7300+ Reasoning for SSC 8527315252, 8527645252, 9868946424, 9268468484, 9268668686

Table of contents :
Cover
Title
Content
ANALOGY AND SIMILARITY
SYMBOLS & NOTATIONS
NUMBER SERIES
MISSING NUMBER
CLASSIFICATION
CODING DECODING
LOGICAL VENN DIAGRAMS
DICE
DIRECTION
R AN K I N G & S I T T I N GAR R AN G E M E N T
CLOCK
CALENDER
CUBE & CUBOID
SYLLOGISM
STATEMENT ARGUMENTS ANDASSUMPTIONS
BLOOD RELATION
ARITHMETICAL PROBLEM
ARRANGEMENT OF WORDSIN LOGICAL ORDER
WORD FORMATION
COUNTING FIGURES
ANALOGY AND SIMILARITY
MIRROR AND WATER IMAGES
COMPLETION OF FIGURE
EMBEDDED FIGURE
PAPER CUTTING AND FOLDING
SERIES
CLASSIFICATION &DEVIATION OF FIGURE
Back Cover

Citation preview

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PRUDENCE

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Maths Special By

Rakesh Yadav Sir th

12 Feb. 2018 05:00 pm – 07:00 pm th

12 March 2018 10:45am – 12:45pm 8527315252, 8527645252, 92-686-686-86, 92-684-684-84 Whatsapp no.: 9868946424

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641, Ground Floor, Dr. Mukherjee Nagar, Delhi-110009

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Coaching Centre

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ONLINE MATHS SPECIAL CLASSES Including Reasoning, G.S. & English 

All Topics of Maths with detailed explanation

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iqLrd ds vkWuykbu daVsaV dks iksiM+h (b) Palace/egy (c) House/ ?kj (d) Bungalow/caxyk

76. Touch : Feel :: Great : ?

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83. Canoe : Boat :: Mansion : ?

LdkiZ : Hkkstu :: de : ? (a) Bread/cszM (b) Tea/ pk; (c) Wine/ 'kjkc (d) Rice/Hkkstu

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: o"kkZ :: ? : jkf=k rwiQku (a) Evening/ 'kke (b) Dark/va/sjk (c) Day/fnu (d) Dusk/laè;k

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82. Thunder : Rain :: ? : Night

75. Scarp : Food :: Less : ?

: ?kqM+lokj :: dkj : ? ?kksM+k (a) Mechanic/ dkjhxj (b) Chauffeur/eksVj pkyd (c) Steering/ijhpkyd (d) Brake/cszd

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ykyef.k: yky :: uhyef.k: ? (a) Blue/uhyk (b) White/liQsn (c) Green/ gjk (d) Silver/pk¡nh

78. Horse : Jockey :: CAR : ?

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: dhV:: exjePN: ? edM+h (a) Reptile/jsaxus okyk (b) Mammal/cPpk nsus okyk (c) Frog/esa 3 or 19 > 3, which is true 33. (a) Using the proper notations in (a), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

7 + 7 – 7 ÷ 7 < 14 or 13 < 14, which is true 26. (a) Using the proper notations in (a), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

8 ÷ 4 × 1 – 2 = 16 – 16 or 0 = 0, which is true 31. (b) Using the proper notations in (b), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

31 + 1 – 2 < 4 + 6 × 7 or 30 < 46, which is true 25. (a) Using the proper notations in (a), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls) 5 × 2 ÷ 2 < 10 – 4 + 8 or 5 × 1 < 18 – 4 or 5 < 14, which is true 19. (c) Using the proper notations in (c), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

14 = 2 × 4 × 2 – 2 ÷ 1 or 14 = 14, which is true 30. (a) Using the proper notations in (a), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

9 + 5 + 4 = 18 ÷ 9 + 16 or 18 = 18, which is true 24. (d) Using the proper notations in (d), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls) 9+9÷9–9×9=9+1–9× 9 = 9 + 1 – 81 = 10 – 81 = –71 18. (c) Using the proper notations in (c), we get the statement as:

10 × 2 > 2 ÷ 1 × 10 ÷ 2 or 20 > 10, which is true 29. (b) Using the proper notations in (b), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

4 + 3 × 8 – 1 = 6 ÷ 2 + 24 or 27 = 27, which is true 23. (b) Using the proper notations in (b), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls) 8×8+8÷8–8=8×8+1– 8 = 64 + 1 – 8 = 65 – 8 = 57 17. (d) Using the proper notations in (d), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

8 – 4 ÷ 2 < 6 + 3 or 6 < 9, which is true 22. (b) Using the proper notations in (b), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls) 52 – 4 × 5 + 8 ÷ 2 = 52 – 4 × 5 + 4 = 52 – 20 + 4 = 56 – 20 = 36 16. (d) Using the proper notations in (d), we get the statement as:

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2 × 5 – 6 + 2 = 6 or 10 – 6 + 2 = 6 or 6 = 6, which is true. 15. (a) Using the proper notations in (a), we get the statement as:

djds] ge dFku izkIr djrs gSA tSls)

2 ÷ 1 + 10 × 1 < 6 × 4 or 12 < 24, which is true 28. (d) Using the proper notations in (lehdj.k eas mfpr fpUg dk iz;ksx djds] (d), we get the statement as: ge dFku izkIr djrs gSA tSls)

1 × 3 ÷ 2 > 1 < 5 > 3 × 1 ÷ 2 or 3 3 >1 , which is true 2 2 21. (c) Using the proper notations in (c), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

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18 ÷ 6 × 7 – 5 + 2 = 3 × 7 – 5 + 2 = 21 – 5 + 2 = 18 14. (d) Using the proper notations in (d), we get the statement as:

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(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

2 < 2 × 4 + 1 × 4 – 8 or 2 < 4, which is true 27. (b) Using the proper notations in (b), we get the statement as:(lehdj.k eas mfpr fpUg dk iz;ksx

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5 × 2 + 1 = 3 × 4 – 1 or 10 + 1 = 12 – 1 or 11 = 11, which is true. 20. (d) Using the proper notations in (d), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls) 4+5×9÷3–4=4+5×3– 4 = 4 + 15 – 4 = 15 13. (d) Using the proper notations in (d), we get the statement as:

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12. (a) Using the proper notations in (a), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

15 ÷ 5 + 3 = 2 × 3 or 6 = 6, which is true 34. (d) Using the proper notations in (d), we get the statement as:

(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

Rakesh Yadav Readers Publication Pvt. Ltd

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6 vkSj4 dks ijLij cnyus 15. (c) On changing – to + and ck;sa i{k ls interchanging 2 and 6, we get ij] gesa izkIr dFku blizdkj gS

2 – 4 = 3 or 3 = 3, 3

izkIr gksrk gS

Given expression = 5 + 6 × 3 – 12 ÷ 2 = 5 + 6 × 3 – 6 = 5 + 18 – 6 = 17 (a) On interchanging × and +, we get:

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5 + 3 × 4 – 6 ÷ 2 = 4 × 3 –10 ÷ 2 + 7 or 5 +12 – 3 =12 – 5 + 7 or 14 = 14, which is true. (d) On interchanging 7 and 6,w e get the statement as :

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which is true.(tks lR; gS) (a) On interchanging ÷ and ×, we get: '÷', vkSj'×' dks ijLij cnyus ij] ges 9.

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(b) On interchanging – and ÷, we get the equation as : '–' vkSj'÷' dks ijLij cnyus ij gesa 5 + 3 × 8 ÷ 12 – 4 = 3

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TYPE-2

or 5 + 3

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30 + 6 ÷ 2 > 4 × 3 or 33 > 12, which is true

izkIr gksrk gS

'–' dks'+' vkSj2 vkSj6 dks ijLij

cnyus ij]gesa izkIr dFku blizdkj gS

4 × 2 + 6 =14 or 8 + 6 = 14 or 14 = 14, which is true.

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cnyus ij] gesa izkIr dFku blizdkj gS

(6 ÷ 3) – 2 = 0 or 2 – 2 = 0 or 0 = 0, which is true. (tks lR; gS)

3 vkSj5 dks ijLij cnyus ij] gesa izkIr 17. (b) Clearly, or 5 – 0 + 3 × 5 = 20, 5 + 3 × 5 = 20 = 5 + 15 dFku blizdkj gS = 20 15 + 5 × 4 – 8 ÷ 2 = 8 × 3 + 16 ÷ 2 – 1 or 15 + 20 – 4 18.(b) Using the operations given in (b), we get the given = 24 + 8 – 1 or 31 = 31, which is true. expression as : '–', vkSj'÷' dks ijLij cnyus ij] ges 11. (d) On interchanging 9 and 5 (b) esa nh gqbZ fØ;kfof/ dk iz;ksx djus izkIr gksrk gS on R .H.S ., w e ge t th e ij] gesa izkIr gqvk O;atd Given expression statement as: 200+100 – 300 + 200 ÷ 10 × = 16 ÷ 8 – 4 + 5 × 2 2 – 40 = 300 – 300 + 20 × 2– nk;sa i{k 9ls vkSj5 dks ijLij cnyus = 2 – 4 + 5 × 2 = 2 – 4 + 10 = 8 40 = 0 + 40 – 40 = 0 ij] gesa izkIr dFku blizdkj gS (c) On interchanging ÷ and –, 19.(c) Using the operations given 6×3+8÷2–1=5–8÷4+ we get: in (c), we get the given 9 × 2 or 18 + 4 – 1 = 5 – 2 + expression as : '÷', vkSj'–' dks ijLij cnyus ij] ges 18 or 21 = 21, which is true. dk iz;ksx djus (c) esa nh gqbZ fØ;kfof/ izkIr gksrk gS 12. (c) On interchanging 9 and 6, ij] gesa izkIr gqvk O;atd Given expression = 9 + 5 – 4 we get the statement as : 1 700 ÷ 10 × + 35 –70 = 70 × 1 9 vkSj6 dks ijLij cnyus ij gesa izkIr 2 ×3÷6=9+5–4× 2 dFku blizdkj gS 1 + 35 – 70 = 35 + 35 – 70 = 0 = 9 + 5 – 2 = 12 8 ÷ 2 × 5 – 11 + 6 = 9 × 2–5 + 2

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Given expression = 2 + 3 × 6 – 12 ÷ 4 = 2 + 3 × 6 – 3 = 2 + 18 – 3 = 17 (b) On interchanging – and ÷, we get:

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the equation as:

16. (d) On changing × to – and 7 vkSj6 dks ijLij cnyus ij] gesa izkIr interchanging 2 and 3, we get dFku blizdkj gS the equation as: 6×2–3+8÷4=5+7×2– '×' dks'–' vkSj2 vkSj3 dks ijLij 24 ÷ 3 or 12 – 3 + 2 = 5 + 14 –

8 or 11 = 11, which is true. '×', vkSj'+' dks ijLij cnyus ij] ges 10. (a) On interchanging 3 and 5, we get the statement as :

izkIr gksrk gS

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(lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls) 7.

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4 ÷ 2 or 4 × 5 – 11 + 6 = 18 –5 + 2 or 15 = 15, which is true. '–', vkSj'+' dks ijLij cnyus ij] ges 13. (c) On interchanging + and × and 4 and 6, we get the izkIr gksrk gS equation as: Given expression = 12 ÷ 2 + 6 '+' vkSj'×' vkSj4 vkSj6 dks ijLij × 3 – 8 = 6 + 6 × 3 – 8 = 6 + 18 cnyus ij] gesa izkIr dFku blizdkj gS – 8 = 16. 4 + 6 × 2 = 16 or 4 + 12 = 16 (c) On interchanging + and, ×, we or 16 = 16, which is true. get the equation as : '+', vkSj'×' dks ijLij cnyus ij] ges 14. (a) On interchanging + and ÷ and 2 and 3,we get the equation izkIr gksrk gS as: given expression = 10 × 10 ÷ '+' vkSj'÷' vkSj2 vkSj3 dks ijLij 10 – 10 + 10 = 10×1 – 10+10 cnyus ij] geas izkIr dFku blizdkj gS = 10 – 10 + 10 = 10 (2 + 4) ÷ 3 = 2 or 6 ÷ 3 = 2 or (c) On interchanging 6 and 4 on 2 = 2, which is true. L.H.S., we get the statement as: (b) On interchanging – and + we get:

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24 ÷ 3 × 2 = 2 × 8 or 16 = 16, which is true 35. (a) Using the proper notations in (a), we get the statement as:

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6. (lehdj.k eas mfpr fpUg dk iz;ksx djds] ge dFku izkIr djrs gSA tSls)

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EXERCISE

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PREVIOUS YEAR QUESTIONS

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TYPE-I (II)

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In the following equation, select correct combination of mathDirection:- In the following question 4. If X stands for addition, V stands ematical sings to replace * signs for subtraction, U stands for you have to identify the correct and to balance the equation response from the given premises equal to', stands for division, 16 *4*5*9*1 stated according to following symbols:  stands for multiplication,  fuEufyf[kr lehdj.k esa * ds LFkku ij dkSu ls :- fn;s x;s iz'uksa esa lgh mÙkj pqus tks fn;sstands funsZ'k x;s for greater than and a xf.krh; fpUg dk la;kstu lehdj.k dks larq"V djus fpUgksa ds vk/kj ij lgh gksA stands for less than. State which ds fy;s lgh gksxkA expression is true. 1. If '+' stands for division, '÷' stands 16 * 4 * 5 * 9 * 1 for multiplication, multiplication ;fn X dk vFkZ '+', gS V dk vFkZ '–', gS U (a) + ÷ = × (b) ÷ + = × stands for addition; then which dk vFkZ'=', gS dk vFkZ'÷', gS  dk vFkZ (c) × = + – (d) + × = ÷ one of the following equation is gS'×',  dk vFkZ'>' gSvkSja dk vFkZ' vkSj< dks Øe'k% '÷' dk vFkZ '–' '×' rks gSvkSj'+' dk vFkZ gS

8 * 5 * 9 * 31 * ds LFkku ij dkSu ls fn;s x;s iz'u esa xf.krh; fpUg dk Øe lgh gksxk\ 8 * 5 * 9 * 31 (a) – × = (b) – = × (c) = × – (d) × – = Select the correct combination of mathematical signs to replace * signs and to balance the given equation. 4 * 6 * 6 * 2 * 20 * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

fuEufyf[kr esa ls dkSu&lk lehdj.k lgh gS\ δ , , γ , η , ω , β vkSj ls n'kkZ;k (a) 16 + 5 – 10 × 4 ÷ 3 = 9 tkrk gS] rks fuEufyf[kr esa ls dkSu&lk fodYi dk la;kstu lgh gksxk tks fn;s x;s lehdj.k (b) 16 – 5 × 10 ÷ 4 + 3 = 12 lgh gS\ dks larq"V djs\

(c) 16 + 5 ÷ 10 × 4 – 3 = 9 (d) 16 × 5 ÷ 10 ÷ 4 – 3 = 19 18. If '×' means '–', '–' means '×', '+' means '÷' and '÷' means '+', then (15 – 10) ÷ (130 + 10) × 50 = ? ;fn '×' dk vFkZ'–', gS '–' dk vFkZ'×', gS '+'

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(a) 3 γ 6 η 2 δ 8  4 ω 5 (b) 3 η 6 γ 2 δ 8  4 β 5 (c) 3 γ 6  2 δ 8 η 4  5 (d) 3 δ 6  2 γ 8 η 4 ω 5

4 * 6 * 6 * 2 * 20 (a) + ÷ = ÷ (b) × – + = (c) + – = ÷ (d) – + = ÷ 28. Select the correct combination of mathematical signs to replace

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fuEufyf[kr esa ls fpUgksa vkSj vadks dk dkSu&lk ijLij cnyko fn;s x;s lehdj.k dks larq"V djsxk\

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dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\

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(5 + 2) × 2 – 10 = 16 (a) + and × (b) 5 and 10 (c) + and – (d) 5 and 2 Which interchange of signs will make the following equation correct? 30 – 6 ÷ 4 + 2 × 3 = 7

fuEufyf[kr esa ls fpUgksa dk dkSu&lk ijLij cnyko fn;s x;s lehdj.k dks larq"V djsxk\

30 – 6 ÷ 4 + 2 × 3 = 7 (a) + and × (b) – and + (c) – and ÷ (d) + and – 40. Which of the following interchanges of signs would make the given equation correct? 5 + 6 ÷ 3 – 12 × 2 = 17

dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\

18 * 6 * 3 * 12 * 24 (a) ÷ – = × (b) × ÷ – = (c) + ÷ × = (d) × = ÷ +

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15 * 3 * 5 * 20 * 2 (a) + – = ÷ (b) × – = × (c) + = + × (d) × – = ÷ 30. Select the correct combination of mathematical signs to replace * signs and to balance the given equation. 2*3*2*4*8 35. * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

6 * 15 * 10 * 3 * 12 (a) ÷ + = × (b) + ÷ × = (c) × ÷ + = (d) + – = ÷ 39. Select the correct combination of mathematical signs to replace * signs and to balance the given equation. 18 * 6 * 3 * 12 * 24 * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

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dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\

dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\

(a) + and × (b) × and – (c) ÷ and + (d) – and ÷ Which of the followings interchange of signs or numbers would make the given equation correct? (5 + 2) × 2 – 10 = 16

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(a) = × + ÷ (b) × = + ÷ (c) × + = ÷ (d) + × = ÷ 29. Select the correct combination of mathematical signs to replace * signs and to balance the given 34. equation. 15 * 3 * 5 * 20 * 2 * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

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ww 16 * 4 * 3 * 4 * 13 (a) ÷ × – = (b) × – + = equation. (c) + = ÷ × (d) – × ÷ = 38. 33. Select the correct combination of 8 5 2 72 4 mathematical signs to replace * fpUg dks gVkus ds fy;s fdl xf.krh; signs and to balance the given fpUg dk la;kstu lgh gksxk tks fn;s x;s equation. 6 * 15 * 10 * 3 * 12 lehdj.k dks larq"V djs\ * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg 8 5 2 72 4 signs and to balance the given

TYPE-III

Which of the following interchange of signs would make the given equation correct? (12 ÷ 6) + 3 × 7 = 42

fuEufyf[kr esa ls fpUgksa dk dkSu&lk ijLij cnyko fn;s x;s lehdj.k dks larq"V djsxk\

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5 + 6 ÷ 3 – 12 × 2 = 17 (a) ÷ and × (b) + and × 2*3*2*4*8 (c) + and ÷ (d) + and – fuEufyf[kr esa ls fpUgksa dk dkSu&lk ijLij (a) + – + = (b) × – + = 41. Which interchange of signs or cnyko fn;s x;s lehdj.k dks larq"V djsxk\ numbers will make the following (c) – + × = (d) × + – = 31. Select the correct combination of (12 ÷ 6) + 3 × 7 = 42 equation correct? mathematical signs to replace * (7 + 2) × 3 × 4 – 1 = 20 (a) + and × (b) 6 and 7 signs and to balance the given (c) ÷ and + (d) 12 and 3 fuEufyf[kr esa ls fpUgksa dk dkSu&lk ijLij equation. 36. Which interchange of signs will cnyko fn;s x;s lehdj.k dks larq"V djsxk\ 16 * 2 * 24 * 3 * 6 make the following equation (7 + 2) × 3 × 4 – 1 = 20 * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg correct? (a) 2 and 3 (b) × and – dk la;kstu lgh gksxk tks fn;s x;s lehdj.k 35+7×5÷5–6 = 24 (c) 7 and 3 (d) + and ×

dks larq"V djs\

16 * 2 * 24 * 3 * 6 (a) + = – ÷ (b) × – + = (c) + ÷ = ÷ (d) – – ÷ = 32. Select the correct combination of mathematical signs to replace * signs and to balance the given equation. 16 * 4 * 3 * 4 * 13

fuEufyf[kr esa ls fpUgksa dk dkSu&lk 42.ijLij Which interchange of signs will cnyko fn;s x;s lehdj.k dks larq"V djsxk\ make the following equation cor-

35+7×5÷5–6 = 24 (a) + and – (b) + and × (c) ÷ and + (d) – and ÷ 37. Which of the following interchanges of signs would make the given equation correct? 24 + 6 × 3 ÷ 3 – 1 = 14

* fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\ 48

rect? (16 – 4) 6 ÷ 2 + 8 = 30

fuEufyf[kr esa ls fpUgksa dk dkSu&lk ijLij cnyko fn;s x;s lehdj.k dks larq"V djsxk\ (16 – 4) 6 ÷ 2 + 8 = 30 (a) ÷ and – (b) 4 and 2 (c) – and + (d) 16 and 6

fuEufyf[kr esa ls fpUgksa dk dkSu&lk ijLij TYPE-I (I), 2012 cnyko fn;s x;s lehdj.k dks larq"V djsxk\ 43. If – stands for ÷, + stands for ×, ÷ 24 + 6 × 3 ÷ 3 – 1 = 14

stands for – and × stands for +

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dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\

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16 * 4 * 64 * 4 (a) ×, , ÷ (c) ÷, >, × (d) ×, >, + 54. Select the correct combination of mathematical signs to replace * signs and to balance the given equation : 28 * 4 * 9 *16 * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\

28 * 4 * 9 *16

dqN lehdj.k dks ge ,d fuf'pr i¼fr ls (a) ÷ + = (b) + ÷ = gy dj ldrs gSA rks bl vk/kj ij fcuk gy (c) – × + (d) – = × fd;s gq, iz'u dk lgh mÙkj Kkr djsa\ 55. Select the correct combination of mathematical signs to replace * ;fn 8 + 8 = 72, 5 + 5 = 30 vkSj7 + signs and to balance the given 7 = 56, rks6 + 6 dk eku D;k gksxk\

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(a) 5 – 2 + 12 × 6 ÷ 2 = 27 (b) 5 + 2 – 12 ÷ 6 × 2 = 19 (a) 40 (b) 42 (c) 5 + 2 – 12 × 6 ÷ 2 = 16 (c) 30 (d) 36 (d) 5 ÷ 2 + 12 × 6 – 2 = 3 46. If × stands for –, + means ×, ÷, 51. Some equations are solved on the basis of a certain system. Find means + and – means ÷, then out the correct answer for the unwhat is the value of the given solved equation on that basis. expression? If 3 ÷ 5 = 5, 4 ÷ 7 = 8, 8 ÷ 7 = 6 175 – 25 ÷ 5 + 20 × 3 + 10 = ? then, what should 9 ÷ 6 be ? ;fn '×' dk vFkZ '–', gS '+' dk vFkZ '×', gS dqN lehdj.k dks ge ,d fuf'pr i¼fr ls '÷', dk vFkZ'+' gSvkSj'–' dk vFkZ '÷' gSrks

equation. 16 * 6 * 4 * 24 * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

dk la;kstu lgh gksxk tks fn;s x;s lehdj.k dks larq"V djs\

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16 * 6 * 4 * 24 (a) ÷ = × (b) × = ÷ (c) = ÷ ÷ (d) × ÷ = 56. Choose the appropriate combination of signs to solve. gy dj ldrs gSA rks bl vk/kj ij fcuk gy 16 * 8 * 1 * 8

fn;s x;s O;atd dk eku Kkr djsa\

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50. Some equations are solved on the basis of a certain system. Find out the correct answer for the unsolved equation on that basis. If 8 + 8 = 72, 5 + 5 = 30 and 7 + 7 = 56, what is 6 + 6 = ?

