Math Concept King - All Formulas and Theorums

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Symbol

Table of Symbols ����

implication

Reference equal to not equal to

�

negation, equivalence, relation

�, �

quantifier

�

congruent to

�

approaches, ray

�


proportional to less than not less than greater than not greater than less than or equal to greater than or equal to much less than much greater than

�

infinity

��or �

sigma

% + – �

percentage plus, positive minus, negative plus or minus

�

� — �

acute angle

perpendicular parallel

�

triangle square

logba log10a loge a or ln a

logarithm (to base b) common logarithm natural logarithm

�

conjunction (and)

�

disjunction (or)

= 1 square centimetre = 1 square decimetre = 1 square metre = 1 square decametre = 1 square hectometre = 1 square kilometre = 10000 square metres Conversion of Volume

millimetres centimetres decimetres metres decametres hectometres

1000 1000 1000 1000 1000 1000

= 1 cubic millimetres centimetres = 1 cubic decimetres = 1 cubic metres = 1 cubic decametres = 1 cubic hectometres = 1 cubic Conversion of Capacity

cubic cubic cubic cubic cubic cubic

millilitres centilitres decilitres litres decalitres hectolitres

= = = = = = Conversion of

10 milligrams 10 centigrams 10 decigrams 10 grams 10 decagrams 10 hectograms 100 kilograms 10 quintals or 1000 kg

rectangle

1

= 1 centimetre (cm) = 1 decimetre (dm) = 1 metre (m) = 1 decametre (dam) = 1 hectometre (hm) = 1 kilometre (km) of Area

100 square 100 square 100 square 100 square 100 square 100 square 1 hectare

10 10 10 10 10 10

therefore since line segment

� ��

millimetres (mms) centimetres decimetres metres decametres hectometres Conversion

at io n

pi

on

Pr am

division

Ga

�a � b � �a / b

multiplication

Ch

�a � b � �a � b

ga n

�

10 10 10 10 10 10

set empty set, void set, null set Conversion of Units Conversion of Length

Si

approximately equal to

p

�

a Pu ta bl

identity

r

{} �

�

ic

Symbol = �

Appendix

= = = = = = = =

centimetre decimetre metre decametre hectometre kilometre

1 centilitre 1 decilitre 1 litre 1 decalitre 1 hectolitre 1 kilolitre Weight 1 centigram 1 decigram 1 gram (g) 1 decagram 1 hectogram 1 kilogram (kg) 1 quintal 1 metric tonne

Conversion of Time = 1 minute = 1 hour = 1 day = 1 week = 1 fortnight = 1 month = 1 year = 1 year = 1 leap year = decade = silver jubilee = golden jubilee = diamond jubilee = radium jubilee or platinum jubilee = century = 10 centuries or 1 millennium

Equivalents of Units Units of Lengths

60 seconds 60 minutes 24 hours 7 days 15 days 28, 29, 30 or 31 days 12 months 365 days 366 days 10 years 25 years 50 years 60 years 75 years

= = = = = = = =

1 feet (ft) = 0.348 metres 1 yard (d) 0.9144 metres 1 chain 0.621 mile or 103 metres 1.6093 kilometres or 1760 yards 2.54 centimetres 2.471 acres Units of Area

Si

1 square metre 1 square yard 1 square kilometre 1 square mile

a Pu ta bl

1 acre

on pi

Ch

am

ga n

Pr

ic

at io n

100 square metres 100 acres

Ga

= = = = = = = = = = = = =

r

1 square feet

p

100 years 1000 years

12 inches 3 feet 1 yard 22 yards 1 kilometre 1 mile 1 inch 1 hectare

2

Symbol

144 square inches 0.0929 square metres 1.196 square yards 0.836 square metres 0.3861 square miles 1000 hectares 2.59 square kilometres 640 acres 4840 square yards 4046.86 square metres 1 acre 1 hectare 10,000 square metres

Geometry

Geometry (Line & Angle) Supplementary angle of an angle is 90° more than complementary angle.

Line and Angle Point: Zero dimension figure or a circle with zero

90°

�

�

Parallel lines: two or more line that never intersects L ��� M

Line segment: A line with a fixed length.

L ���M

Q

Ray: A line with uni-direction length. A

B

L

p

P

�

Transversal Line : A line which inte rests (touches) two or more lines at distinct point is called transversal lines of the given lines.

Line: One dimension figure line is a set of points having only length with no ends. _________ _________

�

�

r

radius.

Si

�

d

d

C

�

Angle

�4 + �5 = 180°

+ = 90°

�3 + �6 = 180° �

12° �

AB �� CD A

47°

137°

78°

+90

168°

90° – �

�

� � �+� = 180° C

Complementary Supplementary

+90

If AB �� CD then find the value of �+�+�?

180º

+90

�� �3 = �5, �4 = �6

Alternate Angles

�

43°

�� �1 = �5, �4 = �8 �2 = �6, �3 = �7

Supplementary Angle : If sum of two angles is 180º then they are supplementary to each other.

Supplementary Angles/

D

Corresponding angles

am �

B

6 5 7 8 F

C

on

n

Ch

ga

Ga

Complementary Angle/

�

EF

E 2 1 3 4

A

Complementary Angle : If sum of two angles is 90º then they are Complementary to each other. 90º

�

ic at io n

A B

AB ���CD

Pr

�ABC = �

�

AB ���CD and EF is transversal line

Angle: inclination between two sides is called angle.

pi

�

a Pu ta bl

M

180° – �

A

�

+

B O

�

D

+ = 360° a

c C b x

D b=a+c+x

3

�+�+��

B

�

A ��

LHS

Geometry

B

�

�

A

RHS

B n C

RHS = LHS +

a

b

F

AB DE m = = BC EF n

p F q

c

r

BE =

G

H

D

an+bm m+n

r

b

Si

B C

E

E

A

p

a:b:c=p:q:r

on pi

am Ch

ga

n

Pr

ic at io n

a Pu ta bl

p a = p+q+r a+b+c

Ga

�

D

m

C D Sum of angle on RHS = LHS

+ =

a

4

Geometry

Types of Triangles Triangle

A

A triangle is a 3-sides polygon that consists of three edges and three vertices.

c

3

B

3

3

three unequal angles

3

�A��B��C & a�b�c

By Angle

A

h3

C

All three angles < 90°

a Pu ta bl

�A+�B+�C=180°

B

n

Type of Triangle

Ga

Equilateral Triangle

�

One angle > 90° �

�B

AC = b = A c B

60°

b a

C

Let �C = smallest angle

�C =

� side AB = c = smallest side

3

AB = c =

Isosceles Triangle

�

Let �B = largest angle

> 90°

� side AC = b = largest side

Equilateral triangle has 3 equal sides, each angle 60° �

Obtuse Angle Triangle

Ch

By side

C

One angle is 90°. �B = 90° and �A + �C = 90°

am

By Angle

ga

By side

A

Pr

1 1 1 h1 : h2 : h3 = : : a b c

< 90°

Right Angle Triangle

ic at io n

Corresponding height. � ah1= bh2= ch3 = constant

�

�

1 1 1 1 ×a×h 1 = bh 2 = ch 3 = × Base × 2 2 2 2

on

Area �

Acute Angle Triangle

p

a

b

pi

h2

B

�

h1

c

C

a

three unequal sides

3 sides, 3 vertices, 3 altitudes, 3 angles 3

b

r

�

Scalene Triangle

Si

�

�

�

Inequality of triangle The triangle inequality states that for any triangle the sum of the lengths of any two sides must be greater than the length of the remaining side.

�

two equal sides two angle same

5

Geometry

|b–c| < a < b+c

II.

A c

A b

B

C

a

C

� 4 + 9 < 15

2

2

III. Obtuse Angle Triangle

� 5 + 10 = 15

7, 12, 15 � is possible

c

c =a +b 2

5, 10, 15 � not possible

A

� 7+12> 15

c

b

OR 7+15 > 12 OR 12 + 15 > 7

C

a

B

Si

A

�C = largest

b

�C =

side c = largest

C

a

�C =

side c = largest

eg� 4, 9, 15 � not possible

B

B

a

�C = largest

Inequality of Triangle

c

c

b

|a–c| < b < a+c |a–b| < c < a+b �

Right Angle Triangle

r

�

c

c2 > a2 + b2

p

Sum of any two sides is always greater than 3rd side.

Pr

�

19

I.

Ga

on pi

n

2× –1 � 2×10–1 = 19 Relation between 3 sides of Triangle 3

Acute Angle Triangle

2

Ex: 1�

c B

a

�C = largest

12 3

�C =

side c = largest

c

c < a +b 2

2

on these triplets will

?

3.5 3

C

(7,24,25), (11,60,61), (13,84,85), (33,56,65), (65, 72, 97),

� 2 � (10, 24, 26) (5,12, 13) ��� (3,4,5) � (6,8,10), (9,12,15), (12,16,20), (15,20,25)

A b

b

c

C a 2 b =c +a (3,4,5), (5,12,13), (8,15,17), (9,40,41), (12,35,37), (16,63,65), (20,21,29), (28,45,53), (39,80,89), (36,77,85), (20, 99, 101) multiplication and division also result in triplets.

Possible values of x = 2×small side –1 � 2×10–1 = 19 x

A

2

am

� 19 �'s possible

19

Triplets

B

Ch

ga

xtotal = 19 values possible

�

Take ratio of sides 11.7 : 16.9 : 23.4 9 : 13 : 18 2 2 2 18 > 9 + 13 �� is obtuse angle triangle.

If 10, 17, x are sides of a �, x � integer Then 7 < x < 27 � x � {8, 9, 10, .... 26} xmin = 8, xmax = 26

11.7, 16.9, 23.4.

ic at io n

a+b > c � b > c–a b+c > a ��b > a–c c+a > b � � |c–a| < b < c+a Difference of any two sides is always less than 3rd side. �

sides of triangle : 11.7, 16.9, 23.4. which type of � it is?

a Pu ta bl

�

2

6

7 �÷2

24 �÷2

25 �÷2

3.5

12

12.5 � 3.5 3 , 12 3 , 12.5 3

Geometry

Ex: 2�

Angle Bisector �

?

14

A

E

D

32 � 22 = 13 21 7 ×2 Ex: 3�

7 ×3

BE � �ABC 2� + 2� = 180° + = 90° � Angle between internal angle bisector and external angle bisector of an angle is 90°.

6 � 7 � 7 � 3 � 6 7 � 21

42 3

?

C B BE � exterior angle bisector of �ABC

7 × 13 = 7 13

6 7 �3

18 7

3rd side = 6 7 × 21 � 9 = 6 7 × 12 = 12 21

?

9.6

9.6 : 18

p

a Pu ta bl

sum of all exterior angles = 360° Types of Angles

n

= 360°

B

A

Right Angle

�

Obtuse Angle

�

Ga

�

A B

C

B

90°

C

0°
h > 3 h1 � h 2 h1 � h2

23

'BC'

Geometry

Centroid Centroid �

A

Centroid is the intersection point of all 3 medians of triangle.

2

3 �

2

Median divides the triangle into two equal areas.

G 1

1

Centroid always lies inside the triangle.

B

1

E 1

�

6

�

� BC2 � 2 AD + � � AB + AC = 2 4 � � 2

Ch

Ga

2

B C Centroid always divides median in the ratio 2 : 1.

C

D

cos (180°–�) = – cos �

am

ga n

G 2

2

180–�

�

B

pi

�ABC

1 Area 3

on

Area �AGC = Area �AGB = Area � BGC =

A

ic

1 Area of �ABC 6

Pr

Area of each triangle �=

To find length of median.

at io n

6 triangle made by 3 medians have equal area.

A

1 AD 3

a Pu t bl a

C

D

3 �

GD =

Apollonius theorem

1

1

B

�

1 G

2 AD 3

p

1

F

AG =

Si

A

D

r

�

2

A

�

2:1 F

A 1 2

F

B

B

G

1 1

E

1 D

G

E

C

D �

BC2 �

�

AC2 �

2 AB2 + AC2 = 2 � AD � 4 � � �

C

AG BG CG 2 = = = GD GE GF 1

2 AB2 + BC2 = 2 � BE � 4 � � �

24

Geometry

�

AB � 2

F, E are mid points ��AO = OD =

2 AC2 + BC2 = 2 � CF � 4 � � �

OG = AG – AO = 4 – 3 = 1 unit

BC2 AC2 2 (AB2 + BC2 + AC2) = 2AD2 + + 2BE2 + + 2 2

� AO : OG : GD = 3 : 1 : 2

AB2 2CF + 2

Area of

2

GEF =

B

p

GEF 1 = ABC 12

at io n

a Pu t bl a

Area � 1 : 4 Area �ABC = Ar �BGC × 3 = 4 × 3 = 12

2 CF 3

�

ic

Area �EFG = Area �DFG = Area �DEG = 1 � Area �DEF = 1 + 1 + 1 = 3 unit

Pr

If two medians of a � intersect each other at 90°

on

�

Area DEF 3 1 = = Area ABC 12 4

�

90° A

Ch

am

ga n

�

pi

2 (AD + BE + CF) > AB + BC + CA 3

4 (AD + BE + CF) > 3 (AB + BC + CA) AB + AC > 2AD AB + BC > 2BE AC + BC > 2CF 2 (AB + BC + CA) > 2 (AD + BE + CF) AB + BC + AC > AD + BE + CF

Ga

AB + BC + AC >1 AD + BE + CF

F AB 2

A

�

F

B

C

D2

�EFG ���BCG Side � 1 : 2

2 AD 3

2 (AG + BG + CG) > AB + BC + CA

�

G

Si

4

2 BG = BE 3

�

E

1

r

GB + GC > BC GA + GC > AC AG + BG > AB

2×

1

F

AB + BC + CA 4 1< < AD + BE + CF 3

CG =

of ABC

A

4 AB2 � BC2 � CA 2 = 3 AD2 � BE2 � CF 2

AG =

1 Area 12

�

3 (AB2 + BC2 + CA2) = 4 (AD2 + BE2 + CF2)

�

6 = 3 unit 2

O

D

E

x G y 90°

2y

B

AC 2 2x

C

AB2 + AC2 = 5BC2

E G C

AO : OG : GD = 3 : 1 : 2 �AFE � �ABC Let AD = 6 unit ��AG : GD = 2 : 1 ��AG = 4 unit and GD = 2 unit

x2 + 4y2 =

AB2 4

y2 + 4x2 =

AC2 4

5 (x2 + y2) =

AB2 � AC2 4

5 (4x2 + 4y2) = AB2 + AC2 � 5BC2 = AB2 + AC2

25

(Property)

Geometry

�

If D, E are mid points, AD and BE are medians,

Special case in isosceles

then DE =

AB = AC ��BE = CF AB2 + AC2 = 5BC2

D, E

2AB2 = 5BC2 5 2

AB2 + AB2 4

AD2 + BE2 =

5AB2 4

r

Si

��4 (AD2 + BE2) = 5 × R2

AD2 + BE2 = 5 × 4R2 AD2 + BE2 = 5R2

p a Pu t bl a �

at io n

:1

�ABC

G

Pr

C

Orthocentre AO = OC = OB = R

Example

on

pi

=

If D, E are two points on BC and AC. D, E, BC

15, 36, 39

1 × 15 × 36 = 270 2 4 �1

�

D

A

F A

B

E G D

BE = R =

2 2 2 2 CD� � CE � BC ��� � � CA ���� � � AD2 + BE2 = �� DE2

ABC

�

Area of �ABC (main �) = 3 � � 2 � 15 � 36 � = 360 cm2 � �

AC

C E AD2 = CD2 + CA2 BE2 = CE2 + BC2

:

Area of triangle made by 15, 36, 39 (triplets)

Ch

H H , GO = 3 6

4 × 3

3

am

ga n

BG =

Ga

H 2

=

Length of 3 medians 15, 36, 39. Then find the area of triangle ABC.

BO = Hypotenuse median = Shortest median

=R=

4 × Area of triangle made by using 3

ic

2

Area of �ABC =

length of all 3 medians.

Circumcentre

B

B

AB =R 2

5 AD2 � BE2 = 4 AB2

A

BO

DE =

��CB = 2R

Right Angle Triangle

O

BE

��4 (AD2 + BE2) = 5AB2

B C D If AG = BC or AD = 1.5 BC ���BGC = 90°

�

AD2 + BE2 =

A

G

�

AD

AD2 + BE2 = DE2 + AB2

2

5 AB � AB � �� � �� = � = 2 BC � BC �

�

AB =R 2

AC 2

AD2 + CF2 = 5.R2 AG2 + CG2 = 5BG2

AB2

� AD2 + BE2 = DE2 + AB2

26

C

Geometry

If we make incircle in � ABC with radius. r then r

Exocentre

=

�

� S

� ABC

r O2

A O3

r2

r3 c b a B r1

r · r1 · r2 · r3 = C

r · r1 · r2 · r3 = Example

O1

� � � � �4 · · · = 2 S S � a S � b Sc � 2

:

13

15

r1 =

� S�a

Si

r

14

r2 =

� S�b

Find r = ? � = 84 cm2, S = 21

r3 =

� S�c

a Pu t bl a

p

r

84 = 12 cm 21 � 14

on pi

Ch

Ga

am

ga n

Pr

ic

at io n

r=

27

r=

� S

Geometry

Equilateral triangle �

Equilateral Triangle �

All sides are equal. A 30° 30° a

�

a

R

:

� a

:

� r

2

:

2 3

:

1

Area � 4�

:

3 3

:

�

Circum circle : area

: incircle : square

Area

:

I, O, G, H

C

a 2

3 a 2

�

BC

B

ga n ,

Ga

Area circumcircle 4 = Area incircle 1

�

All 3 medians

�

Side �

3

Height �

2 cm

3 cm

2 3 cm

2k

3 k

2 3k

Circumcircle radius

C

Isosceles Triangle �

Triangle in which any two sides are equal. A � �

Area �

3 a 2

6 6

D

3 = a = Height 2

a

28 cm

�

2 3

1 r= h 3

am

(r) =

a

2 h, 3

B

Ch

Inradius R 2 = r 1

R=

on

a , 3

Circumradius (R) =

A

pi

All centre lies at same point.

3 2 a 4

Pr

Area =

I H G O

a

3 2 a 4

14 3

B

�

3· 3 6

Side of equilqteral �

D b

b 2

a

b 2

C

AB = AC All 4 centre lies on line AD.

196 3

3 ·3 6 � 9 2

2

Circle : Square A � � : 2 or 11 : 7

ic

�

3 a 2

:

C

a

a Pu t bl a

3 a 2 �ADB � �ADC

3 3

a

p

a

AD �

h=

:

A

AD � median, angle bisector, altitude, � bisector of BC =

4

r

60° D

a 2

Si

60°

at io n

B

�

2

4

� 54 3

In radius

AD =

28

AD a2 �

b2 = 4

4a 2 � b2 = Height 2

Area =

Isosceles Right Angle Triangle

b 4a 2 � b2 4

A

�

45° 17 B

17

15

8

16

8

H 2

1 H 2

C

45° 1

Area of � ABC = 15 × 8 = 120 cm2

r

H H 1 H2 × × = 2 2 2 4

Si

Area =

H 2

on pi

Ch

Ga

am

ga n

Pr

ic

at io n

a Pu t bl a

p

Perimeter

29

= H

�

�

2 �1

Geometry

Geometry

Right angle Triangle Right Angle Triangle �

How to Find Triplet?

Triangle in which one angle is 90º.

�

Odd number

90º Right angle triangle is inscribed in a semi-circle.

3

32 = 9

no.

2

4 5

(3, 4, 5)

(no.)

r

Make two factors at a diff. of 1.

r

�

a Pu t bl a

17

(8, 15, 17)

on

r=

8 10

(6, 8, 10)

P�B�H (P � B � H) = –H 2 2

r = S – H = S – 2R

��S = r + 2R

am

pi

�

36 2 = 18

R O

Circumcentre R

G

R

Ch

A

H P

C

Orthocentre

GO =

15

at io n

ic

Pr H 2

ga n

P+B 2

6

B

2 H R= 3 3

�=r·S

R H = 3 6

� = r · S = r (r + 2R) ����= r2 + 2rR

��= S (S – 2R) � Area of right angle triangle � 1 PB = r × S = (S – 2R) = r2 + 2rR. 2

30

84 85

(no.) 2 Break in two parts at the diff. of 2.

R = BO = Shortest median =

BG =

169

2

no.

2r = P + B – H 2r + H = P + B 2r + 2R = P + B 2r + 2R = P + B

B

82 2 = 32

8

P�B�H 2

2

13

Even number

p

B Orthocentre

:1

5 = 25

13

I

r+R=

2

5

2r

r=

12

Si

H = 2R

P

Ga

�

A x

��

x

D r r

r y C Area of ABC = x × y s=r+x+y ��= r · (r + x + y) = xy (x + r)2 + (y + r)2 = (x + y)2 x2 + r2 + 2rx + y2 + r2 + 2yr = x2 + y2 + 2xy

�

r Si

y-a

a

c

p

D p

�

Pr

ab 1 1 ab = c.p ��p = 2 2 c

90-�

�ADB � �ABC �

ay = xy – ax a (x + y) = xy B c

on

x C

am

�

Ch

Ga

B

ga n

-� 90

�

D

C

Q

a

pi

1 1 1 c c2 a 2 � b2 = �� 2 = 2 2 �� 2 = p p p ab a b a 2 b2

A

a

y �a a = y x

B

1 1 1 2 = 2 + p a b2

P

R

at io n

� a

a

a Pu t bl a

B

�ACB � �CDB � �ADC

�

S

ic

-� 90

y

90-�

C

xy

Side of square (a) = x + y �ABC � �ASR A

2r (r + x + y) = 2xy � r (r + x + y) = xy A

�

AB × BC AC

Maximum Area of a Square Inscribed in a Triangle

B

b

BD AD = � BD2 = AD × CD CD BD

BD =

y

r

�

Geometry

�ABD ~ �BDC

x b

Side of square (x) = �

C

C

AB2 = AD × AC

ab a+b

B

a

AB AD = AC AB

A

p c

y

b

A

BC2 = CD × AC

ab c� cp c = abc y= c�p = c2 � ab ab c� c

AD AB2 2 = CD BC

Side of square (y) =

BC CD �BDC � �ABC � = AC BC

31

abc ,x>y a2 + b2 + ab

�

Geometry

AP = x and QC = y

�

Shaded part

Side of square

=a=

xy

A

�ASP � �RCQ a y = ��a = x a

xy

A

B

P Q

= �ABC +

� 2 � 2 z � z = �ABC 2 2

R

r

C

�

All 3 side are unequal.

B

C

(P) = a + b + c =

Area of �ABC =

S � S � a �� S � b �� S � c �

ic

Area of �ABC =

on pi

Ch

Ga

am

ga n

Pr

� ABC

x2 + y2 = z2

at io n

Area of �ABC = r × S

Area of shaded part = Area of � ABC

32

a �b�c 2

Semi-perimeter

p

2z 2y

–

Scalene Triangle

�

Perimeter

2x

C

Area of shaded part = Area of ABC

y

a 90-�

A

�x 2 �y 2 �z 2 � � 2 2 2

Si

a

B �

a

2y

= �ABC +

a Pu t bl a

S

a

B

C

x

�

2x +

+

abc 4R

2z

Geometry

Square and Rectangle

A closed figure with all 4 sides equal and all angles 90º. 90º

45° 45°

a

45° 45° 90°

90° O

C a

a

90°

�

= a2 =

d2 2

� Perimeter

= 4a

� AC = a 2 = BD

AC BD = 2 2

ga n

r

a

A

a

1

1

1

Q

B

P

Area of

Area of ABCD 2

PQRS =

�

am

R

C

1

A

PQRS

Ch

C

pi

90°

D

S

on

� Diagonal bisect each other at 90° � �DOC � �AOB � �AOD � BOC

R

D

Pr

� Diagonal bisect the vertex angle.

� AO = BO = CO = DO =

ABCD

P, Q, R, S

at io n

� Area

�

B

ic

a

: 2 11 : 7 P, Q, R, S are mid points of sides of square. PQRS is also a square.

p

45° 45°

a Pu t bl a

A

(2)2 2

� (1)2 :

90° 45° 45°

a

1

r

a

1

Si

D

Ga

�

Area of circle : Area of square

�

Square

B

• Radius of incircle

=r=

• Radius of circum-circle = R= = R=

a d = 2 2

a 2

Bigger : Outer : Medium : Medium : Small : Small circle

circle

:

a d = 2 2

R 2 = r 1 Area of circumcricle 2 = Area of incircle 1

square

Area

8�

: :

16 1 2

88

33

: 56

square :

4�

:

: 8

1 2

:

circle

44

1 2

:

or 28

square

:

:

: 2 � : 4... 1 2

: 22

: 14 ......

�

Largest square inscribed in semi-circle � �

r

AC = BD = l 2 � b2

a

�ABC � �CDA

o

AO = OC = BO = DO = a2 = r2 4

�

Diagonals bisect each other but not at 90° 90°

�

�DOC � �BOA and �AOD � �COB

�

BO is median of �ABC � Ar �AOB = Ar �BOC Similarly Ar �AOB = Ar �BOC = Ar �COD = Ar �AOD In square/rectangle/parallelogram/rhombus ABCD�

�

2 a= r 5

�

P, Q, R, S are mid points of respective sides.

r

5 2 4 a = r2 � a2 = r2 4 5

�

AC BD = 2 2

Si

a2 �

ABCD

p

P, Q, R, S D

a Pu t bl a

C

A

3 B

P

ga n

ABCD � Square/Rectangle 1 Area 2

Rectangle

�

�

�

1 �

�

o 1 l

B

l

Shaded Area 1 = � Area (x + y) = Area z 2 Area ABCD

Ch

�

D

C y P

z

�

1

b

A

l

z

British Flag Theorem (For Square/Rectnagle)

90º

D

b

b

1 �l �b Area � APB 1 2 = = l�b 2 Area � ABCD

ABCD

A type of quadrilateral that has its parallel sides equal to each other and all the four vertex angles are equal to 90º.

Ga

�

y

A

am

Area (1 + 2) = Area (3 + 4) =

�

C

x

ic

o

Pr

1

Q

on

2

S

P

at io n

4

pi

R

D

Geometry

Area = Length × breadth = l × b Perimeter = 2 (l + b) Diagonal do not bisect vertex angle.

w

x

C A

�

B

Note � AC and BD are not diagonals. 1

��AC

b �

�

BD

P is any point inside

P

PA2 + PC2 = PB2 + PD2 x2 + z2 + w2 + y2 = x2 + w2 + y2 + z2

� B

34

Geometry

Parallelogram/Rhombus/Trapezium

Parallelogram It is a quadrilateral with opposite sides equal and parallel.

4� i.e., Ar �AOB = Ar �BOC = Ar �COD = Ar �DOA =

D

Ar

C

ABCD

r

�

180-�

d12 + d22 = 2 (a2 + b2)

B

�

� AB ��� CD and BC ��� AD

180°

ga n

D � b � A

�

a

d1

1

d2

O

1

�

1

�

a

�

1

C

�

p

Ch

�

h2 a = h b 1

�

P is any point

P

b

a

D

� B

C y

b

�AOD � �COB

AO = CO =

1 ab sin � = ab sin 2

Ar of Parallelogram = Base × Height � ah1 = bh2

�COD � �AOB

� BO = DO =

1 ab sin � = Ar �BDC 2

� Ar of Parallelogram = 2 ×

am

� Opposite �'s are congruent

B

a

Ar of �ABD =

on

AC � BD

h1

Area of Parallelogram = ab sin �

pi

� �ABC � �CDA

b

b

A

Pr

� AB = CD and BC = AD

C h2

at io n

� Sum of two adjacent angles is 180°

� ��� �, � � r

a

ic

� Opposite angles are equal

a Pu t bl a

�x:y:x:y

� Diagonal

Area of Parallelogram D

� �A = �C, �B = �D � Angle ratio

� d1 > d2

Si

d 22 = a2 + b2 – 2ab cos �

A

1 4

d12 = a2 + b2 – 2ab cos (180° – �) = a2 + b2 + 2ab cos �

�

180-�

Ga

�

� Area of all 4 triangle is same.

BD 2

A

AC 2

z

w x h1 a

b h2 B

Area (x + y) = Area (z + w) = ABCD

� Diagonal bisect each other but not at 90°.

1 1 ah1 = bh 2 2 2

90°

35

1 Area of Parallelogram 2

Geometry

�

P

D

C

D

C

d2 2

y

x z

O

1 Area � ABCD 2

Ar (x + y) = Ar(z) = �

B

d12 d22 � = a2 4 4

1 Area � ABCD 2

Q

d12 + d 22 = 4a2

C

Area �BOC =

1 d1 d2 � � 2 2 2

Si

r

D

A

a

d2 2

d1 2 a

B

A

Ar �APB =

d1 2

Area of rhombus = 4 �

P

Perimeter = 4a Figure Made After Joining Mid Points

B

p

A Ar � APQ 3 = 8 Ar � ABCD

a

Ga

� 180-� � A B a � All sides are equal � �A = �C and �B = �D

� Square

Rectangle

� Rhombus

Rhombus

� Rectangle

at io n

am

�

ic

Trapezium

It is a quadrilateral with one pair of parallel opposite sides. The parallel sides of a trapezium is called bases and non-parallel sides are called legs.

