+m+ m(m+1)/2 = 4 or 16 30. (d) Since n= 16 hence there are 16 points or 16 friends. 31. (c) Since m = 5 since m = 2 will give n = 4 but number of friends are at least 5 hence ruled out. Concept Deviator (CD) Solutions (32 to 35) 32. (d) Given condition is diet must contain Fat -10% and Protein ≥ 30% We will eliminate option one by one to get the correct combination from the given options.

Option 1- P and Q is mixed, to get Fat 10% we have to mix them in the ratio of 4:1 and then the protein content would be 110/500 60%



Cost per unit = (200+200+100)/6 =500/6



Bust cost in option 4 (i.e 700/6) is more than that of option 5 (500/6) Option e.

34. (e) The given condition is Carbohydrate ≥ 30% Protein ≥30% Fat ≥ 25% Minerals ≥ 5%

Now we will evaluate and eliminate the options one by one.



Option (1) protein content = (30+20)/200 x 100 =25% Q4 (8) > Q3 (7) > Q1 (5) zz Using statements (ii), (iii) and (vi), the following can be deduced zz Q2 was correctly answered by A, B, C, D, E, F, G, H and I. zz Q4 was correctly answered by A, C, D, F, G, H, I and J zz Q3 was correctly answered by B, E, F, G, H, I and J. zz Q1 was correctly answered by A, B, C, D and E. 25. (a) Q2 was correctly answered by 9 students and they were A, B, C, D, E, F, G, H and I. 26. (c) Both B and E corrected Q1 and Q3. 27. (d) F, G, H, I, and J did not answered Q1 correctly. 28. (c) E did not answer Q4 correctly.

EBD_7743

MT-24