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TYPE-I (III)

53. Select the correct combination of mathematical signs to replace * signs and to balance the given equation. 16 * 4 * 64 * 4 * fpUg dks gVkus ds fy;s fdl xf.krh; fpUg

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(a) 6 (b) 8 (c) 10 (d) 12 49. X stands for +, Z stands for ÷, Y stands for –, and P stands for ×, then what is the value of 10 P 2 X5Y5? gS Z dk vFkZ '÷', gS Y dk X dk vFkZ '+', P dk vFkZ'×', vFkZ'–', gSvkSj gS rks10 P 2 X 5 Y 5 dk eku D;k gksxk\ (a) 10 (b) 15 (c) 20 (d) 25

;fn '–' dk vFkZ'+', gS '+' dk vFkZ'×', gS '÷' dk vFkZ'–' gSvkSj'×' dk vFkZ'÷' gSfuEufyf[kr lehdj.k esa dkSu&lk ,d xyr gS\

175 – 25 ÷ 5 + 20 × 3 + 10 = ? (a) 77 (b) 160 (c) 240 (d) 2370 47. If '+' means '÷' ; '÷' means '–' ; '–' means '×' ; '×' means '+', then 8+2÷3–4×6=? ;fn '+' dk vFkZ'÷' gS; '÷' dk vFkZ'–' gS, gS, '×' dk vFkZ '+', gS rks '–' dk vFkZ '×' 8+2÷3–4×6=? (a) – 12 (b) –2 (c) –10 (d) –15

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(a) 49 × 7 + 3 ÷ 5 – 8 = 16 (b) 49 ÷ 7 × 3 + 5 – 8 = 26 (c) 49 + 7 – 3 × 5 ÷ 8 = 20 (d) 49 – 7 + 3 ÷ 5 × 8 = 24 44. If '+' means '×', '–' means '÷' '×' means '+' and '÷' means '–' then 25 × 5 – 3 ÷ 2 + 5 = ? ;fn '+' dk vFkZ'×', gS '–' dk vFkZ'÷' gS'×' dk vFkZ '+' gSvkSj'÷' dk vFkZ '–' gSrks 25 × 5 – 3 ÷ 2 + 5 = ? (a) 20/3 (b) 50/3 (c) 30/7 (d) 40/7 45. If '–' stands for addition, '+' for multiplication, '÷' for subtraction and '×' for division, which one of the following equation is wrong?

16 P 24 M 8 Q 6 M 2 L 3 dk

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fd buesa ls dkSu&lk ,d lgh gS\

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find out which one is correct. ;fn '–' dk vFkZ'÷', gS'+' dk vFkZ'×', gS '÷' dk vFkZ'–' gSvkSj'×' dk vFkZ'+' gSKkr djsa

fd;s gq, iz'u dk lgh mÙkj Kkr djsa\ ;fn 3 ÷ 5 = 5, 4 ÷ 7 = 8, 8 ÷ 7 = 6 rks9 ÷ 6 dk eku D;k gksxk\

(a) 4 (b) 9 (c) 5 (d) 6 52. Some equations are solved on the basis of a certain system. On the same basis find out the correct answer for the unsolved equation. If 8 × 2 = 61 ; 8 × 5 = 04, what is 8 × 10 = ?

TYPE-I (II) 48. If L denotes ×, M denotes ÷, P denotes + and Q denotes –, then find the value of 16 P 24 M 8 Q 6 M 2 L 3 = ? ;fn L dk vFkZ'×', gS M dk vFkZ'÷', gS P dk vFkZ '+' gSvkSjQ dk vFkZ '–', gS rks

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dqN lehdj.k dks ge ,d fuf'pr i¼fr ls gy dj ldrs gSA rks bl vk/kj ij fcuk gy fd;s gq, iz'u dk lgh mÙkj Kkr djsa\ ;fn 8 × 2 = 61 ; 8 × 5 = 04 rks8 × 10 dk eku D;k gksxk\ (a) 80 (c) 8

(b) 08 (d) 0

fpUgksa dk lgh Øe Kkr djs\ 16 * 8 * 1 * 8 (a) = – ÷ (c) ÷ – =

(b) – ÷ = (d) ÷ = –

TYPE-III 57. Which one of the four interchanges in signs and numbers would make the given equation correct? 6 × 4 + 2 = 16

fn;s x;s lehdj.k esa fpUgksa vkSj la[;kvksa esa pkj esa ls ,d dkSu&lk cnyko lehdj.k dks lgh djsxk\ 6 × 4 + 2 = 16 (a) + and ×, 16 and 4 (b) + and ×, 2 and 4 (c) + and ×, 4 and 6 (d) None of these

49

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12 P 6 M 15 T 16 B 4 ? (a) 70 (b) 75 (c) 83 (d) 110 Identify one response which would be a correct inference from the given premises stated according to the following symbols: 'A' stands for not greater than 'B' stands for equal to 'C' stands for less than 'D' stands for not less than 'E' stands for not equal to 'F' stands for greater than Premises (2 M B N) and (2N A 3K)

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fuEufyf[kr esa ls dkSu&lk cnyko fn;s gq,(BODMAS fu;e ckè; ugha gksxk ) lehdj.k dks lgh djsxk\ 25 A 37 C 2 B 4 R 1 = ?

5 + 3 × 8 – 12 ÷ 4 = 3 (a) – and ÷ (b) + and × (c) + and ÷ (d) + and – 61. Put the correct mathematical signs in the following equation from the given alternatives. 33 ? 11 ? 3 ? 6 = 115

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(a) 32 (b) 35 (c) 30 (d) 27 66. If P denotes '÷' Q denotes ×, R denotes + and S denotes –, then 12 Q 15 P 3 R 4 S 6 = ? ;fn P dk vFkZ'÷' gSQ dk vFkZ'×', gS R dk 70. vFkZ '+' gSvkSjS dk vFkZ '–' gSrks fn;s gq, fodYiksa esa ls dkSu&lk xf.krh; 12 Q 15 P 3 R 4 S 6 = ? O;atd lehdj.k dks larq"V djsxk\ (a) 70 (b) 57 33 ? 11 ? 3 ? 6 = 115 (c) 58 (d) 68 (a) –, ×, + (b) +, –, × 67. If A stands for +, Q stands for –, (c) ×, ÷, – (d) ÷, ×, × V stands for ×, R stands for ÷, 62. If '×' means '+', ÷ means '–', + then what is the value of the means '÷' and '–' means '×' then given equation? what should be the value of the 225 R 5 A 64 Q 13 V 6 = ? given equation? 14 × 4 ÷ 70 + 10 – 2 = ? ;fn A dk vFkZ '+', gS Q dk vFkZ'–', gS V ;fn '×' dk vFkZ'+', gS '÷' dk vFkZ'–', gS'+' dk vFkZ'×', gS R dk vFkZ'÷', gS rks fn;s x;s dk vFkZ'÷' gSvkSj'–' dk vFkZ'×' gSrks fn;s lehdj.k dk eku D;k gS\

gq, lehdj.k dk eku Kkr djsa\ 225 R 5 A 64 Q 13 V 6 = ? 14 × 4 ÷ 70 + 10 – 2 = ? (a) 376 (b) 31 (a) 33 (b) 15 (c) 476 (d) 576 (c) 30 (d) 4 68. If 'P' denotes' multiplied by', 'T' 63. If '+' means '–' ; '–' means '×' ; '×' denotes 'subtracted from', 'M' demeans '÷' and '÷' means '+', then notes 'added to' and 'B' denotes 25 × 5 ÷ 30 + 8 – 2 = ? divided by' then : what should be ;fn '+' dk vFkZ'–' gS, '–' dk vFkZ'×' gS, the correct response of 12 P 6 M 15 T 16 B 4 ? '×' dk vFkZ'÷' gSvkSj'÷' dk vFkZ'+' gSrks ;fn 'P' dk vFkZ '×', gS 'T' dk vFkZ '–', gS fn;s gq, lehdj.k dk eku Kkr djsa\ 50

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TYPE-I (I), 2013 25 × 5 ÷ 30 + 8 – 2 = ? 58. If '÷' means '–', '–' means '×', '×' (a) 54 (b) 15 means '+' and '+' means '÷' then (c) 18 (d) 19 20 × 60 ÷ 40 – 20 + 10 = ? TYPE-I (II) ;fn '÷' dk vFkZ'–', gS'–' dk vFkZ'×', gS '×' dk vFkZ '+' gSvkSj'+' dk vFkZ '÷' gSrks 64. If 'P' means '+' 'Q' means '×' 'R' 69. means '÷' and 'S' means '–', then 20 × 60 ÷ 40 – 20 + 10 = ? 44 Q 9 R 12 S 6 Q 4 P 16 = ? (a) 80 (b) 60 (c) 40 (d) 0 ;fn 'P' dk vFkZ'+' gS'Q' dk vFkZ'×' gS'R' 59. If – stands for addition, ÷ for dk vFkZ'÷' gSvkSj'S' dk vFkZ'–', gSrks fn;s multiplication, × for subtraction, gq, lehdj.k dk eku Kkr djsa\ and + for division, then which of 44 Q 9 R 12 S 6 Q 4 P 16 = ? the following is correct? (a) 25 (b) 112 ;fn '–' dk vFkZ'+', gS '÷' dk vFkZ'×', gS '×' (c) 36 (d) 124 dk vFkZ gS gS rks '–' vkSj'+' dk vFkZ '÷', 65. If 'R' stands for '–', 'A' stands for fuEufyf[kr esa ls dkSu&lk lehdj.k lgh gS\ '+', 'B' stands for '÷' and 'C' stands (a) 25 – 15 + 5 ÷ 4 × 16 = 21 for '×', then what is the value of (b) 25 + 11 – 4 ÷ 10 × 6 = 20 the given equation? (BODMAS (c) 25 × 12 – 14 ÷ 4 + 6 = 16 rule will not be applicable) (d) 25 – 12 + 14 ÷ 2 × 4 = 15 25 A 37 C 2 B 4 R 1 = ? 60. Which of the following inter;fn 'R' dk vFkZ '–', gS 'A' dk vFkZ '+', gS change of signs would make the 'B' dk vFkZ'÷' gSvkSj'C' dk vFkZ'×', gS rks given equation correct? 5 + 3 × 8 – 12 ÷ 4 = 3 fn;s x;s lehdj.k dk eku D;k gksxk

fn;s x;s iz'uksa esa lgh mÙkj pqus tks fn;s x;s fpUgksa ds vk/kj ij lgh gksA 'A' dk vFkZ gS > 'B' dk vFkZ =gS 'C' dk vFkZ Vdk

(b) 75 – 35 (d) 39 – 61 (b) 13 – 156 (d) 16 – 176 (b) 96 (d) 80 (b) 83, 6 (d) 97, 9 (b) 28 (d) 56 (b) 69 (d) 66 (b) 91 (d) 225 (b) 194 : 12 (d) 224 : 14 (b) 3649 (d) 5025 (b) 625 (d) 2401 (b) 2, 3, 5, 8 (d) 4, 2, 3, 9 (b) 24, 25 (d) 20, 21 (b) 64 (d) 98 (b) 15, 3, 4 (d) 23, 4, 5 (b) 840 (d) 475 (b) 56 – 64 (d) 80 – 93 (b) 19 – 27

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(c) Black/dkyk (d) Green/gjk

316. (a) 64 – 36 (c) 57 – 43 317. (a) 12 – 144 (c) 15 – 180 318. (a) 36 (c) 16 319. (a) 41, 4 (c) 74, 7 320. (a) 24 (c) 42 321. (a) 63 (c) 65 322. (a) 108 (c) 144 323. (a) 187 : 11 (c) 195 : 13 324. (a) 1625 (c) 6481 325. (a) 512 (c) 1296 326. (a) 1, 2, 4 , 7 (c) 3, 4, 6, 9 327. (a) 9, 10 (c) 12, 13 328. (a) 83 (c) 56 329. (a) 45, 6, 7 (c) 35, 5, 6 330. (a) 725 (c) 632 331. (a) 14 – 16 (c) 77 – 88 332. (a) 13 – 21

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294. (a) Red/yky

fodYiksa esa ls fo"ke la[;kvks ds tksM+s Kkr djsa

348. (a) 1, 2, 4, 8 (b) 2, 6, 10, 14 (c) 3, 9, 15, 21 (d) 7, 21, 35, 49 349. Find the wrong number in the series:

fn, x, Js.kh esa xyr la[;k Kkr djsaA 30, 27, 36, 45, 72 (a) 30 (b) 27 (c) 36 (d) 72 350. Which one number is wrong in the series?

bl J`a[kyk esa xyr uEcj dkSu lk gS\ A B C D E 225 256 121 289 324 (a) 225 (b) 256 (c) 121 (d) 324 351. Find out the false date of birth given in the series.

nh xbZ J`a[kyk esa xyr tUe frfFk Kkr djsa\ 1. 15 2. 16 3. 18 4. 30 5. 31 6. 31 (a) 4 (c) 3

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02 -1969 03 - 1969 04- 1969 12 - 1969 02 - 1970 12 - 1969 (b) 6 (d) 5

TYPE -I (2015) 352. (a) Tulip/V~;wfyi (b) Lotus/dey (c) Marigold/xsans dk iQwy (d) Rose/xqykc

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379. (a) MOUTH/eq¡g(b) NOSE/ukd

(c) PAINT/isaV (d) SKETCH/LdSp

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(b) Lawyer/odhy

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381. (a) SWING/>wyk (b) Kingfisher/eNyh idMus okyk i{kh (b) SWEATER/LosVj (c) Crane/Øsu (c) UMBRELLA/Nkrk (d) Humming bird/gfEeax cMZ (d) GLOVES/nLrkus

(d) Governor/jkT;iky

368. (a) Thick/eksVk

358. (a) Parallel/lekukarj

382. (a) Swing/Lohax

(b) Fathom/Fkkg ysuk

(c) Obtuse/dqafBr

(c) Plump/LFkwy

(d) Right/lgh

(d) Solid/n`hy vkSj rykc] ikuh dks laxzg 5 × 5 × 5 = 125 432 dks NksM+dj] nwljh lHkh fo"ke gSA djus okys gSA 89. (d) 162  16 ÷ 2 = 8 97. (d) Except in 24 – 121, in all oth80. (d) Except Rain, all other are waers both the numbers have a 405  40 ÷ 5 = 8 ter bodies. common factor. 567  56 ÷ 7 = 8 o"kkZ dks NksM+dj]vU; lHkh ikuh dks laxzg 24 – 121 dks NksM+dj nwljh nksuks la[;kvksa 644  64 ÷ 4 = 16 djus okys gSA ds xq.ku[k.M gSA 90. (a) Except 156, all other number 7 × 2 = 14; 7 × 7 = 49 81. (c) King, Queen and Prince are are odd numbers. 8 × 2 = 16; 8 × 8 = 64 Royal persons. 156 dks NksM+dj] vU; lHkh la[;k;sa 10 × 2 = 20; 10 × 10 = 100 jktk] jkuh vkSj jktdqekj 'kkgh yksx gSA fo"ke la[;k gSA 98. (b) 36 is a Perfect Square. 82. (a) There is only one seed in 115 90 ,d iw.kZ oxZ gSA 36 mango, while others have 91. (a) = 10.45; = 9; 36 = 6 × 6 11 10 many seeds.

fliQZ vke esa ,d cht gksrk gS]tcfd vU; esa cgqr lkjs cht gksrs gSA 83. (b) Steel is a strong hard metal made of a mixture of iron and carbon. Except steel, all others are elements (metals).

72 56 8; 7 9 8

92. (a) Except in the number pair 120 – 560, in all other number pairs both the numbers are multiples of 13.

99. (a) 0 +3

3 +5

8

9 +3 12 +4 16 17 +3 20 +4 24 51 +3 54 +4 58

100. (b) The number 64 is a perfect square as well as a perfect cube. la[;k64 iw.kZ oxZ ds2lkFk& iw.kZ?kuSA Hkh g 64 = 8 × 8 = 4 × 4 × 4

LVhy dBksj /krq gS tks fd yksgs vkSj 120 la[;k tksM+s esa– 560, dks NksM+dj dkcZu dk feJ.k gSA LVhy NksM+dj] vU; 13 dh xq.kt gSA nwljh lHkh la[;k;sa lHkh rRo gS(/krq dh) 84. (c) Except 'Article' all others are printed reading material. Article is a write up.

Rakesh Yadav Readers Publication Pvt. Ltd

13 × 7 = 91; 13 × 23 = 299; 13 × 6 = 78; 13 × 13 = 169; 13 × 8 = 104; 13 × 33 = 429;

101. (d)

515 635  5;  5; 103 127

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745 856  5; 4 149 214

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B +1 C +14 Q –3 N A +1 B +2 D +2 F 47 – 73, la[;k tksM+s dks NksM+dj] nwljs M +1 N +2 P +2 R

lHkh esa vadks dh fLFkfr cny pqdh gSA

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D C G H

111. (d) A B Y Z Pairs of Opposite Letters

la[;k 5462 dks NksM+dj] nwljh lHkh foijhr v{kjksa ds tksM+s la[;k;sa11 dh xq.kt gSA 363 = 11 × 33; 484 = 11 × 44; 1331 = 11 × 121;

G T [V is not opposite pair] H V [V foijhr tksM+k ugha ] gSA

105. (d) Except in the number pair 7 – 50, in all other number pairs, the second number is square of the first number

R +1 S +2 U U +2 W +2 Y J +2 L +2 N E +2 G +2 I

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8 × 8 = 64; 6 × 6 = 36; 9 × 9 = 84; 50 = 7 × 7 + 1 106. (d) In the number pair 8 – 90, both the numbers are even number.

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8–90 la[;k tksM+ks esatksM+s esa nksuks la[;k,a le la[;k gSA

120.(d) K –1 J +3 G –1 F +3 T –1 S +3 Z +1 A +1

122.(d) B

B

N

R

E

I

la[;k 35 dks NksM+dj] lHkh vHkkT; la[;k,a gSA

L H U Y

C

X

V

F

U

–1

O

T

M

H

S

–1

–3

F

L

J

A +5 F –2 T –1 S +4 W –1 I –2 G +5 L –2 O –2 M +5 R –2

O

I

R

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–2 114. (b) C –2

–1 –1 –1 –3

+1

–2 L

M I V B

–1

–2 H

R Q U T

+1

G

24esa – 64 dks NksM+dj la[;k tksM+ks nwljs lHkh tksM+s iw.kZ oxZ gSA

Y

–2 Q

+5 +5 +5 +4

+1

–2 K

M L P P

gS

–2 T

+4 +4 +4 +3

I H L M

118. (b) A and U are Vowels. A vkSj U nksuks Loj gSA 119. (c) Except letter ANHU, All other is are vowels. ANHU dks NksM+dj] vU; lHkh esa ,d Loj

–1

107. (c) Except in the number pair 24 – 64, in all other number pairs both the numbers are perfect squares.