AB ���CD If AD = BC then it is called isosceles trapezium.

Ch

O

a

�

on

�

�

�

ga n

180-�

C

Square

pi

a

D

Pr

Rhombus is a type of parallelogram with all sides equal and diagonals bisect each other at 90º 90º

AD = BC

� Diagonal bisect vertex angle.

D

�

� Diagonal bisect rhombus into two equal areas.

P

A1

� AC � BD

� O

�

� All 4� made by two diagonals are congruent.

�AOB � �COD

� Diagonal bisect each other at 90°.

PQ =

90°

36

Q

B1

B

AC � BD

AC BD and BO = OD = 2 2

C

�

A

4� � AO = OC =

� Parallelogram

Scalene quadrilateral

a Pu t bl a

�

Rhombus �

d1 × d2 1 d1 d2 � � = 2 2 2 2

AB + CD 2

�

b

D

�

C

D

Q

P

m Q:

:P n

n

n

A

B

a

Area � DCQP m = Area � PQBA n

B

a

PQ =

ma � nb m�n

Isosceles Trapezium

ma 2 � nb2 m�n

PQ =

C

m

m

A

Geometry

b

D

�

C

180-�

b

�

b

AC = BD

am

B

�

�

A

� P

a

AP = QB =

Area ��AOD = Ar �BOC

�APD � �BQC

D

x O

� ADP

k

y A

Q

B

Shift � ADP near �BQC ��trapezium converted to rectangle.

C

k

b

b �a 2

�AOB � �COD

�

C

�

Ch

Ga

a

a

D

on

C

h

A

1 × (a + b) × h 2

ga n

D

Pr

Area of trapezium ABCD =

�

or if a trapezium is inscribed in a circle it must be a isosceles trapezium.

B

ic

a

at io n

d2

d12 + d22 = c2 + d2 + 2ab

�

r

d

A

�

B AB ���CD ; AD = BC �A + �C = �B + �D = 180º � Each isosce le s trapeziu m is a c ycli c quadrilateral.

a Pu t bl a

c d1

�

A

C

p

D

pi

�

180-�

O

Si

If m : n = 1 : 1 Then PQ =

a 2 � b2 2

�BQC

��

D

C/A

P

Q

p1

B

k×k=x×y k=

xy

37

B/D

Geometry

Circle Circle �

Circle � A round plane figure whose boundary consists of points equidistant from a fixed point (the centre).

O

�

r

r

Chord � line touches circumference of circle at two points. �

Secant Line

Radius � It is a straight line from the centre to the circumference of the circle.

�

= �r2

M

B

Secant PBA

A

A perpendicular drawn from the centre of a circle to a chord bisects the chord.

OC � AB � AC = BC or If AC = BC then OC � AB

Ch

PQ =

am

PQ = diameter = biggest chord of circle.

pi

= 2�r

ga n

Circumference

�

on

Q

P

ic

Pr

r

Area

The

at io n

P

�

a Pu t bl a

=r

A straight line that intersects a circle in two points is called secant line of circle.

p

�

PQ = radius of circle

Q

Si

P

Ga

Major arc AB

O

Major segment AMB

A

r

B Minor segment AOB

O Minor are AB

A �

Tangent Line �

r c

B

1. Equal chords are equidistant from the centre. 2. Equal chords make equal angle at the centre.

A line that touches the circle at only one point is known as tangent of the circle.

If AB = CD then �AOB = �COD or if �AOB = �COD then AB = CD

�OPQ = 90°

38

Geometry

C

Alternate Segment Theorem �

�

�

O

B D Angle made by an arc on centre is double the angle made by the same arc on the circumference of centre.

B

� O

�

A

r Si

B

O B

�

�

a Pu t bl a

Angle made by an arc on same side of circle are equal.

�

Pr

�

pi

Angle made in semi-circle is right angle.

90–

A

�

O

Ch

Ga

am

�

B

�

ga n

A

�

on

�

ic

�

B

� A

at io n

2�

B

p

2�

O

�

P

A

�

A

90 -�

�

Angle made by a chord and tangent is equal to the angle made by the chord in other segment of the circle.

90 -�

A

C

A

O

180-�

� 2

D

� 90+

B

�ADB = 90° +

� 2

�ACB = 90° –

� 2

P

� 2

�APB = � �

AB and CD two chords cuts each other internally at point P. Then � AB

CD

�APD = �BPC =

Angle made in quarter circle is 135°.

P �AOD � �BOC 2

�APC = 180° – �APD

135°

A O

135°

P

270°

C B

39

D

�

�

Geometry

AB and CD two chords cuts each other externally at point P. Then � AB

CD

B A

�

P

�APC = �BPD =

� � �

P

��� 2

r2

r1 r 01 1 r2

02

C

D

A B

0102 = r1 + r2

�

r1 1 � sin � � r2 1 � sin �

P D �

Si

2

�

E

PCD

A

�

Pr

B

= 2PA = 2PB

ga n

C P

��PC + CE + ED + PD

A

D

AB + CD = BC + AD Two chords AB and CD cuts each other at point P internally. AB

CD

P

C � �

��PA + PB

�

B

x

P

A

��PC + CA + BD + PD

��2PA or 2PB ( � PA = PB)

w

x

PB PD = PC PA

Ch

Ga

D

Q

PA × PB = PC × PD

am

E

y

w

P

A

Perimeter of PCD

C

S

D

B

y

p

90°- �

a Pu t bl a

C

�

R

z

A

O

z

D

�AOB 2

on

�COD =

� 2

at io n

�COD = 90� �

pi

�

If a circle is inscribed in a quadrilateral or a quadrilateral is circumscribing a circle.

r

C

ic

�

O

P

D �

� � B

Two chords AB and CD cuts each other at point P. AB

CD

P

PB×PA = PD×PC

B

A

C

B P

AD = BD = CD

D

�ACB = 90°

C

40

�

Geometry

From an external point a secant and tangent is drawn. Then �

�

When two circles touch each other externally and common tangent is given �

�

PT2 = PB × PA

� �PAC � �PBD

T

A �

� �

A

P

B

� P �

�

� TPB � �APT

� B

r

C

PB PT � PT PA

When a circle is drawn between two parallel lines and a tangent is given intersecting these two parallel lines.

�

Si

PA PC = PB PD

Two circles intersect each other at two distinct points and two common tangents are given then �

p

�

D

a Pu t bl a

AB = CD

�COD = 90°

r

�

� � �

B

E b

b D

�

Q C

S

D

AB × AC + AE × DE = AE2 �ABD � �AEC

Ch

� AB×AC = AD × AE

�PAC � �PBD

A

B

�

B P

AD AB = AC AE

�

�

R

ic

A

RS2 = AB2 + PQ2

When two circle touch each other internally and common tangent is given �

Ga

�

a

on

O

C

am

r

a

pi

A

Pr

ab

ga n

r=

at io n

2� + 2� = 180° � + � = 90°

� �

A �

�

�

�

P

B

C

� D � E

D PA PC = PB PD

41

C

�

Geometry

If two chords intersect each other at 90°. 90° A a x+y 2 x x-y z 2 C y

O

r w z+w 2

�

D

B

If two circles do not touch each other, there can be maximum 4 common tangents. Direct common tangent = 2 Transverse common tangent = 2.

x �y �z �w 2 2

2

2

4

r

r=

2

x× y=z×w

Si

A

2

r=

a2 � y2 x 2 � y 2 � z 2 � w2 = 2 2

Extrenally Touch

Pr

F

D

D

=4

Intersecting circles

on

A

B

AB � common chord

O1PO2 � straight line

Maximum common tangent = 2

Maximum common tangent = 3 AB = CD = EF

Internally touch �

C

pi

am

O2

Ga

C

B

r1

�

Ch

r1 P

ga n

O1

E

Q

DCT = AB & CD TCT = PQ & RS Maximum common tangent = 4

When two circles touch each other externally. Then distance between their centres is sum of their radii.

A

R

B S

at io n

z 2 � w 2 � 2zw � x 2 � y 2 � 2xy 4

ic

r2 =

p

P

a Pu t bl a

2

�z � w � �x �y � r2 = � � � +� � 2 � � 2 �

�

O2 r2 P r1

O1

2 �

When two circles touch each other internally then distance between their centres is difference of their radii.

O1O2P � straight line O1O2 = d = r1–r2 Maximum common tangent = 1 =1

42

If one circle is made inside other circle and they do not touch each other then no common tangent is possible.

Geometry

Direct Common Tangent �

Length of DCT

Divide by

DCT

1 1 1 = + r3 r1 r2

d � �r1 � r2 �

2

2

A

r1–r2 C r1

�

B d

O1

r2

d

r1r2 r3

(Result)

When two circles intersect each other at two distinct points then length of their common chord AB is � Common chord

O2

C

D

A

CB ���O1O2

r1

r2

r

P

2

O1

O2

Si

d2 � � r1 � r2 �

DCT = AB = CD =

�

AB

Transverse Common Tangent

B

�

Length of TCT

TCT

��AB = 2AP = 2PB

r2 O2

�

Q

R

If two circles pass through each others centre.

at io n

d

O1

1 � O1O2 � AP 2

Ar ��O1AO2 =

a Pu t bl a

r1

S

p

AP = PB

P

Length of common chord AB is �

B

r1

r2

O1

d=r1+r2

DCT = �

� r1 � r2 �

2

2

ic

4r1r2 = 2 r1r2

r1

C r3

B r2

2 r1r2 � 2 r1r3 � 2 r2 r3

43

O2

r r

AB = AC + BC A

r

r B

� � r1 � r2 � =

3r

A

O1

Ch

Ga

O2

3 r×2= 2

r

am

A

�

AB

on

AB = DCT = 2 r1r2

ga n

�

AB =

pi

DCT > TCT

2

Pr

d2 � � r1 � r2 �

TCT = PQ = RS =

Co-ordinate geometry

Co-ordinate Geometry +y

+y �

II quadrant

(0,3) I quadrant

P ��(x, y) ��(r cos�, r sin�)

(0,2)

r

(0,1) (1,0) (2,0) (3,0) (0,–1) IV quadrant (0,–2) (0,–3) –y

+x

–y

sin � =

O (origin) � (O, O)

p

Ordinate

tan ��=

Equation of x-axis � y = 0

�

�y= 0

Equation of y-axis ��x = 0

Ga

am

Ch

O

C

B(x2,y2) � x1 � x 2 y1 � y 2 � , � 2 � � 2

Distance formula �

R(10,–8)

Pole � Reference point in polar co-ordinate system the co-ordinates is called a pole.

Distance between two points is the length of the line segment that connects the two points in a plane.

A � (x1, y1)

Polar co-ordiantes of a point When each point on a plane of a 2D co-ordinate system is decided by a distance from a reference point and an angle is taken from a reference direction. It is known as the polar co-ordinate system.

mid point

Mid point C = �

pi

P(7,4)

�

A mid point is the middle point of a line segment which is equidistant from both the end points.

A(x1,y 1)

on

ga n

Q(–8,5)

�

y x

ic

Representation of points

S(–5,–5)

x r

Mid point

�x=0

Pr

�

cos ��=

Mid Point Formula

a Pu t bl a

abscissa

y

y r

x2 + y2 = r2

P ��(x, y)

x

+x

r

III quadrant

O

Si

(–3,0) (–2,0) (–1,0)

y

� x

at io n

–x

–x

� AB =

� x 2 � x1 �

2

� (y2 � y1 )2

B � (x2, y2) �

Intersection formula �

(A) Internal division m1 : m2 A (x1, y1)

2D

P

B (x2, y2)

� m1x 2 � m2x1 m1y 2 � m2y1 � , � m1 � m2 � � m1 � m2

P �� �

44

Co-ordinate geometry

(B) External division

(O, C)

AP : PB = m1 : m2

C

m1 P

B m2 (x2,y2)

� m1x 2 � m 2 x1 m1y2 � m2y1 � , � m1 � m2 � � m1 � m 2

�

P �� �

Slope of a Line

y=

The slope of a line is a measure of its steepness. It is the change in y co-ordinate with respect to the change in x co-ordinate.

y = mx + c

y �

(m) = tan ��(slope)

a Pu t bl a

��m1 = m2 � a1 �a 2 � b1 b2

at io n

(x2, y1) (y2–y1)

ic

Pr

on

�

pi

�

a1 x + b1 y + c1 = 0 m1 = tan �

Ch

Ga

�a b

a2 x + b2 y + c2 = 0 m2 = tan �

am

Standard equation of a line � ax + by + c = 0 slope =

�

a1 b1 � a � b 2 2

� x2–x1

ga n

�

If two lines are parallel to each other their slope

p

y 2 � y1

�

– x of coff.

will be equal � m1 = m2

tan ��= x � x 2 1

(x1–y1)

–a

m = b = y of coff.

compare

x Slope of a line/

�ax c � b b

r

�

ax + by + c = 0 by = – ax – c

Si

A (x1,y 1)

� O (0,0)

Point from of a line

�

If two lines are perpendicular to each other �

Slope of line AP = slope of line AB. AP A (x1, y1)

AB

P (x, y)

a2 x + b1 y + c1 = 0 m1 = tan �

B (x2, y2)

y � y1 y 2 � y1 y2 � y1 � (x � x1 ) x � x1 x 2 � x1 � y – y1 = x 2 � x1

�

a2 x + b2 y + c2 = 0 m2 = tan (90 + �) = – cot �

y – y1 = m (x – x1) �

Slope

=m

Intercept on y-axis

90 + �

y-

=c

y – c = m (x – 0) y = mx + c

� m1m2 = tan ��(– cot �)

m1m2 = – 1

45

Co-ordinate geometry

Intercept form of a line �

Concurrent Lines

Intercept at x-axis = a x-

�

Lines passing through a single point

�

Intercept by line

a

Intercept at y-axis = b =b

yx y + =1 a b (0, b)

–b y – 0 = a ( x – a)

(a, 0)

r

x y � =1 a b

=

x y � a b

a 2 � b2

(O, b)

ax + by =

k

Equation of line perpendicular to line ax + by + c = 0 is bx – ay = k.

bx – ay =

Pr

ax + by + c = 0 k

�

Ga

(a, O)

Line ax + by = 0 always passes through origin. ax + by = 0

on

��� acute angle

Let angle between two lines = � and their slopes are m1 and m2 then

am

ax2 + b2 y + c2 = 0

m1 = m2

No intersecting point

�

a1 b1 c1 � � a 2 b2 c 2

�

m1

Ch

a1 x + b1 y + c1 = 0

O

Angle Between Two Lines

pi

ga n

(A) If two lines are a1 x + b1 y + c1 = 0 and a2x + bz y + c2 = 0 are parallel to each other then m1 = m2 and a1 b c � 1 � 1 a 2 b2 c 2

a 2 � b2

at io n

a Pu t bl a

ax + by + c = 0

p

Equation of line parallel to line ax + by + c = 0 is ax + by = k.

ic

�

Si

�y x � �1 b a

1=

x y � = 1 between both axis a b

m2 m1 � m2

tan ��= 1 � m m 1 2

� No solution possible

(B) If two line are co-incidents on each other then a1 b c � 1 � 1 a 2 b2 c 2 and infinite solutions

��= tan–1

a1 b1 c1 � � a 2 b2 c2 ,

�

(C) If two lines intersect at single point then m1 � m2 a1 b � 1 a 2 b2 , unique solution possible

Distance of line ax + by + c = 0 from a point (x 1, y1) (x1, y1) ax + by + c = 0 ��

�

m1 � m2 1 � m1m2

ax1 � by1 � c a 2 � b2

Distance of line ax + by + c = 0 from origin ax + by + c = 0

m1 � m2, a1 b1 � a 2 b2

=

46

C a � b2 2

�

Co-ordinate geometry

� Reflection of point (x, y) across the line y = x is

Distance between two parallel lines:-

(y, x). (x, y)

ax + by + c1 = 0 ax + by + c2 = 0 distance

=

5. C1 � C2

(y, x)

y=–x�

y

a 2 � b2

Reflection �

y=x

Reflection over y = – x � Both co-ordinates change their place and sign as well.

� Reflection of point (x, y) across the line y = – x is (–y, –x)

A reflection is the mirror image of the shape.

(x, y)

y=–x

(–y, –x)

Centroid and Incentre

Types of reflection in co-ordinate system

A � (x1, y1)

Si

p

r

�

c

h

G

b

I

p

Mid point

I � Incentre

If reflection is asked about origin change only signs

Pr

ic

� x1 + x 2 � x 3 y1 � y2 � y3 � , G� � � 3 3 � �

pi

on

� ax1 � bx 2 � cx 3 ay1 � by2 � cy 3 � , I� � � a�b�c a�b�c � �

Equation of Circle �

am

ga n (3, –2)

Ch

Ga � x-

r �

Equation of circle (x – a) + (y – b) = r 2

y

2

2

r

Reflection over y-axis � y co-ordinate remains same but x co-ordinate changes its signs.

(a, b)

� y

y

Where (a, b) are co-ordinates of center of circle and r is the radius. (a, b)

Reflection over x-axis � x-co-ordinate remains same but y co-ordinate changes its signs. x-

3.

at io n

G � Centroid

–3, + 2

2.

C (x3, y3)

a Pu t bl a

1

p

1.

a

B (x2, y2)

h

x 4.

Reflection over y = x � When a point is reflected across the line y = x, the x-co-ordinates and y co-ordinates change their places. y=x�

y = x x-

if centre is origin x2 + y2 = r2

y

�

Area enclosed by �x�+�y� = a ��2a2 �x�+�y� = a

47

��2a2

2D

2 Dimension Mensuration 2 �

Zero-Dimensional figure

� (point)

One-Dimensional figure

A

�

only length.

Two-Dimensional figure (

r

Si

B

18

= K2

B

b

Perimeter

p

on

H 2

4.

H2 4

Equilateral triangle A a

1 1 1 bcsinA = absinC = acsinB 2 2 2

B

h

abc 4 � Area

Inradius

48

a C

a

Side/

Area S

Circumradius (R) =

Area(�) =

Perimeter = 2H+ 2H = H( 2 +1)

s � s � a ��s � b ��s � c �

Inradius (r) =

H H 2

=a+b+c

a+b+c Semi-Perimeter (s) = 2

Area (�) =

H Hypotenuse = 2 2

Isosceles right angle triangle

C

a

Area(�) =

3.

Ch

c

P�B�H 2

Circumradius (R) =

am

A

1 × PB 2

pi

ga n

Scalene triangle

Ga

1.

B

ic

=K

Triangle

Area(�) =

Inradius (r) =

=K

Area = K2 times

H

Perimeter = P+B+H

If each corresponding length of any 2D figure = K times. Then, perimeter = K times

(Result)

Right angle triangle P

�

Pr

�

21

1 1 1 : : h1 h2 h3

at io n

15

6

2.

�

a Pu t bl a

7

1 1 1 1 ×Base×Height = ah1 = bh2 = ch3 2 2 2 2

�a:b:c=

For Any 2D figure 5

C

� ah1 = bh2 = ch3 = (constant)

H Length, Breadth & Height

�

b h3

a

Area (�) =

Three-Dimensional figure

L

h2

B

)

B Length and Breadth

L

h1

c

=a

(r) =

a h or 3 2 3

2D

a 2h or (R) = 3 3

Circumradius

(�)�=

Area

� �k

Area 3

� �k

� �k2

3K

3K

Find the Area of � with side � 36, 29, 25 29, 20, 21 15, 20, 25 (use the common triplet) 15 + 21 = 36 (Acute �) triplet

� 19,

p ic

1 1 3 2 1 x = xa + xb + xc 2 2 2 4

37

12 16 12 35

3 2 x 3 x = (a+b+c) � x = a+b+c 2 4 2

� H = a+b+c

x=

2 (a+b+c) 3

37

Area of �ABC =

20 16

30° 19 cm

D

b B

�� �

(Result) 1 (a+b+c)2 3

A

�

1 = ×19×12 = 114 2

� Area of original

��

P

H = a+b+c Area �ABC = �BPC + �APC + �APB

pi

20,

H

a,b,c �

Ch

3

x

a,b,c � � (Perpendicular) from point P on each side

35–16 = 19 (obtuse �)

Area

C

x

P � Any point inside triangle./

1 = ×36×20 = 360 cm2 2

Ga

Simplify by 3

B

x

b Pa

ABC �

on

C

57, 60, 111 � Find Area

12

c

x

= 2a + b

A

�

1 length

am

21 D 15

A

1 2 a · sin � 2

Perimeter

Pr

25

20

Area of triangle �

b 4a 2 � b2 4

ABC � Equilateral � with side x and height H.

ga n

B

=

B

A 29

Area

=

C

4a 2 � b2 2

=

at io n

common

�

Area

a Pu t bl a

90°

b

b 2

Si

Heronian Triangle: All sides and area is integer.

2

a

Height

(Result)

2

o h

� b 2

B

3

2K

� � 2 2

a

3 2 a 4

height

2

�

3a 2

(H) =

a(side)

A

= 3a

Height

�

Isosceles triangle

r

Perimeter

5.

x a

P

b a

b C

Q

a x

b x

Area of �ABC = ax = Area of �PQR

114 × (3)2 � 114 × 9 � 1026 cm2

49

R

�

If regular polygon have same perimeter then figure with more number of sides have greater area.

�

2D

Area of quadrilateral made by joining the mid points of all sides of given quadrilateral is half of the original quadrilateral.

� Area of square > Area of ��with same perimeter. �

Square

�

>

� Infinite sides

� circle

�

a

D

� circle has more area then any other closed

r

figure. a

�

a

O r

Quadilateral

a

Si

Area of �AOB = Area of �BOC = Area of �COD =

Square Rectangle Parallelogram Rhombus Trapezium

Area of �AOD =

Quadrilateral : A closed shape and a type of polygon that has 4 sides, 4 vertices and 4 angles.

Perimeter

4

Area

a Pu t bl a

= a2 =

B

h2 h1

B

Ga

A

Area = =

1 × BD × (h1 + h2) 2

ic

�

am

C

R : r = 2 :1,

a 2

(R) =

a d = 2 2

Area of circumcircle 2 � Area of incircle 1

If we make circle inside a square and again make a square inside the circle and so on... the area will becomes half and so on.

Ch

D

(r) =

Circumradius

on

Area of Quadrilateral

ga n

�

pi

�A + �B + �C + �D = 360°

Inradius

Pr

A

d2 2

=a 2

at io n

diameter

D

a2 4

= 4a

p

4 C

B

r

A

Quadrilateral

4

C

1 × diagonal × (Sum of perpendiculars drawn 2

on this diagonal) = �

1 × 2

×

In any Quadrilateral D A4 A

A3 A1

C

Area of largest square

:

middle : smallest

4

:

2

A2 B

� Largest : smallest

A1 × A 3 = A 2 × A4

50

:

1 �4:1

2D

Rectangle �

A B

�

H

D 1

G

D

E

B

1

O 1 L

A

C

C

1

B

Perimeter

F

(P) = 2(L + B)

(A) = L×B

Area Diagonal

Area of A, B, C, D, E, F, G, H is equal

= Area of �AOD =

a

=

a C a –

+

a

ga n

� � 2 2 �� 4 a –a = � � 1� a2 = a2 2 2 7 � �

Ga

r3

C

a

r

b

a , 16

2

3a , 8

= 2 (a + b) 2

Area

= Base

× Height

If length of one diagonal is d Then Area

d

= 2 s(s � a)(s � b)(s � d)

Where

s=

a�b�d 2

Rhombus �

D

a d1

a

r3=

B

2

B

r2=

b

a

A

r1=

C

AC + BD = 2 (a + b2)

r2

A

a

D

Perimeter

Ch

r1

D

=a

Breadth 2

Parallelogram

A

am

If side of square

Breadth 2

at io n

=a A a

Pr

Area of leaf

�

=

ic

B

a

4 2 a 7

a Pu t bl a

Area of leaf

on

a

p

C

a A

inside rectangle =

3a 2 = 14

pi

D

LB 4

Radius of maximum size circle that can be put

Si

�

Shaded Area

L2 � B2

Area of �AOB = Area of �BOC = Area of �COD

�

�

= AC = BD =

a 6

C

O d 2 a

a B

Perimeter

= 4a

4a2 = d12 � d22

� a 3a a � : � × 48 = 3 : 18 : 8 r1:r2:r3 = � : �16 8 6 �

(Property) 1 � d1 � d2 2

area

=

Area

= Base

× Height

Area of ��������� Area of � BOC = Area of COD = Area of �AOD

51

Trapezium �

b

D h

d A

O

�

C d1

r

c

d2 B

a

Perimeter

=a+b+c+d

Area

circumference = constant =�� diameter

1 (Sum of parallel sides 2

=

C =�� � Circumference 2r

) × distance between them

d12 � d22 = c2 + d2 + 2ab

(Property)

Path Around or Inside a Rectangle

p a Pu t bl a

at io n

x

ic

Pr

Area of crossing road

= lx + bx –x2

= 2(l + b – 2x)

Ga

Area of path

am

�=

l = r× b x

Area of path Perimeter of path

�c 180

�� 180

l = r�c or �c =

= 2x (l + b + 2x) = 4 (l + b + 2x)

Where is x is the width of path

=

�° = � ×

x l x

circle)

x

Path Outside a rectangular field

x

� (sector is what part of 360�

� 360�

θ Length of AB = l = × 2�r 360°

= 4 (l + b – 2x)

Where x is the width of the path.

l

θ lr × �r 2 = 360° 2

= 2x (l + b – 2x)

Perimeter of path

B

Area of sector

Ch

x b x l

pi

Path inside a rectangle x

3.

on

ga n

Perimeter of path

x

r

� = central angle =

= x(l+b–x)

2.

O

r

A

l

� 24.5=7×3.5

�

b

x

A=�r2 154 154K2 154×(3.5)2 154×16

Sector of a circle

Crossing road inside rectnagle

�

c=2�r 44 44K 44×3.5 44×4

radius 7 7K 24.5 28

1 (a + b) × h 2

r

=

Si

Area

= 2�r

= �r2

Area

1.

2D

Circle

x

52

l r

OAB =

Segment �

�

Minor segment

A

22

7 Radius 7

O 7 120° 7 30° 30° 7 3

B Minor Segment

Si 7

Circumference

a Pu t bl a

p

Radius

7

Perimeter

�

Area

Pr

Area of major segment

Major Segment

Length of string = 2�r + 2 × number of circles

Ga

Ch

Area of sector (210°) + Area of �OAB (210°) + �OAB

210� 1 × 100��+ ×10×10×sin150° 360� 2

�r 2 4

=

on

am

B

�

2 �r �r = 2r + 4 2

pi

ga n

10 150° 10

=r+r+

ic

Major segment

Area 77 = 38.5 2

25

at io n

� 154 49 3 � � �� 3 � 4 �� cm2 � �

��

11

7

154 49 3 � – × 3 2 2

��

�r 2

r

�

120� 1 � × 154 – × 7 × 7 × sin120° 360� 2

7 3

= �r + diameter

For a quadrant

�OAB

210° O

Area 77

2

Area of minor segment = Area of sector – Area of �OAB

A

7 Circumference 36

Perimeter Area =

�

2D

For semi-circle

= 2�r + 2 × �

r

A

r

r

7 1 1 ×100��+ ×10×10× 12 2 2

B r

r

r

O

C

� 175� � � 25 � cm2 �� � � 3 �

�ABC = equilateral �

Perimeter of segment

OB = circumradius Let radius of small circle = x

�� � �� � sin � = 2r � 2� � 360 �

x OB =

53

side of equilateral� 3

�

2r 3

2D

�x=

2r –r 3

Derived Figure

�2- 3 � �×r x = �� 3 �� �

15

If we draw a big circle around these 3 circles, then radius of that big circle � y =

2r +r 3

radius of shaded part

3

� 3

Radius of each small semi-circle =

�2� 3 � � 3 �� �

y = r × ��

Si

(Result)

Two circles of equal radius are touch each other as shown in figure, a square of side 'a' is placed between two circles. One side of square is on direct common tangent of both circles, find r/a?

A

a

a Pu t bl a

r

R–r

p

�

AB

n

+r

R 15 = 6 6

n semi-circle (even) drawn on diameter AB

r

2r

�y=

=

R n

R +r n

R

O 2

=

R n

B

2

�R � �R� � � r � = � � +(R–r)2 �n � �n�

at io n

r/a

R 3

on

am

� radius of shaded part

= �

R 4

2� 8 8 = 2� 3 3

Find A/B = ? A

Derived Figure

24

A B

B

B

B

A

12

radius of shaded part

nR r = 2 � n +1�

8 Here, n = 2, R = 8

Base Figure

R

Ga

R 2

R 6

pi

ga n

r 5 = a 2

�

radius of small circle

Pr

O2

Ch

O1

ic

After solving this

A=B

R 12 = = =4 3 3

A =1:1 B

54

A

(Property)

2D

Polygon �

Polygon: closed figure of 3 or more sides.

2

(4–2) × 180° = 360°

1

3

or 2��= 2 × 180° = 360° ....... Polygon

2

3

� (5–2) × 180° = 540°

concave/

or 3��= 3×180 = 540° � Each interior angle of a regular polygon � � � n � 2� � 180� n

any one angle is more than 180° Any one diagonal or more will be outside.