49 – 100  (7)2 – (10)2 81 – 144  (9)2 – (12)2 9 – 36 (3)2 – (6)2 108. (a) Except the number 35, all other are Prime Numbers.

W

+3 +3 +3 +3

B G +2 I –1 H M +2 O –1 N P +1 Q +1 R

–2 D

F E I J

121.(d) A +2 C –1

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112. (a) P +2 S +2 la[;k tksM+ks esadks NksM+dj nwljs 750 +2 lHkh tksM+ks esa nwljh la[;k igyh la[;k H C +2

122

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EF U V

But

+2 +2 +2 +2

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C D W X

5462 = 496.54 11

+3 H

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23. A, B, C, D, E, F and G sitting on Note: tc iz'u esa double row gks vkSj ,d dk eqa[k] G mldk fudVre iM+kslh D gSa R a bench facing East. C is imNorth facing v kS j , d South D ds nk,¡ rhljs LFkku ij dh vksj ughaB, gSA mediate right of D. B is sitting facing gks rc arrangement esa ge cSBk gSA gSA dk E, B ds ,dne nk,¡ cSBk G on any end and E is neighbour North facing okys dks T;ks dk R;ks eqa[k P dh vksj ugha gSA fuEu C dk esaeqa[k ls of G. G is between E and F. D arrange djsaxsa vxj left cksys rks left fdldh vksj gSA is 3rd to the south where is D right cksys rks Right ysfdu South Sol. sitting. facing o k ys Row e ss s a tc F E B C G D ( ) arrangement esaleft cksys rks ge viusiafDr&1 A, B, C, D, E, F rFkk G csUp ij iwjc ( ) Right Shift djsaxs iafDr&2 C, DgSaA fn'kk esa eqa[k djds cSBs ds rqjUr S Q R P U T 25. Twelve People are sitting in two nk¡;h vksj B, gSA fdlh vafre Nksj ij cSBk gS parallel rows containing six 26. Five boys standing in a row and E rFkkF ds rFkkE, G dk iM+kslhG, gSA facing towards South. Anand is People each, in such a way that standing immediate right of eè; esa gSA D, nf{k.k fn'kk ls rhljs LFkku ij there is equal distance between Ramu. Chandan is between D dgk cSBk gS\ adjacent persons. In row-I B, C, gSA rks Ramu and Sohan. No one D, E, F and G are seated and all (a) C and F (b) G and C standing immediate Right of of them are facing south. In (c) A and E Babu. Who is standing in exrow-2 P, Q, R, S, T and U are (d) None of these/buesa ls dksbZ ugh act middle of Row. seated and all of them are Sol. (a) ik¡p yM+ds ,d iafDr es nf{k.k dh vksj eaq[k facing north. Therefore, in the North Left/ North djds [kM+s gSA vkuUn] jkew ds rqjUr nk¡;h vksj given seating arrangement, B gSA pUnu] jkew rFkk lksgu ds eè; gS ckcw ds each member seated in a row E West East faces another member of the rqjUr nk¡;h vksj dksbZ ugha gSA iafDr ds fcYdqy other row. (Please note: None eè; esa dkSu [kM+k gSA G of the information given in Sol. North South F necessarily in the same order). R sits second to the right of S. D West East Only one person sits between C R and U. C sits to the South South Facing immediate left of the one who A faces U. Only two people sit Babu Anand Ramu Chandan Sohan Right/South between C and F. Q sits fourth Answer to the left of T. G is an DOUBLE ROW BASED QUESTION TYPE–4/VkbZi&4 immediate neighbour of the nks iafDr;ksa ij vk/kfjr iz'u Hex agon based Question:one who faces T. D does not face Six people A, B, C, D, E and F 24. Six friends P, Q, R, S, T, U sitR. B sits third to the right of D. are seating on the ground in a ting in a row some facing toE is an immediate right of B. G hexagonal shape. All the wards North and some towards does not face P. Who amongst shapes of hexagon so formed South the following is facing C? are of same length. A is not N% nksLr P, Q, R, S, T, U ,d iafDr esa adjacent to B or C. D is not ad12 O;fDr] nks lekarj ikafDr;ksa esa] 6 O;fDr cSBs gSa muesa dqN dk eqa[k mRrj dh vksj gS jacent to C or E. B and C are rFkk dqN dk eqa[k nf{k.k dh vksj gSA izR;sd iafDr esa ,d nwljs ls leku nwjh ij cSBs adjacent F is the middle to D gSaA iafDr&1 esa] B, C, D, E, F rFkkG cSBs (i) Q is facing towards North but not and C ? gS rFkk lHkh dk eqa[k nf{k.k dh vksj gSA sitting next of S. A, "kVHkqt ij vk/kfjr iz'u%& N% O;fDr Q dk eqa[k mÙkj dh vksj gS fdUrq S ds og P, Q, R, S, T rFkkU cSBs gS iafDr&2 esa] B, C, D, E, rFkk F fdlh "kVHkqt ds izR;sd

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(a) 14 (b) 15 (c) 16 (d) 17 (a) Ravi (b) Mahima I n a row of 16 boys, when (c) Girdhar (d) Rama Prakash was shifted by two places towards the left, he 10. Five people are sitting in a row facing you. Y is at the left of X, viuh fLFkfr dks vkil esa cny nsrs gSa rksbecame 7th from the left end. W is sitting at the right of Z. V JhukFk vc ck¡;s22 ls ok¡ gks tkrk gSA iafDr What was his earlier position is sitting at the right of X and from the right end of the row? esa dqy fdrus yM+ds gSA W is sitting at the left of Y. If Z 16 yM+dksa dh iafDr esa] ;fn izdk'k ds nks (a) 19 (b) 31 is sitting at one end who is in LFkku ck¡;h vksj foLFkkfir djrs gSa rks ogthe vc middle? (c) 33 (d) 34 7 ck¡;h Nksj ls ok¡ gks tkrk gS mldh iwoZ Suresh is 7 ranks ahead of 5 yksx ,d iafDr esa rqEgkjh vksj eq[k djds cSBs fLFkfr nk¡;h Nksj ls D;k Fkh\ Ashok in the class of 39 Y, X ds ck¡;s gS] W, Z ds nk¡;s cSBk V, gSA gSaA (a) 7th (b) 8th students. If Ashok's rank is X ds nk¡;s gS rFkk W, Y ds ck¡;s cSBk gSA ;fn 17th from the last. What is (c) 9th (d) 10th Z ,d fljs ij cSBk gS rks eè; esa dkSu cSBk gS\ Suresh's rank from the start? 8. In a classroom, there are 5 (a) V (b) X (c) Y (d) Z 39 Nk=kksa dh d{kk esa lqjs'k] 7 LFkku v'kksd ls rows. and 5 children A, B, C, D 11. A, B, C, D and E are 5 schools and E are seated one behind vkxs gSA ;fn v'kksd dk LFkku17 varok¡ ls facing towards north. A is in the gS] rks lqjs'k dh izkjaEHk ls fLFkfr crkb,\the other in 5 separate rows as middle of E and B. E is to the follows. right of D. If C and D are at two (a) 16th (b) 23th A is sitting behind C, but in ends. Which school is on the (c) 24th (d) 15th front of B. left side of C? Sudheesh ranks seventh from C is sitting behind E, but D is A, B, C, D rFkk E ik¡p fo|ky; gS rFkk the top and 28th from the sitting in front of E. ftudk eq[k mÙkj dh vksj A,gSA E o B ds bottom. How many students The order in which they are are there in the class? E, D C D nksuksa eè; gSA ds nk¡;s gSA ;fn rFkk sitting from the first row to the lq/h'k Åij ls 7 ok¡ rFkk uhps 28lsok¡ gS] fljksa ij gSaA ds rks ck¡;s dkSu lk fo|ky; gS\ C last is rks ml d{kk esa fdrus Nk=k gSa\ (a) E (b) A (c) D (d) B ,d d{kk esa5 iafDr gS rFkk 5 cPpsA, B, (a) 34 (b) 35 C, D vkSj E dks ,d nwljs ds ihNs 5 vyx 12. Six persons A, B, C, D, E and F sit in 2 rows. 3 in each. If E is (c) 28 (d) 21 iafDr;ksa esa cSBk;k x;k gSA not at any end, D is second to Raju ranks 10th from the top A, C ds ihNs cSBk gSBijUrq ds lkeus cSBk gSA left of F, C is neighbour and Ravi ranks 21st from the C, E ds ihNs cSBk gS]DijUrq , E ds lkeus opposite to B and B is bottom. If there are 3 students os yksx igys ls vkf[kjh iafDr esa fdl izdkj neighbour of F, who will be between them, how many opposite to B? students are there in the class? cSBs gSa\ N% yksx A, B, C, D, E, rFkkF nks (a) DECAB (b) BACED jktw lcls Åij ls 10 ok¡ gS rFkk jfo lcls 3 esaA ;fn E fdlh iafDr;ksa esa cSBs izR;sd gSaA (d) ABEDC 3 ds chp(c) ACBDE uhps ls21 okW gSA ;fn bu nksuksa F ds cak, nwljkC, fljs ij ugha gSAD,rks gS] 9. AgS\ group of friends are sitting in Nk=k gSaA rks d{kk esa dqy fdrus Nk=k ds iM+kslh ds foijhr gS rFkk dk B B, F an arrangement one each at the (a) 34 (b) 33 corner of an octagon. All are B rks iM+kslh gS ds foijhr fn'kk esa dkSu gS\ (c) 31 (d) 32 facing the centre. Mahima is (a) A (b) E If you are 9th person in queue sitting diagonally opposite Rama, (c) C (d) D starting from one end and 11th who is on Sushma's right. Ravi 13. Of the six members of a panel from another end, what is the is next to Sushma and opposite sitting in a row. E is to the left number of persons in the queue? Girdhar. who is on Chandra's left. of B, But on the right of A. F is ;fn ,d fljs ls fxuus ij iafDr esa vkidk Savitri is not on Mahima's right on the right of B but is on the LFkku ukSaok gS rFkk nwljs fljs ls fxuus butijopposite Shalini. Who is on left of G who is to the left of C. Shalini's X;kjosa LFkku ij gks] rks iafDr esa O;fDr;ksa dh right? Find the members sitting right ,d fe=kksa dh lewg dks v"VHkqt ds vkdkj esa la[;k fdruh gS\ in the middle. izR;sd dksus esa ,d Nk=k ds vuqlkj cSBk;k (a) 20 (b) 19 ,d iSuy ds N% yksx ,d iafDr esa cSBs E, gSA (c) 21 (d) 18 tkrk gS lHkh dk eq[k dsUnz dh vksj gSA efgek] B ds ck¡;s gS fdUrq A ds nk¡;s gSA F, B dh

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(a) 32 (b) 42 (c) 45 (d) 46 row. T is on the right of Z, M is (a) 12 (b) 14 23. In an examination, Rahul got on the left of Z but M is on the (c) 15 (d) 16 the 11th rank and he was 47th right of L, T is on the left of Q 19. In a row of trees, A is 7th from from the bottom among those who is sitting first from the left? left end and 14th from the right who passed. 3 students could T, Z ds ik¡p Nk=k ,d iafDr esa cSBs gSaA nka;s end. How many trees are there not appear for the exam.1 gSA M, Z ds ck¡;h vksj gS fdUrq M, L ds in the row? student failed. What is the total nk¡;h vksj T, gSA Q ds ck¡;h vksj gSA ck¡;s ls isM+ksa dh iafDr esa] fljs ls 7 ok¡ gS rFkk number of students? A ck¡;s igyk dkSu gS\ nk¡;s fljs 14 ls ok¡ gSA rks bl iafDr esa fdrus ,d d{kk esa jkgqy11usokW LFkku izkIr fd;k (a) Z (b) Q isM+ gSa\ 47 esa uhps ls rFkk og mÙkh.kZ Nk=kksa dh lwph (c) T (d) L osa LFkku ij gSA rhu Nk=k] ijh{kk ugha nsrs gSa (a) 18 (b) 19 At a college party five girls are rFkk ,d vuqÙkh.kZ gks tkrk gSA d{kk esa dqy (c) 20 (d) 21 sitting in a row. P is to the left fdrus Nk=k gSa\ of M and to right of O. R is 20. Sita is elder than Swapna. sitting to the right of N, but to Lavanya is elder than Swapna (a) 60 (b) 62 left of O. Who is sitting in the but younger than Sita. (c) 59 (d) 61 middle? Suvarna is younger than both 24. Five birds, Crow, Pigeon, Little Hari and Swapna. Swapna is ,d dkWyst ikVhZ esa ik¡p yM+fd;k¡ ,d iafDr Pigeon, Big Crow and Eagle fly elder than Hari. Who is the esa cSBhP,gSaA rFkk ds M ds ck¡;h vksj gS O one after other from a tree youngest? branch. Big Crow flew after Crow nk¡;h vksjR, gSA N ds nk¡;h vksj gS] fdUrq O but is ahead of Eagle. Pigeon is lhrk] Loiuk ls cM+h gSA yoU;k Loiuk ls ds ck¡;h vksj gSA eè; esa dkSu cSBk gS\ between cM+k gS fdUrq lhrk ls NksVk gSA Lo.kkZ] gjh o Crow and Big Crow. (a) O (b) R Little Pigeon is before Crow. (c) P (d) M Loiuk nksuksa ls NksVk gSA Loiuk] gjh ls cM+h Which bird is the last? Five birds are sitting on a tree. gSA lcls NksVk dkSu gS\ ik¡p i{kh] dkSvk] dcwrj] NksVk dcwrj cM+k The Pigeon is to the right of the (a) Sita/lhrk Parrot. The Sparrow is above dkSvk rFkk phy ,d ds ckn ,d 'kk[kk ij (b) Lavanya/ yoU;k the Parrot. The Crow is next to tkrs gSaA cM+k1dkSvk] dkS, ds ckn gS ijUrq the Pigeon. The Crane is below (c) Suvarna/ Lo.kkZ phy ls igys gSA dcwrj dkS,s rFkk cM+s dkS, the Crow. Which bird is at the ds eè; esa gSA NksVk dcwrj dkSvk ls igys gS (d) Hari/gjh centre? rks var esa dkSu lk i{kh gS\ 21.rksrs In ds a row of girls, Kamla is 9th ik¡p ia{kh ,d isM+ ij cSBs gSaA dcwrj] the left and Veena is 16th (a) Pigeon/ dcwrj nk¡;h vksj gSA xkSjS;k] rksrs ls Åij gSA from dkSvk] from the right. If they (b) Big Crow/cM+k dkSvk dcwrj ls vxyk gSA lkjl] dkSa,s ls uhps gSA interchange their positions, dkSu lk i{kh fcYdqy eè; esa gS\ (c) Eagle/phy Kamla becomes 25th from the (a) Crow/dkSvk left. How many girls are there (d) None of these/buesa ls dksbZ ugha (b) Pigeon/ dcwrj in the row? 25. P, Q, R and S are four friends. P, (c) Parrot/rksrk yM+fd;ksa dh iafDr esa] deyk]9 ck¡;s oha ls is shorter than Q but taller than (d) Sparrow/xkSjs;k gS rFkk ohuk nk¡;s ls gSA ;fn os viuh R who is shorter than S. Who is 16 oha the shortest among all? fLFkfr vkil esa cny ysrs gSa] rks deyk vc YEAR-2011 P, Q, R vkSj S pkj fe=k gSA P, Q ls NksVk ck¡;s ls25 oha gks tkrh gSA rks bl iafDr esa B is twice as old as A but twice R ls cM+k gS tksfd S ls NksVk gSA gS fdUrq fdruh yM+fd;k¡ gSa\ younger than F. C is half the lcls NksVk dkSu gS\ (a) 34 (b) 36 age of A but is twice older than (a) P (b) Q D. Who is the second oldest? (c) 40 (d) 41 (c) R (d) S B, A ls nqx quk cM+kijarq gSF ls nqx quk NksVk gSA 22. In a class Rajan got the 11th 26. In a row of trees one tree is the rank and he was 31st from the quk C, A dh vk/h mez dk gS fdUrq D ls nqx 7th from both end of the row. of the list of boys passed. cM+k gSA mez esa nwljk lcls cM+k dkSubottom gS\ How many trees are there in Three boys did not take the the row? (a) B (b) F examination and one failed. (c) D (d) C isM+ksas dh iafDr esa ,d isM+7nksuksa ok¡ Nksj ls What is the total strength of Ramesh ranks 13th in a class gSA iafDr esa fdrus isM+ gS\ the class? of 33 students. There are 5 (a) 11 (b) 13 ,d d{kk esa jktu us 11ok¡ LFkku izkIr fd;k students below Suresh rank (c) 15 (d) 14 wise. How many students are 31 esa uhps ls rFkk og mÙkh.kZ Nk=kksa dh lwph

Ra

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16.

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15.

17.

18.

282

Rakesh Yadav Readers Publication Pvt. Ltd

yo

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;fn jke] ';ke ls de rst nkSM+rk gS rFkk ik¡p ftyksa ds lewg esa vdcjiqj] iQrsgiqj ls NksVk gSA /uckn] ikykew ls cM+k gS rFkk ckjkcadh ';ke] yky ds cjkcj xfr esa nkSM+rk gS ijarq iQrsgiqj ls cM+k gS fdUrq mruk cM+k ugha gS VkWe ls de rst nkSM+rk gSA lcls rst dkSu ftruk ikykewA lcls cM+k ftyk dkSu lk gS\ nkSM+rk gS\ (a) Lal/yky (a) Akbarpur/vdcjiqj (b) Shyam/ ';ke vf[kys' k] 'khcw ls yEck gSA veu]' vf[kys k (b) Fatehpur/iQrsgiqj ds ftruk yEck ugha gSA fdUrq sftUn og j r (c) Tom/ VkWe (c) Dhanbad//uckn ls yEck gSA 'khcw] veu ds ftruk yEck (d) Tom and Lal/VkWe rFkk yky (d) Palamu/ikykew ugha gSA fdUrq 'khcw] sftUn rj ls yEck gSA31. F has less money than H but 35. There are five houses A, B, C, D lcls yEck dkSu gS\ more than G. E has more than and O in a row. A is right side of

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ur

27. Akhilesh is taller than Sheebu. Aman is not as tall as Akhilesh but is taller than Tejinder. Sheebu is also not as tall as Aman but Sheebu is taller than Tejinder. Who is the tallest?

ah

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(a) Akhilesh (b) Sheebu (c) Aman (d) Tejinder 28. There are five friends-Satish, Kishore, Mohan, Anil and Rajesh. Mohan is tallest. Satish is shorter than Kishore but taller than Rajesh. Anil is little shorter than Kishore but Anil is little taller than Satish. Who is taller than Rajesh but shorter than Anil?

B and left side of C. O is in the right side of A. B is right of D. Which house is in the middle? ik¡p ?kjA, B, C, D rFkkO ,d iafDr esa C ls ck¡;k gSA O, gSA, B ls nk¡;k gS rFkk A ls nk¡;k gSA B, D ls nk¡;k gSA eè; esa dkSu

Si

r

F but less than H. Who is the poorest? F ds ikl H ls de iSls gSa ijUrq G ls vf/ d iSls gSaA E ds ikl F ls vf/d gSA ijUrq H ls de gSA lcls xjhc dkSu gS\ (a) F (b) E (c) H (d) G 32. I f Anil runs less fast than Sunil and Sunil runs as fast but not faster than Suraj. Who is 36. ik¡p fe=k gSa lrh'k] fd'kksj] eksgu] vfuy vkSj the faster than runs Suraj ?

lk ?kj gS\

Ya d

av

(a) O (b) A (c) B (d) D Four persons M, N O and P are playing cards. M is one the jkts'kA eksgu lcls yEck gSA lrh'k] fd'kksj;fn vfuy] lquhy ls de rst nkSM+rk gS rFkkright of N and P is one left of O. Then which of the following are ls NksVk gS ijUrq jkts'k ls yEck gSA vfuy lquhy] lwjtdscjkcj nkSM+rk ijarqgS lwjt ls rst partners?

Ashish and Rakesh. Shailendra is shorter than Keshav but taller than Rakesh. Madhav is the tallest. Ashish is a little shorter than Keshav and little taller than Shailendra. Who is the shortest?

ke

29. There are five friends Suresh, Kaushal, Madhur, Amit and Ramesh. Suresh is shorter than Kaushal but taller than Ramesh. Madhur is the tallest. Amit is a little shorter than Kaushal but little taller than Suresh. If they stand in the order of their heights who will be the shortest?

sh

fd'kksj ls NksVk gS ijUrq vfuy lrh'k ls dqNughankSM+rk gS rks ls rst lwjt dkSunkSM+rk S\ g yEck gSA dkSu O;fDr jkts'k ls yEck gS fdUrq (a) As fast as Anil vfuy ls NksVk gSA (b) Faster than Sunil (a) Anil/vfuy (c) Faster than Anil (b) Kishore/ fd'kksj (d) Less fast than Anil (c) Rajesh/ jkts'k 33. There are five friends 37. (d) Satish/ lrh'k Shailendra, Keshav, Madhav,

P iÙks [ksy jgs gSaA pkj O;fDrM, N O rFkk M, N ds nk¡;s gSaP,rFkk O ls ck¡;s gSA rks buesa ls dkSu lkFkh gSa\ (a) P and O (b) M and P (c) M and N (d) N and P Six friends A, B, C , D, E and F are sitting in a row facing East. 'C' is between 'A' and 'E'. 'B' is just to the right of 'E' but left of 'D'. 'F' is not at the right end. Who is between 'B' and 'C'?

Ra

N% fe=k A, B, C , D, E rFkk F ,d iafDr A ogSaA esa iwjc dh vksj eqa[k djds C, cSBs fdUrq E ds eè; gSA B, E ds rqjUr nk¡;k gSD ik¡p fe=k gSa 'kSysUnz] ds'ko] ek/o] vk'kh"k F, nk¡;s Nksj ls var esa ugha cSBk ds ck¡;s gSA rFkk jkds'kA 'kSysUnz] ds'ko ls NksVk gS fdUrq B rFkkC ds eè; esa dkSu gS\ gSA rks jkds'k ls yEck gSA ek/o lcls yEck gSA (a) A (b) D lqjs'k] dkS'ky] e/qj] vfer rFkk jes'k ik¡p vk'kh"k] ds'ko ls dqN NksVk gSA rFkk 'kSysUnz (c) E (d) F fe=k gSaA lqjs'k] dkS'ky ls NksVk gSA fdUrq ls dejes'k yEck gSA lcls NksVk dkSu gS\ 38. Six friends A, B, C , D and E are ls yEck gSA e/qj lcls yEck gSA vfer] (a) Rakesh/ jkds'k sitting in a row facing East. C dkS'ky ls dqN NksVk gSA ijarq lqjs'k ls dqN is between A and E. B is just to 'kSysUnz (b) Shailendra/ yEck gSA ;fn os viuh yEckbZ ds Øe esa [kM+s the right of E but left of D. F is (c) Ashish/ vk'kh"k gksrs gS rks lcls NksVk dkSu gS\ not at the right end. Who is at the left end? (a) Amit/vfer (d) Keshav/ ds'ko N% fe=k (b) Madhur/e/qj A, B, C , D, rFkk E ,d iafDr esa 34. I n a group of five districts iwjc dh vksj eqa[k djds cSBs oE C, AgSaA Akbarpur is smaller than (c) Ramesh/ jes'k Fatehpur, Dhanbad is bigger ds eè; gSA ds rqjUr nk¡;k gS fdUrq B, E D (d) Kaushal/dkS'ky than Palamu and Barabanki is ds ck¡;s gSA F, nk¡;s Nksj ls var esa ugha cSBk 30. I f Ram runs less fast than bigger than Fatehpur but not as gS rks ck¡, Nksj ij dkSu cSBk gSA Shyam and Shyam runs as fast as Lal but less fast than Tom. Who runs fastest?

Rakesh Yadav Readers Publication Pvt. Ltd

big as Palamu. Which district is the biggest?

(a) A (c) C

(b) F (d) B

283

yo

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rksE ds nka, fdrus O;fDr cSBs gS\

sm

(b) 2 (d) 4

ah

bo

(a) 1 (c) 3

Jhfu] vuyq ls yEch gSA jkxw] pUnzq ls yEck gS fdUrq o`Unk ls NksVk gSA Jhfu] pUnzq ls Nks gSA lcls yEck dkSu gS\ (a) Srini/ Jhfu (b) Ragu/jkxw

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YEAR-2012

ss re

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(d) Brinda/o`Unk

or

(c) Chandru/pUnzq

43. Umesh is taller than Satish, A, B, C , D, E rFkk F ,d iafDr N% fe=k Suresh is shorter than Neeraj but taller than Umesh. Who is C, A gSaA esa iwjc dh vksj eq[k djds cSBs o fdUrq the tallest among them? E ds eè; gSA B, E ds rqjUr nk¡;k gSD

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ob

39. Six friends A, B, C, D, E and F are sitting in a row facing East. C is between A and E. B is just to the right of E but left of D, F is not at the right end. Who is at the right end?

44. K is more beautiful than B. B is not as beautiful as Y. J is not as beautiful as B or Y. Whose beauty is in the least degree?

(d) FD

(c) Y

41. Six friends A, B, C, D, E and F are sitting in a row facing East. C is between A and E. B is just to the right of E but left of D.F is not at the right end. Who is to the left of A?

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49. Kathir is senior of Ganesh. Ganesh is senior than Apparu. Apparu, is junior of Raju. Raju K, B ls vf/d lqUnj gSA B mruk lqUnj ugha is junior of Ganesh. Who is the gS ftrukY gSAJ mruk lqUnj ugha gS ftruk most senior? dkfFkj] x.ks'k ls lhfu;j gS] x.ks'k] vik: ls B ;k YA lcls de lqUnj dkSu gS\

Ya d

(c) EB

r

(d) Neeraj/ uhjt

Si

(c) Satish/ lrh'k

(b) J

(d) K

45. Age of Naren is equal to Naveen as they are twins. Nakul is younger than Naren. Priyanka is younger than Balaji but elder than Naveen. Who is the eldest of all?

sh

(b) FA

dyk] uhyk ls cM+h gSA uhyk] ckyk ls NksVh gS] lcls cM+k dkSu gS\ (a) Rita/jhrk (b) Kala/dyk (c) Bala/ckyk (d) Nila/uhyk

(b) Suresh/ lqjs'k

N% fe=k A, B, C , D, E rFkk F ,d iafDr A ogSaA esa iwjc dh vksj eqa[k djds C, cSBs E ds eè; gSA B, E ds rqjUr nk¡;k gSD fdUrq ds ck¡;s gSA F, nk¡;s Nksj ls var esa ugha cSBk (a) B gSA rks D dh rjiQ dkSu lh tksM+h cSBh gS\ (a) CE

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40. Six friends A, B, C, D, E and F are sitting in a row facing East. 'C' is between 'A' and 'E'. 'B' is just to the right of 'E' but left of 'D', 'F' is not at the right end. Which pair is sitting by the side of, 'D'?

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48. Bima is younger than Rita. Rita is younger than Kala. Kala mes'k] lrh'k ls yEck gSA lqjs'k] uhjt ls is elder than Nila. Nila is F, nk¡;s Nksj ls var esa ugha cSBk ds ck¡;s gSA younger than Bala. Who is the NksVk gS fdUrq mes'k ls yEck gS] lcls yEck gSA rks nka, Nksj ij dkSu cSBk gS\ eldest of all of them? dkSu gS\ (a) D (b) B chek] jhrk ls NksVh gSA jhrk] dyk ls NksVh gSA (a) Umesh/ mes'k (c) E (d) C

lhfu;j gSA vik:] jktw ls twfu;j gSA jktw] x.ks'k ls twfu;j gSA lcls lhfu;j dkSu gS\ (a) Ganesh/ x.ks'k (b) Raju/jktw (c) Kathir/dkfFkj (d) Apparu/vik:

50. Four kids P, Q, R and S are up

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ujsu dh vk;q uohu ds cjkcj gS D;ksafd oson the ladder. P is further up tqM+ok gSA udqy] ujsu ls NksVk gSA the fiz;adk] ladder than Q. Q is between ckykth ls NksVh gS fdUrq uohu ls cM+hPgSA and R. If S is further up than N% fe=k A, B, C , D, E rFkk F ,d iafDr P. Who is the third from the lcls cM+k dkSu gS\ esa iwjc dh vksj eqa[k djds C, cSBs A ogSaA bottom? (a) Naren/ ujsu (b) Balaji/ckykth E ds eè; gSA B, E ds rqjUr nk¡;k gSD fdUrq pkj cPpsP, Q, R rFkk S ,d lh 'kSysUnz >jkds'k 34.(c)

Fatehpur > Akbarpur

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nksuks dFkuksa ls] 32.(c)

F is to left of A. F, A ds ck¡, gS 42.(b)

LEFT F A C E B D

LEFT F A C E B D RIGHT

RIGHT

B and D are to the right of E. B vkSj D,E ds nk¡, gSA 43.(d) Umesh>Satish mes'k >lrh'k Neeraj>Suresh>Umesh uhjt>lqjs'k >mes'k Neeraj>Suresh>Umesh>Satish >>mes'k >lrh'k uhjt>lqjs'k 44.(b) K>B Y>B B,Y>J K,Y>B>J 45.(b) Naren = Naveen> Nakul ujsu= uohu> udqy Balaji > Priyanka > Naveen ckykth> fiz;kadk > uohu Clearly, Balaji is the eldest.

Li"Vr% ckykth lcls cM+k gSA

vdcjiqj

(F) (A) Dhanbad > Palamu

Rakesh Yadav Readers Publication Pvt. Ltd

46.(a)

W C

t >X>B

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RIGHT

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H>F>G ......(i) H>E>F .....(ii) From both the statements,

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Tom>Lal = Shyam>Ram

F A C E B D

RIGHT

LEFT F A C E B D

dp

> vfer > lqjs'k > jes'k e/qj > dkS'ky

H>E>F>G Sunil = Suraj > Anil

LEFT

RIGHT

38.(b)

or

Kaushal>Suresh>Ramesh

dkS'ky >lqjs'k >jes'k

31.(d)

LEFT F A C E B D

>fd'kksj >vfuy>lrh'k> jkts'k eksgu

>yky = ';ke >jke VkWe

P

N vkSj P lgHkkxh gSA

Mohan > Kishore > Anil > Satish > Rajesh

30.(c)

B and E are sitting by the side of D. B vkSj E, D ds ,d rjiQ cSBrs gSa 41.(d)

M N and P are partners.

37.(c)

Madhur > Kaushal > Amit > Suresh > Ramesh

RIGHT

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N

>vfuy>lrh'k fd'kksj Now,/vc]

dkS'ky >vfer >lqjs'k

C/O O/C

R I G H T

O

36.(d)

Kishore>Anil>Satish

Kaushal>Amit>Suresh

D B A

T

>lrh'k>jkts'k fd'kksj

29.(c)

L E F

ob

35.(b)

= 7+7–1 = 13 Akhilesh>Sheebu vf[kys'k >f'kcw Akhilesh> Aman>Tejinder vf[kys'k > veu >rsftUnj Aman>Sheebu>Tejinder >rsftUnj veu >f'kcw Akhilesh> Aman >Sheebu> Tejinder > veu >f'kcw > rsftUnj vf[kys'k Kishore>Satish>Rajesh

bo

28.(d)

> iQrsgiqj (F)

Now,/vc] D>P>B>F>A

iafDr esa dqy isM+ks dh la[;k 27.(a)

F A C E B D

ah

(P)

> ckjkcadh (B)

LEFT

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26.(b)

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25.(c)

(3) (4) (5) Q>P>R S>R Clearly, R is the shortest. R lcls NksVk gSA Li"Vr% Total number of trees in the row

Q R

59.(d)

Lalit>Prakash, Kishore

yfyr >izdk'k]fd'kksj

Ra

Lalit>Mukesh>Rakesh >jkds'k yfyr >eqds'k Rakesh>Prakash>Kishore jkds'k >izdk'k >fd'kksj Now,/vc] Lalit > Mukesh >Rakesh > Prakash > Kishore yfyr > eqds'k> jkds'k> izdk'k>

60.(b)

fd'kksj

52. (b) K e s h a >Ashish>Shailendra Madhav is the tallest. ds'ko > vk'kh"k> 'kSysUnz

v 61.(a)

292

r Si 64.(c)

Therefore, Lokita scored the lowest.

vr%] lcls de vad izkIr djrh gSA

L>O M>O Thus,/;|fi] L > O > M > N

Praveen

D C B is sitting on the left of E. 65.(b) B, E ds cka;h rjiQ cSBk gSA N>M X>Y>M N>X Purnima is good in Physics, Now, N>X>Y>M History and Dramatics. N is the most intelligent. iwf.kZek HkkSfrdh] bfrgkl vkSj ukV~;dyk esa N lcls T;knk cqf¼eku gSA vPNh gSA Roshan, Susheel>Hardik 66.(b) >gkfnZd jks'ku] lq'khy Right Hardik>Niza>Harry F gkfnZd >uh”kk >gSjh B Roshan>Susheel jks'ku>lq'khy C Roshan > Susheel > Hardik D > Niza > Harry A jks'ku > lq'khy> gkfnZd > uh”kk > gSjh E Therefore, Roshan is the Left tallest. F is at the right end. vr% jks'ku lcls yEck gSA F nka;h rjiQ gSA M>GQ 67.(d) M N O P Q H is the most lean person Bus O is the middle of the five. in the group. O ik¡pksa ds chp esa gSA H bl lewg esa lcls nqcyk O;fDr gSA Subbu Total number of students in 68.(a) the row Pappu Revathi Subject

Person

Dramatics

Computer Science

Madhavi







×

×

Shalini Anjana

 ×

×

Purnima



  ×

Physics

History

 

  

Maths

 ×

Nimala

×

×







×

iafDr esas dqy fo|kfFkZ;ksa dh la[;k

ek/o lcls yack gSA

53.(d)

Raju

B

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51.(a)

E

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P

B

av

A E

S

D

Ya d

50.(c)

Mohan

Amith

Balu C

E F T

Rakesh < Shailendra < Ashish < Keshav < Madhav Raju>Vasant>Manohar

62.(b)

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L

Priti>Rahul izhrh >jkgqy Rahul>Yamuna>Divya jkgqy >;equk >fnO;k Manju>Manju >eatw eatw Divya>Manju >eatw fnO;k Now, Priti > Yamuna = Divya > Manju > Lokita > vc] izhrh> ;equk= fnO;k> eatw

yksfdrk

57.(a)

58.(c)

63.(d)

.c

R I G H A T

vr%]lyhe lHkh esa lcls NksVk gSA

L E F T

ss re

56.(d)

dp

R I G H T

or

55.(a)

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Li"Vr%] dkfFkj lcls ofj"B gSaA

ob

Kathir>Ganesh>Apparu dkfFkj >x.ks'k >vik: Ganesh>Raju>Apparu >jktw >vik: x.ks'k Clearly, Kathir is the most senior

bo

Li"Vr% jktw bl lewg esa lcls yack gSA CABDE

lEHkor% bu lHkh esa dkyk lcls cM+k gSA 49.(c)

ah

Kala>Rita>Bima dkyk>jhrk>chek Kala>Nila dkyk>uhyk Bala>Nila ckyk >uhyk Most probably Kala is the elder of all of them.

sm

Manohar>Jayant>Dutta euksgj >t;ar >nÙkk Clearly, Raju is the tallest in the group. 54.(d)

Age of Raju = x + 1 year = x +1 o"kZ jktw dh vk;q Age of Smith = x + 2 years fLeFk dh vk;q = x +2 o"kZ Age of Vini = x + 3 years Therefore, Salim is the youngest of all.

ur

jktw >o'kar >euksgj

vr%]c`ank lcls cM+h gSA 48.(b)

yo

w.

ww

47.(d)

Therefor, B is the poorest. vr%B lcls xjhc gSA Srini>Anlu Jhfu > vYuw Brinda>Ragu>Chandru c`ank > j?kw > pUnzq Chandru>Jhfu pUnzq > l`fu Therefore, Brinda is the tallest.

= 7+11–1=17 Suppose the age of Salim is x years x o"kZ gSA ekuk lyhe dh mez

Nisha

Aruna Keerthana

Rakesh Yadav Readers Publication Pvt. Ltd

yo

w.

ww

75.(d)

L E B F T

Son

Father

D E/F

R I A G H T

E/F

C

'D' is sitting on the right of 'B' D, B ds nka;s cSBk gSA

Ya d

Grandfather

Mother

Daughter

79.(b)

72.(b) Subject

r

(45–20) + 1 = 26 88. (b) R > S > T > P > Q 89.(a) th

12 Eng.

L

th

16 16+6

Math nd

22

R

= 22 ck,a Nksj ls xf.kr dh iqLrd dk LFkku = 12 nk,a Nksj ls xf.kr dh iqLrd dk LFkku

Computer Science

Physics

History

Maths

Madhavi







×

×

iafDr esa iqLrdksa dh dqy la[;k

Shalini Anjana

 ×

×

= (22 + 12) – 1 = 33



 

  

×

Purnima

  ×

 ×

Nimala

×

×







P Q T S R

90.(b)

Q, R ds ck,a rhljs LFkku ij cSBk gSA

rkjd dk nksuksa Nksjksa = ls 18LFkku

91.(c)

ek/oh HkkSfrdh ukV~;dyk vkSj dEI;wVj foKku esa vPNh gSA

rks dqy la[;k= (18 + 18) – 1 = 35 92. (a) Book 1 = 2(Book 2)

80.(a) L G D E F B C A E

Right

Tapti

Sudha

Padma

Krishan

Rama

LEFT

Ra

ke

Madhavi is good in Physics, Dramatics and Computer Science.

73.(b)

F T

R I G H T

81.(d) C

D

Kanna>Malik/ Ñ'k>ekfyd E

Krish>De, Malik

Ñ'k>ns]efyd

B F

A

Krish >Veena, Malik

E is to the left of F.

fØ'k>ohuk >efyd

E, F ds cka;s gSA

Kanna

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Shena

Krish>Dev>Veena>Malik

Meena

nso >ohuk

Gayathri

82.(a) Malini

Dev>Veena

Anjana

74.(d)

ck,a Nksj ls uhacw ds o`{k dk LFkku

Dramatics

sh

Person

nk,a Nksj ls uhacw ds o`{k=dk 20LFkku

Si

= 17+22–1 = 38

A Now, A is to the left of D. 71.(b)

85. (c) 86.(a) Unity towers = Cyber towers > pearl towers > sky towers 87.(a) iafDr esa dqy o`{k = 45

iafDr esa dqy fo|kfFkZ;ksa dh la[;k

F D

av

F D

om

78.(d)

B

C

A and C are setting together. A vkSjC ,d lkFk cSBs gSaA

.c

= 45 –15 +1 = 31st Total number of students in the line

A

ss re

vafre ls jkE;k dh fLFkfr

E B

C

D

dp

77.(c)

or

70(c)

E B

84.(a)

= 20+35–1 = 54 The rank of Ramya from the last  46–22+1 = 25th Rank of Neha from the last

76.(d)

= 11+9–1 = 19

.w

la[;k

iafDr esa dqy vknfe;ksa dh la[;k

ob

9th Total number of people in the row/ iafDr esa yksxksa dh dqy

bo

11th B

X>Z>Y X>Z>W Clearly, X is the eldest. Li"Vr% X lcls cM+k gSA

ah

7th A

69.(c)

ehuk vafre ls nwljs ij gSA 83.(a)

sm

20th K M 35th 30th Total number of men in the row

'kCcw iIiw ds cka, gSA

ur

Subbu is to the left of Pappu.

Meena is second from the last.

Book 3 =

2 Book 2 = Book 2 2

Book 4 = Book 2 + 60 Book 4 = Book 1 - 60 (ii)

lehdj.k i vkSjii dks gy djus ij Book 2 = 120 Book 1 = 2 × 120 = 240 Book 4 = 120 + 60 = 180 Book 3 = 120 Book 1 > Book 4 > Book 2 > Book3 93. (c) 8

Js; 10 euq7 8 + 10 + 7 – 1 = 24 94.(c) eksVks > jsM> cqy> ,uthZ> ykW;u 95. (c) R > Q > P > S > T S ls nwljk mQ¡pkbZ ds ?kVrs gq, Øe esa vafre

gSA

293

yo

w.

ww sm

ur ah

11 bo

ob

.w

CHAPTER ss re

dp

or

CLOCK 

2. Angle/dks.k

9 'kCn ty esa leku gSA

3. Time/ le;

5 + 4 (9) B , C , D , E, K, O, X, I, H

4. Error/v'kqf¼ Image/Mirror/Water Image

B B

r

(1)

(6)

Si

niZ.k esa cuus okyk izfrfcEc ,oa ty esa cuus okyk izfrfcEc (1)

av



(7)

Ya d

In mirror, left Right (change) Top and bottom (same)

niZ.k esa nk,a ,oa ck,a fgLls dks fn'kk cny & mirror and water fiq. jgh gS tcfd Åij ,oa uhps vkd`fr esa dksbZ niZ.k rFkk ty vkd`fr ifjorZu ugha gSA (8)

ke

Ra

Z

vkd`fr dk niZ.k izfrfcEc o ty izfrfcEc ty esa Åijh fn'kk esa & uhps fn'kk esa • ck;ka & nk;ka (3)

Z

Z × 4 Letters in mirror and water image is same. i.e. O, X, I, H 4 'kCnksa ds fy, niZ.k rFkk ty ,d leku

P Q

Z ×

jgrs gSA

F

(10)

L (4)

H

P Q

F L

O

I

I

O

X

X

E K E K 11 Letters in Mirror image is s ame 11 'kCn niZ.k esa leku gSA

A A MM T T U U V V

Clock Mirror/?kM+h niZ.k

HR : MIN 12 : 00 or 11 : 60 To know the mirror image subtract from Clock 12 : 00 or 11 : 60

niZ.k ?kM+h 12 : 00 or 11 : 60 esa (&) djs nsxsaA eg : 8 : 40

I Fig

H Fig

O

294

H

D D

9–3 10 – 2 12 – 12 7–5 6–6

(9)

In water, Top – bottom (change) Left and Right (Same)

C C

WWY Y 

sh

(2)

9 Letters in water image is s ame

X

eg : 2 : 47

HR : MIN 12 : 00 – 8 : 40 3 : 20 11 : 60 –2 : 47 9 : 13

eg : 22 : 13 or 10 : 13

Rakesh Yadav Readers Publication Pvt. Ltd

om

(5)

.c

1. Image/izfrfcac

yo

w.

ww ur



ss re

& from 2 P.M ––– 4 P.M

If in Options 10 : 10 is not given, than reduce 1 hour

2 cts ls ––– 4 cts rd

vxj fodYi esa10 : 10 ugh fn;k jgs rks 1 ?kaVk de dj nsaA

?kaVs okyh––lqbZ 2?kaVk =120feuV

eg :

2 : 53

1 hr less

17 : 90 – 2 : 53 15 : 37 3 : 37 2 : 37

Sol.