*

�

180°

180°

Always a convex polygon

ga n

* each side is equal

ic

on

Regular polygon

� Each exterior angle of a regular polygon =

pi

* each interior and exterior angle is equal

am

= No. of sides

Ch

Ga

Regular Hexagon: 6 sides/

�I + �E = 180° × n (n–2)180° + �E = 180° × n

180°n – 360° + �E = 180° n �E = 360° I + E = 180° Internal angle + External angle = 180°

Pr

Ex:

at io n

�

�

6 vertices/

=

=

6 exterior angles/

angles, n exterior angles.

�

n

n

n

a

Sum of all interior angles of a polygon with n sides

F

a

n

� (3–2)×180° = 180°

A

55

C

a a 60° a

a

� (n–2) × 180° 1

2

a

a

a

60°

a

60°

a B

n �n – 3� 2

n � n � 3�

Regular Hexagon a E D

� n sides polygon have: n vertices, n interior

360° n

360° E

No. of diagonals in a polygon =

6 integior angles/

�n

n

� 360°

'n'

*

� n � 2� � 180�

Sum of all exterior angles of a 'n' sided polygon � 360°

a Pu t bl a

�

�

�

p

* each interior angle is less than 180° All diagonals lins inside

Si

convex/

r

1

360° n

Each Interior angle

= 120° = 60°

Each exterior angle

�

Total diagonals

=9

Large diagonal

= FC = AD = BE = 2a

Regular Hexagon = 6 equilateral � = 3 Rhombus 3 2 3 3 2 a a ×6 = 4 2

P

= FD = DB = BF =

3a

B A P, Q, R are mid points P, Q, R

(R) = a (side) D 1

P

a 1

1

F 1

1

1

1

1 a

B A a 6 equilateral ��formed. 6

� �

Pr

Area of each ��is same.

�

3 Rhombus of equal area in a regular hexagon.

Ga A E

�

on

1 a

a

a

a

a

a

a a

= 2

�

�

2 � 1 a2

Perimeter

= 8a

Inradius (r)

(r) =

a 2 2�2

a 1 1

F

=a

Area

B D

a

Octagon

Regular figure with 8 sides

a

Ch

C

1

Circumradius (R)

C a

= 135° Each exterior angle = 45°

1 B A a ��EAC = equilateral ��of side ��EAC =

3a

(R) =

Each interior angle = 135°

1 a

3a 2

Let side of octagon = a

am

ga n

D

F

B

8

pi

3

E

2a � a 3a = 2 2

3 9 2 � a Area � PQR 3 4 4 � = = Area ABCDEF 3 2 8 6� a 4

1

at io n

1

PQ =

�PQR = equilateral � with side

a Pu t bl a

a

a

A

C

Q

p

1

C

Si

1 a

2a

F

ic

Circumradius a E

Q

r

r = Short diagonal

�

C

F

� =3

6

�

D

= 6a

Perimeter

Area =

2D

Area � EAC 1 = Area ABCDEF 2 R E

= 45° 3a

Number of diaonal = 20

�

= 20

56

a 2� 2

=

2� 2 a 2

3D

3 Dimension Mensuration 3 3 Dimension: Length, breadth, height

Cube �

3 Lateral surface Area

(LSA):

surrounding area except (of figures having flat surface) top and bottom

�

a

(TSA) :

a

a

Si

Total surface area

A Solid figure in which length, Breadth and height are equal.

r

�

Area of all surfaces of a figure (LSA) + area of bases (top/bottom)

6 faces

(LSA) +

8 vertices

p

12 edges

Curved surface area (CSA) : of figures having curved surfaces like cylinder, cone etc. (CSA)

at io n

a Pu t bl a

LSA = 4a2 Volume = a3 TSA = 6a2 Diagonal =

Euler's Theorem: for any 3D flat surface figure

Ga

volume = 1×1×1 = 1 m3 1 km = 1000 m 1 m = 100 cm 1m3 = 1000 Liter � 1m3 = 103L � 1L = 10–3 m3 1L = 10–3 × 106 cm3 � 1L = 103 cm3 �

Pr

When a cube of maximum size put inside a hemisphere.

pi

on

�

am

1m

V+F–E=2

a

Ch

1m

1m

:

ga n

Volume capacity

3

ic

CSA : Area of only curved surfaces except top and bottom. CSA

3a

a

r a 2

a

r

a=

r

2 r 3

a2 + a2 = r2 2

3 �� 2 a2 = r2

1m = 100 cm a (side of cube) =

1m = 1000 mm 1 decimetre = 10 cm 1 km = 100 decametre

57

a 2

2 r 3

�

Cuboid � H

� B

L

Face/

=6 = 12

Edge/

x

x

=8

Vertices/

3D

If x is the thickness of a cuboid, then volume of the hollow cuboid = lbh – (l – 2x) (b – 2x) (h – 2x) x lbh – (l – 2x) (b – 2x) (h – 2x) Making open rectangular box by cutting 4 corners of a retangular sheet. 4

= LH, BH, LB

Adjacent faces

x

x

x

x x

b

x l

LSA = 2(bh + lh) = 2(L+B) × H = Area of 4 walls

r

Volume of rectangular box = (l – 2x) (b – 2x)x.

Diagonal

=

L2 � B2 � H2

�

x, y, z

xyz

x = lb y = bh 2 2 2 xyz = l b h

Volume = �r2h = Base Area = × Height CSA = 2�rh = Base Perimeter × Height TSA = CSA + 2 × Base Area = 2�rh + 2�r2 = 2�r(r+h)

z = hl

Pr

xyz = lbh = volume

l 2 +b2

y=

b �h

z=

h2 � l 2

l=

2

2

2

l2+b2+h2 =

2

2

2

on

Folding of rectangular sheet to form a cylinder

b l

A

Folding along length

�x

2

2

x2 � y2 � z2 2

D = l 2 � b2 � h2 =

x2 � y2 � z 2 , b= 2

Volume =

�

CSA h = TSA r + h

x +y +z =2(l +b +h ) 2

Ga

x=

am

L

x, y, z

B

Ch

x

H

ga n

y

�

Ratio ��

If x, y, z are diagonals of three adjacent faces of a cuboid

z

r

at io n

=

ic

volume

�

h

a Pu t bl a

3

after opening

p

If area of 3 adjacent faces of a cuboid are x, y, z respectively.

r

pi

�

Cylinder

= L×B×H

Volume

= (l – 2x) (b – 2x)x.

Si

TSA = 2(LB+BH+HL)

h=b

x2 � y2 � z2 2

x2 � y2 � z 2 y2 � z2 � x2 , h= 2 2

� y 2 � z 2 �� x 2 � y 2 � z 2 ��y 2 � z 2 � x 2 �

2�r = l B

r=

l 2�

Folding along breadth

2 2

Longest rod that can be put inside a cuboid (Room) = Diagonal =

h=l

l 2 + b2 + h 2

2�r = b l 2 + b 2 + h2

58

r=

b 2�

3D

Hollow Cylinder r

�

�

R

=

Height of cylinder

=a

Ratio ��

h

a 2

Radius of cylinder

vol. of cube 14 � vol. of cylinder 11

Maximum size cylinder inside a cone

Thickness

thickness

=t = �R2h – �r2h

volume of metal

= �(R2 – r2)h =H

Si

Height of cone

Height of cylinder

=h

Radius of cone

A maximum size cone inside a cylinder

=R

Radius of cylinder

=r

p

�

r

Volume = �(R+r) (R–r)h CSA = 2�(R+r)h TSA = 2�(R+r)h + 2�(R2–r2) = 2�(R+r) (h+R–r)

h

r

Cone Apex semi-vertex angle l (slant height)

h r �l=

l2 = h2 + r2 volume =

am

Ch

Ga

h = 2r

on

A cylinder encloses a sphere

ga n

�

pi

vol. of cylinder 3 � Ratio �� vol. of cone 1

Pr

r

ic

�

(Property)

at io n

a Pu t bl a

R H � r H�h

1 2 �r h 3

CSA = �rl TSA = �rl + �r2 = �r (r + l) �

Height of cylinder = Diameter of sphere

If H, C and V are the height, curved surface area and volume of a cone. Then find the value of 3�VH3–C2H2+9V2? H, C

V

vol. of cylinder 3 Ratio �� vol. of sphere � 2

�

h2 � r 2

3�VH3–C2H2+9V2

A maximum size cylinder inside a cube

Let

r = 1, h = 1

Volume = a

1 �, 3

3�VH3 – C2H2 + 9V2 = 3�× � �2 – 2�2 +��2 = 0

59

�l=

C=

2 2�

1 1 ��– 2�2 + 9 �2 3 9

�

3D

If S denotes the area of the curved surface area of a right circular cone of height and semivertical angle � then S equals?

2�r = 2�R ×

S

r=R

h �

S

� 360�

� 360�

slant height of cone = radius of sector, h = R 2 � r 2 l

r tan��= h

r

h = R 2 � r2

h cos��= l

Cutting of Cone �

r = htan� l = hsec� S =��rl = h2sec tan When a cube of maximum volume is cut from a cone

l1

h1

h2

r1

l2

r

Si

�

r

�

h

R

r1 h1 l1 � � r2 h2 l2

a Pu t bl a

a 2

at io n

2 rh

a (side of cube) = �

2rh 2r � h

Ga

a(h + 2 r) =

2

am

2 ra = ah

�

1

3 4

Ch

2 rh –

on

a h�a = 2r h

a

2

1 2 �r1 h1 3 3 3 � r1 � � l1 � � h1 � small cone volume 3 = 1 2 =� � =� � =� � Big cone volume �r2 h2 � r2 � � l2 � � h2 � 3

Pr

a r

ga n

a 2

2

ic

a a

2

small cone CSA �r1l1 � r1 � � l1 � � h1 � = �r l = � � = � � = � � Big cone CSA 2 2 � r2 � � l2 � � h2 �

pi

h

p

r2

h–a

Frustum

5

When a sector is folded to make a cone: O R R

R

CSA of 5 parts � 12 : 22 – 12: 32 – 22 1 : 3 : 5 Volume of 5 parts � 13 : 23 – 13: 33 – 23 1 : 7 : 19

h r

l circumference of base of cone = arc l

arc l

60

: 42–32 : 52–42 : 7 : 9

(Ratio)

: 43 – 3 3 : 53 – 4 3 : 37 : 61

(Ratio)

�

When a cone is cut parallel to its base, lower portion is called frustum.

A b

c

Or

B

B

C

a

C

r2

Sum of vol. of 2 cones 2 h

h

3D

A

Frustom of Cone

l

=

1 (ac)2 � 3 b

Sphere � r

r1 –r2

r

r1

Si

CSA = �(r1+r2)l TSA = �(r1+r2)l + � � r12 � r22 �

2

Rotation Direction

r=c h=a Rotation along perpendicular BC

Ga

B

C

TSA of each part =

Rotation Direction

C

Second cut

8�r 2 = 2 r2 4

(Quarter sphere) If we make 3 cuts at x, y, z axis

l=b AC

x, y, z

3

3 cut � 8 parts 3 cut � 6 circle area 8 parts TSA = 4�r2 + 6�r2 = 10 r2

b a

2 3 �r 3

First cut r

l=b

A

B

=

2 cut (4 pieces) 1 cut � 2 circle area ��(Increase) 4 parts TSA = 4�r2 + 4�r2 = 8 r2

r=a h=c Rotation along hypotenuse AC

c

r

�

A

C

r

r

CSA of Hemi-sphere = 2�r2 TSA of Hemi-sphere = 2�r2 + �r2 = 3�r2 TSA of both parts = 4�r2+2�r2 = 6�r2

BC

B

Hemi-Sphere

at io n

C

am

A

a

ga n

Rotation along base a

B

Sphere

Volume of hemisphere

on

C

a

Ch

B

Pr

b

c

Cutting of Sphere

ic

A

A

�

Rotation of right angle triangle to form a cone

pi

�

h 2 � � r1 � r2 �

4 �R3 �� V�R3 3

CSA = TSA = 4�R2 � Area ��R2

p

l=

1 2 2 � � r1 � r2 � r1r2 � × h 3

a Pu t bl a

volume =

Volume =

Rotation Direction

TSA of each part

61

TSA =

10�r 2 5 2 = �r 4 8

�

A maximum size sphere inside a cube

3D

Prism �

A prism is a solid figure with identical ends, flat faces and same cross section all along its length.

� Base and Top of the prism is same

a Diameter of sphere

=a

� Radius of sphere vol. of cube

Surface of prism is lateral and not curved.

a = 2

Cube, cuboid are prism but cylinder is not prism

21

Ratio � vol. of sphere � 11

Triangular base prism

r

A maximum size cube inside a sphere

Si

�

a

b

p

c

a Pu t bl a

h

a

b

c

at io n

Diagonal of cube = Diameter of sphere

Volume of prism = area of base × height

vol. of sphere 11 3 � vol. of cube 7

LSA = perimeter of base × height

Pr

Ga

r = radius of base of cone r =

hr =R= l�r

Hollow Sphere

R

LSA

�

�

A B

TSA = nah + 2 × C

r

=

na 2 � cot 4 n

Volume = Base area × Height = × na 2 � cot � h 4 4

Here , n � no. of sides of regular polygon n� a � side length of regular polygon a�

4 �(R3–r3) 3

h � height of prism/

TSA = 4�R2 Thickness

3

ah + bh + ch = (a + b + c) × h TSA = LSA + 2 × Base Area General formulae for a prism:– – CSA = Base perimeter × Height = na × h TSA = CSA + 2 × Base area

Volume =

Volume of metal

×

� 3 rectangles formed if we open it

on

Ch

l = slant height of cone l =

Radius of sphere

pi

am

h = height of cone h =

�

ic

A maximum size sphere inside a cone

ga n

�

×

(t)= R–r

62

3D

Pyramid �

� 4×

Apex

height (dotted)

�

1 × base perimeter × slant height 2

�

1 × 2

14

× �

In square pyramid

Square Pyramid

�a � l =h + � � �2� 2

Triangular Pyramid

Height � Apex to centre of base �

2

� a � � � 2�

at io n

ic

a

l

�a � � �

2

e2 = l2+ � � 2

a/2

a

e 2 = h 2 + R2

pi

am

ga n

e

on

Pr

slant height

��

In triangular pyramid

a Pu t bl a

�

�

Ch

l h

�

l2 = h2 + r2

r

Ga

1 ×perimeter of base × slant height 2 1 × 2

×

TSA � LSA + Base Area Volume �

1 ×base area × height 3

1 � × 3

h

e

r= ×

R

After opening square pyramid � LSA = 4�'s � LSA = 4�'s

63

2

= e 2 = h2 + �

p

�

slant height � Apex to side of base

for a regular pyramid/

�a � =e �e =l + � � �2�

If slant edge

slant edge � Apex to vertex of base

a

2

2

r

14

LSA =

1 × 4a × slant height 2

Si

r=7

LSA =

�

slant height

slant edge

1 × a × slant height 2

a 2 3

,R=

a 3

2

2

3D

Tetrahedron

LSA =

�

1 (P +P ) × l 2 1 2

P1, P2 � Perimeter of bases � a

a

A1, A2 � Area of bases �

l a

a

Volume =

a a

Pentagonal Prism

Slant height (l) =

�

3 a 2

h

3 2 a ×3 4

r

Surface area of pentagonal =

Si

3 2 TSA = a ×4= 4

a

3a

2

Ga

p ic

a

Surface area of hexagonal � �

3 3 2 a =2.5981a2 2

3 3 2 a =2.5981a2 2

Lateral surface area = 6a × h = 6ah = 6a × h = 6ah

a–b 2

Total surface area = 6ah + 3 3 a2 = 6ah + 3 3 a2 Volume

a

�a � b� h2 � � � � 2 �

3a 2 � h � 3a 2h

h

Ch

l

=

Hexagonal Prism

on

Pr

�

am

ga n

b/2

l=

= 5ah + 2 3a 2

Volume

b

a/2

= 5a × h = 5ah

at io n

a Pu t bl a

Frustum of a Pyramid

h

3a 2

Total surface area = 5ah + 2 3a 2

1 a3 2 3 2 Volume = × a × a= 3 6 2 4 3

�

=

3a 2

Lateral surface area = 5a × h = 5ah

2 a 3

height =

pi

LSA =

1 (A +A + A1 � A 2 ) × h 3 1 2

2

64

=

3 3 2 a h = 25981a2h 2

Pentagonal Pyramid

3D

Hexagonal Pyramid

�

�

s s

h

h a

a

Lateral surface area =

1 5 ×5a×s= as 2 2

s = slant height Lateral surface area =

Si

r

1 5 = ×5a×s= as 2 2

5 Total surface area = as+ 3a 2 2

p at io n

a Pu t bl a

1 1 2 2 a = � 3a � 3 3

on pi

Ch

am

ga n

Pr

ic

Volume

Ga

1 ×6a×s=3as 2

Total surface area = 3as +

5 = as+ 3a 2 2

Volume

=

1 ×6a×s=3as 2

65

= 3as + =

3 3 2 a 2

3 3 2 a 2

3 2 ah 2

3D

Solid figure

Figure a

Cube

a

a

H

Cuboid

B

L

Volume

CSA/LSA

TSA

a3

4a2

6a2

L×B×H

2 (L + B) H

2(LB + BH + HL)

2 rh

2 r (r + h)

r

r Si

1 2 rh 3

r

1 (R2 + r2 + Rr)h 3

p

h

rl

R

4 3 r 3

r

a

Prism

h

4 (R3 – r3) 3

r (r + l)

(R + r)l + (R2+ r2)

4 r2

4 R2

Ch

am

pi

r

r

Ga

Hemi-sphere

R

ga n

Hollow sphere

4 r2

on

Pr

Sphere

(R + r)l

at io n

Frustum of cone

R r

ic

h

Cone

r2h

r

a Pu t bl a

Cylinder

h

2 3 r 3

2 r2

3 r2

Base area × Height

Base peri. × Height

LSA+ 2 × Base area

1 × Base area × H 3

1 × Base peri. × Slant h. LSA + Base area 2

b c

a

b c

e

Pyramid

a

l

a/2 a

66

Number System

Number system Decimal Numbers

Numbers

Real number

Imaginary numbers

Rational numbers

Terminating Non-Terminating decimal Repeating decimal

Non-terminating Non-repeating decimal

1 2

0.73 = Whole numbers

73 100

81 125

p

Negative numbers

0.333333...=

1 3

0.565656...=

56 99

Si

0.5 =

Integers

Non-integer rational numbers

r

Irrational numbers

0.648 =

0.137137137...=

2 =1.414....

�

137 999

Irrational

a Pu t bl a

Rational Numbers

Natural numbers

Classification of Numbers

on

�

a+ib

�5 ,

Ex.

5+3i

�

Irrational Numbers

which can be written in

can not be written in

5 13 �8 , , , 0.5555, 3 1 1 22 7

� 0 � Neither positive nor negative Whole Numbers

–1 = i2

Rational Numbers

p q

Non-negative Integers

{–�, ....–4, –3, –2, –1} {0,1,2,3,4,....�}

–1 = i

Real Numbers

p form (q�0) p, q � Int q

Integers

Negative Integers

pi

�3 ,

Ch

am

�7 ,

5 , 11

Ga

0.0675,

can not be denoted on number line.

ga n

19 5 Ex. +3, –7, 5, , 13 7

All integers are rational no.

ic

Imaginary Numbers

which can be denoted on number line.

Integers

�p� � � where q = 1 �q �

Pr

Real Numbers

�

at io n

Zero

Numbers

�

Natural numbers {1,2,3,4,5,....�} Integers

Even

p form. q

Odd

Even � which are divisible by 2. (2K form) �

p q

2

2K

{–8,–6, –4, –2, 0, 2, 4, 6, 8) Odd � which are not divisible by 2. (2K�1 form)

0.1342607532 ....

�

2

{–5, –3, –1, 1, 3, 5, 7, 11}

q ������ = 3.141592 ....

67

2K�1

Number System

Odd × odd � odd

�

odd × even � even odd�odd � even even � even � even odd ± even � odd

Ex. 4, 6, 8, 9 etc. 1 � Neither prime nor cocmposite

a – b = even a + b = even a – b = odd

a�

even � even � even 2

a�

4� 9 � Smallest odd composite number. 9� �

even � odd odd � � natural 2 2

Natural numbers

�

1

2, 3, 5, 7, 11, 13, 23, 61, 67, 97 etc. 2 � even prime no. & smallest prime no. 2 �

�

1-50

� 15

�

pi

on

3, 5, 7

Smallest 3 digit prime � 101 3

= 101

largest 3 digit prime � 997

am

ga n

50-100 � 10 1-100 � 25 1-200 � 46 1-1000 � 168 Each prime number can be written in (6k±1) form. (6k±1)

only pair of prime no. with a gap of 2 is 3, 5, 7.

ic

Pr

Prime no.

2

2

3,5,7 � only pair of consecutive odd prime no. 3,5,7 �

Two prime numbers with a

(3, 5) (5, 7) (11, 13)

a Pu t bl a

Only two factors 1 & itself.

Twin-prime numbers gap of 2.

p

Composite numbers

Prime Numbers

HCF = 1

(25, 19) (16, 9) (2, 3) (11, 13) �

Prime numbers

Relatively prime/co-prime numbers Two numbers in which nothing is common i.e. their HCF = 1

r

a + b = even

4 � Smallest composite number.

odd � odd � even 2

Si

a – b = odd

more than two factors.

1�

a�

But every (6k±1) form may not be necessarily prime no.

Ga

�

Composite Numbers

at io n

a + b = odd

Ch

�

3

� 997

Perfect Numbers

If the sum of all the factors (excluding that no.) is equal to that number then it is called a perfect no.

6 � 1,2,3,6 (factors) ����� 1+2+3 = 6 (Smallest perfect no.) � 6 is perfect no. 28 � 1,2,4,7,14,28 (1+2+4+7+14) = 28

(6k±1)

Perfect numbers

13 � 6×2+1 (prime)

25 � 6×4+1 (not prime)

68

� 6, 28, 496, 8128 ....

Number System

Divisibility Rules � If difference is divisible by 7, 11, 13 then

Divisibility Rules

number will be divisible by 7, 11, 13 respectively.

1 is not divisible by any number except 1 but 1 is a universal factor.

7, 11, 13

1

1, 1

7, 11, 13 7�

Divisibity Rule of 2, 4, 8, 16

005922 � 922 – 5 = 917

2�

r

2� Last digit should be divisible by 2.

11 × 13 ×

6489 � 489 – 6 � 483

7� 11 × 13 ×

8� Last 3 digit should be divisible by 8.

Si

2

8�

380247 � 380–247 �133

7� 11 × 13 ×

4� Last 2 digit should be divisible by 4. 4 8

p

3

16� Last 4 digits should be divisible by 16 4 16 Divisibility Rule of 3 and 9

�

73×101 = 7373

�

3� Sum of digits should be divisible by 3. 3�

ga n

5. � Last digit should be 0 or 5 0

� �

on

Pr

9 Divisibility Rule of 5, 25, 125

5. �

ABCABC � divisible by 1001 687×1001 = 687687

3

9� Sum of digits should be divisible by 9. 9�

ABAB � divisible by 101

a Pu t bl a

16�

at io n

2

5

7×11×13 = 1001 (Remember) Divisibility Rule of 11

ic

4�

If the difference between the sum of the digits at odd places and sum of the digits at even places is zero or multiple of 11.

25 � Last two digit should be divisible by 25.

am

25

11

pi

25 �

Add even place digits

125. �

Ch

125. � Last 3 digit should be divisible by 125.

Add odd place digits

3 125 Divisibility Rule of 6

Ga

�

166452 � 1+6+5 = 12 6+4+2 = 12 diff. = 0 � div. by 11 7945938 � 28–17 = 11 Divisibility Rule of 12

6. � 6=2×3 (co-prime factors) � If a number is divisible by 2 & 3 both, that number will also be divisible by 6 �

2

3

6

Divisibility Rule of 7, 11, 13

12 = 4 × 3 �

5922 � Make pair of 3 digits from RHS 5922 � RHS

Take diff.

If diff. is O or multiple of 11 then no. will be divisible by 11

If a number is divisible by 4 and 3 both then that number will also be divisible by 12. 4

3

� Add alternate pairs & take difference �

69

3

12

Number System

Remainder Theorem Fermat's Theorem

Remainder �

The remainder is the value left after the division if the dividend is not completely divided by the divisor.

�

Fermat's Theorem �� �

Fermat's

a P-1 =1 P

p = prime number

If dividend is completely divided by the divisor then in that case remainder will be zero.

a, p � co-prime

r

�

a P-1 = 1 (Remainder) P

Si

5016 �R=1 17

Remainder Theorem

Wilson's Theorem

�

� P � 1� !

quotient

Remainder

�

72 = 13 × 5 + 7

Dividend = Divisor × Quotient + Remainder +

206 ,R=8 11

n

on

� Remainder = 1n = 1

a n

a

� ax � 1� a

� Remainder = (–k)n

n

� Remainder = (–1)n

am

an + bn (n=odd) � a3+b3=(a+b) (a2–ab+b2)

a

� ax � k �

pi

ga n

a, b, n � natural numbers

n

� ax � 1�

Remainder is always less than divisor. �

� Remainder = –1 or (P–1)

� ax � k � = Remainder � Kn

ic

×

Pr

=

P

a3+b3=(a+b) [a2b0–a1b1+a0b2] � (a+b) [a2–ab+b2]

n = even 1

Ch

start

+ then – goes on)

Ga

(+ – + – + – (+

(a5 + b5) = (a+b) [a4b0–a3b1 + a2b2 – a1b3+a0b4]

No. of the form

�

a �(N) � R=1 N

�(N)=Toient function of N

a, N � co-prime

Div. by (a–b)

1. an+bn (n � odd)

�

×

N � Natural number /

2. a +b (n � even)

×

×

3. a –b (n � odd)

×

�

How to find 72 = 23 × 32

4. an–bn (n � even)

�

�

n

n

�

Div. by (a+b)

n

n

If Power is odd

�

�

a +b +c +d is divisible by (a+b+c+d)em n

n

n

n = odd –1

Euler's Theorem

= (a+b) (a4 – a3b + a2b2 – ab3 + b4)

� an + bn � n odd � (a+b) is a factor always.

P=

at io n

5

a Pu t bl a

13 72 65 7

Divisor

P = any prime number

p

Dividend

1� � 1� � �(72) � 72 × �1 � � × �1 � � 2� � 3� �

n

= 72 ×

70

(N)

1 2 × = 24 2 3

Number System

�

100 � 22 × 52

n

1� 1� � � 1 4 �100 � 100 × �1 � � × �1 � � � 100 × × � 40 2 5 � � � � 2 5

�

�(P) = P–1 where P = prime number � �(N) � N

�

N

4 � Rem = 4 6

�

10n � Rem = 4 6

Unit digit (UD)

co-prime

Product of any 'n' consecutive (+ve) numbers is always divisible by n! 'n'

�

(+ve)

�

n!

5 × odd � U.D = 5 5 × even � UD = 0 one zero = one pair of 5 × 2 875 × 64 � 5 × 5 × 5 × 7 × 2 × 2 × 2 × 2 × 2 × 2

15 � 16 � 17 � 18 � 19 � 5 numbers 5!

0� �

8697 511 85 19 17 27 17 10

ga n

77 15 5 6 27 25 2

Ga

OR

Final Quotient

am

Ch

successive remainders

15 2 12 3

� 4, 2, 3

Successive remainders

UD = same

odd

(132)25 �

25 , R=1 4

� (132)1 � UD = 2

31 32 33 34 35 36 37 38 39 310 � � � � � � � � � � UD � 3 9 7 1 3 9 7 1 3 9

Some Important points

r1

� 865 × 865 × .... 99 time UD = 5 Any power of 5 � UD = 5

99

(4) = UD � 4 (4)even = UD � 6 9 � (9)odd � UD = 9 (9)even � UD = 1 10. Rule of 2, 3, 7, 8 21 22 23 24 25 26 27 28 29 210 � � � � � � � � � � UD � 2 4 8 6 2 4 8 6 2 4 UD repeat after every power 4 � cyclicity = 4 � UD � 2n = 2n+4 �

If two numbers are divided by same divisor the remainders are respectively r1 and r2. If sum of these two numbers are divided by the same divisor the remainder is r3. Then divisor is � divisor = r1 + r2 – r3

� 371 × 371 .... 108 times

Any power of 0, 1, 5, 6

4�

Final quotient

�

unit digit same

UD = 1

p

Divide 620 by 8, 5, 6 successively

8 620 4 5 77 2 6 15 3 3 2

(371)

a Pu t bl a �

on

Successive Division

620 77 5 56 60 56 4

power

108

6 � (106)357 � 106 × 106 × .... 357 times UD = 6

= 1, 2, 10

Pr

Divisor � HCF [85, 17, 17] � 17

8

0, 1

5 � (865)

consecutive remainder

8.