Ra

r

Si

60  5 – 11 40 2

=

300 – 440 140 = = 70° Ans. 2 2

av

eg :

225 =112.5° 2

Hr : min 12 : 20 =

Q.

360 = 6° 60

Min hand = 6°/min Hour Hand/feuV okys lqbZ 12 hr/?kaVk –– 360°

360 1 hr/?kaVk –– = 30° 12

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=

11 12 1 10 2 9 3 8 4 7 6 5

Total/dqy=

Minute Hand/feuV okys lqbZ

190 = 95° 2

=

Ans. 

18 : 30 – 10 : 27 08 : 03 Ans. 60 min/feuV––– 360°



225 min/ feu V =

60  5  1110 300  110 = 2 2

60 (H) – 11(min) 2



1 2

feuV 10

eg : 5 : 40 NksVk dks.k Kkr dhft,A

12 :00

?kaVs okyh= lqbZ 1 feuV–

:

or 360° – 95° = 265° Ans. eg : 5 : 40 find smaller Angle

1 Hour hand = 1 min – 2

eg : 22 : 27 ––– 10 : 27 24 hr 12 hr



=

Dff 3 hr 45 min 3 : 45 3 × 60 + 45 = 225 min/feuV

[18:30 vkSj17: 90] nksuksa ,d leku gSA

1 min/feuV–––

CGL – 2015 12:00 Noon to 3 : 00 : 45 min 12:00 nksigj ls3 : 00 : 45 feuV

Note : [18:30 or 17: 90] both are the s ame



120 = 60° Ans. 2

Ya d

18 : 30 – 1 : 30 24 hr 17 : 00 12 hr 5 : 00 Ans.

1 : 30

Q.

=

sh

eg :

8 : 30

ke

eg:

18 : 30 – 8 : 30 10 : 00 Ans. 9 : 00 Ans.

1 2

120 min/feuV =

?kaVk :

5



eg

420 – 330 90 = = 45° Ans. 2 2

or 360° – 45° = 315° Ans. eg.: Hr min

Hour hand –– 2hr =120min

 1 min/feuV =

om

=

.c

60(H) – 11(min) 607 – 1130 = 2 2

30°/ ?kaVs ?kaVs okyh––lqbZ

18 : 30 9 : 00 – 9 : 00 09 : 30 Ans. 8 : 30 Ans.

dp



1 ?kaVs okyh––lqbZ/feuV 2

Hour hand –– 30°/ hr.

18 : 30 Ans. – 8 : 20 10 : 10

eg: 8 : 20

?kaVk : feuV

or

?kM+h dk ty izfrfcEc Kkr djus ds fy, fn, x, le; dks 18 : 30 esa ls ?kVk nsa

Hr : min

.w

Subtract from 18 : 30 to know Clock water Image

7 : 30 find Angle/dks.k Kkr dhft,

ob



1 /min 2

60(H) –11(min) 2

bo

eg:



Hour hand ––

Angle/dks.k:

ah

30 1 = 60 2

60 min/feuV––

sm

11 : 60 –10 : 13 1 : 47

12  60 – 11 20 2

720 – 220 500 = = 250° 2 2

360 – 250° = 110° Ans. or 12 : 20 let 12 be 00

360 = 30° 12

What is the angle between both the hands at 7 : 30

0  60 – 11 20 220 = = 110° 2 2 eg:

Hr

:

min

7 : 30 cts nksuksa lqbZ;ksa ds chp dks.k crkb;s?kaVk : feuV 360° – 45° = (30°+15°) 3 : 24 315 ° or 45° Ans. 3  60 – 11 24 = 11 12 1 2 10 2 7

45°

9 8

3 7

30

4 6 5

=

180 – 264 84 = = 42° Ans. 2 2

295

ob

eg 12 : 00 & 1 : 00 ds chp250° dk

dks.kdc cusxk\

r

7 min = 42°

Sol. 12:00 1:00

420 – 510 90 = = 45° Ans. 2 2

2 (0× 30 + 250°) 11

At what time between 7:00 and 8:00 are the hands at angle 45°

:

12 : 45

2 (A +A2) 11 1

Ya d

le; 45° dk dks.k cukrh gSA

5 11

= 250°/110° Ans. eg: 5 : 30 & 6: 00 A2 : 70° A1 = [Smaller number/NksVk la[;k × 30 5 : 00 & 6 : 00

2 : (210° + 45°) 2nd 11

time/ le;

Ra

Time/le;

2 510 4 × 255 = = (46 ) 11 11 11 2 (210 – 45°) 11 Ist time/(igyh ckj)

2 × 165° = 30° = (7 : 30) Ans. 11 Ist time  –    2nd time    igyh ckj –    nwljh ckj   

296

2 2 (A1 + A2) ( 5× 30 +70°) 11 11

ke

eg: 7 : 00 to 8 : 00 A1 7×30 = 210° A2 45°

Q.

=

2 × (150 70) 11

=

2 × 220 = 40° = 5 : 40Ans. 11

When in between 3:00 and 4:00, minute hand leg by 4 minute from hour hand in a clock? 3:00 vkSj 4:00 ds chp dc feuVdh lqbZ ?k.Vsa dh lwbZ ls ihNs gksxh\ 4 feuV

Sol. 1 min –– 6°

4 min : 24°

2 (3 × 30 – 24°) 11 2 × (90 – 24) 11 2 × 66 = 12 11 3 : 12 Ans.

3 : 14 min At what time between 8 : 00 and 9 : 00 clock are the nanda opposite to each? 8 : 00 rFkk9 : 00 ds chp fdl le; ij

lqbZ;k foijhr fn'kk esa gksxh\ Half : 180°

A1 = 8 × 30 = 240 A2 = 180°

2 × (A1 + A2) 11 2 (240° + 180°) 11

sh

A2 = Given angle/fn;k x;k dks.k

2 × 132 = 24 11

Sol. 8 : 00 & 9 : 00 total : 360°

2 500 5 × 250° = = 45 11 11 7 : 00 rFkk8:00 ds chp nksuksa lqbZ;ka fdl11

Time/le;

Q.

av

Q.

A2

Si

510 510 =7: 11 11

2 (90° + 42°) 11

2 2 A2 = 110° (0 + 110°) = × 11 11 110 = 20° & 12 : 20 Ans. eg when angle between 12 : 00 & 1 : 00 is 250°

om

4 4 46 11  4 = 46 + = 11 11 11

2 (A + A2) 11 1

.c

4 dks.k Kkr dhft, 11

A1 = 3 × 30 = 90°

ss re

= 5 :10 Ans. eg: 12 :00 & 1 : 00 & 110° = ? A1 = 00 × 30 = 0

60  7 – 11 510  = 2 11 =

dp

=

or

46

4 find angle 11

.w

7 : 46

bo

7 : 46

3:00 vkSj4:00 ds chp dc feuVdh lwbZ 7 feuV ?kaVs dh lwbZ ls vkxsa gksxh

2 × 55 = 5 11

A2 = 95° eg:

2 (150 – 95°) 11

ah

180 – 132 48 = = 24° Ans. 2 2

When in between 3:00 and 4:00, minute hand ahead by 7 minute from hour hand in a clock?

sm

=

A1 = 5 × 30 = 150

Q

ur

eg :

yo

w.

ww

eg: 5 : 00 & 6 : 00 ; 95° = ?

3  60 – 1112 3 : 12 = 2

2 840 4 ×420 = = 76 but 11 11 11 2 (240 – 180) 11 2 120 10 × 60 = = 10 11 11 11 It is greater than 60 mins. 60 feuV ls T;knk 8 : 10

10 11

Hence angle 180° comes once in an hour

vr% 180° dk dks.k ,d ?kaVs es fliQZ ,d ckj gh cusxkA

CLOCK Number of times hands of a clock makes 180°/Straight line

Rakesh Yadav Readers Publication Pvt. Ltd

From 11 AM to 5 PM

CLOCK ERROR

11 AM ls 5 PM rd

,d ?kM+h tks fd yxkrkj rst gks jgh gSa jfookjlqcg 8%00 cts 5 feuV ihNsFkh] ;fn ;g eaxyokj lqcg 8%00 cts 7 feuV vkxsgS]rc blus lgh le; de fn[kk;k gksxk\

av

TYPE - I Q.

If in a clock minute hand cross hour hand in 65 minute, then how many times clock in 24 hr goes slow down or fast

Sol.

Sun day– 8 Am––– –5 times 48hr

Ya d

Sol. 11Am –– 5 Pm –– Diff = 6 hr. 12:00 0° ––– ? 6 –– 1 = 5 times Ans. 6 : 00 180°––– ? 6 times Ans. Q. From Sunday 5 pm to Monday 7 pm how many times hands of clock are in straight line.

A clock which moves continuously fast, It lags 5 minute on Sunday 8 am, it is ahead 7 minute on Tuesday 8 am then find when clock shows right time?

r

le; Q.

TYPE - II Q.

Si

(6 : 00) –– 180° 6 –– 1 = 5 times/

om

Ans.

2 pm ls 10 pm rd Sol 2 pm –– 10 pm –– (8Hr) 12:00 0° –– 8 times Ans. 6 : 00 180° – 8 – 1 = 7 times Ans. 3 : 00, 9 : 00 –– 90° 8 × 2 = 16– 2 = 14 times.

60 ×22 = +2 Fast 11 60

.c

Sol. 5pm –– 11pm –– Diff = 6 hr. (12:00 o’ clock) 0° –– 6 time Ans.

=

ss re

5 p.m ls 11 p.m rd

dp

From 5 pm to 11 pm

or

0° = 12 o’clock/cts

5 5 – 60 65 11 11 ×22 × 22 = 60 60

.w

24 ?kaVk – 22 ckj

=

65

ob

24 hr –22 times

bo

2 ?kaVk –– 11 ckj

(12:00 ct s) 0° –– 6 le;

ah

1 pm ls 5 pm rd Sol. 1 pm ––– 5p.m. – (4hr) 12 :00 0° – ? 4 times Ans 6 : 00 180° ? 4 times Ans. 3:00 /9:00 – 90° ? 8 – 1 = 7 times Ans Q, from 2 pm to 10 pm

12 hr–– 11 times

Q.

;fn ,d ?kM+h esa feuV dh lwbZ ?kaVs dh lwbZ dks60 min esa ikj dj ysrh gSA rc ?kM+h 22 hr esa fdruh /heh vFkok rst gks tk,xh\

– 44 ckj Ans. ,d fnu – 24 ?kaVs from 1 pm to 5 pm

sm

Q.

ur

1 ?kaVk –– 1 ckj

yo

w.

ww

1 hr –– 1 times

Tuesday 8 Am – +7 times ;fn ,d ?kM+h es feuV dh lwbZ ?kaVs dh lwbZ dks65 min esa ikj dj ysrh gSa rc ?kaMh slow / fast formula:-1st × total (24 hr) ,d fnu esa fdruh /heh vFkok rst slow  fast jfookj 'kke 5 cts ls lkseokj 'ke 7%00 cts rd ?kM+h dh lwbZ;k fdruh ckj ,d lh/h gks tk,xh\ 5 time = × 48 = 20 Hr js[kk esa gksxh\   5 12

24 hr –– 22 times 22

ke

jfookj5 pm ls lkseokj7 pm rd Sunday/jfookj – 5 p.m. 180°

– Time 65   11  Total time(Hr /days)  Time    

Ra

Monday 5 pm/lkseokj5 pm

1 2 Monday/lkseokj7 pm + 23 24 = 47 times/le; Q.

Number of times hands of a clock make 90º 90° Right angle – 3: 00 o’clock 9:00 o’clock 1 Hr/?kaVs –– 2 times/ckj 12 Hr/?kaVs –– 22 times/ckj 24 Hr/?kaVs –– 44 times/ckj How many times hands of clock makes right angle.

from 65

65

5 5 = 11 11

5 – 65 5 11  24 = 65 11 es ls 65 65

5 11

?kVk nsus ij & vk;sxk 5 24 = 1165

24 fast 143 Hrs.

Q ,d fnu esa ?kM+h dh lwbZ;k fdruh ckjIf in a clock minute hand cross hour hand in 60 minute, then ledks.k cukrh gSA how many times clock in 22 Sol. One days– 24 Hr – 44 times

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lw=k :-1st

  5 – le; 65   11   ( / ) dqy le; ?kV a s fnu   le;    

5 65 – 65 11 Sol.  24 = 65 subtract 65

90° ledks.k– 3: 00 cts, 9:00cts

Q.

5 /hek/ rt s × le; = × 48 12 /hek rt s = 20 ?kaVk

sh

Sol. from sunday 5 pm to Monday 7 pm

hr goes slow down or fast

Sunday 8 am + 20 hr = Monday 4 am

jfookj 8 am + 20 ?kaVk = lkseokj Q.

4am A clock which moves continuously fast, It lags 9 minute on Sunday 10 pm, it is ahead 3 minute on Tuesday 10 pm then find when clock shows right time?

,d ?kM+h tks fd yxkrkj rst gks jgh gS] jfookj lqcg 10% 00 cts 9 feuV hNs i Fkh] Fkh ;fn ;g lkseokj jkr 10%00 cts 3 feuV gks vkxs rc bl us lgh le; dc fn[kk;k gksxk\ Sol. 24 + 12 = 36hr

Sun – 10 am –– –9 min Mon – 10 pm –– +3 min

9 ×36 = 27 hr 12 Sunday – 10 am + 27 hr = Monday = 1pm

297

Sun 7 am + 72 hr Wed 7 Am fd;k x;k gks rc ml le; lgh le; D;k = 3 days Ans. gksxkA tc ?kM+h eaxyokj lqcg 11%00 cts Clock we read fn[kkbZxhA ?kM+h dk le; Sol. Right Wrong lgh xyr Image/izfrfcEc– (a) 12 : 00 or 11

sm

ur

ah

bo

(4)

0° – 12h – 11 times/ckj

Tuesday 9:00 AM

Ans.

24 hr –– 4 hr slow Right wrong

24 20 ––

24 – 4 = 20 hr

W R 20 24

Tuesday 7AM

r

Error

S a me

Sun 7 AM

180° – 12hr – 11 times/ckj

av

Q.

2 (A1 + A2) 11

Si

24 ×50 = 48 HR (Right) 25

48 Hr + 12 hr = 60hr

Ya d

A clock fast 1 hour in each 24 hour. If clock shows correct time on Sunday 9 am then what is the correct time, when clock, shows 11:00 am Tuesday?

time/ty :

=

65

65

5 – time 11 × T.T time 5 – time 11 ×T.T time

= Wrong + Right (?) = xyr + lgh (?)

ke

sh

,d ?kM+h gj 24 ?kV+s esa ,d ?kVsa rst gks tkrh 24 = × 60 = 72 hr gS ;fn blls jfookj lqcg 9%00 cts lgh 20

=

Ra

By The Team of Best Faculties of Mukherjee Nagar

Batches for

STATE POLICE EXAM. 298

Rakesh Yadav Readers Publication Pvt. Ltd

om

50

TYPE - III •

(3)

Tues – 11:00 AM

10 ×60 = 40 hr 15 Sun 7 am +24 hr Mon -7 Am +12 h Mon 7 pm + 4n = Monday 11 pm Ans.

Angle

50 Hr

Sun – 7 am –– –10 min Tue – 7 pm –– –5 min

60hr – 11min 2

(2)

.c

48 + 12 = 60hr

ss re

Sol.

: 60 Water – (b) 17 : 90

25 hr R 24

dp

,d ?kM+h tks fd yxkrkj rst gks jgh gSA24hr W – 10 min vkxs Fkh] jfookj lqcg 7%00 cts ;fn eaxyokj 'kke 7% 00 cts 5 feuV ihNs 25 – Fkh rc blds lgh le; dc fn[kk;k gksxk\ Sun – 9:00 AM

or

.w

ob

A clock which moves continuously fast, It lags 10 minute on Sunday 7 am, it is ahead 5 minute on Tuesday 7 am then find when clock shows right time?

yo

w.

ww

Q.

yo

w.

ww ur sm

EXERCISE – I

ah bo

3 (c)   10 

0

10  (d)   3

What angle is made by minute hand in 59 sec ?

r

(a) 66º (b) 44º (c) 54º (d) 60.5º 59 }kjk ?kM+h dh feuV lqbZ lsdaM esa cuk;k 10. Time in a clock is 3 : 13, what 17. What is the angle between dks.k D;k gksxk\ time will be appear in water? minute and hour hand at 9 : 53? (a) 6º (b) 5º : 13 gSA rks ty izfrfcac ;fn ?kM+h esa3le; feuV vkSj ?kaVs dh lqbZ9 ds chp cts : 53 (c) 5.9º (d) 4.9º esa le; D;k gksxk\ cuk dks.k D;k gksxk\ What angle is made by second (a) 3 : 17 (b) 2 : 17 (a) 121.5º (b) 21.5º hand in 15 sec.? (c) 130º (d) 68.5º (c) 3 : 23 (d) 2 : 13 ?kM+h dh feuV okyh lqbZ }kjk esa11. Time appears in water is 5 : 47, 18. What is the angle between 15 lsda.M cuk;k dks.k D;k gskxk\ minute and hour hand at 12 : 46? what will be correct time in (a) 15º (b) 1.5º watch. 12ds: 46 feuV vkSj ?kaVs dh lqbZ chpcts cuk dks.k D;k gksxk\ ty izfrfcEc esa le;5 : 47 gS rks okLrfod  1 º (c)  (d) 90º   le; D;k gksxk\ (a) 97º (b) 107º 8  (c) 153º (d) 7º (a) 12 : 47 (b) 12 : 43 TYPE -II 19. What is the angle between (c) 1 : 47 (d) 1 : 43 What time is shown by the mirminute and hour hand at 7 : 09? 12. Water image of a clock is showror if the real time is 9 : 27 : 09 feuV vkSj ?kaVs dh lqbZ7 ds chp cts ing time 4 : 42 what is the real ;fn okLrfod le; 9 : 27 gS rks niZ.k time? cuk dks.k D;k gksxk\

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TYPE -III

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;fn ?kM+h esa okLrfod12le;: 23 gS rks 16. What angle is made by minute niZ.k izfrfcac esa le; D;k gksxk\ and hour hand at 4 : 12? (a) 12 : 33 (b) 11 : 37 4 : 12}kjk feuV vkSj ?kaVs dh lqbZ ij fdruk dks.k cuk;sxh\ (c) 12 : 37 (d) 1 : 23

(b) 3º

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(a) 120º

izfrfcac esa D;k le; fn[ksxk\

6.

(a) 3 : 33 (b) 2 : 33 (c) 3 : 27 (d) 2 : 23 Time appears in the mirror 11 : 09. Then what time will be appear in clock?

(a) 1 : 48 (b) 2 : 48 (c) 12 : 48 (d) 12 : 42 13. If time in a clock is 7 : 35 then at what time will be appear in 11 : 09 gS rks ;fn niZ.k izfrfcac esa le; water?

?kM+h esa D;k le; fn[kkbZ iM+sxk\

7.

?kM+h ds ty izfrcEc dk le; 4 : 42 gSA rks okLrfod le; D;k gksxk\

(a) 1 : 51 (b) 12 : 09 (c) 12 : 51 (d) 1 : 09 Time appears in the mirror 6 : 00. Then what is the correct time in clock?

11ds: 10 feuV vkSj ?kaVs dh lqbZ chpcts cuk dks.k D;k gksxk\ (a) 265º

(b) 175º

(c) 85º (d) 95º ;fn ?kM+h esa7le; : 35 gSA rks ty esa le; 21. What is the angle between D;k gksxk\

(a) 11 : 35 (b) 10 : 35 (c) 10 : 05 (d) 10 : 55 14. If time in a clock is 9 : 11, then ;fn niZ.k izfrfcac esa 6le;: 00 gS rks ?kM+h at what time will appear in water? esa D;k le; fn[kkbZ iM+sxk\

Rakesh Yadav Readers Publication Pvt. Ltd

(a) 120.5º (b) 160.5º (c) 49.5º (d) 19.5º 20. What is the angle between minute and hour hand at 11 : 10?

minute and hour hand at 3 : 56?

feuV vkSj ?kaVs dh lqbZ3 ds chp cts : 56 cuk dks.k D;k gksxk\ (a) 152º (c) 360º

(b) 228º (d) 142º

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9.

?kaVs dh lqbZ36}kjk lsda.M esa cuk;k x;k dks.k D;k gksxk\

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2.

(a) 1740º (b) 2.9º (c) 29º (d) 260º What angle is made by hour hand in 36 sec?

(a) 9 : 19 (b) 8 : 11 (c) 8 : 19 (d) 9 : 30 ;fn ijkofrZr le; 3 : 43 rks ?kM+h 15. dk If time in a clock is 8 : 52, then at what time will appear in water? okLrfod le; D;k gksxk\ ;fn ?kM+h esa8le;: 52 rks ty izfrfcEc (a) 3 : 17 (b) 7 : 17 esa le; D;k gksxk\ (c) 8 : 17 (d) 8 : 43 (a) 8 : 38 (b) 9 : 52 If the time in clock is 12 : 23. (c) 8 : 22 (d) 9 : 38 What is the time in the mirror? TYPE -IV

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8.

29 lsda.M esa cuk;k x;k feuV dh lqbZ }kjk dks.k D;k gksxk\

;fn ?kM+h esa9le;: 11 rks ty izfrfcEc esa le; D;k gksxk\

(a) 5 : 58 (b) 6 : 60 (c) 6 : 00 (d) 6 : 01 If reflecting time is 3 : 43 then the real time of clock is?

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What angle is made by minute hand in 29 sec.?

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TYPE -I

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9 minute loss 11

(d) 6 : 16

4 6 , 6 : 54 11 11

(b) 32

8 minute gain 11

(c) 33

9 minute gain 11

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(c) 3 : 49

8 11

(d) 6 : 5

5 11

24. At what time between 10 to 11 O’ clock minute and hour hand will coincide or makes 0° angle?

1 11

10 (d) 10 : 10 11

(c) 6 : 00

5 (d) 6 : 5 11

29. At what time between 8 to 9 O’ clock the minute and hour will apart 7 minutes to each other?

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3 (b) 2 : 27 11

(b) 8 : 36, 8 : 51

3 11

10 (c) 2 : 10 11

4 (d) 2 : 16 11

(c) 8 : 09, 8 : 47

4 11

300

(b) 6 : 43

7 1 , 6 : 49 11 11

117 minute gain 121

(d) 11

117 minute loss 121

32. A clock is set right at 8 a.m. on

,d ?kM+h dks jfookj lqcg 8 cts ij lsV fd;k tkrk gSA ;g feuV rst gks 24 ?kaVs 8 esa 9 cts tkrh gSA tM+h vkus okys jfookj ds jkr D;k le; n'kkZ;sxh\

9 (d) 8 : 17, 8 : 28 11

TYPE -VI 30. The minute hand of a clock overtakes the hour hand at intervals of 64 minutes of correct time. How much a day does the clock gain or lose?

6 esa ?kM+h vkSj7 ds chp fdl le; ?kaVs vkSj 90° dk dks.k cuk;sxhA feuV dh lqbZ;ka

2 7 , 6 : 43 11 11

(c) 11

time when the clock indicates 9 p.m. on upcoming sunday?

3 (a) 8 : 42, 8 : 51 11

8 (a) 2 : 32 11

(a) 6 : 38

109 minute loss 121

8 esa ?kM+h vkSj9 ds chp fdrus cts ?kaVs vkSj sunday. It gains 8 minutes in feuV dh lqbZ ,d nwljs7 ls feuV nwj gksxh\ 24 houus. What is the correct

?kM+h vkSj3 ds chp fdl le; ?kaVs vkSj 2 esa 90° dk dks.k cuk;sxh\ feuV dh lqbZ;ka

26. At what time between 6 to 7 O’ clock minute and hour hand will be at right angle or makes 90° angle?

(b) 11

(b) 6 : 60

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25. At what time between 2 to 3 O’ clock minute and hour hand will be at right angle to each other or makes 90° angle

,d ?kM+h dh feuV okyh lqbZ ?kaVs okyh lqbZ dks 66 feuV ds varjky ij ikj djrh gSA rks iwjs fnu esa ?kM+h fdruk lqLr ;k rst gksxh\

?kaM+h esa7 ds chp esa fdrus cts ?kaVs (a) 11 109 minute gain 6 vkSj 121 vkSj feuV dh lqbZ;ka ,d nwljs ds foijhr ;k 180° dk dks.k cuk;sxh\

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6 (c) 10 : 54 11

6 11

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2 (b) 10 : 38 11

(d) 3 : 54

tervals of 66 minute of correct time. How much a day does the clock gain or lose?

11

28. When did the minute and hour hand makes 180° angle between 6 to 7 O’ clock?

10esa ?kaM+h o 11 ds chp fdrus cts ?kaVs vkSj feuV dh lqbZ ,d nwljs ds Åij ;k 6 (a) 6 : 54 0° dk dks.k cukrs gS\ lkikrh gksxh rFkk 11

7 (a) 10 : 43 11

(b) 3 : 38

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(c) 6 : 32

11

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7 (b) 6 : 43 11

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(a) 3 : 43

2 (a) 6 : 38 11

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3 vkSj ?kaM+h esa4 ds chp esa fdrus cts ?kaVs 8 vkSj feuV dh lqbZ;ka ,d nwljs ds foijhr ;k (d) 32 11 minute loss 180º dk dks.k cuk;sxh\ 6 o esa 7 ds chp fdrus cts ?kaVs vkSj feuV ?kaM+h 31. The minute hand of a clock dh lqbZ ,d nwljs ds Åij ;k lkikrh gksxhA overtakes the hour hand at in7 2

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23. At what time between 6 to 7 O’ clock minute and hour hand will coincide?

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27. At what time between 3 to 4 O’ clock minute and hour hand are opposite to each other?

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(b) 110º (d) 20º

1 4 , 6 : 16 11 11

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(a) 260º (c) 120º

(c) 6 : 49

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feuV vkSj ?kaVs dh lqbZ chpcts 12ds: 20 cuk dks.k D;k gksxk\

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22. What is the angle between minute and hour hand at 12 : 20?

(a) 9 P.M

(b) 8 : 30 P.M

(c) 8 P.M

(d) 10 P.M

33. A clock is set right at 10 A.M on sunday. It loses 8 minutes in 24 hours. What is the correct time when the clock indicates 9 P.M on next sunday?

,d ?kM+h dks jfookj lqcg 10 cts ij lsV fd;k tkrk gSA ;g feuV /heh 24 ?kaVs8 esa 9 cts gks tkrh gSA ?kM+h vxys jfookj ds jkf=k D;k okLrfod le; n'kkZ;sxh\ ,d ?kM+h dh feuV okyh lqbZ ?kaVs okyh lqbZ dks (b) 9 A.M 64 feuV ds varjky ij ikj djrh gSA rks iwjs (a) 9 P.M (d) 10 P.M fnu esa ?kM+h fdruk lqLr ;k rst gksxh\ (c) 10 A.M Rakesh Yadav Readers Publication Pvt. Ltd

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esa okLrfod le; dc n'kkZ;k Fkk\

,d ?kM+h tks leku :i ls lqLr gksrh gSA xq:okj lqcg6 cts 3 feuV rst gks tkrh gS rFkk vkus okys cq/okj5'kke cts 3 feuV 12 lsd.M lqLr gks tkrh gS rks ?kM+h us okLrfod le; dc n'kkZ;k Fkk\

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(a) 2 A.M Thursday (b) 6 P.M Wednesday (c) 1 A.M Wednesday (d) 6 P.M 35. A watch which loses uniformaly ,d ?kM+h tks leku :i ls rst gksrh gS jfookj lqcg 9.00 cts 4 feuV lqLr gks tkrh gSA is 3 minute fast at 6 A.M on Thursday and is 3 minute 12 9.00 rFkk vkus okys 'kqØokj jkrcts 4 sec. slow at 5 P.M on upcoming feuV15 lsda.M rst gks tkrh gSA bl ?kM+hwednesday. When was it correct?

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(a) 9 P.M on Sunday

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34. A watch which gains uniformaly is 4 minute slow at 9 A.M on Sunday, and is 4 minute 15 sec. fast at 9 P.M on upcoming Friday. When was it correct?

(b) 9 A.M on Monday

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(c) 9 A.M on Sunday (d) 8 A.M on Sunday

ANSWER KEYS 5. 6. 7. 8.

(b) (c) (c) (c)

9. 10. 11. 12.

(b) (b) (b) (a)

13. 14. 15. 16.

(d) (c) (d) (c)

17. 18. 19. 20.

(b) (b) (b) (c)

21. 22. 23. 24.

(d) (b) (c) (c)

25. 26. 27. 28.

(b) (c) (c) (c)

29. (b) 30. (b) 31. (b)

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(b) (c) (c) (d)

32. (c) 33. (d)

34. (c) 35. (c)

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1. 2. 3. 4.

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(Exercise- I)

EXERCISE 1.

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PREVIOUS YEAR QUESTIONS By looking in a mirror it appears that it is 6 : 30 in the clock. What is the real time?

(b) 34

vafdr gSa rFkk ?kM+h dks ,d niZ.k ds lkeus j[kk tkrk gSA rFkk ,d O;fDr niZ.k izfrfcEc 10:20 n'kkZ jgk gSA okLrfod le; crkb;s\ esa

8 minutes past 6 11

niZ.k esa ns[kus ij yxrk gS fd6:30 ?kM+h 8 (c) 30 minutes past 6 n'kkZ jgh gSA ?kM+h dk okLrfod le; crkb;s\ 11 (b) 5 : 30 (d) 5 : 00

(d) 32

(SSC 31.07.2005)

2.

After 9’O clock at what time between 9 p.m and 10 p.m will the hour and minute hands of a clock point in opposite direction?

5 minutes past 6 7

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(a) 6 : 30 (c) 6 : 00

(a) 07 : 10 (c) 04 : 50

(b) 02 : 40 (d) 10 : 20 (SSC CGL 05.05.2002 )

6.

(SSC CGL26.06.2011)

A clock only with only dots marking 3, 6, 9 and 12 positions has been kept upside down in front of mirror. A per9 cts ds i'pkr] jkr 9 cts rFkk jkr10 son reads the time in the rects ds chp fdl le; ij ?kaVs okyh lqbZ rFkk flection of the clock as 4.50. 5 feuV] ,d ?kM+h v/Zjkf=k esa igys ?kaVs lqLr feuV okyh lqbZ foijhr fn'kk esa gksxh\ What is the actual time? gks tkrh gSA nwljs ?kaVs 10 ds feuV vUr esa rFkk rhljs ?kaVs ds vUr esa vkSj vkxs (a) 15 minutes past 9 15 feuV ,d ?kM+h3, esa 6, 4, 9, 12 ij fpÉ vafdr 6 ?kaVs Hkh blh izdkj lqLr gks tkrh gSA ckn (b) 16 minutes past 9 gSa rFkk ?kM+h dks ,d niZ.k ds lkeus j[kk tkrk

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4.

A clock goes slow from midnight by 5 min. at the end of the first hour, by 10 min. at the end of the second hour, by 15 min. and the end of the 3rd hour and so on. What will be the time by this clock after 6 hours?

(c) 16

gSA rFkk ,d O;fDr niZ.k izfrfcEc 4:50 esa n'kkZ jgk gSA okLrfod le; crkb;s\

4 minutes past 9 11

(a) 08 : 10 (c) 04 : 50

1 (d) 17 minutes past 9 11 (SSC CGL 19.06.2011 )

3.

(SSC Matric Level 05.05.2002)

5.

At what time are the hand of clocks together between 6 and 7? 6 rFkk7 cts ds chp fdl le; ij ?kaVs o

feuV okyh lqbZ;ka lEikrh gksxh\

(a) 32

(b) 01 : 40 (d) 10 : 20

8 minutes past 6 11

Rakesh Yadav Readers Publication Pvt. Ltd

A clock only with dots marking 3, 6, 9, and 12 O’ clock postions has been kept upside down in front of a mirror. A person reads the time in the reflection of the clock as 10 : 20. What is the actual time?

,d ?kM+h 3, esa6, 9 vkSj12 ij fpg~u

?kM+h D;k le; n'kkZ;sxh\

(a) 6 : 00 am (c) 6 : 30 am

(b) 5 : 30 am (d) 5 : 15 am (SSC CGL 05.05.2002 )

7.

A clock goes fast by one minute during the first hour, by two minutes at the end of the second hour, by 4 minutes at the end of 3rd hour, by eight minutes by the end of 4th hour, and so on. At the end of which hour, will it be fast by just over sixty minutes?

,d ?kM+h igys ?kaVs eas rst gks tkrh 1 feuV 301

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10 : 30 a.m. What was the time gSA nks feuV vxys ?kaVs esa rst gks tkrh gS ,d rFkk ?kM+h 3, esa6, 9 vkSj12 ij fpÉ equiry clerk told this? rhljs o pkSFks ?kaVs esa Øe'k% rFkk vafdr gSa rFkk ?kM+h dks ,d Åijh niZ.k when ds 4 feuV 8 feuV rst gks tkrh gSA rFkk ;gh Øe vkxslkeus j[kk tkrk gSA rFkk ,d O;fDr niZ.kfdlh cl fMiks ls izfr30 feuV esa psUubZ ds pyrk gSA fdl ?kaVs ds vUr 60 esafeuV og izfrfcEc esa 6 : 10 n'kkZ jgk gSA okLrfod fy, cl NwVrh gS iwNrkN fyfid us ;k=kh ls 10 feuV igys dgk fd psUubZ ds fy, cl rst gks tkrh gS\ le; crkb;s\ NwV pwdh gSA rFkk vxyh 10:30 cl lqcg (a) Fifth (b) Sixth (a) 06 : 50 (b) 12 : 40 cts fudyrh gS tc iwNrkN fyfid esa ;gka (c) Seventh (d) Eighth (c) 11 : 20 (d) 6 : 10 rd le; D;k gks jgk Fkk\

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(a) 10 : 20 a.m. (b) 10 : 10 a.m. (c) 10 : 00 a.m. (d) 09 : 50 a.m.

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(SSC LDC 28.10.2012)

12. If 50 minutes ago, it was 45 minutes past four O’ clock, how many minutes is it until six O’

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A clock with only dots marking 3, 6, 9 and 12 positions has been kept upside down in front of a mirror. A person reads the time in the reflection as 9.50. What is the actual time?

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(SSC 12.05.2002)

10. A clock with only dot markings 3, 6, 9 and 12 positions has been kept upside down in front of a mirror. A person reads the time in the reflections of the clock as 12 : 30 the actual that will be

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8.

(SSC CGL 5.05.2002 )

(a) 2 : 15 (c) 8 : 50

(b) 8 : 40 (d) 4 : 15

(a) 12 O’ clock (b) 12 : 30 (c) 6 O’ clock (d) 03 : 45

(SSC CGL 05.05.2002 )

(SSC 12.05.2002 )

A clock with only dots marking 3, 6, 9, and 12 positions has been upside down in front of a mirror. A person reads the time in the reflection as 6 : 10 The real time is:

11. The bus for Chennai leaves every 30 minutes from a bus depot. The enquiry clerk told a passenger that the bus for Chennai left 10 minutes ago, and the next bus will leave at

(c) 25

(d) 35

(SSC Constable (GD) 04.05.2002 )

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,d ?kM+h 3, esa6, 9 vkSj12 ij fpÉ clock? ,d ?kM+h 3, esa6, 9 vkSj12 ij fpg~u vafdr gSa rFkk ?kM+h dks ,d Åijh niZ.k ds vafdr gSa rFkk ?kM+h dks ,d Åijh niZ.k ;fn ds 50 feuV igys pkj ctdj 45 feuV lkeus j[kk tkrk gSA rFkk ,d O;fDr niZ.k lkeus j[kk tkrk gSA rFkk ,d O;fDr niZ.kct jgk FkkA 6 ctus esa fdrus feuV 'ks"k gSa\ izfrfcEc esa 9 : 50 n'kkZ jgk gSA okLrfod 12 : 30 n'kkZ jgk gSA okLrfod izfrfcEc esa (a) 45 (b) 15 le; crkb;s\ le; crkb;s\

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ANSWER KEY

PREVIOUS YEAR QUESTION

1.

3. 4.

(a) (b)

(b) as we know,

5.

(a)

6.

(b)

7.

(c)

8.

(b)

9.

(c)

tSlk gesa Kkr , gS Minute hand in 1 minute = 6°

Minute hand in 60 second = 6°

2.

= rFkk29 lsda.M esa

12. (c)

 3 º 1 In 36 Sec = × 36 =   120 10 

6 × 29 = 2.9° 60

(c) as we know hour hand in

= 6°

Then in 1 Sec/rks1 lsda.M esa

6 = 60

1 36 lsda.M esa = × 36 = 120

 3 º   10 

In 60 minute = 30°

feuV okyh lqbZ 60 feuV esa dks.k ?kwerh gS 60 feuV esa = 30°

302

11. (b)

SOLUTION

1 feuV esa dks.k ?kwerh 1 hour = 30° feuV okyh lqbZ = 6° kaVs okyh1 lqbZ ?kaVs esa ?kwerh = 30° gSa

and In 29 sec =

10. (c)

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(b) (c)

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1. 2.

3.

(c) Minute hand in 1 sec. =

In 3600 sec = 30° 3600 esa ls= 30°

30 1 = 3600 120 30 1 rks1 lsda.M esa = = 3600 120

feuV okyh lqbZ 1 lsda.M=

1 10

1 10

Then, In 1 sec. =

6 × 29 = 2.9° 60

Then in 59 sec. =

1 × 59 = 5.9° 10

Rakesh Yadav Readers Publication Pvt. Ltd

9.

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18 : 30

14.

(c) In this time 9 : 11, the hour hand of the clock is between 6 : 00 to 12 : 00 clockwise and minute hand is between 12 : 00 to 6 : 00 clockwise. Hence we subtract that time from 17 : 30.

Si

–3 : 13 15 : 17 –1 14 : 17

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(b) In the time 5 : 47 hour hands of the clock is between 12 : 00 to 6 : 00 and the minute hand of the clock is between 6 : 00 to 12 : 00 clock wise hence we subtract that time from 5 : 90

9 : 11, ij ?kaVs okyh6lqbZ : 00 rFkk 12 : 00 ds chp gS rFkk nf{k.kkorZ gS rFkk 12 : 00 rFkk 6 : 00 feuV okyh lqbZ

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12.

11 : 60 – 3 : 43 = 8 : 17 (b) Because the time 12 : 23 lies between 11 : 00 to 1 : 00, Hence we subtract that time from 23 : 60

D;ksafd 12 : 23, 11 : 00 rFkk 1 : 00 ds eèk vkrk gS blfy, bl le; dks 23 : 60 esa ls ?kVkrs gSaA 10.

6 or

5 : 47 le; ij ?kM+h dh ?kaVs okyh lqbZ ds chp gSA rFkk nf{k.kkorZ gSA vr% bl 12 : 00 rFkk 6 : 00 ds chp gS rFkk feuV le; dks ge 17 : 30 esa ls ?kVkrs gSA okyh lqbZ 6 : 00 rFkk 12 : 00 ds chp gS 17 : 30 – 9 : 11 = 8 : 19 blfy, bl le; dks ge 5 : 90 esa ls 15. (d) In this time 8 : 52, both ?kVkrs gSaA the hands of clock are between 6 : 00 to 12 : 00 clockM wise. Hence we subtract that time from 17 : 90 H

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D;ksafd le;3 : 43, 1 : 00 rFkk11 : 00 ds eè; vkrk gS blfy, bl le; dks 11 : 60 esa ls ?kVkrs gSA

17 : 90 – 7 : 35 = 10 : 55

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8.

11 : 60 – 6 : 00 = 5 : 60 5 : 60 Means (6 : 00) (c) Because the time 3 : 43 lies between 1 : 00 to 11 : 00 Hence we subtract that time from 11 : 60

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D;ksafd le;6 : 00, 1:00 rFkk11 : 00 ds eè; vkrk gS blfy, ge le; dks 11 : 60, esa ls ?kVkrs gSaA

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D;ksafd le;11 : 09, 11:00 rFkk1 : 00 ds eè; tkrk gS blfy, ge bl le; dks23 : 60 esa ls ?kVkrs gSA 23 : 60 – 11 : 09 = 12 : 51 (c) Because the time 6 : 00 lies between 1 : 00 to 11 : 00, hence we subtract that time from 11 : 60

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D;ksafd le; 9 : 27, 1 : 00 ds 11 : 00, chp vkrk gS blfy, ge bl le; dks 11:60 esa ls ?kVkrs gSaA 11 : 60 – 9 : 27 = 2 : 33 (c) Because the time 11 : 09 lies between 11 : 00 to 1 : 00, Hence we subtract that time 11. from 23 : 60

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7.

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6.

(b) Because the time 9 : 27 lies b/w 1 : 00 to 11 : 00, hence we subtract this time from 11:60,

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5.