(1370) � UD = 0 No. of zero at the end = 189

1�

pi

17

� UD repeat after power 4 � cyclicity = 4 �

334

333337

�

337334 � 1334 � 1 4

� 3331 � UD = 3

r2 r3

�

= r1 + r2 – r3

3

189

Si

Consecutive Remainder �

r

� 3 zero at the end of product

15 � 16 � 17 � 18 � 19 ,R=0 120

at io n

�

3 pair of 5 & 2

ic

�

�

Similarly cyclicity of 7 & 8 is also 4. 7

71

8

4

Number System

Number of Factors Factors �

Factors are the positive integers that can divide a number exactly.

�

1� 25 � 1� 2 �1

×

1 � 34 � 1� 3 �1

×

� r � 1�

1� 52 � 1� 5 �1

�

2160 = 24×33×51

r

1 is a factor of every number. 1

Even factors � minimum 21

2.

Every natural number is a factor of itself.

3.

Apart from 1 all natural numbers have atleast two factors.

NOF = 5×4×2=40 (20 21 22 23 24) × (30 31 32 33) × (50 51) Even NOF � 4 × 4 × 2 � 32 OR 2160 = 24×33×51 � 2(23×33×51) ���� 4×4×2 = 32

NOF of 12 � 6

NOF � 2

perfect square NOF Even NOF of 12 � 4 perfect cube NOF � 1

NOF = 1.

20 21 22 23

Ga

30 31 32

Si

p

on

am

Direct: NOF = (power + 1) × (power + 1) .... NOF of 72 � (3+1) × (2+1) � 4 × 3 � 12 N = ax × by × cz �

�

2160 � 2 ×3 ×5 4

3

Sum of reciprocal of all factors =

sum of factors number

8 � 1, 2, 4, 8

Sum of factors (SOF) 72 ��[20+21+22+23]×[30+31+32] = 15 × 13 = 195

�

2

4×2×2= 16

NOF of N = (x+1) (y+1)(z+1) �

1

18=2 ×3 � minimum 21 and 32 required for multiple of 18. � 4 × 2 × 2 = 16 OR 2160 = 24 × 33 × 51 � 2×32[23×31×51] ��������

� NOF � 12

a,b,c � prime number

3

18 1

OR 4 × 3 = 12 combinations

�

2160 = 2 ×3 × 5 4

(20 21 22 23 24) (30 31 32 33) (50 51) Number of factors which are multiple of 18 � 18=21×32

Ch

72 � 23 × 32 (write in prime base)

� Absence of

� 40 × 6 = 240

Sum of odd factors

pi

ga n

�

No. of odd factors 2 � 2160 = 24 × 33 × 51 � 4 × 2 = 8

�

odd NOF of 12 � 2

NOF of multiple of 3 � 3

� 40–32 = 8

No. of odd factors

Pr

12 � 1, 2, 3, 4, 6, 12 (factors)

21

� 30×40×6 �

Sum of even factors 7200

at io n

12 � 12, 24, 36, 48, 60 .... (multiples)

�

ic

Number of Factors (NOF)

a Pu t bl a

1

�

� GP

� 31 × 40 × 6 = 7440

Properties of factors 1.

�

OR Sum of factors

a ��r n � 1��

Sum of reciprocal �

1

SOF � (20+21+22+23+24)×(30+31+32+33)×(50+51) ������ 31×40×6 � 7440

=

72

Sum of factors number

1 1 1 1 8 � 4 � 2 �1 + + + = 8 1 2 4 8

Number System

�

10800 � 24×33×52

�

Only perfect square number has odd number of factors.

�

Perfect square number upto 100 � 10

Number of factors which are perfect squares � �

a

2n

= perfect square number

100

a6n = perfect square as well as perfect cube

� Even number of factors �

Perfect square of a prime number has exactly 3 factors. 3

�

1

3

p

make pairs from 1st & last, NOF = 12

at io n ��3

Pr

ic

= 726

NOF 2

Ch

am

pi

on

= � N�

ga n

N

=4

only 1 factor = 1 (1 has only one factor) � More than 3 factors

a Pu t bl a

4×18 = 72 6×12 = 72 8×9 = 72

Ga

more than 3 factors between (1–100) = ? exactly 3 factors = 4

72 � 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

� Product of factors of N

Si

�

� 3×2×2 � 12

r

49 7 (1–100) � 22, 32, 52, 72 � 4 numbers

�(2+1) (1+1) (1+1)

Product of factors

72 � 2 + 1 = 3

49

Power � Integer � 1 2

1×72 = 72 2×36 = 72 3×24 = 72

�

100–10 = 90

(20 22 24) × (30 32) × (50 52) � 3 × 2 × 2 = 12 OR No. of factors (perfect squares) �

�

= 10

a3n = perfect cube number

73

��100 – 5 = 95 numbers

Number System

Sequences and Series Sequence �

A sequence is an arrangement of numbers in definite order according to some rule.

�1 1 � 4 1� � �= � 5 25 � 25

Progression

1 1 1 1 + + + + 1 � 4 � 7 4 � 7 � 10 7 � 10 � 13 10 � 13 � 16

�

A sequence whose terms follow a certain pattern

r

�

1 1�1 1 � 1 = ? �� � � � � 13 � 16 � 19 6 � 4 16 � 32

Si

is called a progression.

� 1 1 � 1 � � � diff.of 1st & 3rd no.in den. � 1st 2no. last 2 no. �

�

Series

Pr

Step 1 �

Multiply

Step 2 �

9

on

= 999 zero = 000

Step 3 � Diff. of 999 – 991 = 8

AP �

=1

6�5 7�6 8�7 9�8 25 � 24 + + + +.... 5�6 6�7 7� 8 8� 9 24 � 25

n

If a is the first term, d is the common difference

a

d

l

AP

a, a+d, a+2d, a+3d,..., l, (a)

nth term is given by an = a+(n–1)d an = a+(n–1)d

n

nth term of an AP from the last term is given by a 'n = l – (n–1)d

1 1 1 1 1 + .... + + + + 8×9 24×25 7×8 6×7 5×6

common difference

d=

and l is the last term of an AP, i.e. the given AP is a, a+d, a+2d, a+3d,..., l, then

(b)

� 999000 – 8 = 998992

a=

nth Term (or General Term) of an AP

Ch

991 ×999 = ? 999

Ga

999

�

Where a = first term and d = common difference.

am

If denominator is same as multiplier

pi

ga n

Special series

AP

Ex. a, a + d, a + 2d,... is an AP

A series having infinite number of terms is called infinite series.

�

called the common difference of AP.

A series having finite number of terms is called finite series.

�

regularly by a fixed number. This fixed number is

a1+a2+a3+...+an+...

at io n

a1,a2,a3,...,an,... �

A sequence in which terms increase or decrease

�

a Pu t bl a

expressed as a1+a2+a3+...+an+... is called a series.

Arithmetic Progression (AP)

p

If a1,a2,a3,...,an,... is a sequence, then the sum

ic

�

� � 1 1 1 � � � common difference �1st no.of den. Last no.of den. �

OR

n

AP

a 'n = l – (n–1)d

Note (i)

an+ a 'n = a + l i.e. nth term from the beginning + nth term from

1 1 1 1 1 1 1 1 1 1 1 1 – + – + – + – +.... – + – 5 6 6 7 7 8 8 9 23 24 24 25

the end = first term + last term n

1 1 4 – = 5 25 25

74

+

n

=

+

Number System Properties of Arithmetic Progression

Note

(i)

(i)

If a constant is added or subtracted from each term

A sequence is an AP if the sum of its first n terms

of an AP, then the resulting sequence is also an

is of the form A 2n + Bn, there A and B are constants

AP with same common difference.

and common difference in such case wil be 2A.

AP

AP

AP

A

(ii) If each term of an AP is multiplied or divided by a is also an AP, with common difference kd or

(ii) an = Sn – Sn–1 i.e. nth term of AP = Sum of first n terms – Sum of first (n – 1) terms an = Sn – Sn–1

given AP.

AP

AP d k

+4 +4 +4

p

(d) = 4

an AP, then 2an+1 = an + an+2.

Tn = nth term = a+(n–1)d = n

(iv) If the terms of an AP are chosen at regular AP AP

ic

on

�n=

linear expression in n, i.e. its nth term is given

An+B

Ga

B

n A=

(i)

�

2

3

3

3

n � n � 1�� 2n � 1� 6 2

3

Sum of 1st n even natural numbers

2 + 4 + 6 + 8 + .... +2n =

(iii) 5 terms � (a – 2d), (a – d), a, (a + d), (a + 2d) AP

n [2+2n] = n(n+1) 2

Sum of 1st n odd natural numbers

n

1st n

Sum of first n terms of AP, is given by AP

n � n � 1�

1st n

(ii) 4 terms ��(a–3d), (a–d), (a+d), (a+3d)

�

l –a +1 d

� n � n � 1� � �n = 1 +2 +3 + .... +n = � � 2 � � 3

3 terms � (a–d), a, (a+d)

Sum of First n Terms of an AP

l �a =n–1 d

�n2 = 12+22+32+ .... +n2 =

A

Any terms in AP can be taken as AP

2

AP n [2a+(n–1)d] 2

OR

n

Selection of Terms in an AP AP �

�a � l �

�n = 1+2+3+ .... +n =

am

n

Ch

AP

pi

by An+B, where A and B are constants and A = common difference.

� n

Tn = l = a + (n–1)d �

If a sequence is an AP, then its nth term is a

ga n

(v)

Sum

Pr

intervals, then they form an AP.

(a) = 7

Tn = nth term = a+(n–1)d = Last term for n term AP

2an+1

at io n

AP

1st term

a, a+d, a+2d, a+3d, ....a+(n–1)d

a Pu t bl a

(iii) If an, an+1 and an+2 are three consecutive terms of an+2

(n – 1)

Ex. 7, 11, 15, 19, 23..... common difference

d =

AP

an, an+1 = an + an+2

=

Si

k

kd

B

2A

d respectively, where d = common difference of k

AP

A 2n + Bn

r

non-zero constant k, then the resulting sequence

n

n

1 + 3 + 5 + 7 +.... +(2n–1) =

n n [2a + (n – 1) d] = [a + l] 2 2 When l = last term

Sn =

75

n [1+2n–1] = n2 2

Number System

(iv) If

Sp = q and

If a, m and b are in AP, then m is called the

(v)

Sp = Sq, then

arithmetic mean of a and b and it is given by

(vi) If

Arithmetic Mean (AM) (i)

m=

a�b 2

If

b, AP

m m =

a

b

a2, b2 and c2 are

r

Si

at io n � n �a

= r,

n>1

If a is the first term, r is the common ratio and l is the last term of a GP, then the GP can be written as a, ar, ar2, ....., arn–1, .... l. a

r

on

GP

a, ar, ar , ....., ar

4, 12, 36, 108

a2

nA

A=

a�b 2

a�b 2

GP

If r

ap = q and

aq = p, then

(ii) If

pap = qaq, then

(iii) If

ap = q and

1

1

n If a is the first term, r is the common ratio and l is the last term, then

(i)

nth term of a GP from the beginning is given by an = arn–1

a

r

a an = arn–1

ap+q = 0, ar = p+q–

n

l

l

' (ii) nth term of a GP from the end is given by a n �

ap+q = 0 aq = p , then

12

�

Important Results on AP AP (i)

, .... l

nth Term (Or General Term) of a GP

i.e. A1 + A2 + A3 + ... + An = nA, where A = A1 + A2 + A3 + ... + An = nA,

GP

r= a = = 3 4 1

Ch

n AM's

Ga

b

n–1

r = common ratio

(b) Sum of n AM's between a and b is nA a

l 2

am

b�a n �1

pi

ga n

and

a � nb n �1

�

(r)

ic

Pr

:

d�

ratio (r).

i.e. a n

n �1

A n � a � nd �

throughout. The constant ratio is called common

a Pu t bl a

na � b A1 � a � d � n �1

:

first term) to its just preceding term is constant

p

(iii) If a, A1, A2, A3, ..... An, b are in AP, tern (a) A1, A2, A3, ..... An are called n arithmetic mean between a and b, where B

n �1 a1a n

A sequence in which the ratio of any term (except

�

a, A1, A2, A3, ..... An, b, AP

�

Geometric Progression (GP)

a � a 2 � .......a n A� 1 n

� n � 1� a � 2b � a � 2d �

AP

1 1 1 1 � � � ..... � a1a 2 a 2a 3 a 3a 4 a n �1a n

AM

A

AP

a1, a2, ...., an

a1 � a 2 � .......a n n

(a) A1, A2, A3, ..... An

a b c , , both b�c c�a a �b

(vii) If a1, a2, ...., an are the non-zero terms of an AP, then

a�b 2

a1, a2, a3, ......, an

:

in AP, then

are also in AP

(ii) If a1, a2, a3, ......, an are n numbers, then their AM is given by, A �

Sp+q = –(p+q)

Sp+q = 0

1 1 1 , , and b�c c�a a �b

a, m

A2

Sq = p, then

apq = 1

GP

76

n

a'n �

l r n�1

l r n�1

GP

n

m

arm–n

Number System

Sum of First n Terms of a GP a GP (i)

×

n

=

×

ana = al ' n

Properties of Geometric Progression If all the terms of GP are multiplied or divided by same non-zero constant, then the resulting sequence is also a GP with the same common ratio.

(ii)

Sn �

a � lr lr � a , r < 1 or Sn = ,r>1 1� r r �1

Si

(i)

n

� a �1 � r n � � ,if r � 1 � 1� r � a rn � 1 � ,if r � 1 � � Sn � � � r �1 �na,if r � 1 � ��

ana'n = al n

Sum of first n terms of a GP is given by GP

(iv) ana'n = al i.e. nth term from the beginning × nth term from the end = first term × last term

n

r

(iii) The n th te rm from the e nd of a fini te GP consisting of m terms is arm–n.

where, l = last term of the GP.

GP

l = GP

GP (ii) The reciprocal of terms of a given GP also form a GP. GP

GP GP

on

ga n

GP

S� does not exist.

If a, G, b are in GP, then G is called the geometric

Ga

am

G=

GP

1

� b � n�1 G1 � ar � a � � , �a�

b2 = ac

2

� b � n �1 G2 � ar � a � � �a � 2

:

a 3 terms �� , a and ar. r

a a , , a, ar, and ar2. r2 r

GM

(a) G1, G2, G3, ....., Gn, are called n GM's between a and b, where

GP

(iii) 5 terms ��

a1, a2, a3, ......., an 1/n

a, G1, G2, G3, ...., Gn b

Any terms in a GP can be taken as

a a , , a and ar3. r3 r

B

( ii i) If a, G1, G2, G3, ...., Gn b are in GP, then

Selection of Terms in a GP GP

(ii) 4 terms ��

A

G=

(a1a2...an)

(vi) If a, b and c are three consecutive terms of a GP, then b2 = ac. GP

G

n

Ch

GP

a, G, b GP

ab .

ab (ii) GM of n positive numbers a1, a2, a3, ......., an are given by G = (a1a2...an)1/n

pi

If a1, a2, a2, a3, ....., an are non-zero and nonnegative term of a GP, then log a1, log a2, log a3, ....., log an are in an AP and vice-versa. a3, ....., log an AP

(i)

a 1� r

Geometric Mean GM

ic

Pr

GP

c

|r|�1, then

S� =

mean of a and b and is given by G =

(iv) If the terms of a GP are chosen at regular intervals, then the resulting sequence is also a GP.

a, b

|r| < 1, then

(ii) If (i)

a1, a2, a2, a3, ....., an

If

at io n

(iii) If each term of a GP is raised to same power, then the resulting sequence also forms a GP.

(v)

(i)

a Pu t bl a

GP

p

Sum of Infinite Terms of a GP a GP

:

: n

� b � n �1 Gn � ar � a � � and �a� n

1

� b � n �1 r �� � �a�

(b) Product of n GM's, G1×G2×G3×....Gn=Gn, where G =

77

ab

Number System

Harmonic Progression �

Harmonic Mean

A sequence a1, a2, a3, ....., an, .... of non-zero

(i)

If a, H and b are in HP, then H is called the

numbers is called a Harmonic Progression (HP),

harmonic mean of a and b and is given by H=

1 1 1 1 if the sequence a , a , a , ....., a , ... is in AP. 1 2 3 n

a, H

b HP

a1, a2, a3, ....., an, ....

H=

1 1 1 , , , ....., a1 a 2 a 3

(HP)

Si

p

� 1 1 1� � � n � 1� � � � a1 � a 2 a1 �

ga n

1

1

H�a H�b � �2 H�a H�b

1.

If A, G and H are arithmetic, geometric and harmonic means of two positive numbers a and b, then A, G

pi

am

1

(c)

Properties of AM, GM and HM between Two Numbers AM, GM HM

a1a 2l = a a � l � n � 1� � a � a � , where l is the last term. 1 2 1 2 1

1 1 1 1 � � � H�a H�b a b

at io n

Pr

n

1 � 1 1 1� � � n � 1� � � � l � a 2 a1 �

(b)

ic

(ii) nth term of the HP from the end

1

mn mn , amn=1, ap= p m�n

(ii) If in a HP, ap=qr and aq=pr, then ar = pq (iii) If H is HM between a and b, then (a) (H–2a)(H–2b) = H2

1

a1a 2 = a � n �1 a � a � �� 1 2� 2

a 'n �

If in a HP, am=n and an = m, then am+n=

n

HP

r

(i)

a Pu t bl a

an �

2ab a�b

Important Results on HP HP

nth term of the HP from the beginning HP

b

1 1� 1 1 1 1 � � � � � � ... � � H n � a1 a 2 a 3 an �

on

(i)

a

(ii) Harmonic Mean (HM) of a1, a2, a3, ..., an is given by

1 , ... AP an

nth Term (or General Term) of Harmonic Progressio n

H

H

A

(ii) A � G � H (iii) G2 = AH and so A,G,H are in GP.

Ch

1 (iv) an= a � n � 1 d , if a, d are the first term and � �

Ga

(i)

common difference of the corresponding AP. a, d

AP

(iv)

A=

a

HP =

a�b 2ab , G= ab , H = 2 a�b

n �1

n �1

�b a n � bn

Note There is no formula for determining the sum of harmonic series. AP = 2, 5, 8, 11 ....

B

1

(iii) a � a ' � a � l � First term of HP � Last term of HP n n

1 an= a � � n � 1� d

2ab a�b

� A,if n � 0 � 1 � � �G,if n � � 2 � ��H,if n � –1

Exponential Series �

The sum of the series 1 �

1 1 1 1 � � � � ...� is 1! 2! 3! 4!

denoted by the number e.

1 1 1 1 , , , 2 5 8 11

1�

78

1 1 1 1 � � � � ...� 1! 2! 3! 4!

e

Number System

1 1 1 1 � � � � ... 1! 2! 3! 4!

2 2 2 2 2 (viii) � n � 1 � 2 � 3 � ... � n � r �1

(i) e lies between 2 and 3. (ii) e is an irrational number. (iii) e x = 1 �

(v)

� n � n � 1� � 3 3 3 3 3 (ix) � n � 1 � 2 � 3 � ... � n � � � 2 r �1 � �

x x2 x3 � � � ...�, x � R 1! 2! 3!

(iv) e –x = 1 �

(x)

x x2 x3 � � � ...�, x � R 1! 2! 3!

For any a > 0, ax = e x log

n

�n

4

� 14 � 24 � 3 4 � ... � n 4 �

30

(xi) Sum of n terms of series n 1 – 2 + 32 – 42 + 52 – 62 + 72 – 82 + ... 2

2

3

x x (logea)2 + (logea)3 +...�, x � R 2! 3!

Case I when n is odd =

Si

(vi) Sum of first n even natural numbers. n

I

i.e. 2+4+6+...+2n=n(n+1) (vii) Sum of first n odd natural numbers.

n

p

II

on pi

Ch

am

ga n

Pr

ic

at io n

a Pu t bl a

2

Ga

=

n � n � 1� 2

n � n � 1� 2

Case II when n is even =

n i.e. 1+3+5+...(2n–1) = n

2

n � n � 1� � 6n 3 � 9n 2 � n � 1�

r �1

a e

2

= 1+x(logea) +

n � n � 1�� 2n � 1� 6

n

r

� e = 1�

n

79

n

=

�n � n � 1�

�n � n � 1� 2

2

Number System

LCM & HCF

The least common multiple (LCM) is defined as the smallest multiple that two or more number have in common. (LCM)

z

�

Find LCM of 55, 66, 60.

Find LCM of 13, 29, 41.

Si

Ch

Ga

�

Find LCM of 24, 30, 36. 24 � 12 × 2 30 � 6 × 5 36 �

In other words HCF is the largest positive integer that divides each of the given integers.

HCF

�

Let LCM = 36

40 � 8 × 5 64 � 8 × 8 42 � 6 × 7

6, 12 included

Ratio

no. common co-prime

Find LCM of 55, 66, 60. 55 � 11 × 5

LCM = 66×5×2 = 660

66 � �

60���5 × 6 × 2 �

Co-prime � HCF = 8

Co-prime no. � HCF = 6 66 � 6 ×11 co-prime no. HCF

� 36 × 5 × 2 = 360

in 36 so skip them �

HCF

HCF is the greatest common number which divide all the given numbers.

13 � LCM (prime no.) � their product 29 � � LCM = 13×29×41 41 �

HCF

Ex. Find HCF of 40, 64. 40 � (1, 2, 4, 5, 8, 10, 20, 40) 64 � (1, 2, 4, 8, 16, 32, 64) common-factors of 40, 64 � [1,2,4, 8 ] HCF = 8 40, 64 � [1,2,4, 8 ] HCF = 8

am

�

ga n

LCM = 22×3×5×11 = 660 55 � 5×11 66 � 2×3×11 60 � 22×3×5

on

= 8 × 9 × 5 = 360

pi

1

HCF is used to find the highest common factors of any two or more given integers.

p �

�

Pr

2

(d) = a–x = b–y

HCF (Highest common factor)

LCM=2 ×3 ×5 (Take max. power of each) 3

x, y,

common difference (d) = c–z number = LCM (a, b, c)k – d

a Pu t bl a

24 � 23×31 30 � 21×31×5 36 � 22×32

The number which when divided by a, b, c respectively gives remainder x, y, z such that a, b, c

LCM of the any two number is the value that is completely divisible by the two given numbers.

Find LCM of 24, 30, 36.

'r'

= LCM(a,b,c) × k + r �

Ex. Find LCM of 36, 45. 36 � 36, 72, 108, 144, 180, 216, 252 .... 45 � 45, 90, 135, 180, 225, 270 .... LCM (36, 45) � 180 (Least common multiple of both) �

a, b, c

at io n

�

The number which when divided by a, b, c leaves remainder 'r' in each case = LCM(a,b,c) × k + r

ic

�

�

r

LCM (Least Common Multiple)

Product of co-prime number = their LCM LCM

80

40 : 64

42 : 66

5:8

7 : 11

HCF of two no. = H Numbers = Hx, Hy x, y � co-prime LCM = Hxy

Ratio of no's = Hx : Hy x:y

Number System

How to find HCF

Long Division Method

Factorization Method

�

Find HCF of 168, 294, 420 168 � 23 × 3 × 7

HCF = 63 693 945 1 693 252 693 2 504 189 252 1 189 Final divisor 63 189 3 = HCF 189 ×

Take minimum power of common in all � HCF � 2×3×7 = 42

Difference Method HCF of two numbers = H Numbers = Hx, Hy x, y � co-prime

Diff. = Hx – Hy = H(x–y) � HCF = diff. factor

1,2,1,3 � successive quotient �

If any one number is odd their HCF can not be even.

Diff. = H(x – y)

2

If(x–y)=1

If(x–y)>1

HCF=diff.

HCF=factor of diff.

�

HCF

d = 85 � 17 × 5 1518 1840 D = 322

161×2

ga n

�

23 ×7×2

Ga

323, 456, 703 133

� LCM = Hxy

p ic

�

HCF = 6 All will be divisible by HCF. HCF

Hx × Hy = Hxy × H I × II = LCM × HCF H2xy = H2xy

�

I × II = LCM × HCF ��1st no. × 2nd no. = LCM × HCF

�

'K'is the largest number which when divide a,b,c gives same remainder 'r' & quotients are x,y,z respectively a,b,c

'K'

19 ×7

k

a r

x k

x,y,z b r

y

k

c

z

r

a–b, b–c, c–a � ��� ��� ��� k(x–y) k(y–z) k(z–x) a = kx + r b = ky + r � K = HCF [a–b, b–c, c–a]

1008, 1323, 1722 d=315 63 ×5 3× 21 HCF of (1008, 1323) = 63

1722 not divisible by 9 � Remove 9 from 63 Hence not divisible by 63

18, 30

'r'

19 divide all 3 � HCF = 19 �

HCF

If HCF of 2 numbers = H

LCM=90 Diff.=12 Sum=48

Ch

Both are even number � HCF = 23×2 = 46

23

am

� Both number divisible by 23

�

�

on

306, 391 � HCF = 17

HCF

HCF, LCM

Difference=1

pi

�

48 = 8 × 6 48 � Difference=8=HCF 56 � 56 = 8 × 7 � HCF = difference = 8

Pr

�

3

Diff. = H(x–y) Nos. � Hx, Hy x,y � co-prime Sum = H(x+y) � HCF is present in LCM, difference & sum of the numbers.

a Pu t bl a

HCF of 2 numbers can not be greater than their difference. 2

Note: If 2 or 3 numbers are even their HCF will be even.

r

�

Si

294 � 2×3×72 420 � 22×3×5×7

693, 945

at io n

�

�

� a c e � HCF � a,c,e � HCF � , , � � LCM b,d,f � � �b d f �

63

� 1722 divisible by 21 1722 21 � HCF (1008, 1323, 1722) = 21

LCM � a,c,e � �a c e� LCM � , , � � HCF b,d,f � � �b d f �

�

81

HCF [an �1, am �1] = aHCF(n,m) 1

c = kz + r

Ch on

pi

am

ga n

Ga

at io n

ic

p

a Pu t bl a

Pr Si

r Number System

82

Ch on

pi

am

ga n

Ga

at io n

ic

p

a Pu t bl a

Pr Si

r Number System

83

Number System

Calculation & simplification Conditions to be a perfect square number

If unit digit is 5 then ten's digit always 2. 5

2

672 =

1 (67+17)��172 � 42�289 � 4489 2

= 1

24

2192 = 2(219+19)� 192 = 476�361�� 47961 1942 = 2(194–6)� 62 = 37636

at io n

1822 = 2(182–18)� 182 = 328�324 � 33124

n2=(2k)2=4k2 � multiple of 4

n2=(2k+1)2=4k2+4k+1�4(k2+1)+1

100 �

Ga

1022 = 102 + 2� 22 = 10404 1072=107+7��72 = 11449

�

1522 =

3 (152+2)� 22 �� 23104 2

1412 =

3 (141–9)�92 � 19881 2

6142 = 6(614+14)�142 � 6×628� 196 ���376996 7932 = 8(793–7)�72 � 8×786�49���628849

�

Last 2 digits of 18 ,

32 , ��� ����(50–18)2 = 24 2

Ch

Base

am

Find squares

on

� 67 � ,R � 3 � � � 47 �

ga n

51767 � Not a perfect square

pi

� n2 � 4K, 4K + 1

Pr

R=1 after divisible by 4

1132 = 113+13��132 = 126�169 � 12769

�

2 68 , 82 , �� (100–18)2 2

2

1182, 3682 ����� (350+18)2

2 �

x(x+a) (x+2a)(x+3a) +k is a perfect square then k=? k=?

x(x+a) (x+2a)(x+3a) +k

922 = 92–8� 82 = 8464

� (x2+3ax) (x2+3ax +2a2)

832 =83–17� 17 2 = 66�289 � 6889 Base

2

� x2, (50k � x)2, (100k � x)2 � Last 2 digits same.

1212 = 121+21��212 = 142�441 � 14641 972 = 97–3�� 32 = 9409

3 × 100 2

150 �

Base

ic

2K(even)

208 = 2(208+8)� 82 = 43264

p

Last 2 digit of a perfect square number = last two digits from the squares of numbers between (1 to 24).

n

200 � 2 × 100

2

2K+1(odd)

�

1 (38–12)��122 � 13�144 � 1444 2

Base

9

2

�

382 =

Si

No number can be a perfect square unless its digital root is 1, 4, 7 or 9. 1, 4, 7

�

1 (44–6)��62 � 1936 2

a Pu t bl a

�

442 =

If a no. ends with 2,3,7,8 it can't be a perfect square 2,3,7,8

�

1 (48–2)��22 � 2304 2

r

�

482 =

1 50 � ×100 2

�

2 2 (x2+3ax)2 +2×a × (x +3ax)

a2

+2 a b

+ b2 = (a+b)2

� to make perfect square add (a2)2 = a4

1 592 = (59+9)��92 � 3481 2

�

(a2)2 = a4

� (x2+3ax +a 2) 2 & k = a4

1 542 = (54+4)��42 � 2916 2

84

Number System

If I, II, III, IV

d

9999800001 + 1111088889 – 4444355556 =? 99999 + 33333 – 66666 = 66666

�

n3

AP

Then I × II × III × IV + d = (I × IV +d2)2 4

�

�

To make perfect square what should add from:

� Gap between these consecutive numbers = 7

= 7 � 7 = 2401 should be added. 4

To make perfect square what should subtract from:

� 841 × 846 × 851 × 856 �

N2

p �

Ga

99992 = 99980001

N�

even no.of digits 2

17 � 1 = 9 digits 2

on

N2 � 24 digits

N�

24 = 12 digits 2

Square Mirrors �

6 is repeated n times

92 = 81

odd no.of digits � 1 2

N2 � even no. of digits

N2 � 17 digits

N�

I � Before 3 (n–1) times 4 II � Before 6 (n–1) times 5 eg. � 62 = 36 662 = 4356 6662 = 443556 6666662 = 444443555556 �

571787 = 83

N2 � odd no. of digits

eg. � 32 = 09, 333332 = 1111088889 332 = 1089 3332 = 110889 �

3

= 512 = 83

�

Ch

I � Before 0 (n–1) times 1 II � Before 9 (n–1) times 8

3

N�

am

If 3 is repeated n times

�

� 571

pi

ga n

�

1 4 9 6 5

� �

at io n

N

�

83

a Pu t bl a

Unit digit of N2

�

� smaller than 571 perfect cube number = 512 =

Pr

Unit digit of N

�

571787 � UD = 3

�

841×846×851×856+625 + 75 should be � Perfect square subtracted

N=1, 9 N=2, 8 N=3, 7 N=4,6 N=5

3

1 8 7 4 5 6 3 2 9

� �

� ignore last 3 digit

54 =

625

C.

�

�

Si

841 × 846 × 851 × 856 + 700 � 841 × 846 × 851 × 856 � to make square 54 = 625 should be added

1 8 27 64 125 216 343 512 729

1 23 33 43 53 63 73 83 93

1119 × 1126 × 1133 × 1140

�

UD

3

r

If I, II, III, IV are in AP with common difference d

ic

�

14 + 87 = 782 + 412 152 +752 = 572 + 512 172 + 842 = 482 + 712 262 + 972 = 792 + 622 272 + 962 = 692 + 722 Non-terminating repeating decimal 2

2

Let x = 0.55555 .... (–) 10x = 5.55555 .... 9x = 5 � x = 5 9

�

992 = 9801 9992 = 998001

7 0.77777 .... = 0.7 � (no. of bar = no. of 9) 9 83

0.838383 .... = 0.83 � 99

85

Number System

514

0.514514514 .... = 0.514 � 999 6823 � 68 6755 � 9900 9900

11.4325 � 11 +

(no. of non bar digit = no. of zero) OR

5186 7.518651865186 ....= 7+ 9999

0.866666 .... = 0.86 =

�

86 � 8 78 13 = = 90 90 15

0.531313131 .... = 0.531 =

B

O

D

M

A

S

�

�

�

�

�

�

div. multiply Add subtract

�

531 � 5 526 = 990 990

small

()

middle { }

Si

Larger [ ]

of means multiplication

p

8169 � 8 8161 0.8169169 .... = 0.8169 = 9990 = 9990 5816 � 581 5235 �7 9000 9000

on pi

Ch

am

ga n

Pr

ic

at io n

a Pu t bl a

7.5816 � 7 +

4325 � 43 4282 2141 ��� ��� 9900 9900 4950

114325 � 1143 113182 = 9900 9900

�Brackets of

437 � 43 394 0.43777777 .... = 0.437 = = 900 900

Ga

�

75816 � 7581 68235 = 9000 9000

r

0.6823232323 ....= 0.6823 �

OR

86

Number System

Surds & Indices na

Law of Indices �

am×an×ap = am+n+p (a � 0)

1 (n > m) a n �m

=1

(m = n)

�

(a ) = am×n = an×m = (an)m

�

(abc) = a ×b ×c

�

� a � an � � = n (b � 0) �b� b

�

(am)n � a m

�

� a�

= a n = n am

�

� a�

= an = a

�

n m

m n

n

n

n

n

�

n

4

38 � 316

�

p

� 1q � 1 = �� a �� = � a p � q � �

�

a = a

�

If am = an then m = n

Pr

1 �p q

�

a–1 =

1 (a � 0) a

�

a–n =

1 1 & an = � n an a

�

�a � �b� � � =� � b � � �a �

�

(–1)n = +1 (n = even)

on

a° = 1

�

Ga

ab =

�

a n a � a �n n = = b n b �� b ��

n

1

�

�2 � 3 �

2

(a+b)2=a2+b2+2ab

28 � 10 3 = 10 3 � 2×5 3 52 + 32 =28

�

� a–b � (a>b) b–a � (b>a)

�

�

24 3= 2 ×12 3 4 3 3 16+27=43

�

�

99 � 70 2 = 5 2 � 7

�

70 3= 2 ×35 2

x = 5+ 2 6 1 1 1 5�2 6 5�2 6 = � × � x 5�2 6 5�2 6 5�2 6 2 5 � 2 6

�

1

=

1

a × n b = a n × b n = � ab � n 1

�

= 2� 3

43 � 24 3 = 3 3 � 4

�

1 n

n

2

2 7 � 4 3 = 2 + 3 + 2 ×2× 3 = ab

a2+b 2–2ab

Laws of Surds

�

n � � � mno � � = xyz a � �

7 5 2 49+50=99

= –1 (n = odd)

a =a

� a�

m

(a–b)2 = (b–a)2 = a2+b2–2ab

�

�m

n

0

x

Ch

am

�

�

1

1

� 5� 3

pi

ga n

If am = bm then a = b

m

n

p �

�� 32

p q

n

a = n a m = a mn

� �z �y � �� �

a Pu t bl a

2 4

n

m

Find Square Root

n

�3 �

m

n

at io n

=

order of surd

r

am = am–n (m > n) an

Radicand

Si

a × a × a × .... n times = an

ic

�

�

87

5 � 2 6 5–2 6 = 1 25 � 24

1

sign change diff. of square

x=7+4 3 � = x �� ������ 49 48

7�4 3 1

�

2

Number System

1

x = 5 3 +4 2 � = x

5 3 �4 2 43

a � a � a....� = 3 a

����� ���������

75 �

�

32

(a+b+c) = a2+b2+c2 + 2ab + 2bc + 2ca

4a � 1 � 1 =x 2

a � a � a � ....� =

4a � 1 � 1 =y 2

2

(a–b+c)2 = a2 + b2 + c2 – 2ab – 2bc + 2ca (a+b–c)2 = a2 + b2 + c2 + 2ab – 2bc – 2ca �

a � a � a � ....� =

x–y = 1 xy = a OR Take two factors of a whose difference is 1 like

(x–1) (x+1) = x –1 2

1 � � x � 1�� x � 1� = x

� 1 + (x – 1) (x + 1) = x2 �

Componendo & dividendo a c = b d

1

8 7

56

8 9

72

Then x = Larger factor

x=

Si

�

a

r

�

Apply C & D (Add and subtract den. in number)

y = smaller factor

D

x+y=

p

a�b c�d = a �b c�d

at io n

a � b � �a � b � c � d � �c � d � a � b � �a � b� = c � d � �c � d �

y=

Pr

2a 2c = 2b 2d

a c = � b d

x+y =

�

x� y

on

am

2�x � y �

Ch

x � y

= x �y � � x� y

Ga

x � y

+

D

pi

ga n C

�

�4 xy x � y – x � y = x �y

�

x� y

� –�

x� y

� =4 2

xy

�

x=

a � a � a � ....� =

y=

a � a � a � ....� =

1� a a a....ntimes = a 2n = a

3

a 3 a 3 a....� = a

n

2n �1

4a � 3b2 � b 2

a � b a � b a � ....� =

4a � 3b2 � b 2

2,

3

3,

4

1 1 1 5 Powers � 2 , 3 , 4

LCM = 12

� 2� , � 3� , � 5� 12

3

12

4

4

12

5 >33 > 2

� � � 64 81 125

2n

a n a n a....� = n�1 a

4a � 3 � 1 2

xy = 2(a–1)

� 26, 34, 53 � 1

4a � 3 � 1 2

a � b a � b a � ....� =

�

a a a a....� = a

�

4a � b2 � b 2

4a � b2

x–y = 1

x – y

2

a � b a � b a – ....� =

4a � b2 � b 2

x–y = b xy = a

If we apply C & D two times on a fract ion, same fraction is achieved.

x� y

x = a � b a � b a � ....� =

a Pu t bl a

�

Apply again

�

4a � 1

ic

C

y=

�

If x + y = 12 (constant) (xy)max = diff. of x & y should be min. = 6 × 6 = 36

88

Number System

(xy)min = diff. of x & y should be max. = 1 × 11 = 11 �

a>b>c 1 1 1 < < a b c

�

a × b = 16 (constant) � (a+b)min = 4 + 4 = 8

1×16� 16+1=17

(a +b )min = 4 +4 =32 2

2

2

2

* a×b×c = 125 (constant) for min a=b=c=5 (a2+b2+c2)min= 25+25+25 = 75 Approx Root Value

r

13 9 3

16 4

Si

�

� 3< 13

am

x–

If x
b

189 =9 21

92

(b–a) � if b > a

Algebra

Concept of Degree 5.

3x3 – 7x2y + 8zx ² ��Degree = 3 (highest powers) 8x + 7 � Degree = 1 25x2 – 10x3 + 19z26x1 + 48w1 Degree = 27 Multiply Divide

1 1 1 + + 2 =0 a 2 + b 2 – c 2 b2 + c 2 – a 2 c + a 2 – b2

Then

2 � a 4 + b4 + c 4 �

6.

Then a 2b2 + b2c2 + c 2a 2 = 4 � �

7.

Then

power add power subtract

a+b 2 b+c 2 c+a (a + b2 – c2) + (b + c2 – a2) + ab bc ca

xy � Degree = 2

(c2 + a2 – b2) = 0

x8 ��Degree = 8 – 3 = 5 y3

8.

2a 2 2b2 2c 2 Then � b2 +c 2 –a 2 � + � a 2 +c 2 –b2 � + � a 2 +b2 –c2 � = –3

Degree of each term is same on both sides.

�

a + b + c, a2bc + b2ca + c2ab

(x + ax + bx + ab) = (x + a) (x + b)

�

1 + A + B + AB = (1 + A)(1 + B)

a 3 � b3 = a2 – ab + b2 � Both side degree is 2 a�b

�

(1+a) (1+b) (1+c) = 1 + a + b + c + ab + bc + ca + abc

�

If xy = 1 or x = y then + n = 1 1� xn 1� y

Symmetric Function

= ab + bc + ca

�

a3 � �b � c� b3 � � c � a � c3 � �a � b � + + � a � b �� a � c � � b � c �� b � a � � c � a �� c � b �

2

Pr

2

=a+b+c

If x = a+

�

If x + y = 2z, then the value of =

�

If x + y = 2z, then

pi

�

Ch

Ga

y x + y �z = 2 x �z

�

1. 2.

a2 b2 c2 + 2 + 2 =2 Then 2 a – bc b – ca c – ab

3.

Then

a2 b2 c2 + 2 + 2 =1 2a + bc 2b + ca 2c + ab

4.

Then

�a – b � + � b – c � + �c – a �

2

a 2 + b2 + c 2 2

2

2

=

r

Si

a 2 + b2 + c 2 �

1

1

1

�

�

p2

q2

r2

�

If pq + qr + rp = 0, then � p 2 – qr + q 2 – rp + r 2 – pq � � � �

If �

�a

2

+b2 +ab � �

�

�a

2

+b2 –ab � � = 1, then (1 – a2) �

3 4

��

�

2 2 If x � 1 � x y � 1 � y = 1, where x and y are

real numbers, then (x + y)2 = 0 �

If a + b + c = 0, then following results follows 1 1 1 Then (a � b)(b � c) � (a � c)(b � a) � (c � a)(c � b) = 0

+ s2

If xy + yz + zx = 0, then � x 2 � yz � y 2 � zx � z 2 � xy � � �

(1 – b2) =

x 4 � y 4 � 2x 2y 2 = 4 y z +y �z =0 x �z

2

=1

am

1 1 and y = a – then a a

�

on

�

ga n

�

x z x m � ym � z m � wm m/2 If y = then � m = xyzw � x � y �m � z �m � w �m � w

2

= 0 (x, y, z � 0)

a � �b � c� b � �c � a � c � �a � b� + + � c � a �� a � b � � a � b �� b � c � � b � c �� c � a � 2

� s – a � + � s – b � + �s – c � 2

If a + b + c = 2s, then =1

�

= ab + bc + ca �

p

a 2 � b � c � � b2 � c � a � � c 2 � a � b �

1

1

at io n

�

a 3 � b2 � c 2 � � b3 � c2 � a 2 � � c 3 � a 2 � b2 �

1

ic

�

Some other results

2

a Pu t bl a

�

x4 � x � y �� x 2 � z 2 � + 2

2

y4 �y � x ��y 2 � z 2 � 2

2

z4 + z 2 � x 2 z 2 � y 2 =1 � �� �

�

If a + b + c = abc, then

�1 – a ��1 – b � + �1 – b ��1 – c � + �1 – c ��1 – a � = 4 2

ab

�

2

2

2

bc

If bc + ca + ab = abc, then b+c c+a a+b + + bc � a–1� ca � b–1� ab � c–1� = 1

1 3

93

2

ca

2

�

� x �y

z�x �

y�z

If xy + yz + xz = 1, then � 1 � xy � 1 � yz � 1 � xz � � �

2

2

5.

If a +b +c –ab–bc–ca=0 � a=b=c 2 2 2 OR a +b +c =ab+bc+ca

6.

a3+b3+c3–3abc = (a+b+c)(a2+b2+c2–ab–bc–ca)

1

= xyz If x �

= a �b b �c c �a ,y � ,z � , a �b b �c c �a

a 3 � b 3 � c 3 � 3abc

�a � b �

�1 � x ��1 � y ��1 � z �

2

=

a �b �c 2

If x2 + y2 = z +1, y2 + z2 = x + 1, z2 + x2 = y + 1, then

a3+b3+c3–3abc= 7.

(a + b + c) (ab + bc + ca) – abc = (a + b) (b + c) (c + a)

�

(a + b + c) (ab + bc + ca) = a2b + b2a +b2c + bc2 + a2c + a2c + 3abc

�

(a + b + c) (ab + bc + ca) – 3abc = a (b + c) + b (c + a) + c2 (a + b)

If a+b+c=0 then a3+b3+c3–3abc=0 OR a3+b3+c3=3abc 8.

2.

(a+b+c) = a +b +c +2(ab+bc+ca) 2

a2+b2+c2 = (a+b+c)2 –2(ab+bc+ca)

� a+b+c �

2

– � a 2 +b2 +c 2 �

ga n 2

(a–b+c)2 = a2+b2+c2–2ab–2bc+2ca

1 [(a–b)2+(b–c)2+(c–a)2] 2

Ga

a2+b2+c2–ab–bc–ca =

Ch

(a+b–c)2 = a2+b2+c2+2ab–2bc–2ca

��������

a2+b2+c2–ab–bc–ca =

If a, b, c are +ve integers & a3+b3+c3 = 3abc then a = b = c

9.

(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c) (c + a) (a + b + c)3 = a3 + b3 + c3 + 3 [a2 (b + c) + b2 (c + a) + c2 (a + b)] + 2abc (a + b + c)3 – a3 – b3 – c3 = 3 (a + b) (b + c) (c + a) 1

1

1

10. If x + y = a, y + = b, z + =c z x 1

then xyz + xyz = abc – (a + b + c) 1

1

1

11. If x – y = a, y – = b, z – =c z x 1

1 [(a–b)2+(b–c)2+(c–a)2] 2

If a, b, c are in A.P. with comon difference d +d +d

If a, b, c are distinct integers & a3+b3+c3 = 3abc then a+b+c=0

If a, b, c are in A.P. then a3+b3+c3–3abc = 9bd2

1 [2a2+2b2+2c2–2ab–2bc–2ca] 2 1 2 2 [a +b –2ab+b2+c2–2bc+a2+c2–2ca] 2

a=b=c

�

am

ab + bc + ca =

�

on

2

�

pi

2

a+b+c = 0

Pr

If x2 + y2 + z2 = 0, then x = 0, y = 0, z = 0

at io n

1 1 1 � b � � c � (where a� b�c) then abc is b c a equal to = ��1 3 Variable Formulae

1.

1 (a+b+c) [(a–b)2+(b–c)2+(c–a)2] 2

If a3+b3+c3–3abc = 0 OR a3+b3+c3 = 3abc

2

If a �

2

a3+b3+c3–3abc=

a Pu t bl a

2

a3+b3+c3–3abc=(a+b+c)[(a+b+c)2–3(ab+bc+ca)]

p

�

1 (a+b+c)[3(a2+b2+c2)–(a+b+c)2] 2

r

1 8

ab(a – b) + bc (b – c) + ca(c – a) = (b – a) (b – c) (c – a)

4.

2

a3+b3+c3–3abc = (a+b+c) [(a+b+c)2 – 3(ab+bc+ca)]

�

3.

� �b � c � � �c � a �

then 1 � x 1 � y 1 � z = 1 � �� �� �

xyz = 1 or �

�

2

ic

�

1 (a+b+c)[(a–b)2+(b–c)2+(c–a)2] 2

Si

�

Algebra

}

2

then xyz – xyz = abc + (a + b + c) 12. If a + b + c = x and

1 2 2 [d +d +(2d)2] = 3d2 2

1 1 1 � � =y a b c

Then

94

a b c a b c � � � � � = xy – 3 b a a c c b

Algebra

Theory of Equations

f(x)

Polynomial �

An

alg e braic

e xpre s si on

of

th e

fo rm Quadratic Equation

a0+a 1x+a2x +...+anx , where n � N, is called a 2

f(x)=0

n

polynomial. It is generally denoted by p(x), q(x), f(x),

A quadratic polynomial f(x) when equated to zero

�

g(x) etc.

is called quadratic equation.

a0+a1x+a2x +...+anx 2

n

f(x)

n � N,

r

p(x), q(x), f(x), g(x)

Si

i.e. ax2+bx +c =0, where a,b,c�R and a � 0. Real Polynomial

Roots of a Quadratic Equation

Let a0,a1,a2,...,an be real numbers and x is a real variable, then, f(x) = a0+a1x+a2x2+...+anxn is called

The values of variable x which satisfy the quadratic

a real polynomial of real variable x with real

p

equation is called roots of quadratic equation.

coefficients. x

a Pu t bl a

a0,a1,a2,...,an f(x) = a0+a1x+a2x2+...+anxn Degree of a Polynomial

1.

Pr

Let ax2 + bx + c = a(x–�) (x–�) = 0. Then, x = � and x

complex is a polynomial of degree n, if an � 0.

= � will satisfy the given equation.

f(x)=a0+a1x+a2x2+a3x3+...+anxn,

an � 0

n

on

Linear Polynomial A polynomial of degree one is

ax2 + bx + c = a(x–�) (x–�) = 0.

2.

Direct Formula Quadratic equation ax2 + bx + c = 0 (a � 0) has two roots, given by

Ch

ax2 + bx + c = 0 (a � 0)

(ii) Quadratic Polynomial A polynomial of second

Ga

x=�

x=�

am

known as linear polynomial.

pi

(i)

ga n

Some Important Deduction

Factorisation Method

ic

A Polynomial f(x)=a0+a1x+a2x2+a3x3+...+anxn, real or

at io n

Solution of Quadratic Equation

x �

x

degree is known as quadratic polynomial.

(iii) Cubic Polynomial A polynomial of degree three is known as cubic polynomial.

�=

�b � b2 � 4ac , 2a

�=

�b � b2 � 4ac 2a

or � =

(iv) Biquadratic Polynomial A polynomial of degree four is known as biquadratic polynomial.

�b � D �b � D ,�= 2a 2a

where, D = � = b2 – 4ac is called discriminate of the equation. D = � = b2 – 4ac

Polynomial Equation

Above formulas also known as Sridharacharya formula.

If f(x) is a polynomial, real or complex, then f(x)=0 is called a polynomial equation.

95

Algebra

Nature of Roots (i)

b Then, ��� = �+ � + � = � a

Let quadratic equation be ax2 + bx + c = 0, whose discriminate is D.

���� = �� + �� + �� =

Nature of Roots ����= �

If D = 0

If D > 0

3.

If D < 0

d a

Biquadratic Equation If �,�,� and � are the roots of the biquadratic equation

Roots are imaginary

Roots are real and equal

�,�,�

D = Perfect square

D = Not square

Roots are real, unequal and rational

Roots are real, unequal and irrational

c a

�

ax + bx + cx2 + dx + e = 0, then 4

3

r

S1 = � + � + � + � = �

b , a

Si

S2=��+��+��+��+��+��

(ii) Conjugate Roots The irrational (complex) roots of a quadratic equation, whose coefficients are rational (real) always occur in conjugate pairs. Thus,

=(–1)2

c c = a a

p

or S2=(�+�)(�+�)+��+��=

c a

� + i�

If one root be � + �–

� , then other root will be �+

�.

�

�–

or S3 = ��(�+�)+��(�+�)= �

�

Relation between Roots and Coefficients

and S4 = ��������= (–1)4

am

�

�

�

=S=�+�=

e e = a a

If �1,�2,�3,...,an are the roots of an nth degree equation, then the equation is xn – S1xn–1 + S2xn–2– S3xn–3+...+(–1)nSn=0, where Sn denotes the sum of the products of roots taken n at a time. �1,�2,�3,...,an

�b a

Ch

Ga

Sum of roots

pi

If roots of quadratic equation ax2+bx+c = 0 (a � 0) are � and �, then ax2+bx+c = 0 (a � 0)

d a

Formation of Polynomial Equation from Given Roots

on

Quadratic Equation

ga n

1.

d d =� a a

ic

(b)

� – i�

=(–1)3

at io n

If one root be � + i�, then other root will be � – i�.

Pr

(a)

a Pu t bl a

S3 = ���+���+���+���

x – S1xn–1 + S2xn–2–S3xn–3+...+(–1)nSn=0 n

Sn

coefficient of x

= � coefficient of x 2

1.

Product of roots = P=����=

If � and � are the roots of a quadratic equation, then the equation is x2–S1x+S2=0, where S1 = sum of roots and S2= product of roots

c constant term = 2 a coefficient of x

�

2.

x 2– S1 =

S2=

Cubic Equation If �, � and � are the roots of cubic equation, then the equation is

2. Cubic Equation If �,� and � are the roots of cubic equation ax3 + bx2 + cx + d = 0. �

�

S1x+S2=0

D

Also, |�–�| = a

�,�

Quadratic Equation

�, �

�

x3–S1x2+S2x–S3=0 i.e. x3–(�+�+�)x2+(��+��+��)x –���� = 0

ax3 + bx2 + cx + d = 0

96

Algebra

3.

Biquadratic Equation

Inequality

If �,�,� and � are the roots of a biquadratic equation, then the equation is �,�,�

�

�

x –S1x +S2x –S3x+S4 = 0 i.e. x4–(�+�+�+�)x3 + (��+��+��+��+��+��)x2 – (���+���+���+���)x + ���� = 0 Maximum and Minimum Values of Quadratic Expression 4

(i)

3

2

>, 2, 8 < 21 (ii) Literal inequality An inequality which have variables is called literal inequality.

(ii) If a < 0, quadratic expression has greatest value

at io n

b � This greatest value is given by 2a

e.g. x < 7, y � 11, x – y � 4 (iii) Strict inequality An inequality which have only is called strict inequality.

on

< >

e.g. 3x + y < 0, x > 7 (iv) Slack inequality An inequality which have only �or� is called slack inequality. ��

Ch

Ga

am

pi

4ac � b2 D �� 4a 4a

ga n

=

b 2a

Pr

x= �

ic

4ac � b2 D �� � But their is no least value. 4a 4a

a < 0,

��

Types of Inequalities

But their is no greatest value.

at x = �

>

e.g. 5 0, quadratic expression has least value at

a > 0, x =

A statement involving the symbols >, (greater than), � (less than or equal to) and �(greater than or equal to) are known as symbol of inequalities.

�

e.g. �3x + 2y � 0, y � 4

97

Trigonometry

Trigonometry Basic Triplets Theory

1.

Hypotenuse (H)

P

�

B

�

120( 3–1)

P

B P � Perpendicular

9, �×3 27

B � Base

a, b

c

x

2

12.5 : 32.5 5, 12, 13 � Triplet 5 : 13 � x = 12 × 2.5 = 30

p

32 + 4 2 = 5 2

a Pu t bl a

� 3, 4, 5 are triplets

3.

Some Pythagorean Triplets

(3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97) (20, 99, 101) In a triplet largest side is hypotenuse

�

Ga

a2+b2

triplets

2ab

:

on

8×2 :

�

4.

2

2

1.

2.

(3, 4, 5) ×2 (6, 8, 10) ×2 (12, 16, 20)

�3

2,4 2,5 2

:

8 9 � 4 = 8 13

�

sin �=

P H

cot�=

B P

cos�=

B H

sec�=

H B

Tan�=

Multiplication and division of basic triplets results into other triplets

(3, 4, 5) × 2

8×3

H

P

Triplets form

(5, 12, 13) ×2 (10, 24, 26) ×1.5

x �

B

2

a –b , 2ab, a +b 2 2 x –1, 2 x, x +1 a–b, 2 ab, a+b

:

Basic Trigonometry Ratios

(a2–b2) 2

24

pi

am

4

24

16

32 � 42 � 25 � 5

4

x

ic

Pr

3

Triplet

16

Ch

3

ga n

5

32.5

12.5

at io n

a +b =c 2

Si

A Pythagoras triplet is a set of Positive integers a, b and c that fits the rule : a2 + b2 = c2 2

41 � Triplet �×3 123

40, �×3 120

r

H2 = P2 + B 2 ��Pythagoras Theorem 2.

123( 3–1)

27( 3–1)

H

�

Cosec�=

98

1 � sin × cosec = 1 sin�

Sec�=

1 cos �

� cos × sec = 1

Cot�=

1 t an �

� tan × cot = 1

(15, 36, 39)

�

P H cosec�= B P

Trigonometry

3.

sin � sec � = cos � cos ec�

Tan �= Cot�=

2.

(sin2�+cos2��= 1) � cos2� tan2�+1 = sec2� tan2� = sec2� – 1 � tan2�=(sec�+1)(sec�–1) �

cos � cos ec� = sin � sec �

�sec � � 1� Tan � � tan � �sec � � 1�

Basic Trigonometric Identities �

sec2 – tan2 = 1 (sec��– tan�) (sec� + tan�) = 1

H

P

�

1 (sec� – tan�) = �sec � � tan � �

B

P2 + B 2 = H 2 2

sin2�=1–cos2� ��

r

Si

sin2� + cos2� = 1 sin � 1 � cos � � 1 � cos � sin �

3.

a Pu t bl a

1+cot2� = cosec2� cot2 = cosec2 –1 �

cot � cos ec� � 1 � cos ec� � 1 cot �

at io n

cosec2�–cot2��= 1 (cosec�+cot�)(cosec�–cot�)=1

ic

1 (cosec� + cot�) = � cos ec� � cot � �

Pr

(sin2� + cos2�)2 = 12 sin4� + cos4� + 2sin2�cos2� = 1 sin4 + cos4 = 1–2sin2 cos2 (sin2� + cos2�)3 = (1)3 sin6� + cos6� + 3sin2�cos2� = 1 sin6 +cos6 =1–3sin2 cos2

(sin2� + cos2� = 1) � sin2�

p

cos � 1 � sin � � cos2� = 1–sin2� �� 1 � sin � cos �

cosec4 + cot4 = 1+2cosec2 cot2 cosec6 – cot6 = 1 + 3 cosec2 cot2

Angle (In Radians)

0

� 6

on

pi

am

Degrees)

Ch

Angle (In

ga n

Trigonometry Ratio Table

Ga

1.