7 : 35 ij nksuksa6lqbZ : 00 rFkk12 : 00 ds chp gS rFkk nf{k.korZfy,gS bl le; dks 17 : 90 esa ls ?kVkrs gSA

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rks15 lsda.M esa = 15 × 6 = 90°

5 : 30 – 3 : 13 = 2 : 17 12

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lsds.M1 sec. = 6° Then in 15 sec = 15 × 6 = 90°

(d) In this time 7 : 35 both hands are between 6 : 00 to 12 : 00 clockwise. Hence we subtract that time from 17 : 90

sm

(d) Second hand in 1 sec. = 6°

13. ty izfrfcEc esa le; 3 : 13 ij nksuksa 12 : 00 rFkk6 : 00 ds eè; gS lqbZ;ksa rFkk nf{k.korZ gS vr% bl le; dks 5 ge : 30 esa ls ?kVkrs gSaA

ur

4.

yo

w.

ww

1 rks59 lsda.M esa = × 59 = 5.9° 10

23 : 60 – 12 : 23 = 11 : 37 (b) In water Image the time 3 : 13 has the both hands between 12 : 00 to 6 : 00 clockwise hence we subrtract that time from 5 : 30

Rakesh Yadav Readers Publication Pvt. Ltd

8 : 52, ij ?kM+h dh nksuksa lqbZ;ka 6 : 00 rFkk 12 : 00 ds chp gS rFkk nf{k.kkorZ

5 : 90 – 5 : 47 = 00 : 43 0 : 43 means 12 : 43 (a) In the time 4 : 42 hour hand of the clock is between 12 : 00 to 6 : 00 and the minute hand of the clock is between 6 : 00 to 12 : 00. Hence we sustruct that time from 5 : 90 12 : 00 rFkk6 : 00 ds 6 :lqbZ 00 rFkk eè; gSA rFkk feuV okyh 12 : 00 ds eè; gSA vr% ge 5 : 90 esa

gS blfy, bl le; dks 17 : 90 esa ls ?kVkrs gSA 17 : 90 – 8 : 52 = 9 : 38

M H

4 : 42 le;

ls bl le; dks ?kVkrs gSA

5 : 90 – 4 : 42 = 1 : 48

16.

(c) Formula for Angle/dks.k ds fy, lw=k= H × 30 = x° M×

M

11 = y° 2

(Subtract smaller from larger)

H

dks.k ds fy, lw=k Hence/vr] H = 4, M = 12 303

yo

w.

ww

H = 3, M = 56 3 × 30° = 90°

11 56 × = 308° 2 308° – 90° = 218° But 218° is more than 180° hence we subtract this Angle from 360°

25.

304

r Si 12 11

  12  12  6:  45   , 6 : 15   11 11     6:

540 180 ,6: 11 11

6 : 49 27.

1 4 , 6 : 16 11 11

(c) By unique formula /lw=k ls] H = 3, Angledks.k = 180° Note: Hands are opposite means 180°

3 : (15  30) ×

6 10 : 54 11 (b) By unique Formula Here H = 2, Angle = 90°

3 : (15 + 30) ×

H = 2, dks.k= 90° lw=k ls]

×

 90  12  2:  2  5  6  11 

3 : (45) ×

2 : (10  15) ×

360° – 218° = 142°

2 : (25) ×

(b) By Angle Formulalw=k ls H = 12, M = 20 12 × 30 = 360°

2:

12 , 6 : (30 – 15) 11

 180  12  3:  3  5  6  11 

600 11

ij 218°, 180° ls vf/d gS vr%] 22.

av

Ya d

sh

ke

Ra

(d) By Angle Formula/lw=k ls

12 11

12 11

12 , 3 : (15 – 30) 11

12 11

3:

12 12 , 2 : (–5) × 11 11

300 – 60 2: 11 11

12 11

: lqbZ;ka foijhr fn'kk esa gS bldk vFkZ gS uksV 180°

12 10 : (50  0) × 11

ij 275°, 180° ls vf/d gS blfy,

om

 0  12 10 :  10  5   6  11 

.c

(c) By Unique Formula Here H = 10, Angle = 0

10 :

×

360 8 = 6 : 32 11 11

;gk¡H = 10, dks.k= 0

330° – 55° = 275° But 275° is more than 180° hence we subtract this angle from 360°

21.

12 11

6 : (30 + 15) ×

ss re

24.

11 10 × = 55° 2

360° – 275° = 85°

6 : (30  15) × 

dp

=6:

(c) By Angle Formula/lw=k H = 11, M= 10 11 × 30 = 330°

or

 90  12  6:  6  5  6  11 

= 6 : (30  0) ×

210° – 49.5° = 160.5°

.w

360° – 250° = 110° (c) “Coincide means O° Angle” By Uniqe Formula

 0  12 =6:  6  5   6  11 

11 9× = 49.5° 2 20.

H = 6, Angle/dks.k = 90°

Angle/ dks.k= 0°, h = 6 then

(b) By Angle Formula/lw=k H = 7, M= 9 7 × 30 = 210°

(c) By unique Formula/lw=k ls]

ij 250°, 180° ls vf/d gS] blfy,

 Angle  12 = H : H × 5  6  ×   11

360° – 253° = 107° 19.

26.

vr% lkekU; lw=k ls

11 = 253° 2

3 11

ob

46 ×

: 27 ;g laHkao ugh2gS

“lEikrh dk vFkZO° ” gSdks.k

(b) By Angle Formula/ lw=k = H = 12, M = 46 12 × 30 = 360°

3 11

bo

23.

291.5° – 270° = 21.5° 18.

ah

(b) By Angle Formual/lw=k ls] H = 9, M = 53 9 × 30 = 270°

11 53 × = 291.5° 2

This is not possible 2 : 27

360° – 110° = 250° But 250° is more than 180° hence subtract this angle from 360°

120° – 66° = 54° 17.

11 = 110° 2

sm

20 ×

11 12 × = 66° 2

ur

4 × 30 = 120°

12 , 11

3 : (–15) ×

12 11

 –180  540  , 3:   11  11 

angle/dks.k= 180° 3 : 49

1 11

Rakesh Yadav Readers Publication Pvt. Ltd

ss re

r

Si

av

Ya d

= 32

= 180 ?kaVs lgh ?kM+h ds The correct time would be = 9 p.m – 1 hour = 8 p.m

(b) Total loses in 66 minutes is

rks lgh le; = 9.00 – 1.00 ?kaVk = jkr 8 ctsA

66 feuV esa dqy le; dh gkfu

5 720 = 66 – 11 11

726 – 720 6 = 11 11 In 1 minutes/,d feuV 1 esa =

24 15 181 362

= 180 hours of correct clock

8 minutes/feuV 11

sh

ke

Ra

=

16  24  60 360 = = minutes 11 64 11

6 = 11 66

33.

(d) Total hours of clock from 10 A.M Sunday to 9 P.M on following Sunday = 179 hours 24 hour – 8 minutes = 23 hour 52 minutes

jfookj lqcg10 cts ls vxys jfookj jkr 9 cts = 179 ?kaVs 24 ?kVs – 8 feuV= 23 ?kaVs 52 feuV

In 24 × 60 Minute

12 12 , 8 : 33 × 11 11

24 × 60 feuV esa

564 396 ,8: 11 11

= = 32.

(b) Normal watch overtakes in = 65

om

31.

12 12 8 : (40 + 7) × , 8 (40 – 7) × 11 11

30.

181 hour of this clock

vr%24 × 60 feuV esa ?kM+h vkxs fudyrh = 180 ?kaVs lgh ?kM+h ds gS=

= 66 – 65

3 8 : 51 , 8 : 36 11

15 ?kaVs 362

lgh ?kM+h ds

In 24 × 60 Minute clock gains

By unique formula/lkekU; lw=k ls]

8:

15 hour of correct clock 362

1 ?kaVs bl ?kM+h = 24ds×

16 = 11 64

uksV ?kaVs rFkk feuV okyh lqbZ;ka 5 rFkk 6 6 o 7 ds ds chp 180° cukrh gSa rFkk chp Hkh 180° ugha cukrh gS vr% ;g dFku 6:00 dks lgh cukrk gSA

8 : 47 ×

 24 ?kaVs lgh ?kM+h ds 1 hours of this clock = 24 ×

vr% 1 feuV esa ?kM+h vkxs fudyrh gS

Note: Minute and hour hand does not makes 180° Angle between 5 to 6 and 6 to 7 O’ clock . It makes it correct at 6 O’ clock.

12 8 : (40  7) × 11

 24 hours of correct clock

“In one day = 24 × 60 minutes” Then in 1 minute clock gains

720 , 6 : 00 11

12 8 : (8 × 5  7) × 11

5 5 16 – 64 = 1 = min. 11 11 11

.c

= 65

“,d fnu esa= 24 × 60 feuV ”

Angle (b) = minutes/feuV 6

dp

12 12 , 6 : (30 – 30) × 11 11

Not possible/laHkao ugha gS

29.

362

64 feuV esa ?kM+h vkxs 12 hours of this clock/bl ?kM+h bldk vFkZ gS fudyrh gSA ds

12 12 6 : (60) × , 6 : (0) × 11 11 6:

or

12 11

.w

It means In 64 minutes the clock gains

362 hours/?kaVs 15

ob

6 : (30 + 30) ×

=

bo

6 : (30  30) ×

bl ?kM+h esa lqbZ;ka vkxs =fudyrh 64 gS feuV esa

8 2 360  2 = 24 + = 60 15 15

ah

 180  12 6  5  6:   6  11 

24 +

sm

H = 6 Angle/dks.k= 180°

This watch overtakes in = 64 minute

ur

(c) By unique formula/lw=k ls]

yo

w.

ww

28.

5 minute 11

6  24  60 1440 = 11 66 121

109 minutes loss/feuV gkfu 121 (c) Total hours form 8 a.m. Sunday to 9 p.m. following sunday = 181 hours 24 hours + 8 minutes 11

jfookj lqcg8 cts ls vxys jfookj jkr9 lkekU; ?kM+h esa lqbZ;ka vkxs = fudyrh gS = 181 ?kaVs cts rd dqy ?kaVs 5 feuV esa 65 24 ?kaVs + 8 feuV 11 Rakesh Yadav Readers Publication Pvt. Ltd

23 +

52 345  13 358 = = hr. 60 15 15

358 hours of this clock = 24 15 hour of correct clock 358 bl ?kaVs ?kM+h d = 24 ?kaVs lgh ?kM+h 15 1 hour of this clock = 24 ×

15 358

hour of correct clock

305

yo

w.

ww 31 minutes lose in = 155 hr. 5

132 × 4 33

or

.w

ob

179 ?kaVs bl ?kM+h ds

bo

1 minute gains in =

= 24 × 6 + 11 = 144 + 11 = 155 hour

ah

?kM+h ds 179 hour of this clock

31 feuV gkfu gksrh = gSa 155 ?kaVs esa 5

ss re

= 180 ?kaVs lgh ?kM+h ds The correct time would be = 9 p.m. + 1 hour = 10 p.m.

1 minutes lose in =

132 × 4 × 4 = 64 hour 33 9 A.M Sunday + 64 hours 9 lqcg jfookj+ 64 ?kaVs 9 A.M Sunday + 2 dyas 16 hours = 1 A.M Wednesday 9 lqcg jfookj+ 2 fnu 16 ?kaVk = 1 lqcg

155 1 feuV gkfu gksrh = gS × 5 31

cq/okj

av

vxys 'kqØokj 9 cts – 4 feuV4 feuV 15 lsda.M

(c) 6 A.M Thursday 5 P.M upcoming Wednesday +3 minutes – (3 minutes + 12 sec.)

xq:okj lqcg6 vxys cq/okj5 'kke 5 cts +3 feuV– (3 feuV+ 12 lsda.M )

15 1 33 = 8+ =8+ = min. 60 4 4

Total loses = 6 min. + 12 sec

Total hours = 5 days + 12 hr.

dqy gkfu= 6 feuV+ 12 lsda.M

= 5 fnu + 12 ?kaVs dqy ?kaVs = 120 + 12 = 132 hours

75 hour = 3 day + 3 hour 75 ?kaVs = 3 fnu + 3 ?kaVs Thursday 6 A.M + (3 day + 3

hour) = Sunday 9 A.M xq:okj6 A.M + (3 fnu + 3 ?kaVs ) = 9 jfookj

12 1 31 =6+ = min. 60 5 5

Total hour = 6 days + 11 hour

Ra

ke

33 min. gains in = 132 hour 4

155 × 5 × 3 = 75 ?kaVs 31

sh

=6+

155 × 5 × 31

3 = 75 hour 3 feuV esa =

Ya d

35.

3 minutes lose in =

Si

jfookj lqcg9 cts

r

=

= lgh le; gksxk= 9.00 + 1.00 ?kaVk 10 jkr ds nl cts (c) Sunday 9 A.M

155 × 5 31

DSSSB 306

om

4 pkj feuV ykHk gksrk gS

.c

4 minute gain in

= 180 hour of correct clock.

Upcoming Friday 9 P.M – 4 minute 4 Minute 15 sec. Total gains = 8 min. 15 sec

dp

132 1 feuV esa ykHk = ×4 33

24 15 179 = 358

34.

= 6 fnu + 11 ?kaVs dqy ?kaVs

33 132 ?kaVks esa ykHk gksrk=gSA 4

sm

15 lgh 358

ur

1 ?kaVk bl ?kM+h = 24dk×

Rakesh Yadav Readers Publication Pvt. Ltd

yo

w.

ww ur sm

SOLUTION PREVIOUS YEAR

ah bo

180 Not Possible/laHkao ugha , 9 gS : 11

sh

By unique formula/lw=k H ls = 6

4.

17 : 30 17 : 30 – 10 : 20 = 7 : 10 (b) It goes a low In 1 hour = 5 mts

6.

gS 1 ?kaVs esa lqLr gksrh = 5 feuV then in 6 hour = 5 × 6 = 30 minutes

12 11

= 5 × gS 6 = 30 rks6 ?kaVs esa lqLr gksrh fefuV

Ra

 0  12 6:  6  5   6  11  6 : (30  0) ×

r

;gka ijkorZu ty izfrfcEc dh rjg dk;Z djrk gSA

360 8 6: = 6 : 32 11 11 (b) Here the reflection works as a water image.

In this time 9 : 50 the both hand of clock are lies between 6 to 12 clockwise. Hence we subtract that time from 17 : 90

bl le; 9 : 50 ij ?kaVs okyh6lqbZ : 00 rFkk12 : 00 ds eè; gS rFkk feuV okyh lqbZ rFkk nf{k.kkorZ gSA vr% bl le; 17 dks : 90 esa ls ?kVkrs gSA

bl le; 10 : 20 ij ?kaVs okyh6lqbZ :® 00 rFkk 12 : 00 ds eè; gS rFkk feuV okyh 6 lqbZ : 00 17 : 90 – 9 : 50 = 8 : 40 rFk12 : 00 ds eè; gS rFkk nf{k.kkorZ gSA9. vr% bl (c) In this time 6 : 10 the hour hand lies between 6 : 00 to 12 : le; dks 17 : 30 esa ls ?kVkrs gSA

ke

9 : 16

3.

(b) In this question reflection works as water image

Si

av

Ya d

In this time 10 : 20, the hour hand is between 6 : 00 to 12 : 00 and the minute hand is between 12 : 00 to 6 : 00 clockwise. Hence we subtract that time from

12 12 , 9 : (15) × 11 11

4 11 (a) Together = 0° Angle = 0° dks.k lkFk esa

8.

om

12 11

12 12 9 : (45 + 30) × , 9 : (45 – 30) × 11 11 9 : (75) ×

5 : 90 – 4 : 50 = 1 : 40 (a) Here the reflection works as a water image

;gka ijkorZu ty izfrfcEc dh rjg dk;Z djrk gSA

 180  12  9:  9  5  6  11  9 : (45  30) ×

: 00 bl le; 4 : 50 ij ?kaVs okyh12 lqbZ rFkk6 : 00 ds eè; gS rFkk feuV okyh lqbZ 6 : 00 rFk12 : 00 ds eè; gS rFkk nf{k.kkorZ 64 minute is just over 60 minutes gSA vr% bl le; dks 5 : 90 esa ls ?kVkrs gSA 64 feuV60 feuV ls vf/d gSaA

.c

H = 9, dks.k= 180º lw=k ls]

ss re

11 : 60 – 6 : 30 = 5 : 30 (c) oppsite direction means in 180° Angle/foijhr fn'kk180° dks.k By unique formula H = 9, Angle 5. = 180º

dp

D;ksafd le;6 : 30 1 : 00 rFkk11 : 00, ds eè; vkrk gSvr% 11 : 60 esa ls bl le; dks ?kVkrs gSA 2.

8 feuV rst gksrh =gS4 osa ?kaVs esa ×2 16 feuV rst gksrh gS =5 osa ?kaVs esa ×2 32 feuV rst gksrh =gS6 osa ?kaVs esa ×2 64 feuV rst gksrh gS = 7osa ?kaVs esa

is between 12 : 00 to 6 : 00 and the minute hand is between 6 : 00 to 12 : 00 clockwise. Hence we subtract that time from 5 : 90

or

(b) Because the time 6 : 30 lies between 1 : 00 to 11 : 00, hence we substruct that time from 11 : 60

.w

ob

1.

00 clockwise and hour hand lies between 12 : 00 to 600. Hence we subtract that time from 17 : 30

bl le; 6 : 00 ij ?kaVs okyh6lqbZ : 00 rFkk12 : 00 ds eè; gS rFkk feuV okyh lqbZ 12 : 00 rFk6 : 00 ds eè; gS rFkk nf{k.kkorZ gSA vr% bl le; dks17 : 30 esa ls ?kVkrs gSA

Then after 6 hour the time will be = 5 : 30 a.m.

7.

;gk ijkorZu ty izfrfcEc dh rjg dk;Z djrk gSA

In this time 4 : 50 the hour hand

Rakesh Yadav Readers Publication Pvt. Ltd

17 : 30 – 6 : 10 = 11 : 20 10. (c) In this time 12 : 30 we sub= 5 : 30 lqcg ds rks6 ?kaVs ckn le; gksxk tract that time from (c) Every hour it is double fast 12 : 30 le; ij bl le; dks 17 : 90 esa of given minutes ls ?kVkrs gS

izR;sd ?kaVs esa ;g fn;s x;s feuV dk nksxquk 17 :gks 90 – 12 : 30 = 5 : 60 = 6 : 00 tkrh gSA 11. (b) 10 : 10 a.m. In 8 minute fast = In 4th hour ×2 16 minute fast = In 5th hour ×2 32 minute fast = In 6th hour ×2 64 minute fast = In 7th hour

12. (c) before 50 minutes its 4 : 45 means now times = 4 : 45 + 50 minutes = 5 : 35 50 feuV igys4 : 45 dk vFkZ gS vFkkZr = 4 : 45 + 50 feuV= 5 : 35 = rks6 cts ds fy, = 25 feuV ckdh gSA

307

yo

w.

ww sm

ur ah

12 bo

ob

.w

CHAPTER ss re

dp

or

CALENDER

1186  4 ls foHkkftr ugha gSA vr%

1900  400 ls foHkDr ugha gSA

1943  Not divisible by 4 hence  Ordinary Year

vr% lk/kj.k o"kZ 2000  divilsible by hence  Leap Yaer

1943  4 ls foHkkftr ugha gSA vr%

2000  400 ls foHkDr ugha gSA

lk/kj.k o"kZ

Leap Year = 366 Days = 52 weeks + 2 days Extra

vr% yhi o"kZ

1784  Divisible by 4 hence  Leap Year

yhi o"kZ = 366 fnu = 52 lIrkg + 2 vfrfjDr fnu

av

1784  4 ls foHkkftr gS vr% yhi o"kZ 2012  divisible by 4 hence  Leap Year

Distinction (to know) of leap year  An year which is divisible by 4 completely is called Leap Year otherwise called simple year.

Ya d

2012  4 ls foHkkftr gS vr% yhi o"kZ Century Year Case

'krkCnh o"kZ ds lEca/ esa]

yhi o"kZ dh igpku  og o"kZ tks iw.kZ Ex. In 1700, 1200, 500, 1900, :i ls 4 ls foHkkftr gksrk gSA yhi o"kZ 2000. 2100. Find which is dgykrk gSA vU;Fkk lkekU; o"kZ dgykrk gSA Leap Year and which is ordinary Year?

2100  Not divisible by 400 hence  Ordinary year 2100  400 ls foHkDr ugha gks jgk gSA

vr% lk/kj.k o"kZ

Extra Days/Odd days:- Most of questions from this topic depends on Extra days

vfrfjDr fnu@fo"ke fnu%& bl vè;k; ds vf/drj iz'u vfrfjDr fnuksa ij vk/kfjr gksrs gaSA Methods to Find Extra days:-

iz'u vfrfjDr fnuksa ij vk/kfjr gksrk gSA (a) Extra days in days

sh

But in case of century year. We divide that year by 400, if divided completely will called Leap Year otherwise called Simple Year.