(sec2� – tan2�)2 = (1)2 sec4� + tan4� – 2sec2�tan2� = 1 sec4 + tan4 = 1 + 2sec2 tan2 (sec2� – tan2�)3 = (1)3 sec6� – tan6� – 3sec2�tan2� × 1 = 1 sec6 – tan6 = 1 + 3sec2 tan2

2

� P � � B� � � + � � =1 � sin2 + cos2 = 1 � H� � H�

� 4

� 3

� 2

�

3� 2

2�

sin

0

1 2

1 2

3 2

1

0

�1

0

cos

1

3 2

1 2

1 2

0

�1

0

1

tan

0

1 3

1

3

3

1

1 3

2 3

2

2

2

2

2 3

cot sec cosec

Not Defined 1 Not Defined

99

Not Defined 0 Not Defined 1

0 Not Defined �1 Not Defined

Not Defined 0 Not Defined �1

1 Not Defined 1 Not Defined

Trigonometry

sec2� + cosec2� = sec2�cosec2�

2.

tan2�–sin2� = tan2�sin2� cot2�–cos2� = cot2�cos2�

90°, 270° .... (90°

=sec�cosec�=2cosec2� Range

3

–1 � sin� � + 1

(–1)2 = 1 �

p

a Pu t bl a

0°, 360°

ga n

Ga

Any Acute Angle

All +ve

at io n

ic

�

Sin(180°+�) = –sin� Cos(180°+�) = –cos� Cos210° = cos(180°+30°) = –cos30° =

ASTC

Sin225°=sin(180°+45°) = –sin45° = III tan cot

sin cos tan

+ve

IV cos sec

�

� 3 2

�1 2

Cot240° = Cot(270°–30°) = –Tan30° =

+ve

� sin +ve in I, II

�2 3

Tan135° = Tan(180°–45°) = –Tan45° = –1

I

+ve

�1 2

Sin(180°–�) = sin� cos(180°–�) = –cos� cot(180°–�) = –cot� Sec150°=sec(180°–30°) = –sec30°=

am

270°

270° + � 360° – � 0° – �

Ch

IV

�

on

I 0° + � 90° – �

1 3

3 2

Cos120°=cos(180°–60°) = –cos60° =

pi

90° + �� 180° – ��

180°, 540° III 180° + � 270° – �

cosec

Sin(90°+�) = cos�

Tan150° = Tan(90°+60°) = –cot 60° = �

Pr

90°, 450°

II

cosec(90°–�) = sec�

Sin120°=sin(90°+30°) = cos30° =

Quadrant theory

sin

cot(90°–�) = tan�

Cosec(90°+�) = sec�

sin71° > cos���

II

sec(90°–�) = cosec�

Sec(90°+�) = –cosec�

45° < � < 90° sin� > cos�

1.

tan(90°–�) = cot�

Cot(90°+�) = –tan�

sin19° < cos���

x > x2

4 = cos� 5

Tan(90°+�) = –cot�

sin� < cos�

If 0 < x < 1,

sin(90°–�) =

Cos(90°+�) = –sin�

0° � 90° sin� increases from 0 to 1 0° � 90° cos� decreases from 1 to 0 sin61° > cos32° � cos32° = sin58° � sin61° > sin58° � sin61° > cos32°

3.

�

Si

� 0�� sin2�, cos2� � +1

0° < � < 45°

change)

� cos(90°–�) = sin�

– ��� tan�, cot� � + �

2.

5

90–�

4

P H > P (� Always between –1 and +1) sin�= H

–1 � cos� � + 1

odd multiple

even multiple 180°, 360° .... no change (90° no change) sin � cos, tan � cot, sec � cosec

sin � cos � 1 tan�+cot�= + = cos � sin � sin � cos �

1.

Change of T-Ratio

r

�

1 3

Cos(270°–�) = –Sin� Tan(270°–�) = cot� Cosec(270°–�) = –sec�

� cos +ve in I, IV

Sec240° = sec(270°–30°)= –cosec30°= –2

� Tan +ve in I, III

100

Trigonometry

�

Sin(270°+�) = –cos�

2.

Cos(270°+�) = sin� Tan(270°+�) = –cot�

1C

57°16'22''

� radian = 180°

Sin300° = sin(270°+30°) = –cos30° =

180� � 7 630 � 3� � � 57 � � 22 11 11

1 rad (1c) =

� 3 2

� 3°=180' �

Cot315° = cot(270°+45°)= –Tan45°= –1

180' 4' =16'+ 11 11

Cosec330°=Cosec(360°–30�) = –cosec30°= –2 �

� 4' =240'' �

Sin(360°–�) = –sin�

� 1C=57°16'22''

240 = 22'' 11

Sin(–�) = –sin� Tan780°= Tan(360°×2+60�) = Tan60°= 3 Cosec1125°= Cosec(360°×3+45�) = Cosec45°= 2 3.

cot(–�) = –cot�

�� AB = L = 2�r × 360�

p

cos180° = –1

a Pu t bl a

sec(–�) = sec� cosec(–�) = –cosec� If A+B = 90° then

L=r×

on

×

ic

1.

If asin � + bcos� = c acos� – bsin� = x (let)

�

�

�

�

P

B

H

a sin��+ b cos��= 1 c c sin��sin��+ cos��cos�� = 1

sin��= c

r

�

2�r � 360° 360� 180� = = 1 radian 2� �

circle

Radius

Arc centre

square and add

If asin� + bcos� = c and a2+b2 = c2

180° c � =1

r�

180°

a2+b2 = c2 + x2 OR x2 = a2 + b2 – c2

r

O

c

Some Important Properties

Ch

Ga

= � (Irrational numer)

°=

am

Radian Angle Theory

pi

ga n

Pr

If A+B=90° � A, B are complementary to each other sinA = cosB � sinAsecB=1 TanA=cotB � TanATanB=1 or cotAcotB = 1 secA=cosecB � cosAcosecB = 1 Tan31×Tan59°=1 sin2A+sin2B=1 � sin2A+sin2(90–A) cos2A+cos2B=1 � sin2A+cos2A=1

circumference = fixed number diameter

��� 180�

at io n

L=r×

tan180° = 0

1.

B

L

tan(–�) = –tan�

5. 6.

r

�°

A

sin180° = 0

1. 2. 3. 4.

r

Si

Cos(–�) = cos�

r

O

�

a b , cos��= c c

a b

If 48sin� + 55cos� = 73 then cot� = �

1 radian

P

Angle

compare

�

�

B

H

55 48

� radian (� ) = 180°

�

(sin�+sec�) +(cos�+cosec�)2 = (1+sec�cosec�)2

1°=60', 1'= 60''

�

(1–sec�+tan�)(1+cosec�+cot�) = 2

c

101

2

2.

Trigonometry

)

If asecA+btanA = c Then atanA+bsecA = x (Let)

Formula form - (A+B)

square and subtract

1.

a2–b2=c2–x2 � x2 = c2 – a2+b2 x= �

sin(A–B) = sinAcosB – cosAsinB cos(A+B) = cosAcosB – sinAsinB

c2 – a 2 + b2

cos(A–B) = cosAcosB + sinAsinB

If asecA – btanA = c

2.

atanA – bsecA= x (let) If asec�–btan� = c and a2=b2+c2 a sec�– b tan �=1 c c sec� sec� –tan�=1

)

compare

2sinAsinB = cos(A–B) – cos(A+B)

a sec� = c

3.

b tan� = c

a

r

4.

sinC + sinD = 2sin

C�D C�D cos 2 2

�

sinC – sinD = 2cos

C�D C�D sin 2 2

p

b c

�

cosC + cosD = 2cos

C�D C�D cos 2 2

(a > b)

�

C�D C�D sin 2 2

a Pu t bl a

a a+b If secx + tanx = then cosecx + cotx = b a–b

=

a 2 � b2 ab

secx =

a 2 � b2 2ab

secx =

a 2 � b2 � H 2ab � B

� P = a2–b2 �����

on

2secx

1.

pi

b a

Ch

2.

� a � b� = � a � b �� a � b� 2

�

sin2A = 2sinAcosA =

2 Tan A 1 � Tan2 A

A A 1 � cos 2A cos = 2 2 2

Sec A.Cos ecA 2

cos2A=cos2A–sin2A=1–2sin2A=2cos2A–1= =

a 2 � b2 � 2ab a 2 � b2

C�D D�C sin 2 2

Multiples of Angle

cosec2A =

H B H�B + = P P P

�

cosC – cosD = 2sin

sinA = 2sin

am

ga n

secx – tanx =

�

cosD – cosC = 2sin

ic

Pr

a b

secx + tanx =

Ga

3.

B

at io n

�

cos(A+B)cos(A–B) = cos2A–sin2B=cos2B–sin2A

Si

P

sin(A+B)sin(A–B) = sin2Acos2B – cos2Asin2B = sin2A(1–sin2B)–(1–sin2A)sin2B = sin2A–sin2B = cos2B–cos2A

�asec�–btan� = c ������ �� ������

H

2sinAcosB = sin(A+B) + sin(A–B) 2cosAsinB = sin(A+B) – sin(A–B) 2cosAcosB = cos(A+B) + cos(A–B)

� a2–b2=c2–x2 �

sin(A+B) = sinAcosB+cosAsinB

cot 2 A � 1 cot2 A � 1

cosA =

1 � cos 2A 2

2Tan A 1 � tan2 A

3.

Tan2A =

4.

Sin3A = 3sinA–4sin3A

a�b a �b

Cos3A = 4cos3A – 3cosA Tan3A =

102

3tan A � tan3 A 1 � 3 tan2 A

1 � tan2 A 1 � tan2 A

Trigonometry

6.

Tan(A+B) =

Tan A � TanB 1 � Tan A TanB

Tan(A–B) =

Tan A � TanB 1 � Tan A Tan B

Cot(A+B) =

cos�cos(60–�)cos(60+�)=

tan�tan(60–�)tan(60+�)=tan3� 10. cos�cos2�cos22�cos23�......cos2n�=

cot A cot B � 1 cot B � cot A

cot A cot B � 1 Cot(A–B) = cot B � cot A

1 � tan � 1 � tan � , Tan(45+�)= 1 � tan � 1 � tan �

1.

Tan(45–�) =

2.

If A+B = 45° or 225° � (1+TanA) (1+TanB) = 2

5 �1 4

cos18°=

10 � 2 5 4

cos36° =

5 �1 4

sin36°=

10 � 2 5 4

r

12. Sin75° = cos15° =

tan(A+B) = tan45° tan A � tanB =1 � tanA+tanB+tanAtanB = 1 1 � tan A tanB

1+tanA+tanB+tanAtanB=1+1

am

� TanATanB+TanBTanC+tanCTanA = 1

Tan15° = cot75°=

1.

2.

on

pi

ga n

sin90� 1 tan A � tanB � tan C � tan A tan B tanC = = cos 90� � 1 � tan A tanB � tanB tanC � tan C tan A

�

3.

Ch

Ga

� 1 � tan � � � � = tan2� � 1 � cot � �

1 � sin � 1 � sin � 1 � sin � � = =sec�–tan� 1 � sin � 1sin � cos �

1 � sin � = 1 � sin �

5.

1 � cos � 1 � cos � 1 � cos � 1 � cos � � = = sin � 1 � cos � 1 � cos � 1 � cos �

6.

cos2� + cos2(60–�) +cos2(60+�) =

7.

cos3A+cos3(120°–A)+cos3(120°+A)=

8.

Tan� –Tan(60–�)+tan(60+�)=3tan3�

1 � 6tan2 � � tan 4 �

1 sin3� 4

term direct

=cosec�+cot�

Tan4��=

sin�sin(60–�)sin(60+�)=

cot A � tan B = tanBcotA cot B � tan A

4.

4tan � �1 � tan2 � �

5.

cot�–cot(60–�)+cot(60+�)=3cot3�

(sin2A+cos2A – sinAcosA) (sinA+cosA)

2

� TanA + TanB + TanC = TanATanBTanC � cotA.cotB + cotBcotC + cotCcotA = 1

Some other formulae

multiply

OR cotA+cotB+cotC = cotAcotBcotC If A+B+C = 180°

3 �1 = 2– 3 3 �1

= sin3A + cos3A

Pr

If A+B+C=90°

9.

3 �1 = 2+ 3 3 �1

ic

tan A � tan B � tan C � tan A tanB tan C Tan(A+B+C) = 1 � tan A tan B � tanB tan C � tan C tan A

3 TanA – tan3 A Put B = C = 'A' � Tan3A = 1 – 3 tan2 A

�

Tan75° = cot15°=

at io n

OR (1–cotA) (1–cotB) = 2 4.

3 �1 2 2

a Pu t bl a

If A+B = 45° or 225° � (cotA–1) (cotB–1) = 2

3 �1 2 2

Sin15° = cos75° =

p

�

� (1+tanA) (1+tanB) = 2

sin 2n�1 � 2n �1 sin �

11. sin18° =

Some special formulae

3.

1 cos3� 4

Si

5.

3 2 3 cos3A 4

6.

cos ecA � 1 1 � sin � = =sec �+tan � cos ecA � 1 1 � sin �

7.

sec � � tan � = sec � � tan �

8.

cot � � cos � = sec�+tan� cot � � cos �

9.

103

2 � sec � � tan � � =sec�+tan�

sin1º.sin2º.sin3º ...... sin180º = 0 cos1º.cos2º.cos3º ..... cos90º = 0 tan1º.tan2º.tan3º ..... tan89º = 1

Trigonometry

Interchange of Trigonometry Ratio Table

1 � sin2 �

cos �

sin �

1 � cos 2 � cos �

1 � sin � 2

1 � sin2 � sin � 1 1 � sin2 �

1 � tan �

1 � cot �

1

cot �

1 � tan2 �

1 � cot2 �

tan �

1 cot �

1 tan �

cot �

1 � tan2 �

1 � cot2 � cot �

2

cos � 1 � cos � 2

1 cos � 1 1 � cos � 2

1 � tan2 � tan �

2

1 � cot2 �

ic

at io n

a Pu t bl a pi

on

Pr am

ga n Ga

Ch 104

sec2 � � 1 sec �

1

p

1 sin �

tan �

1 sec � sec2 � � 1 1 sec2 � � 1

r

1 � cos 2 �

Si

sin �

sec � sec � sec2 � � 1

1 cosec � cosec2� � 1 cose � 1 cosec 2� � 1 cosec2 � � 1 cosec2 � cosec2 � 1 cosec �

Trigonometry

Trigonometric Functions, Identities and Equations Graph of Trigonometric Functions 1.

Graph of sin x

5.

Graph of sec x

1 0.5 O –0.5

2

90° 180° 270° 360°

–1

–1 –2

�

2� 3� 2

x

Si

O –0.5

p

90° 180° 270° 360°

(i) Domain = R � (2n + 1)

(i) Domain = R (ii) Range = [–1, 1] (iii) Period = 2� Graph of tan x

Pr

1

ic

6.

0.5 O –0.5

� ,n�I 2

Ga

(iii) Period = � Graph of cot x

2 1

–� –� 2

–� 2 O –1 –2

–�

2 –� 2 1 O –1 � 2 –2

3� � 2

x

y’ y = cosec x

Ch

(ii) Range = (–�, �)

am

(i) Domain = R � (2n+1)

pi

ga n

–1

x’

on

90° 180° 270° 360°

� ,n�I 2

(ii) Range = (–�, –1] � [1, �) (iii) Period = 2� Graph of cosec x

a Pu t bl a

–1

x’

–� 2

y’ y = sec x

0.5

4.

O

r

(i) Domain = R (ii) Range = [–1,1] (iii) Period = 2� Graph of cos x 1

3.

1

at io n

2.

–�

x’

(i) Domain = R � n�, n ��I (ii) Range = (–�, –1] ��[1, �) 3� 2

2�

x

(iii)Period = 2� Note |sin�| � 1,|cos�|� 1,|sec�|�1,|cosec�|�| for all values of �, for which the functions are defined.

y y = cot x

(i) Domain = R�n�, n � I (ii) Range = (–�, �) (iii) Period = �

105

Trigonometry

Maxima & Minima

1.

–1�sin���+ 1

sin� =

����� min = – a 2 � b2

–1�cos���+ 1

asin2�+bcosec2� min = 2 ab (when a>b)

6.

P Tan� = � (We can take any B

–��� tan����+ �

asin�+bcos� � max = + a 2 � b2

5.

P ,H�P H

acos2�+bsec2�

max = �

value of P and B)

cosec� =

–��tan3�, cot3��� + � +1�sec2�, cosec2��� + � –��sec �, cosec ��� + ��But –1

+1

asin2�+bcos2�� max value = max [a,b]

ga n

3.

sin �cos � n

1 max = 2n min = –1 2n

n

1 max = 2n

min = 0

r

T - ratio sin�,cosθ(odd power)

2. 3. 4.

sin2θ,cos 2θ(even power) tanθ,cotθ(odd power) tan2θ,cot 2θ(evenpower)

5. secθ,cosec θ(odd power) 6. sec2θ,cosec 2 θ(even power)

max value = 1 min value = put ��� 45°

11. atan2��+ bcot2�

Ch

4.

n

Ga

n

1.

10. sin2n��+ cos2n�

am

� min value = min [a, b] 37 sin2� + 45cos2� � maximum = 45 minimum = 37

sin �cos �

If y = cos2x + sec2x then y � 2

ic

0�tan2�, cot2��� + �

Pr

0�sin2�, cos2�� + 1 –1�sin3�, cos3�� + 1

3

9.

at io n

2 = 0 at x = 0 x min

3

sin2m� + cos2n� � max = +1 m, n � natural no.

a Pu t bl a

If x = Real � x2 � +ve

2

8.

p

H B

when n=odd

when n=even

106

a� b

�

min value � asec2�+bcosec2� �����

pi

2.

H ...H ��(B, P) P

on

sec�=

1

�

7.

Si

–��� cot����+ � –��� sec�, cosec����+ ��But –1

min = a+b (when a < b)

min value = 2 ab max value = �

min max �1 �1 0 �� 0

�1 �� �

�� 1

�� �

Trigonometry

Height & Distance Elevation Angle

Change of Angle 1.

Li ne of sig ht

Object

When elevation angle changes from 30º to 45º 30º D

Angle of Elevation

�

Horizontal

1

� Angle of Depression �

2.

2

A

A 3 –1

1:

C

3:

75º

�Tan15°=

75� �

3 �1

3 �1 2 2

B

3 �1

=

45º

60º

B

45º 3–1

D

3

BC AC

A

1

60º

120°

30° 3

4.

When elevation angle changes from 15º to 30º 15º

30º

C

15º 2

A

1 30� 30� 120� � � �

1 30°

C

3

90� �

A 1

3 �1

60º

C

15� �

4.

1

2 2

15º 3 +1

B

2

1 BC = 3 AC

Ga

Tan30° =

90� �

3.

am

30º 3

B

3.

30� 60� � �

2

30º B 3–1=2 C 3 When elevation angle changes from 45º to 60º A

Ch

1

ga n

2.

60º

at io n

1:

ic

1:

C

3

on

45º

C

pi

1

Pr

B

B

3 When elevation angle changes from 30º to 60º

a Pu t bl a

45� 45� 90� � � �

�

60º

45º

D

A

1.

30º

1

p

Object

2

30º 3–1

A

Angle-side Ratio

45º

Si

r

Angle of Depression

1

45º

1:

1:

B

3

C

107

3

30º

D

5.

Trigonometry

When elevation angle changes from ....

2.

A

D

.... A

h

A

�1

B a

asin� B

6.

C

B

� hcot�

A

3.

C

D

When elevation angle changes from �º to 2�� �º

2�

A

y

h

�1

D

A

A

�2

C

d

�1

D

ga n

B

�2

ic

a cot �1 cot �1 � cot �2

D

A H2

am

C

�

�

x

B

Ch

��+ � = 90º

If Angles of elevation are complimentary

Ga

B

H1

pi

d = h (cot�1 – cot�2)

Some other results

5.

on

h

Pr

h=

x = h1h2

6.

A

D

H2

H1

h

B

C

a

at io n

C

�

C

a Pu t bl a

x

2�

D

p

h2 = y2 – x2

A

�2

F a

Si

h

1.

�1

a = h (cot�1 + cot�2) 4.

7.

h

r

B

B

C

a

a = h (cot�1 + cot�2)

h

� acos�

h

�2

a

90–� C x y

�

C

D

1 1 1 = + a H1 H2

h � xy

108

B

Arithmetic

Percentage Fraction percentage chart

Percentage

Fraction

Percentage

Percentage

1

100%

100%

4.76%

4

1 2

50%

50%

1 21

16 % 21

1 3

33.33%

1 33 % 3

1 22

4.54%

4

6 % 11

1 4

25%

25%

4.34%

4

8 % 23

1 5

20%

20%

1 24

4.16%

1 4 % 6

1 6

16.66%

2 16 % 3

1 25

4%

4%

1 7

14.28%

2 14 % 7

1 40

2.5%

2

1 8

12.5%

1 12 % 2

3 8

37.5%

1 37 % 2

1 9

11.11%

1 11 % 9

5 8

62.5%

1 62 % 2

1 10

10%

4 7

57.14%

57

1 11

9.09%

5 7

71.42%

3 71 % 7

1 12

8.33%

1 8 % 3

2 3

66.66%

2 66 % 3

1 13

7.69%

7

4 5

80%

80%

1 14

7.14%

1 7 % 7

3 4

75%

75%

6.66%

2 6 % 3

5 11

45.45%

45

5 % 11

6.25%

1 6 % 4

7 11

63.63%

63

7 % 11

1 17

5.88%

5

10 11

90.90%

90

10 % 11

1 18

5.55%

5 5 % 9

4 9

44.44%

4 44 % 9

1 19

5.26%

5

7 9

77.77%

7 77 % 9

1 20

5%

5%

1 16

Si p ic

at io n

a Pu t bl a

1 % 11

Ch

am

9 % 13

pi

9

on

Pr 10%

ga n

Ga

1 15

1 23

r

Percentage

Fraction

15 % 17

5 % 19

109

1 % 2

1 % 7

Arithmetic

Derived fraction from base fractions �

� 100%-20

1 2 � 14 % 7 7 4 1 � 57 % 7 7 5 3 � 71 % 7 7

40 4 = 4 � � 400% + 44.44% � 444.44% 9 9

�

43 1 � 7 � � 700% + 16.66% � 716.66% 6 6

1 = 20% 5

�

13 6 5 � 1 � � 100% + 85 % 7 7 7

3 � 3 � 20% = 60% 5

�

2 1 = 16 % 6 3

� �

Si

29 2 2 � 9 � � 966 % 3 3 3

� 1 � 71 11 2 �5� 500% + 11 ���8 % ���� � 500% + 91 % � 3 12 12 3

Percentage to fraction conversion

17.5% � 17.5×

1 7 � 40 or 100

� � 7 ��2.5% � 1 �� � 7 � 17.5%= 40 � �� 40 �

pi

on

�

2 2 37 7 � 2� � 200% + 46 3 % � 246 3 % 15 15

164

41

15

125 125 5 5 � % � % � 8 8 800 32

1 1 �6 % 16 4

�

35

5 1 1 5 % � 5×7 % � 5× � 7 7 14 14

1 1 13 =13×6 % = 81 % 16 4 4

�

29

1 1 1 1 7 � % � 25% + 4 % � � 6 6 4 24 24

�

23.33% � 20% + 3.33% �

1 2 � 14 % 7 7

�

78

6 1 2 5 � 1 � =100-14 % = 85 % 7 7 7 7

�

46.66% � 40%+6.66% �

Ch

�

Ga

� 164% � 100 25

or �

35 5 1 1 � 5 � � 500% + 83 % � 583 % 6 6 3 3

�

5 17 1 = 17×2 % = 35 % 48 12 12

�

am

1 1 =2 % 48 12

ga n

11 1 3 =11×6 % = 68 % 16 4 4

�

�

Pr

1 1 �6 % 16 4

1 1 �4 % 24 6

5 � 185 7 %

2 � 591 3 %

1 2 �6 % 15 3 11 2 1 =11×6 % =73 % 15 3 3

�

r

�

2 5 1 = 5×16 % = 83 % 6 3 3

�

5 1 % � 79 % 6 6

p

�

� 1 �� 19 5 � � 1� � 100%-5 ��4 % �� � 6 � 24 24

at io n

�

�

ic

�

11 1 1 2 � 1� � 100% -8 3 % � 91 3 % 12 12

1 � 25% 4 � �3 3 � 75% 4

a Pu t bl a

�

1 1 �8 % 12 3

13 3 3 1 = 1– = 100% – 18 % = 81 % 16 16 4 4

110

7 1 1 � = 5 30 30

1 1 3 1 49 % � 75% + 3 % � � � 3 3 4 30 60 2 1 7 + � 15 5 15

Arithmetic

�

1 2 1 5 % � 100%-16 % � 1- � 3 3 6 6

�

83

�

237.5% � 200% + 37.5% � 2 +

�

3 24 342.84% � 300% + 42.84% � 3+ � 7 7

�

756.25 � 7+

�

� 538.33 � 500% + 30% + 8.33% � 5 + � 60 10 12

�

528.56% � 5+

�

�y = �x x% of y = y% of x � 100 100

�

840% of 62.5 � 62.5% of 840 � � 840 � 525 8

�

7.44% of 3750 � 37.50% of 744 � ×744 � 279 8 99 is what % of 135

323

� Initial

y

�

p

-11

�

ic

�

on

�

am

Ga 135

��

–36

36 �100% 135

4

40 : 31 �9 � 40 -9

�26% x �� �� � � x � x � 26% �

x+x×

� 26 �� 26 � = x ���1 � 100 �� 100

Final value �

37 �� �x 100 ��

�

45 �� �y 100 ��

x

+37%

1.37x = ����1 �

y

-45%

0.55y = ����1 �

z

+45.45%

z+z×

2

5 11

����������������������������� � 5 �� �� � �11��

180

×0.8

100

+7

or

180 × 100% = 144% 125

:

F.V 27

-22.5% �

� 26 % 15 : 4 � 15 3 180 is what % of 125

or 125

�7 20

= x × 1.26 multiplying factor (m.F) Initial value Change

99 is what % less than 135 99

+35% � 35% = 1.V 20

pi

ga n

100 99 ��� 99 � % 135

� 11 25

F.V 14

1.V 25

Pr

100 1 ��� % 135

or

-44% � 44% =

a Pu t bl a

3

11 1 = 73 % 15 3

Initial value Final 32

+7

5

1 2 =6 % 15 3

Change in value

Si

25

135 ���100%

�

1

+7 28% = 25

2 37 � 7 7 x

or 15 : 11

Final 128%

r

3

� multiplying factor

+1.28

9 121 � 16 16

99 1 ×100% = 73 % 135 3

�

Initial 100%

Ch

�

� multiplying factor

3 19 � 8 8

x � �

� �

� x % � K+kx%=K(1+x%)=K �1 � � K ���� 100

at io n

�

Scaling factor/multiplying factor

4 1 33 82.5% � 80% + 2.5% � + � 5 40 40

w

+31.25%

� � ��1 � 5 �� w = 21 w �� 16 �� 16

u

-46 %

2 3

� 7� 8 ���1 � ��� u = u � 15 � 15

�

144

� 7 �� �� � �15 ��

+44%

111

� � ��1 � 5 �� ×z = 16 z � 11�� 11

Net income = Income – Income Tax Rs.700 = Rs.800 – Rs.100 Rate of income tax

=

�

100 ×100% = 12.5% 800

1.

Income Tax

= Total income �100% Net Income = Income – Income Tax 700

=

800

–

Arithmetic

Second Method :Initial : Final 100 : {(100 × 1.2) ×1.3}×0.5 100 : 78

22% Change (Decrease) If the population/cost of a certain town/article, is P and annual increment rate is r% , then P r% � �

100

(i) After 't' years population/cost � P �1 �

+50

–50

't'

Si

650 = 800 – 150 Change in net income & tax will be same

r

�

r � � � P �1 � � � 100 �

(ii) Before 't' years population/cost � Concept of deviation

7+13+5

Total deviation

divide

ga n

 

p 't'

3.

Ga

Ch

am

18% + 0.96% = 18.96% or Rice : Dal : Exp � 7 : 13 : % Change 17% 17+3% Let overall % change = 17% deviation = 13 × 3% + 5 × 2% = 49% divide

Sugar 5 17%+2%

(i)

100

+20%

120

+30%

156

–78 –50%

t

P r � � �1 � � 100 � �

t

If the population of a town /cost of a article is P and it decreases/reduces at the rate of r% annually, then. P r%

r � � After 't' years population/cost � P �1 � � � 100 � r � � � P �1 � � 100 � �

(ii) Before 't' years population/cost �

� 17% + 1.96 � 18.96% Annual increment/decrement � Successive percentage change:-

+36

r � � �1 � � � 100 �

Formula: If we change a number by x% & y% successively Then, net% change x% y%

't'

49 % � overall % change � 17% + 25

P

� xy �� � � ���x � y+ 100���%

pi

� �7 � 26 � 5 �� % �� � 18% + ��� � 25

+20

�

t

at io n

2.

Pr

Deviation Deviation Deviation in dal in rice in sugar (-1% × 7 + 2% × 13 + 1%×5)

Sugar 5 +19% +1% 18%

:

t

ic

Dal 13 +20% +2%

a Pu t bl a

Rice Exp 7 : � +17% % Change � Deviation � –1% Let overall change is 18%

on

�

r � � 100 �

�

't'

4.

r � � �1 � � � 100 � P

r � � �1 � � � 100 �

z%

xy � yz � zx xyz � � � % �� x � y � z � � 100 10000 �� �

–22 net % change = 22%

112

t

P

Net % change of x% , y% and z% is x% , y%

78

t

t

t

If the present population of a town is P and the populat ion increases or decreases at rat e of R1% , R2% and R3% in first, second and third year respectively. P R1% , R2%

(Negative for decrease, Positive for increase) 3.

R �� R �� R � � � P �1 � 1 � �1 � 2 � �1 � 3 � � 100 � � 100 � � 100 �

� Population after 'n' years 'n'

x2 % be a decrease of 100

Ga

x%

r

� Negative sign for decrease � � � � Positive sign for increase �

4.

at io n

If a number is first increased by a% and then again increased by b% , then total increase percent is a% b%

pi

ab � � �a � b � �% 100 � �

6.

x%

x2 % 100

10%

ab � � � �a � b � � % –ve 100 � �

ic

5.

If the cost of an article is increased by A% , then how much to decrease the consumption of article, so that expenditure remains same is given by A% OR If the income of a man is A% more than another man, then income of another man is less in comparison to the 1st man by A%

Ex. Price of an article is increased by 10% and then reduced by 10% . What will be net percentage change? 10%

� � A � 100 � % �� � � �100 � A � �

102 % = 1% � Price will be decrease by 100

2.

If a number is first decreased by a% and then by ab � � b% , then net decrease percent is � �a � b � �% 100 � � (–ve sign for decrease) a% b%

on

Pr am

If an amount is increased by x% and then it is reduced by x% again, then percentage change will

Ch

1.

+30% +35% –48% = +35% –48% +30% Initial × 1.3 × 1.35 × 0.52 = Initial × 1.35 × 0.52 × 1.3 sequence change successive change

ga n

�

ab � � � �a � b � �% 100 � �

a Pu t bl a

R �� R � � R � � � P �1 � 1 � �1 � 2 � ... �1 � n � � 100 � � 100 � � 100 �

Important Points Based on increase/decrease

If a number is decreased by a% and then it is increased by b% , then net increase or decrease percent is a% b%

p

'+' is used when population increases '+' '–' is used when population decreases. '–' The above formula may be extended for n number of years. n

�

b%

ab � � �a � b � �% 100 � �

R3%

then the population of town after 3 years

6.

Arithmetic

a%

Si

5.

If a number is increased by a% and then it is decreased by b% , then resultant change in

7.

ab � � percentage will be � a � b � �% 100 � �

113

If the cost of an article is decreased by A% , then the increase in consumption of article to maintain the expenditure will be? A%

OR If 'x' is A% less than 'y'. then y is more than 'x' by �

�

A

Arithmetic

13. If every side of cube or radius of sphere is increased by a% , then increase % in volume a%

Required% = �� 100 – A � 100 �� % (increase) � �� �

8.

A%

y 'x'

� � A � 100 � % �� � � �100 – A � �

If the length of a rectangle is increase by a% and breadth is increased by b% , then the area of rectangle will increase by a%

� 3a 2 a3 � � � 3a � � �% 2 � 100 �100 � �� �

14. If a% of a certain sum is taken by 1st man and b% of remaining sum is taken by 2nd man and finally c% of remaining sum is taken by 3rd man, then if 'x' rupee is the remaining amount then, a% b%

b% ab � � = �a � b � �% 100 � � If the side of a square is increases by a% then, its area will increase by a%

Initial amount

Required Increase

x�

p

at io n

� a2 � 2a � � �% 100 � �

10. If radius of circle is increased by a% then its area

�

�

ic

on

a2 �

'y' Increased/decreased cost of the article

Ch

Ga

� a2 � � Decrease � � �2a � �% 100 � �

� xy � �� � � 100 � a �

This formula is also applicable for circles where decrease % of radius is given.

12. If the length, breadth and height of a cuboid are increased by a% , b% and c% respectively, then, Increase% in volume a% , b% c%

100 100 100 � � 100 � a 100 � b 100 � c

16. On increasing/decreasing the cost of a certain article by x% , a person can buy 'a' kg article less/ more in 'y' rupees, then x% 'a'

am

11. If the side of a square is decreased by a% , then the area of square will decrease by a%

c% 'x'

Initial amount x�

pi

�

= � 2a � 100 � %

ga n

a%

Pr

� a2 � will be increased by = � 2a � 100 � % � �

100 100 100 � � 100 � a 100 � b 100 � c

15. If an amount is increased by a% and then again increased by b% and finally increased by c% . So, that resultant amount is 'x' rupees, then a% b%

a Pu t bl a

9.

'x'

Si

c%

r

'x' 'y'

%

�

And initial cost

xy

�100 � x � a

[Negative sign when decreasing and positive sign when increasing] 17. If a person saves 'R' rupees after spending x% on food, y% on cloth and z% on entertainment of his income then. x% y% z%

� ab � bc � ca abc � % � �a � b � c � � 2� 100 �� �100� ��

Monthly income

114

'R' �

100 �R 100 � � x � y � z �

1.

Examination based In an examination, a% candidates failed in Maths and b% candidates failed in English. If c% candidate failed in both the subjects, then a% b%

Arithmetic

� Total no. of votes =

Price increase/decrease If the price of an article is reduced by a% and buyer gets c kg more for some Rs. b, the new price per

�

c% (i) Passed candidates in both the subjects

kg of article �

b �

r 10%

Total marks =

Ga

p

at io n

pi

If a candidate got A% votes in a poll and he won or defeated by 'x' votes, then, what was the total no. of votes which was casted in poll ? A% 'x'

�Total number of votes

3.

50 � x � A � 50 �

Ex. A candidate got 55% votes in an election and won by 3600 votes. Find total number of votes? 55%

y% 100 � x � 100% 100 � y

If two numbers are respectively x% and y% more than a third number the first as percentage of second is x% y%

If a number 'a' is increased or decreased by b% , then the new number will be 'a' b% � 100 � b � � ��a � 100 �

�

x%

100 � x � 100% 100 � y

Ch

� B.x � G.y � �� �% � B�G �

am

y%

2.

on

In a certain examination, 'B' boy and 'G' girls participated x% of boys and y% of girls passed the examination, then. 'B' 'G' x%

4.

100(80 � 25) 100 � 105 = = 300 35 35

ga n

3.

Pr

25

100 � x

percentage of second is 100 � y � 100%

ic

80

1.

10 � 300 15 3 � � 3 Rs / kg 4 100 � 8 4

Miscellaneous If two numbers are respectively x% and y% less than the third number, first numbe r as a

a Pu t bl a

Ex. In an examination passing marks are 35% . A person got 80 marks and fail by 25 marks. Find total marks?

300

8

New price per kg =

100 � b � c � a

35%

ab 100 � c

Ex. Price of an article is reduced by 10% and buyer gets 8 kg more for Rs. 300. Find new price per kg?

'c' Total marks �

c

Si

= (a + b – c)% In a certain examination passing marks is a% . If any candidate obtains 'b' marks and fails by 'c' marks then, a% 'b'

ab 100 � c

a%

= 100 – (a + b – c)% (ii) Percentage of candidates who failed in either subject

2.

50 � 3600 = 36000 55 � 50

3600

115

Arithmetic

Profit & Loss

�

CP ` 250

25 : 21

Loss% = �

: : : OR 100% :

Profit = ` 40 40 Profit% = ×100% = 16% 250

�

at io n

ic

on pi �

mark - up(rs) ×100% CP

=

MP – CP ×100% CP

=

800 �500 ×100% = 60% 500

If an object is sold on r% Profit. r%

100 � � � S.P. � � 100 � Profit% � �

Similarly, if an object is sold on r% loss, then r%

76 Loss% = × 100 = 16% 475

�

=

�100 � Profit% � then, S.P. = C.P � � or C.P. 100 � �

SP ` 399 Loss = ` 76

Loss% =

Label price/marked price = printed price on a product

Mark-up%

Loss CP – SP × 100 = × 100% CP CP

CP ` 475

13 20

Mark-up = MP – CP = Rs. 300 MP – CP = Rs. 300

Ch

%

7 � Loss Loss = 35% = 20 � CP

Loss% = �

20

Ga

Loss %

7 � �

CP MP Rs.500 Rs.800

am

ga n

� 3 � 23 For SP � ×1.15 or × ����1 � ���� � 20

�

Pr

3 � Profit 20 � CP

: SP 23 SP 115%

� �

a Pu t bl a

40

`100 ��� ×100 = `16 Profit 250

CP 100%

65%

The cost price of an article is Rs. 500 and its mark price is Rs. 800, then find markup percentage ? 500 800

40

`1 ��� Profit 250

CP 20

SP (20 – 7) 13

p

Profit(Rs) SP – CP ×100% = ×100% CP CP

15% Profit � 15% =

7 � Loss 20 � CP

For SP � × 0.65 or × �1 � 20 � =

`250 ��� `40 Profit

�

35% Loss � CP 20 20

SP ` 290

Profit% =

25 � 21 × 100% = 16% 25

r

1.

Cost price (CP) = Total investment for a product. (CP) SP > CP � Profit = SP- CP SP < CP � Loss = CP-SP SP = CP � No profit No loss Profit % %

Si

�

�100 � Loss% � S.P. � CP � 100 � �� or � S.P. � 100 � ��100 � Loss% ��

Loss CP – SP × 100 = × 100% CP CP

CP SP 475 : 399

116

Arithmetic

Ex. On selling 12 articles the profit earned is equal to selling of 4 articles. Find profit% .

Profit% = 3.

ab � � � �a � b � �% 100 � �

x% x%

� 100 � � 100 � �� � paid by A � x � � � 100 � a � � 100 � b �

B

a%

B

C

%

b%

C, x

4.

A

a Pu t bl a

If a% profit and b% loss occur, simultaneously

Pr

ab � � �a � b � �% 100 � �

ga n

(–ve sign for loss, +ve sign for profit) –ve +ve

R C.P. of items � � y � x � �100 '+' = when one is profit and other is loss. '+' '–' = When both are either profit or loss. '–' Ex. A man sells his items at 10% profit. If he had sold for Rs. 40 more then he would have gained 15% profit. Find cost price of item.

pi

Based on number of article 1. If cost price of 'x' articles is equal to selling price of 'y' articles, then Selling Price = x. Cost Price = y 'x' 'y'

am

10%

y

Hence, Profit and Loss% = =

2.

40

Ch

Ga

x

CP =

x �y � 100 y

5.

x �y � 100 y

'x'

y � 100 x �y

Loss% �

y � 100 x �y

40 � 100 �� Rs. 800 (15 � 10)

If a man purchases 'a' items for Rs. x and sells 'b' items for Rs. y, then his profit or loss per cent is given by x 'a' y � ay � bx � � � � 100% � bx �

'y' =

15%

'b'

On selling 'x' articles the profit or loss is equal to selling of 'y' articles, then Profit%

y%

ic

b%

on

a%

R

at io n

ab � � then overall loss or profit% is � a � b � �% 100 � �

x2 % 100

A man sells his items at a profit/loss of x% . If he had sold it for Rs. R more, he would have gained/ loss y% . Then. x%

p

� 100 � � 100 � � x�� �� � � 100 � a � � 100 � b �

2.

If a man sells two similar objects one at a loss of x% and another at a gain of x% , then he always incures loss in this transaction and loss% is x2 % 100

If A sells an article to B at a% profit and B sells it to C at b% profit and if C paid Rs. x, then amount

A

4 � 100 � 50% 12 � 4

r

then Total Profit

4

12

Si

1.

Successive profit/loss Successive Profits : If A sells an article to B at a% profit and B sells it to C at b% profit A B a% B C b% OR If a% and b% are two successive profits a% b%

1.

y � 100 x �y

117

Dishonest Shopkeeper A dishonest shopkeeper sells his goods at C.P. but uses false weight, then his profit =

Gain%

Arithmetic

[Profit or loss as per positive or negative sign).

True weight � False weight � 100 % = False weight Error

or Gain% = True value � Error � 100 (Positive and negative sign conventions are used for profit and loss.)

Ex. If a vendor used to sell his articles at 13% loss but uses 150 gm instead of 200 gm. Then his profit/ loss% is? 13% 200

Ex. A dishonest shopkeeper sells his goods at CP but uses 750 gm weight instead of 1000 gm. Find his profit percent?

200 � � ��100 � 13 � � 150 � 100 � % � �

4.

ab � � � � �a � b � �% 100 � �

If a% loss and b% profit occur then, total loss/profit is a% b%

Pr

ic

ab � � � �a � b � �% 100 � �

pi

on

(negative sign for loss and positive sign for profit)

Ch

am

ga n

y � � ��100 � x � z �100� % � �

Total loss%

a Pu t bl a

If a vendor used to sell his articles at x% loss on cost price but uses y grams instead of z grams, then his profit or loss% is x% y z

Ga

2.

250 1 × 100% = 33 % 750 3

p

����

If a% and b% are two successive losses then (negative sign shows loss and positive sign shows profit). a% b%

Si

3. 1000 � 750 � 100% Gain% = 750

r

4 � � = � 87 � � 100 � % = (116 – 100)% = 16% profit 3 � �

750

at io n

1000

150

118

Arithmetic

Discount

Discount

18

� Discount = MP – SP � Always calculated on marked price

2 � 25 �100 = 8% profit

� Can not be greater than 100%

100% CP SP MP 500 700 800 Discount = MP – SP = 100

r

Si

Discount ×100% MP MP – SP ×100% MP

Equivalent discount of 10% , 8.33% & 18 3.

2.

If shopkeeper does not allow any discount

3.

MP = SP � Mark-up% = Profit% SP = CP + Profit = MP - Discount 700 = 500 + 200 = 800 - 100

4.

CP

CP 100 – d% = MP 100 + p%

xy � yz � zx xyz � 100 10000

� 35 � 56 � 40 �� 5 � 7 � 8 �� = 20 – ���� �� 100

10000

= 20 – 1.31 + 0.0280 = 18.718% Free Article �

Successive Discount 1.

� x% , y% , z%

Ex:- 5%, 7% , 8% Net discount

am

M �D�

Ga

�

�� % P% = ����M � D � 100 ��

Successive discount

= (x + y + z) –

Net effect (successive) of mark up% and discount% is profit %

Ch

5.

ga n

SP = CP × (100 + P)% = MP × (100-d)% �

100

Net discount

on

-d discount

4.

pi

+P% Profit

MP

32.5% Successive discount � x% , y% � xy �� Net discount- ����x � y � ��� %

Pr

SP

2 % = 11

p

100 �100% = 12.5% � 800

ic

=

2 –18 % 11 MP –10% –8.33% 100 DI 90 DII 82.5 DIII 11 9 × 90× 12 11 D = 32.5% = 67.5

at io n

Discount%=

2.

a Pu t bl a

1.

2 �2 %= 11 11

2 –10% –8.33% –18 11 % 160 100 Markup MP DI 144 DII 132 DIII 108 CP +60%

Buy 4 get 1 free 4 1 Article MP = Rs. 1 D=1 MP =5 D%=

1

:-

1 ×100 =20% 5

Note: Any kind of Discount is calculated only on marked price and not on selling price or cost price.

8% profit �3 5 �1 10% = 10 �1 8.33% = 12

'y' articles (quantity/number) are given free on purchasing 'x' articles. Then, 'x' 'y'

60% =

Discount%

119

�

x � 100 x �y

Ex. If 2 articles are given free on purchase of 8 articles then find discount % . 8

30% 15%

2

2 Discount % = (2 � 8) × 100% = 20%

Miscellaneous If article is sold on D% discount, then D% SP �

MP �100 � D� 100

3.

.

The marked price of an article is fixed in such a way that after allowing a discount of r% a profit of R% is obtained. Then the marked price of the

Si

R%

� r1

Pr Ch

am

pi

on

Ex. A shopkeeper marks his goods 30% above the cost price. If he allows a discount of 15% on the mark price. Then his profit or loss percent is?

ga n

20% 30%

ic

(Positive sign signifies profit and negative sign signifies loss).

Ga

� r +R � ×100 � % � 100 r � �

Ex. The MP of an article is fixed in such a way that after allowing a discount of 20% a profit of 30% is obtained. Then mark-up percent is?

a Pu t bl a

100

255 � 15 ��25.5 – 15 = 10.5% profit 10

p

r1% r � �100 � r1 �

�

r

SP � 100 100 � D A tradesman marks his goods r% above his cost price. If he allows his customers a discount of r1% on the marked price. Then the profit or loss percent is

r%

30 � (100 � 15) � 15 100

� r�R � � 100 � % more than its cost article is � � 100 � r � price. r%

MP �

2.

�

at io n

1.

Arithmetic

120

� 20 � 30

�

Mark-up% = � (100 � 20) � 100 � % � � =

50 × 100% = 62.5% 80

Arithmetic

Simple interest �

Important Points �

Borrowed money is called Principal and it is denoted by 'P'. 'P'

�

Money is borrowed for certain time period, that time is called interest time and it is denoted by 'T' or 't'.

If there are distinct rates of interest for distinct time periods i.e.

Rate for 1st t1 years � R1%

A So, Amount = Principal + Interest � A = P + S.I. � A = P + S.I.

r

�

OR Interest = Amount – Principal ��S.I = A – P ��S.I = A – P When Interest is payable half – yearly

ga n

When interest is payable quarterly

on

Rate will be half and time will be twice

or, S.I �

pi

T�

S.I � 100 R�T

R�

S.I � 100 P�T

S.I �100 P �R A = P + S.I or, S.I. = A – P

� n � 1� � 100% and, T

� n � 1� � 100% R

Ex. If a certain sum becomes 3 times of itself in 4 years on simple interest. Find rate percent per annum. 4

P �R � T 100

P�

T�

am

Principal � Rate � Time 100

Ch

SI =

Ga

�

Rate of Interest

Interest on Rs. 100 in one year is called Rate of Interest. 100 If a certain sum becomes 'n' times of itself in T years on Simple Interest, then the rate per cent per annum is. T 'n'

R% �

Rate will be one-fourth and time will be four times. Simple Interest (SI)

100

ic

�

Pr

�

P � R1t1 � R 2t 2 � R 3t 3 �

p

�

a Pu t bl a

�

Si

'T' 't' The principal becomes Amount when interest is added to it Amount is represented as A.

at io n

�

t1 � R1% Rate for 2nd t2 years � R2% t2 � R2% Rate for 3rd t3 years � R3% t3 � R3% Then, Total S.I. for 3 years S.I.

R= �

�

121

3

(3 � 1) × 100% = 50% 4

If Simple Interest (S.I.) becomes 'n' times of principal i.e. (S.I.) 'n' S.I. = P × n then. RT = n × 100 If an Amount (A) becomes 'n' times of certain sum (P) i.e. (A) (P) 'n' A = Pn then RT = (n – 1) × 100

If the difference between two simple interests is 'x' calculated at different annual rates and times, then principal (P) is 'x'

Arithmetic

�

(P)

�

x � 100 � diff. in rate� � �diff. in time�

T = no. of years A = amount �

x2

� Diff in amount � � 100 � Change in interest Rate � � time

�

S.I. � 2400 n � n � 1� � � deposited amount �

where n = no. of months The difference between the S.I. for a certain sum P1 deposited for time T1 at R1 rate of interest and another sum P2 deposited for time T2 at R2 rate of interest is R1 T1

and

a Pu t bl a

A.t2 � B.t1

P1

at io n

At2 � Bt1 t2 � t1

on

Pr

ic

S.I. �

Ch

am

ga n

P�

� B � A � � 100

r�

pi

R% �

r = Rate of Interest To find the rate of interest under current deposit plan,

p

If a sum with simple interest rate, amounts to 'A' in t1 years and 'B' in t2 years, then. t1 'A' t2 'B'

Ga

�

A � 200 T ��200 � � T � l � r ��

where

If a sum amounts to x1 in t years and then this sum amounts to x2 in t yrs. Then the sum is given by t x1 t

P�

�

Equal installments

r

P�

Installment If a sum is to be deposited in equal installments, then

Si

�

122

R2 P2

P2R 2T2 � P1R1T1 100

T2

Arithmetic

Compound interest Compound interest is the interest you earn on interest If rate of interest in first and 2nd cycle are x% & y% respectively x% y%

7�8

5.

� xy �� � CI2 = ���x � y � �� % 100

If rate is same �

= r%

R = 10% p.a =

r2 � �

� � CI2 = ��2r � 100 ��� %

5% 10.25% 0.25% 10% 21% 1% 15% 32.25% 2.25% 20% 44% 4% 25% 56.25% 6.25% 30% 69% 9% Rate = x% , y% , z% for 3 cycles. �

�

4.

ga n

1.

×1.1

CIII 1100

×1.1

11 or 1.1 10

A3 13310 ×1.1

CIIII 1210 ×1.1

Amount & CII, CIII, CIIII G.P Ratio of CI of 2 years � 2 : 1 3 years � 3 : 3 : 1 4 years � 4 : 6 : 4 : 1 5 years � 5 : 10 : 10 : 5 : 1 Some important points If A = Amount P = Principal r = Rate of Compound Interest (C.I.) (C.I.)

am

P = Rs. 4000 R = 2% , 3% , 5% CI = (10 + 0.31 + 0.0030)% = 10.313% � 4000 × 10.313% = Rs. 412.52 If r% for 3 consecutive years r% , r% , r%

t = no. of years

Ch

Ga

�

�

A2 12100

Si

6.

�

�

CII 1000

a Pu t bl a

Difference between CI and SI

�xy + yz + zx � xyz � + CI = ��� x + y + z � + 100 10000 � % �

×1.1

on

3.

r2 % 100

pi

Rate

� r 2 �� � CI2 ��2r � 100 ��� � �

A1 11000

p

Difference between CI & SI for 2 years =

Pr

2.

P 10000

�

ic

�

1 10

r

�

CI2 - SI2 � % = 0.56% 100 7500 × 0.56% � Rs. 42 P = 10.000 T = 3 years

at io n

1.

P = Rs. 7500 R = 7% , 8%

�

then

, t

� 3r2 r3 �� � CI3 = ��3r � 100 � 10000��� % � �

r � � A � P �1 � � , C.I. � A � P 100 � �

P = Rs. 10, 000 R = 7% p.a. CI3 = ? Eff. Rate = (21 + 1.47 + 0.0343)% = 22.0543% � CI3 � 10, 000 × 22.5043% = Rs. 2250.43 Rate CI 3 CI3 - SI3 5% 15.7625% 0.7625% 10% 33.1% 3.1% 15% 52.0875% 7.0875% 20% 72.8% 12.8% 25% 95.3125% 20.3125% 30% 119.7% 29.7%

t �� � r � C.I. � P ��1 � � � 1� 100 � �� �

2.

Compound interest is calculated on four basis:

Annually Half-yearly Quarterly Monthly

123

Rate r% r % 2 r % 4 r % 12

Time(n) t years t × 2 years t × 4 years t × 12 years

3.

Arithmetic

If there are distinct 'rates of interest' for distinct time periods i.e.

�B � (i) If b – a = 1, then R% � � � 1� ×100% A � � � B

� r1%

Rate for 1st year

� r3% and so on

Rate for 3nd year Then

where n is a whole number. n If a sum becomes 'n' times of itself in 't' years on

8.

r �� r �� r � � A � P �1 � 1 � �1 � 2 � �1 � 3 � ... � 100 � � 100 � � 100 �

r

� 1t � � compound interest, then R% � n � 1� × 100% � �

C.I. = A – P If the time is in fractional form i.e.,

Si

4.

1 � � � B �n � � � � 1 (iii) If b – a = n then, R% ��� A �� � × 100% � �

� r2%

Rate for 2nd year

�

p

t×n

at io n

9.

on

�

am

Ch

Ga

�

�

�

�

�

3 – 1 ×100%

P � 100 � � 100 � � ��� � � 100 � r � � 100 � r �

2

(ii) For n = 3. Each annual installment n =3

81 = 3 � Time = 5 × 4 = 20 years The difference between C.I. and S.I. on a sum 'P' in 2 years at the rate of R% rate of compound interest will be C.I. 'P' 2 S.I.

�

P 2

� 100 � � 100 � � 100 � � ��� � �� � � 100 � r � � 100 � r � � 100 � r �

3

10. The simple interest for a certain sum for 2 years at an annual rate of interest R% is S.I., then R% S.I.

2

R � S.I. � R � � 200 � 100 �

For 2 year,� C.I. � S.I. � P ��

R � � C.I. � S.I. �1 � � 200 � �

11. A certain sum at C.I., becomes x times in n1 year

2

R � � R � � For 3 years, C.I. – S.I. � P � � � �3 � � 100 100 � � � �

1

If on compound interest, a sum becomes Rs. A in 'a' years and Rs. B in 'b' years then, 'a' A ‘b’

1

(i) for n = 2. Each annual installment n =2

3

R%

7.

�

��

pi

ga n

81

3

2 R% = � 3 � 1� � 100%

Ex. A certain sum becomes 3 times of itself in 5 years on compound interest. Then the time it will take to becomes 81 times of itself is � 5

2 R

ic

Pr

A certain sum becomes 'm' times of itself in 't' years on compound interest then the time it will take to becomes mn times of itself is t × n years. 'm' 't' mn

�

Ex. If a sum becomes 3 times of itself in 2 years on compound interest then R% is �

a Pu t bl a

3

r � � r 5� � A � �1 � � � � �1 � � 100 � � 100 7 �

6.

'n'

�

1

�

n

r � � rF � 5 � A � P �1 � � �1 � � e.g. t � 3 yrs, then 7 � 100 � � 100 �

4

't'

t R% = � n -1� × 100

t = nF, then

5.

�

(ii) If b – a = 2, then R% � �� A � 1�� × 100% � �

B

124

1

and y times in n2 years then x n � y n . 1

n1 y

1

1

x n1 = y n2

2

x

n2

Arithmetic

Ratio & Proportion Concept of Ratio �

(i) Part of A �

The comparative relation between two amounts/ quantities of same type is called ratio.

m �R m�n

(ii) Part of B �

Second : Second etc. Let an amount be x and another is y, then the ratio between them is x : y or x ÷ y. x y x:y

(i) Part of A �

5.

Then

where m > n If the ratio of alligation of milk and water in a glass is m : n and in other glass alligation is p : q, then the ratio of milk and water in third glass which contains alligation of both glasses is m:n

� m p � � n q � Ratio � � � � �:� � m � n p � q m � n p �q� � � �

If

a c e a � c � e � ... � � � ...... then each ratio � b d f b � d � f � ...

Definition of different types of ratio

What should be added to all of a, b, c, d (numbers) so that these become proportional respectively? a, b, c, d x

m�n �R m�n

p:q

6.

a a�c � = ac : bc = a : b b b�c

Let x should be added

3.

Ch

Ga

2.

at io n

(iii) The sum of parts of A and B �

� Product of extremes = Product of means. i.e., ad = bc It does not change the ratio, when we multiply or divide antecedent and consequent of the ratio by a same non-zero number as-

e.g. a : b =

n �R m�n

ic

'y' If a : b : : c : d, then a and d are called extremes and b and c are called means. a:b::c:d a d b c

m �R m�n

(ii) Part of B �

a Pu t bl a

x÷y In ratio 1st number i.e. 'x' is called "antecedent" and 2nd number i.e., 'y' is called "consequent". 'x'

r

: Hour

Si

Hour

m–n �R m�n

where m > n If the ratio of A and B is m : n and the difference in their share is 'R' units then, A B m:n 'R'

p

: kg

am

1.

kg

4.

on

�

: Rupees

pi

�

Rupees

(iii) Difference of part of A and B �

Pr

�

Ratio always occurs between units, as-

ga n

�

n �R m�n

:

1.

2.

a�x c�x :: b�x d�x

If an amount R is to be divided between A and B in the ratio m : n then R A B m:n

125

Mixed ratio – Let x : y and P : Q be two ratios, then Px : Qy is called mixed ratio. x:y P:Q Px : Qy Duplicate ratio – The mixed ratio of two equal ratios is called the duplicate ratio as

duplicate ratio of a : b is a2 : b2 a:b a2 : b2

The subduplicate ratio of a : b = a:b

Triplicate ratio – The cube of a certain ratio is called triplicate ratio.

a :3b

r

a�b c�d � a �b c �d To simplify the proportion any one method of componendo , divi dendo, compo nendo and Dividendo can direcly be used.

ic

Concept of Proportion

1.

am

Directly Proportional: If x = ky, where k is a constant, then we say that x is direcly proportional to y. If it is written as x � y. x = ky k

a c � b d

2.

k

Inversely Proportional: If x � y where k is a constant, th en we say that x is inverse ly 1

a b � c d

proportional to y. It is written as x � y .

a : b :: c : d a : c :: b : d 9.

x�y

x y

Ch

Ga

alternendo is a : c :: b : d. i.e alternendo of

a c � b d

a �b c �d � b d

(a + b) : (a – b) :: (c + d) : (c – d) or,

a : b = c : d is b : a = d : c

a c b d � � means then b d a c Alternendo – If a : b :: c : d is a proportion then its

is

�

Pr

ga n

Invertendo – The proportion in which antecedent and consequent quantities change their places, is called invertendo, as -

Invertendo of

a c a c � � �1 � �1 b d b d

Si

a :3b

1 1 �1 1� : or � � : � × (L.C.M of a and b) a b �a b�

,

p

3

It means

11. Componendo and dividendo – If there is a proportion a : b :: c : d then its componendo and dividendo is – a : b :: c : d

Inverse ratio – The reciprocal of quantities of ratio is called its inverse. Reciprocal or inverse ratio of a:b

�

8.

(a – b) : b :: (c – d) : d

at io n

=

3

a:b

7.

c a � b c � d� �a � or, � � 1 � � 1 � b d b d �� � 10. Dividendo – If a : b :: c : d is a proportion, then its dividendo is (a – b) : b :: (c – d) : d a : b :: c : d

The triplicate ratio of a : b = a3 : b3 a:b = a3 : b3 Subtriplicate ratio – The cube root of a certain ratio is called subtriplicate ratio as -

a:b 6.

a: b

a: b

The Subtriplicate ratio of a : b =

a �b c�d � b d

a Pu t bl a

5.

=

a c a�b c�d � � then, b d b d

It means, If

a c � b d

on

4.

Arithmetic

Subduplicate ratio – The square root of a certain ratio is called its subduplicate.

pi

3.

x�

a b � c d

x y

Componendo – If a : b :: c : d is a proportion, then componenedo is (a + b) : b :: (c + d) : d a : b :: c : d (a + b) : b :: (c + d) : d

3.