400

r

lk/kj.k o"kZ

Ordinary year = 365 Days = 52 weeks + 1 day Extra = 365 fnu = 52 lIrkg + lk/kj.k o"kZ 1 vfrfjDr fnu

(ii)

hence  Ordinary Year

Si

(i)

hence  Ordinary Year

om

o"kks± ds nks izdkj

.c

Two Types of years

Ra

ke

1700, 1200, 500, 1900, 2000. 2100 esa dkSu lk yhi o"kZ vkSj dkSu lk days = Remaining days are lk/kj.k o"kZ gSA 7 All are century years hence Extra days ijUrq 'krkCnh o"kZ ds lEca/ esa ge ml o"kZ we check these years dividdks400 ls foHkkftr djrs gSA ;fn ;g fnu ing by 400 (a) vfrfjDr fnu = 'ks"k cps gq;s 400 ls iw.kZr;k foHkkftr gks rks yhi o"kZ 7 400 ge lHkh 'krkCnh o"kZ gSA vr% ls

vU;Fkk lkekU; o"kZ dgykrk gSA

Ex.

Sol.

In 1323, 1726, 1186, 1943, 1784, 2012 Find which is Leap Year and which is Ordinary Year?

1700  Not divisible by 400 hence  Ordinary Year

esa dkSu lk yhi o"kZ vkSj dkSu lk lkekU;1700  400 ls foHkDr ugha gSA vr% lk/kj.k o"kZ o"kZ gSA 1200  divisible by 400 1323  Not divisible by 4 hence  Leap Year Sol. hence  Ordinary year ls foHkDr gSA vr% yhi 1200 400   1323  4 ls foHkkftr ugha gSA vr% lk/kj.k o"kZ

1726  Not divisible by 4 hence  Ordinary Year

308

vfrfjDr fnu gksrs gSA foHkkftr djds irk djsxsa fd dkSu&lk o"kZ Ex. Find Extra days in 77 days, 12 D;k gSA

o"kZ

500  Not divisible by 400 hence  Ordinary Year

1726  4 ls foHkkftr ugha gSA vr%

500  400 ls foHkDr ugha gSA

lk/kj.k o"kZ

vr% lk/kj.k o"kZ

1186  Not divisible by 4

1900  Not divisible by 400

days, 67 days, 41 days, 39 days, 26 days 77 fnu]12 fnu]67 fnu]41 fnu]39 fnu vkSj26 fnu esa esa vfrfjDr fnuksa dh

la[;k Kkr djsa

In 77 days = 77 fnu esa =

77 = 0(Remainder) 7

77 = 0 ('ks"kiQy) 7

hence 0 is Extra day

vr% 0 vfrfjDr fnu gS In 12 days =

12 =5 7

(Remainder)

Rakesh Yadav Readers Publication Pvt. Ltd

= 24 (Quotient)/(HkkxiQy)

Si

r

 53 o"kks± esa] 10 o"kks± esa fo"ke fnuksa dh la[;k Kkr dj

av

In 39 days  4 E.D

esa4 vfrfjDr fnu 39 fnuksa In 26 days  5 E.D

fo"ke fnu=

26 fnuksa esa5 vfrfjDr fnu Here we found that the Extra days always comes 0 to 6.

E.x

ke

Ra

eghus pkj izdkj ds gksrs gSA (i) 28 days = 0 E.D

+2 o"kZ 7

= 'ks"kiQy

10  2 12 = = 7 7

sh

;gk¡ ge ikrs gS fd vfrfjDr fnuksa dh la[;k ges'kkO vkSj6 ds chp vkrh gSA

Months are of Four types

year + 2 =Remainder 7

Ya d

odd days =

= 5 (Remainder)/('ks"kiQy) Find odd days in 46, 99, 53, 76, 83 years respectively?

mnk- Øe'k%46, 99, 53, 76, 83 o"kks± esa fo"ke fnuksa dh la[;k Kkr djsa\  In 46 years/o"kZ

28 fnu= 0 vfrfjDr fnu (ii) 29 days = 1 E.D

29 fnu= 1 vfrfjDr fnu (iii) 30 days = 2 E.D 30 fnu= 2 vfrfjDr fnu (iv)31 days = 3 E.D 31 fnu= 3 vfrfjDr fnu Extra days in Years 

o"kks± esa vfrfjDrfnu Odd days in a Ordinary Year =1

lk/kj.k o"kZ esa fo"ke fnu =1 Odd days in a Leap Year = 2

yhi o"kZ esa fo"ke= fnu 2 Rakesh Yadav Readers Publication Pvt. Ltd

No. of Leap Years =

om

99  24 123 = 7 7 = 4 (Remainder)('ks"kiQy)  In 53 year,

o"ke fnu=

4 Find Odd days in 10 years

blh izdkj&

.c

o"kks± dh la[;k =(HkkxiQy)

99  24 123 = 7 7

ss re

dp

=

odd days =

10 yhi o"kZ = = 2 (HkkxiQy) 4

Similarly,

(c)

or

No. of year =(Quotient) 4

.w

=

ob

vr% 6 vfrfjDr fnu

Extra days in months 

bo

=Remainder/'sk"kiQy

10 Leap years = = 2 (Remain4 der)

hence 6 is E.D

eghuksa esa vfrfjDr  fnu

ah

Ex. (Remainder)

41 41 fnu esa = = 6 ('ks"kiQy) 7

(b)

sm

hence 4 is E.D

41 =6 7

ur

67 67 fnu esa = = 4 ('ks"kiQy) 7

In 41 days =

= yhi o"kks± dh la[;k

Where, Leap Year/tgk¡] yhi o"kZ

(Remainder)

99 4

o"kZ $ bl o"kZ esa yhi o"kks± dh la[;k = lw=k 7

vr%5 vfrfjDr fnu gSA

99 4

No. of Leap Years =

year + no. of Leap Year in these Year 7

hence 5 is E.D (E,D – Extra Days)

67 In 67 days = =4 7

yo

w.

ww

Formula =

12 12 fnuksa=esa = 5 ('ks"kiQy) 7

46 = 11 4

No. of Leap Years =

53 = 13 4

(Quotient)

53 4

yhi o"kks± dh la[;k = = 13 (HkkxiQy) 53  13 66 = = 3 7 7 (Remainder) odd days =

53  13 66 = 7 7  In 75 years,

fo"ke fnu=

 75 o"kZ

No. of Leap Years =

76 = 19 4

(Quotient)

76 4

= = 19 (HkkxiQy) yhi o"kks± dh la[;k odd days/fo"ke fnu = =

76  19 7

95 =4(Remainder)/('ks"kiQy) 7

 In 83 Years

(Quotient)

46 4

 83 o"kZ]

= = 11 (HkkxiQy) yhi o"kks± dh la[;k

No. of Leap Years/yhi o"kks± dh la[;k

46  11 57 odd days = = = 1 7 7 (Remainder)

=

fo"ke fnu =

46  11 57 = = 1 ('ks"kiQy) 7 7

 In 99 years,  99 o"kks± esa

83 = 20 (Quotient)/(HkkxiQy) 4

odd days/fo"ke fnu = =

83  20 7

103 =5(Remainder)/( 'ks"kiQy ) 7

Thus, in 46, 99, 53, 76, 83 Years 1, 4, 3, 4, 5 Extra days Respectively

309

Odd days in 400 years 

1200 + 100 + 22

sm

ah

400 o"kks± esa fo"ke fnuksa  dh la[;k 

ur

0

300 Years + 99 Years + 400 th year (Leap Year)

+ 5 + 5

= 10 days

bo

ob

+

4

+

2

=

In 22 year, 22 o"kZ esa

7

or

1

.w

But 7 can not be odd days,

7

99 years + 100th year (Ordinary Result)

(lk/kj.k o"kZ) 99 o"kZ + 100oka o"kZ

200 o"kks± esa fo"ke fnu

= 0fnu  400 o"kks± esa vfrfjDr

Note:- Every multiple of 400 have odd days = 0

In 100 Years = 5 days ×2

×2

sh

Ex.

Odd days in 1700 Years  1700 o"kks± es fo"ke fnu

ke

But 10 days can never be odd days, hence we will divide it by 7 again and remainder would be odd days

1600 +100

Ra

ijarq10 fnu dHkh fo"ke fnu ugha gksrsA 0 + 5 = 5 days 7 ls foHkkftr fd;k tk;sxk vr% bls iqu% Ex. Odd days in 2100 Years vkSj tks 'ks"kiQy cpsxk oks fo"ke fnu gksaxsA 2100 o"kks± es fo"ke fnu  10 = 3 (Remainder)/('ks"kiQy) 7





Odd days in 300 years 



300 o"kks± esa fo"ke fnuksa dh  la[;k Ex. th

200 Years + 99 Years + 300 year (Ordinary Year) 4

+

1

=



ijarq8 fo"ke fnu ugh gks ldrk 8 = 1 odd days/fo"ke fnu 7

2000 +100 0

+ 5 = 5 days

Odd days in 1745 years 1600 + 100 + 45 0

+ 5 + 0

=5

NOW TYPE-I

To find the day of the week on a particular date when no preference day is given:

,d fuf'pr rkjh[k ij] lIrkg ds fnuksa dks Kkr djuk tc dksbZ ojh;rk okys fnu u fn;s x;s gks When we count no. of odd days on the given particular date. Then we write

fn;s x;s ,d fuf'pr rkjh[k ij ge fo"ke fnuksa dh la[;k irk djrs gSA fiQj ge fy[krs gSA Sunday for



0 odd day

Odd days in 1900 years

jfookj ds fy;s



0 fo"ke fnu

1900 o"kks± es fo"ke fnu

Monday for



1 odd day

lkseokj ds fy;s



1 fo"ke fnu

Tuesday for



2 odd days

1600 + 300

8

But 8 can not be odd days, hen ce

310

= 3 Remainder

vc ge fdlh Hkh o"kZ esa fo"ke fnuks dh la[;k Kkr dj ldrs gS

×2

=5 fnu 200 o"kks± esa

+

7

1745 o"kks± esa fo"ke fnuksa dh la[;k

Now we can find odd days in any no. of years

=5 fnu 100 o"kks± esa

3

10

= 3 ('ks"kiQy) Ex.

uksV%izR;sd400 ds fo"ke fnu= 0

In 200 Years = 10 days

×2

22  4 = 5 ('ks"kiQy) 7

 As in 400 year E.D = 0

Ya d



av

4 fnu+ 1 fnu= 5 fnu vfrfjDr Odd days in 200 years

fo"ke fnu=

10 can not be odd days = 400 o"kks± dk dSys.Mj 400– vkjafHkd nwljs 7 800 o"kks± esa iz;ksx gksxk400 vksj izR;sd 10 o"kks± esa ckj&ckj iz;ksx gksxkA 10 fo"ke fnu ugha gks ldrk =

4 days + 1 day = 5 days Extra 

r

100 o"kks± es fo"ke fnu



7

Initial 400 Years calender used in next 400 to 800 years and repeated again and again in every 400 years

Si

Odd days in 100 years



om

uksV%;fn fo"ke fnuksa dh la[;k 7 ls T;knk 22 yhio"kZ = 4 (HkkxiQy) 7 ls vk jgh gSA rks ge bl fnu dks iqu% ijarq7 fo"ke fnu ugha gks ldrk gSA vr% 5 foHkkftr djsaxs vkSj fiQj tks 'ks"k cpsxk 7 22  4 ogh gekjk fo"ke fnu gksxk\ = 0 fo"ke fnu Odd day= =5 (Remainder)

.c

7 = 0 odd day 7

22 = 4 (Quatiet) 5

ss re

hence =

Leap years

dp

Note: If result of odd days comes more t han 7 t he n w e wi ll again divide these days by 7 and Remainder would be ‘Odd days’

yo

w.

ww

ifj.kke Lo:i] 46, 99, 53, 76,  83 o"kks± es Øe'k vfrfjDr fnu 1, 4, 3, 4, 5 gSA



0

+ 1 = 1days

Ex.

Odd days in 1322 years



1322 o"kks± esa fo"ke fnu

eaxyokj ds fy;s 

2 fo"ke fnu

Wednesday



3 odd days

cq/okj ds fy;s



3 fo"ke fnu

Thursday



4 odd days

c`gLifrokj ds fy;s 

4 fo"ke fnu

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tc ,d fuf'pr fnukWd ij lIrkg dk fnu iwNk tk; vkSj mlls lEcaf/kr fnu fn;k x;k gks]

om

.c

(b) Wednesday/cq/okj

ss re

Ex.2 What was the day of week on 13th April 1723?

dp

,d vfrfjDr fnu ds fy;s  lkseokj

(a) Tuesday/eaxyokj

Ex.3 If the third day of month is Monday. Which of the 13 vizSy1723 dks lIrkg dk dkSu lk following will be the 29th day fnu iM+sxk\ of week?

(c) Monday/lkseokj (d) Sunday/jfookj Sol.

For 1 day Extra  Monday

or

iM+sxk\

14 twu1993 dks dsoy ,d vfrfjDrfnugS

When the day of week is asked on a particular date. And reference day is given

.w

14 twu1993 dks lIrkg dk dkSu&lk fnu

TYPE - II

ob

On this particular date 14 June 1993 we count only one day Extra 

6 fo"ke fnu  Ex.1 What was the day of week on 14th June, 1993?

2 fo"ke fnu ds fy;s  eaxyokj

(c) Odd days in 1992 years  1992 o"kks± es fo"ke fnuksadh la[;k

(a) Monday/lkseokj (b) Tuesday/eaxyokj (c) Wednesday/cq/okj

1600 + 300 + 92 0 + 1 + 3 =4 We take 1992 years. Because 1993rd was running than,

(d) Thursday/c`gLifrokj Sol.

;fn eghus dk rhljk fnu lkseokj gSA 29 rks oka fnu lIrkg dk dkSu lk fnu gksxk\ (a) Friday/'kqØokj (b) Saturday/'kfuokj (c) Sunday/jfookj (d) Monday/lkseokj

r



'kfuokj ds fy;s

For 2 odd days  Tuesday

Si

Saturday

15 15 fo"ke fnu ugha gks ldrk = gSA= 1 7

bo

5 fo"ke fnu 6 odd days

ah

'kqØokj ds fy;s 

sm

5 odd days

ur



yo

w.

ww

Friday

sh

Ya d

av

Write completed years, months days till 13th April 1723. And find odd days in Sol. (b) 3rd = Monday ge 1992 o"kks± dh gh ysrs gSA] D;ksafd these days The,/ rc 1993 okW o"kZ pkyw o"kZ gSA 13 vizSy1723 rd lHkh o"kks± vkSj fnuksa3rd + 7 = 10th  Monday In 92 years = 92 years + 23 iw.kZr;k fy[ks vkSj bu fnuksa esa fo"ke rhljk fnu + 7 = 10oka lkseokj Leap year Kkr djs\ 10th + 7 = 17th  Monday 92 o"kks± = 92 + 23 yhio"kZ esa o"kZ Year Year Year Month (of 1723) days of April 10oka + 7 = 17oka lkseokj 1600 100 22 Jan = 31 = 3 13 115 17th + 7 = 24th Monday = =3(Remainder)/('ks"kiQy) Feb = 28 = 0 7 17oka + 7 = 24oka lkseokj 0 5 6 March = 31 = 3 6 Odd days in Months/eghus fo"ke

fnuks dh la[;k

3

0

3

2

ke

Jan Feb March April May

3 = 11

Ra

1993rd was a ordinary year. Hence its February would be of 28 days  1993 ,d lk/kj.k o"kZ gSA vr% blesa iQjojh28 fnu dh gksxhA  Odd days in days 

fnuksa eas fo"ke  fnu

22 In 22 Year Leap Year = 4 = 5 (Quotient)

24th + 5 = 29th 24oka + 5 = 29oka

 Monday + 5 = Saturday

22 lkseokj+ 5 = 'kfuokj 22 o"kZ esa yhi=o"kZ= 5 (HkkxiQy) 4 Ex.4 If the 26th day of month is Friday. Which of the following Then odd days/rc] fo"ke fnu will be the 5th day of week? 22  5 27 ;fn eghus dk26 ok fnu lkseokj dks iM+rk = = 7 7 gSA rks lIrkg 5dkoka fnu] fdl fnu = 6 (Remainder)/(HkkxiQy)

iM+sxk\

Total odd days/dqy fo"ke fnu

26th – 7 = 19 – 7 = 12– 7= 5th If Friday on 26th day, then Also Friday on 5th day

In June 14 days completed

= 0 + 5 + 6 + 3 + 0 + 3 + 6 = 23

14 twu esa 14 fnu iwjk gSA = =0 7

23 days can never be odd days, so we divide 23 by 7 and remainder would be odd days?

5 oka ;fn 26 oka fnu 'kqØokj dk gSA rks fnu Hkh 'kqØokj dk gh gksxk

15 days can not be odd days,

Ex.5 23rd March of a general year 23 fnu dHkh fo"ke fnu ugh gksaxs blfy;s was Tuesday. Then what was ge 23 dks7 ls foHkkftr djrs gSA vkSj tks the day of the week on 17 July of this year? 'ks"kiQy cpsxk og fo"ke fnu gksxk

15 then = = 1 7

23 days =2(Remainder)/('ks"kiQy) 7

,d lk/kj.k o"kZ esa 23 ekpZ dks eaxyokj 17 tqykbZ dks lIrkg dk Fkk] rks blh o"kZ dkSu lk fnu gksxk\

Total odd days = 4 +11 + 0 = 15

dqy fo"ke fnu = 4 + 11 + 0 = 15

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311

ss re

dp

or

.c

om

April = 30 May = 31 June = 30 Till 17th July = 17

15 iQjojh1790 dks dkSu&lk fnu gksxk\

9th June  Thursday – 1 = Wednesday

= 116

r

(a) Monday/lkseokj (b) Saturday/'kfuokj

Si

9 twu c`gLifrokj – 1 = cq/okj Note: (i) First day and the last day of Every General year are same

17 tqykbZ rd = 17 Total = 116

(c) Tuesday/eaxyokj

(d) Wednesday/cq/okj fdlh Hkh lk/kj.k o"kZ dk igyk fnuSol. (b) 15th Feb, 1789  Friday vkSj vfUre fnu leku gksrk gSA 15 iQjojh1789  'kqØokj

av

116 Odd days in 116 = = 4 (Re7 mainder)

(ii) Last day of a Leap Year precedes one day to the first day of that year

116 116 esa fo"ke fnu = = 4 ('ks"kiQy) 7

Ya d

,d yhi o"kZ dk vfUre fnu mlho"kZ ds igys fnu ls ,d fnu vkxs gksrk gSA

 Add these odd days in the day of 23rd March

sh

 23 ekpZ ds fnu ds lkFk bu fnuksa dh In G.Y. In L.Y. + 4 = 'kfuokj tksM+us ij eaxyokj lk/kj.k o"kZ esa] yhi o"kZ esa Tuesday + 4 = Saturday 1 Jan  Monday Ex.6 9th June of a Leap Year was 1 tuojh  lkseokj on Thursday. Then what was the day of week on 17 Febru- Then/ rc] ary. 31 December  Monday

ke

,d yhi o"kZ esa 9 twu dks c`gLifrokj Fkk] rks 31 fnlEcj  lkseokj 17 iQjojh dks lIrkg dk dkSu&lk fnu gksxkA In leap year (a) Wednesday/cq/okj 1 Jan  Monday (b) Monday/lkseokj 1 tuojh  lkseokj (c) Thursday/c`gLifrokj Then/ rc] (d) Tuesday/eaxyokj 31 Dec.  Tuesday

 31  30  31  9  113 Days

113 7

leku fnukad= 15 Month same  February

 G. Y. (28th Feb. of 1789)

ge ns[krs  gS lk/kj.k o"kZ (28 iQjojh 1789) Hence we forward 1 day Friday + 1  Saturday Ex.8 The Republic day in 1996 was celebrated on Wednesday. On what day was it celebrated in the year 1997?

o"kZ 1996 esa x.krU=k fnol cq/kokj dks 1997 esa ;g fdl fnu euk;k x;k rks o"kZ euk;k tk;sxk\ (a) Thursday/c`gLifrokj (b) Friday/'kqØokj (c) Saturday/'kfuokj (d) Sunday/jfookj

year  Tuesday 1 Jan of next 1 tuojh dks  eaxyokj nwljs o"kZ] year  Wednesday

nwljs o"kZ] 1 tuojh dks  cq/okj Result/ifj.kke 1.

date same = 15

We corss

rc] 31 fnlEcj  eaxyokj

(D;ksafd ;g yhi o"kZ gSA)

15 iQjojh1790  ?

leku eghuk iQjojh

1 Jan of next

 29 – 17 =12 (Because It is a leap year)

15th Feb, 1790  ?

'kqØokj + 1  'kfuokj

Ra

(a) Left days in Feb

iQjojh esa 'ks"k fnu

312

.w

ekpZ eghus ds 'ks"k fnu

Odd days in 113 days =

ob

Left days in March = 31– 23 = 8

March April May June Total

bo

;fn fnukad] eghuk leku gSA vkSj ge ,d yhi o"kZ (iQjojh 29 fnu) dks ns[krs gSA rks ge 2 fnu vkxs c