126

k y

k x�

1 y

Proportion: When two ratios are equal to each other, then they are called proportional as

4.

a : b = c : d, then, a, b, c and d are in proportion. a : b = c : d, a, b, c d or, a : b :: c : d E.g. 2 : 5 = 6 : 15, then we write 2 : 5 :: 6 : 15 Mean Proportion – Let x be the mean proportion between a and b, then a : x :: x : b (Real condition) x, a b a : x :: x : b

7.

x : a :: b : c �

x b � � cx � ab a c

�x �

8.

a x � � � x 2 � ab x b

Arithmetic

First Proportional – Let x be the first proportional of a, b and c. then, x : a :: b : c (Real condition) x a, b c

ab c

Let 'x' be a number which is subtracted from a, b, c and d to make them proportional then 'x' a, b, c d

� x � ab x�

p

ab a�b

x�

b a

�a

b

at io n If

a�b c�d = b d

(iv)

a �b c�d = b d

(v)

1.

bc a

a�b c�d = a �b c�d

If 3A = 4B = 5c find A : B : C step1: Take LCM of (3,4,5) = 60 �A=

� Fourth proportional of a, b and c � � a, b

a b = c d

How to find ratio

a c � � ax � bc b x

�x �

Law of Ratios

(iii)

a : b :: c : x

�

c

d

a c = then b d

(ii)

Fourth Proportional – Le t x be the fourth proportional of a, b and c, then a : b :: c : x (Real condition) x a, b c

Ga

6.

b2 � a

a, b, c

(i) ad = bc

Ch

b2 �Third proportional of a and b � a

�

pi

ga n

�x �

2

Here a, b, c and d should always be in ascending order.

on

a b i.e. � ��ax = b2 b x

bc � ad � a � d � � � b � c�

ic

Pr

Th ird proporti on al – Le t 'x' be the thi rd proportional of a and b then, 'x' a b a : b :: b : x (Real condition)

am

5.

Let 'x' be a number which is added to a, b, c and d to make them proportional, then 'x' a, b, c d

a Pu t bl a

x�

Si

a b � ab If x be the mean proportion between (x – a) and (x – b) then what will be the value of x ? x (x – a) (x – b) x

ad � bc � a � d � � � b � c�

r

So, mean proportion of a and b � ab

60 = 20 3

� A : B : C = 20 : 15 : 12

bc a

bc � a

127

B=

60 = 15 4

C=

60 = 12 5

2nd Method 3A=4B=5C A:B:C = 4×5 : 3×5 : � � forA forB multiply the multiply the coeff. of B & C coeff. of A & C � 20 : 15 : 12 If A:B = 4:5,

3×4 � forC multiply the coeff. of A & C

B:C = 2:5 then find A:B:C = ?

A: B :C 4: 5

1.

If

Arithmetic

a 7 � b 3

then

5a � 3b 5 � 7 � 3 � 3 44 = = 7a – 4b 7 � 7 – 4 � 3 37

2.

2a 2 � 3b2 98 � 27 �25 = = 7 a 2 � 4ab 49 � 84

3.

3a 3 � 4b2 = can not be determined because degree 5a 2 � 2b3

2 :5 8 : 10 : 25

of each term is not same. 3a 3 � 4b2 = 5a 2 � 2b3

B

A : 4×2

B : C : 5×2 2×5 : 5×5

8 :

10 :

Si

B

r

Second methtod:-(B is common, so make B equal)

Note:To solve this type of equations degree of each term should be same.

on pi

Ch

am

ga n

Pr

ic

at io n

a Pu t bl a

p

25

Ga

2.

Concept of Degree

128

Arithmetic

Mixture & Alligation Concentration �

Concentration of milk

�

Concentration is the percentage of a particular quantity in the full mixture

3 Left pure milk � 4 �� �� � = 0.512 = �� 5 �� Total mixture

Ex. In a mixture ratio of milk; water = 60 : 40 2.

= 60 : 40 � Concentration of milk

=

r

Si

–40

= 11 : 5 Then concentration of milk

p

5 ×100% = 31.25% 16

at io n

=

–80L mix +80L water =

Pr

or concentration of water = 100 – concentration of milk

on

pi

am

ga n

Ch

Ga

��1

left quantity of pure milk after 3 times 3 � 400 ×

�1 5

+80L

4 4 4 � � 5 5 5

:

�1 5

252 � �1

400L 252L If from x litre of liquid A, p litre is withdrawn and same quantity of liquid B is added. Again from mixture, q litre mixture is withdrawn and same quantity of liquid B is added. Again from mixture, r litre is withdrawn and same quantity of liquid B is added, then x p B

q B

3

� 4� � 400 ��� ��� �� 5 �

r B

� 400 � 0.512 � 204.8L

water in final mixture 400–204.8 = 195.2L

=

9 7 4 × � = 252L 10 8 5

400

3.

�80L pure milk � � � �80L water �

�1 8

water = 400 – 252 = 148L 2nd method :10 : 9 8 : 7 5 4

�80 �1 � = part of 20% removed each time 400 5

20%

=

Left Quantity of pure milk = 400×

Replacement of mixture When same quantity is replaced each time from 400 L Pure Milk. 400

�1 10

�1 8

ic

–80

= 100 – 68.75% = 31.25%

=

+50L

a Pu t bl a

–50

Concentration of water

1.

40

–50L mix +50L water =

11 ×100% = 68.75% 11 � 5

= 100 –

�1 10

–40L pure milk +40L water =

If ratio of milk : water = 11 : 5

=

= 51.2% Concentration of water = 100% – 51.2% = 48.8% When different quantity is replaced each time. 400L pure milk

60 × 100% = 60% 60 � 40

E.

=

In final mixture, liquid A is

=

� x � p�� x � q �� x � r � x� �� �� � ......... � x �� x �� x �

129

A

Arithmetic

If only one process is repeated n times, then liquid � x � p� � � x �

n

p�

�

A in final mixture is � x �

n

or �1 � � and x� � liquid B in final mixture = x – (liquid A in final mixture) n n

p� � �1 � � x� �

B

=x– A If x is initial amount of liquid. p is the amount which is drawn, and this process is repeated ntimes such that the resultant mixture is in the

: D–M Cheap Quantity

n

p

2.

n a � x �p� �� � a�b � x �

a:b

unit.

% increase/decrease (Male)

a Pu t bl a

Pr

on

�100 � y �

Ch

x%

3.

a �x � y � y

Initial number of males

6500 �

y%

Overall increase in population =

500 50 ×100 = % 6000 6

Male 5%

unit.

Female 9% 50

�

Initial number of females

9%

The amount of acid/milk is x% in 'M' litre mixture. How much water should be mixed in it so that percentage amount of acid/milk would be y% ? x% 'M' y% Amount of water

% increase/decrease (overall population)

6000

�Required quantity of vapourised water �

% increase/decrease (female)

Ex. The population of a town is 6000. If males increase by 5% and females increase by 9% then population will become 6500 after 1 year. Find the initial ratio of males and females?

am

There is x% water in 'a' unit the mixture of sugar and water. The quantity of water vapourised such that decrease in the percentage of water is from x% to y% is given by 'a' x%

Ga

Alligation in population related questions

pi

a �x � y �

ga n

�

M–C Dear Quantity

n

Increase/decrease of mixture 1. There is x% milk in 'a' unit mixture of milk and water. The amount of milk that should be added to increase the percentage of milk from x% to y% is given by 'a' x% x% y% Required quantity of milk

Mixture (M)

p

x

Dear(D)

r

a � x � p� �� � a�b � x �

Cheap (C)

at io n

ratio a : b then,

2.

Cheap object D � M � Dear object M�C

n

ic

4.

M

Si

�x �p� � �

A � x� � x

1.

Alligation The cost of cheap object is Rs. C/kg and the cost of dear object is Rs. D/kg. If the mixture of both object costs Rs. M/kg then C D

6

:

50 �5 6

M �x � y �

4 6

:

20 6

y

1

:

5

130

9�

50 6

%

5% 1

3.

Arithmetic

Alligation in income related questions:

Profit/loss% (A)

5.

Profit/Loss% (B)

overall profit/loss%

% increase/decrease in expenditure

% increase/decrease in savings

Ex. First watch is sold at 10% profit and 2nd watch is sold at 15% profit and the overall profit on both the watches is 12% if cost price of first watch is Rs. 360 Find cost price of second watch ? 10%

Ex. A man spends 75% of his income. If his income is increased by 20% and expenditure increased by 10% . Then find % change in savings.

12% 360

r

75% 10%

Si

20%

Watch 1 Watch 2 10% 15%

% 3 � spends � Exp: saving = 3 : 1 4 � income

p

ga n

x – 20 = 30 x = 50%

cost price A

cost price B

45

B

Rate of Interest % (B)

10,000

5%

x

10%

8% 5%

Type B 45 Rs/Kg

: : :

Alligation in simple interest

Ex. Rs. 10,000 is lent at 5% per annum simple interest and Rs. x is lent at 10% p.a. If overall rate of interest is 8% then find value of x.

x 10%

8%

39 Rs/Kg 45–39 6 2

:

:

39 Type A 36 Rs/Kg

overall discount%

Total rate of interest (A+B)

Ex. Cost price of type A sugar is 36 Rs./Kg and type B sugar is 45 Rs/Kg. In what ratio these type of sugar should be mixed to get a mixture worth 39 Rs./Kg. 36

Discount% (B)

Rate of Interest %(A)

Ch

Ga

cost price of mixture

A

7.

pi

Alligation in profit loss

am

4.

Discount%(A)

6.

x � 20 3 = 10 1

Ratio of cost price 2 ×120 240 Rs.

at io n

10 1

: :

Pr

�

: 3 ×120 360 Rs.

ic

x – 20 3

Savings x%

a Pu t bl a

Expenditure 10%

12%

on

75% =

15%

: 2 ×5000

39–36 3 1

Ratio of principal 3 ×5000

Rs. 10,000 Rs. 15000

131

8.

Arithmetic

Alligation in Average Average 1

11. speed 1

Average 2

speed (Avg)

Total average

9.

Bowling Average (till innings A)

speed 2

Time 1 : Time 2

Speed=

Bowling Average (till innings B)

dis tan ce time

Note � Mean value

Total average

�

:

P%

L%

P%/L% CP2

Si

Average speed (for part 1)

CP1

r

10. Alligation in time and distance Average speed (for part 2)

P% =

p

Average speed (whole journey)

P ×100 CP

on pi

Ch

Ga

am

ga n

Pr

ic

at io n

a Pu t bl a

:

132

respect

ratio

Arithmetic

Partnership

simple partnership

compound partnership

If all partners invest different capital for same time period or same capital for different time period

If all partners invest their different capitals for different time period

�

Profit = Capital × time P=C×T C=

P T

T=

P C

Profit and capital × time

r

�

Partnership

p

I 1, I 2, I 3, I 4 �

am

Ch

Ga

salary from the profit.

on

who only invests money and does not take part in business activities.

ga n

invests money as well as takes part in business activity for which he is paid

Sleeping Partner

pi

Active Partner

Pr

Partners

ic

at io n

a Pu t bl a

T1, T2, T3, T4 �

�

×

Profit = I1T1 : I2T2 : I3T3 : I4T4 where I 1, I 2, I 3, I 4 � Investments by different persons T1, T2, T3, T4 � Time

Si

�

133

Arithmetic

Average Concept of Deviation �

Find Average of

40, 42, 35, 50, 85

�

Traditional method is to add all the numbers and divide by the number of observations but this method is lengthy and calculative. To avoid calculation and save time we solve it by concept of deviation.

r

Step I : Consider any number in the range of these numbers as average.

Si

I Step II : Find the difference of average from each number (deviation)

II Step III: Add the deviation and divide it by total number of observations.

40,

42,

35,

50,

85

Deviation:

0

+2

–5

+10

+45

=

Net deviation � Actual Average

Pr

Let Average = 40

ca

IV Example:

ti on

a Pu t bl a i

p

III Step IV: Add or subtract (according to sign of deviation) the deviation from the average that we considered to get accurate average.

0 � 2 � 5 � 10 � 45 52 = = +10.4 5 5

= 40 + 10.4 = 50.4

2.

= × Average of two or more numbers/ quantities is called the mean of these numbers, which is given by

pi

am

Ga ga

on

Sum of the observation = Average × total number of the observation

n

1.

Ch

Sum of all observation

Average (A) = Total no. of all observation Example:- Weight of 60 students = 40, 42, 35, 50.... Total weight = 2400kg Average = 3.

If the given observation (x) are occuring with certain frequency (A) then, (x) (A) Average =

4.

2400 = 40 kg/student 60

A1x1 + A 2 x 2 + ... + A n x n x1 + x 2 + .... + x n

where A1, A2, A3, ...... An are frequencies If the average of 'n1' numbers is a1 and the average of 'n2' numbers is a2, then average of total numbers n1 and n2 is, Average = 'n1'

n1a1 + n 2a 2 n1 + n 2

a1

'n2'

a2

134

n1

n2

n1a1 + n2a2 n1 + n2

Ex. If average of 10 numbers is 24 and average of 5 numbers is 15. Find the combined average?

Combined average =

10 � 24 � 5 � 15 315 = = 21 10 � 5 15

No. of data � n1 n 2 n 3 n 4 Average � a1 a 2 a 3 a 4 �

Example: � Class A No. of Students � 9 Average weight � 53 kg

B 17 : 59 kg

:

C 14 64 kg

OR, Deviation Method : Student � 9 : 17 : 14 Avg wt. 53 59 64kg � (–6kg×9) 0 (+5kg ×14) Let Average weight of all classes

��54 � 70�

= 59 kg

a Pu t bl a i

= 59 kg +

Si

9 � 53 � 17 � 59 � 14 � 64 2376 = = 59.4 9 � 17 � 14 40

p

Avg.wt. of all class �

n1a1 � n2a 2 � n3n3 � n 4a 4 n1 � n2 � n3 � n4

40

= 59 + 0.4 = 59.4 kg

n

n(n+1) n +1 2 = n 2

on

Pr

The average of 'n' consecutive natural numbers starting from 1 i.e. Average of 1,2,3,....n 1 'n' 1,2,3, n sum of first n natural no. = n

pi

The average of squares of 'n' consecutive natural numbers starting from 1 i.e. 1 'n'

Ga ga

2.

ca

Average of consecutive numbers 1.

r

Net avg/weighted avg

ti on

5.

Arithmetic

2

2

2

Ch

3.

2

am

n(n+1)(2n+1) � n +1�� 2n +1� 6 Average of 1 , 2 , 3 , 4 , .... n = = n 6 2

The average of cubes of first 'n' consecutive natural numbers i.e. Average of 13, 23, 33, .... n3 'n' 13, 23, 33, .... n3 2

� n(n+1) � 2 n � n � 1� � � =� 2 � = 4 n

4.

n(n+1) = (n + 1) n

The average of first 'n' consecutive even natural numbers i.e. Average of 2, 4, 6,.... 2n = 'n'

2, 4, 6, .... 2n

=

n(n+1) = (n + 1) n 2

5.

The average of first 'n' consecutive odd natural numbers i.e. 1, 3, 5, ... (2n – 1) = 2

'n'

1, 3, 5, ... (2n – 1) =

135

n =n n

n =n n

Arithmetic

6.

a+n First no.+Last no. = 2 2

The average of certain consecutive numbers a, b, c, .... n is

a�n First no.+Last no. = 2 2

a, b, c, .... n Ex. Find average of 4, 5, 6 .............. 20.

Average =

The average of 1st 'n' multiples of certain numbers x 10

First 10 multiples of 3 = 3, 6, 9 ..... 30 3

10

3(1 � 10) 33 = = 16.5 2 2

Average of square of 1st n even number

9.

Average of cube of 1st n even number

n

10. Average of square of 1st n odd number

n

n

Pr

11. Average of cube of 1st n odd number

ti on

a Pu t bl a i

n

=

p

8.

ca

Average =

Si

Ex. Find average of first 10 multiples of 3 3

1

x �1 � n � 2

n

2n � n +1�� 2n +1� 3

= 2n(n+1)2 =

n � n +1�� 2n – 1� 3

= n(2n2–1)

=

Last odd no. � 1 2

=

Last even no. � 2 2

am

Ch

2xy

pi

Average speed If A goes from P to Q with speed of x km/h and returns from Q to P with speed of y km/h, then the average speed of total journey is A P Q x Q P y

Ga ga

1.

1

n

13. Average of 1 to n even number

n

on

12. Average of 1 to n odd number

�

'n'

r

7.

4 � 20 24 = = 12 2 2

total distance

Average speed = x + y = total time taken 2.

If a distance is travelled with three different speeds a km/h, b km/h and c km/h, then Average speed of total journey =

3abc km / h ab + bc + ca

a

b

c

3abc km/h ab � bc � ca

Ex. A particular distance is travelled with 2 km/hr, 3 km/hr and 4 km/hr. Find average speed of the whole journey.

Average speed =

3�2�3� 4 72 = 2� 3 � 3� 4 � 4� 2 26

=

36 km / hr 13

136

1.

Average age 't' years before, the average age of N members of a family was 'T' years. If during this period 'n' children increased in the family but average age (present) remains same, then. Present age of n children = n.T – N.t N ‘T’ 't'

5.

Arithmetic

If in a group, one member is replaced by a new members, then. Age of new member = (age of replaced member) � xn where, x = increase (+) or decrease (–) in average n = Number of members.

'n' � xn

x 6.

't'

�x (n + 1). where x = increase (+) or decrease (–) in average age (or income) n = Number of members. x

Then, the age of the new person

N

'T'

on pi

Ch

If the average age of the group decreases by 't' years, then Average age of new students 't'

When change in data happens 1.

�N � � T � � � 1� t �n �

4.

'y'

(cms) 'z'

(cms)

If in any series having common difference 'd' and Average 'k', 'x' numbers are added in forward or backward, then d k x

If the average age (height) of 'n' persons is x year (cms) and from them 'm' persons went out whose average age (height) is 'y' years (cms) and same number of persons joined whose average age (height) is 'z' years (cms) then what is the average age (height) of n persons ? 'n' x (cms) 'm'

ti on

n

am

n

Ga ga

�N � � T � � � 1� t �n �

� x(n – 1) where, x = increase (+) or decrease (–) in average income (or age) n = Number of members. x

'n'

't'

n If a member leaves the group, then income (or age) of left member = Average income (or age) � x(n – 1)

ca

�N � new students � T � � � 1� t �n �

7.

Pr

3.

a Pu t bl a i

p

= T + N.t If the average age decrease by 't' years after entry of new person, then the age of the new person = T – N.t 't' = T – N.t The average age of a group of N students is 'T' years. If 'n' students join, the average age of the group increases by 't' years, then Average age of

n If a new member is added in a group then. age (or income) of added member = Average age (or income) � x (n + 1).

r

n = n.T – N.t If in the group of N persons, a new person comes at the place of a person of 'T' years, so that average age, increase by 't' years N 'T'

Si

2.

=k�

New Average 2.

xd 2

In series of even or odd having Average "k", when we add "x" number in forward or backward, Then k x

3.

n

m �y � z � � � � years � cms � � Average age = � x � n � �

New Average = k �x In series of natural number having Average "k", when we add "x" number in forward or backward, Then k "x" New Average

137

= k � x/2

5.

If average of n observations is a but the average becomes b when one observation is eliminated, then value of eliminated observation = n(a–b)+b n a b =n(a– b) + b If average of n observations is a but the average becomes b when a new observation is added, then value of added observation = n(b–a) + b n a

3.

(n + 1)

b

n

nb

Pr

Si

2.

'm'

pi

am

Ch a, b

Second number

�a � b� n

If the average of n numbers is m but later on it was found that two numbers a and b misread as p and q. n m b

p

q

The correct average

c

= m�

� a +b + c � � =k 2 � �

1.

= �

= 2b – k

n

If the average of n students in a class is a, where average of passed students is x and average of failed students is y, then n a x

y

Number of students passed/ n �a � y �

= x �y � � 2.

Bowling Average

=

Total runs given Total wickets taken

y 3.

(i) Average of remaining numbers

�a � b � p � q �

Miscellaneous

Third number = 2c – k If the average of m numbers is x and out of these 'm' numbers the average of n numbers is y. (or vice ve rsa) the n the ave rage of re maining numbers will be m x 'm' n

mx � ny m�n

'b'

a

= 2a–k

First number

=

'y'

'a'

�m�

Related to numbers If there are 3 natural numbers and average of any two number when added with third number gives a, b, c. Then natural numbers. 3

Sum of number

2.

'l'

'z' mth observation = m(y + z) – nx (m + 1)th observation = nx – m(y + z) When data is misread 1. If average of n numbers is m but later on it was found that a number 'a' was misread as 'b'. The correct average will be n m

on

n

Mathe matical ope ration performed on each observation results in same effect on the average.

Ga ga

1.

'f'

ti on

, a = a1 + a2 + a3 + ....

Note: In this formula, the signs of '+' and '–' depend upon the increment or decrement in the average '+' '–' 7.

'L'

ca

Where/

4.

'F'

f – l = n(F – L) If the average of 'n' observations is 'x' and from these the average of 1st 'm' observation is 'y' and the average of last 'm' observations is 'z' then 'n' 'x' 'm'

a Pu t bl a i

=a

n

p

= n(b–a) + b We have n observations out of which some observations (a1, a2, a3....) are replaced by some other new observations and in this way, if the average increase or decreases by b, then value of new observations n (a1, a2, a3....)

ny � mx (if n > m) n�m If from (n + 1) numbers, the average of first n numbers is 'F' and the average of last n numbers is 'L', and the first number is 'f' and the last number 'l' then �

b 6.

Arithmetic

(ii) Average of remaining numbers

r

4.

Batting Average Total runs scored Total number of innings played

(if m > n)

138

=

Arithmetic

Time & Work 1.

If M1 men finish W1 work in D1 days, working T1 time each day and M2 men finish W2 work in D2 days, working T2 time each day then M1

W1

D1 M2

3.

T1 W2

D2

If A can do a work in 'x' days. B can do the same work in 'y' days, C can do the same work in 'z' days then, total time taken by A, B and C to complete the work together A 'x' B 'y'

T2

C A, B

M1D1T1 M2D 2 T2 = W1 W2

'z'

C

4.

If A can finish

m part of the work in D days. n

Then, total time taken to finish the work by A

p

Ex. 5 men can finish a work in 10 days working 8 hours each day. How many men will be needed to finish the same work in 5 days working 4 hours each day?

Si

r

1 xyz � � 1 1 1 xy � yz � zx � � x y z

B

'y'

1 x

B

�

1

1 y

� Work done by A and B in 1 day �

A

B

�

ti on 4 5

� Time taken to complete the work

5.

1 1 x �y � � x y xy

5 × 20 = 25 days 4

If A and B can do a work in 'x' days B and C can do the same work in 'y' days. C and A can do the same work in 'z' days. Then total time taken, when A, B and C work together �

1 1 x �y � � x y xy

�Total time taken to complete the work by A and � xy � B both � � x � y � � �

A

4 th part of the work in 20 days. 5

In how many days he will complete the work?

Ch

1

Work done by B in 1 day � y

n �D m

Ex. Rahul can finish

pi

�

1

D

A

am

A

n

1 Work done by A in 1 day � x

m n

A

ca

'x'

n � D days m

on

A

Pr

5 × 10 × 8 = m × 5 × 4 m = 20 If A completes a piece of work in 'x' days and B completes the same work in 'y' days, then.

Ga ga

2.

a Pu t bl a i

=

2xyz days xy + yz + zx

A C

B

'x' 'y' A, B �

� xy � �� � �x �y �

139

B C

'z'

B

2 OR �1 1 1� �x � y � z � � �

A

C

2 �1 1 1� � � � � �x y z �

2xyz xy � yz � zx

6.

Arithmetic

If A alone can do a certain work in 'x' days and A and B together can do the same work in 'y' days then B alone can do the same work in A

‘x’

B

1

(E) � No. of days

A

'y'

B

E1 : E2 : E3 �

� xy � � days �� � �x �y�

Ex. If A can do a work in 8 days and A + B together can do the same work in 5 days. Then B alone can do the work in how many days?

E2D2 Ex. 3 persons can complete the work in 3 days, 4 days, 5 days respectively. Find ratio of their efficiency? 3 4 5 3 Persons Time

A+B B

Si

�1

Efficiency �� � 3 �

Time taken by B alone 8�5 40 = days 8�5 3 If food is available for 'x' days for 'A' men at a certain place and after 'y' days 'B' men join, then the remaining food will serve total men for

�

1� � × 60 (LCM) 5�

A �x � y �

� A � B�

100 work in x 100 � R days � �

Pr

�

ti on

'B'

A:B 2x : x A:B t : 2t 11. If A can do a work in 'x' days and B is R% more efficient than A then 'B' alone will do the same

ca

'x'

'y'

Required time days

B A1

Time taken �

C B1

on

pi

'x'

R%

B A 100 x �100 � R �

'B'

B

35(60 � 15) Required time = = 35 days 35 � 10 If A men or B boys or C women can do a certain work in 'x' days then A1 men B1 boys and C1 women can do the same work in

A

A

Ex. A can do a work in 10 days. B is 20% more efficient than A. Then B alone do the same work in? B A A

Ch

10

am

Ga ga

n

Ex. If food is available for 60 days for 35 men. After 15 days 10 new men join, then remaining food will serve total men for? 35 60 15

9.

1 4

�

C 5

20 : 15 : 12 10. If the efficiency to work of A is twice the efficiency to work of B, then, A : B (efficiency) = 2x : x and A : B (time ) = t : 2t A B

a Pu t bl a i

'A' A

8.

B 4

p

=

7.

� A � 3

r

A

1 1 1 E� k : : , or, ED = k or, E1D1 = D D1 D2 D3

B alone will do

'x'

100 100 � 20 100 = 10 × 120 25 �� days 3

= 10 ×

12. A can do a work in 'm' days and B can do the same work in 'n' days. If they work together and total wages is R, then. 'm' B A

C1

x A1 B1 C1 � � A B C

'n'

The comparison of rate of work done is called efficiency of doing work.

R Part of A

A

�

n �R �m � n�

Part of B

B

�

m �R m � � n�

1

Efficiency (E) � No. of days

140

Arithmetic

13. If A, B and C finish the work in m, n and p days respectively and they receive the total wages R, then the ratio of their wages is A, B C m, n p R 1 1 1 : : m n p

14. If A working alone takes x days more than A & B and B working alone takes y days more than A & B. Then the number of days taken by A & B working together to finish a job �� xy days. B B

x y

A

B �� xy

Ch

am

n

4 � 9 = 6 days

Ga ga

=

B

Pr

� Time taken by A and B together

A

ti on

9

ca

B B

on

A A

pi

B

a Pu t bl a i

p

Ex. If A working alone takes 4 days more than A and B and B working alone takes 9 days more than A and B working together. Then the number of days taken by A and B working together to finish a job? A A B 4

r

A A

Si

A B

141

Arithmetic

Pipe & Cistern Ex. If a pipe fills a tank in 20 hours but it takes 4 hours more to fill it due to leakage in the tank. If the tank is filled completely, then in how many hours it will be empty? 20 4 20 � (20 � 4) 4

r

1.

Amount of water released or filled = Rate × time. × Two taps 'A' and 'B' can fill a tank in 'x' hours and 'y' hours respectively. If both the taps are opened together, then how much time it will take to fill the tank? 'A' 'B' 'x' 'y'

= 5 × 24 = 120 hours

� xy � Required time = � � hrs � x +y �

Two taps 'A’ and 'B' can empty a tank in 'x' hours and 'y' hours respectively. If both the taps are opened together, then time taken to empty the

5.

a Pu t bl a i

� xy �

'y'

� xy � �� � �x �y�

when both are opened � xy � �� �:x � y �x �y �

(b) time taken to empty the tank

6.

Ch

where T, is the required time. T Note: Positive result shows that the tank is filling and Negative result shows that the tank is getting empty.

4.

If a pipe fills a tank in 'x' hours but it takes 't' more hours to fill it due to leakage in tank. If tank is filled completely, then in how many hours it will be empty? [due to leakage outlet] 'x' 't'

Required time=

when both are opened � xy � �� �:y � x �y � x �

pi

am

Ga ga

1 1 1 1 � � � .... � x y z T

on

If x, y, z,........... all taps are opened together then, the time required to fill/empty the tank will be: x, y, z,...........

n

3.

'y’

ca

'x’

Pr

'B’

(a)

ti on

tank will be Required time � � x � y � hrs � � 'A’

A tap 'A' can fill a tank in 'x' hours and 'B' can empty the tank in 'y' hours. Then (a) time taken to fill the tank 'A' 'x' 'B'

p

2.

Si

Required time =

x �x + t � t

142

Two taps A and B can fill a tank in x hours and y hours respectively. If both the pipes are opened together, then the time after which pipe B should be closed so that the tank is full in t hours A B x y x t Required time

� � t �� � �y �1 � �� hours x �� � �

Arithmetic

Time, Speed & Distance �

Distance = D, Speed = S, Time = T

�

D D D = S × T, S = , T = T S 1km

If they meet after 't' hours then t = 't'

5 m/sec 18

1 m/sec =

18 km/hr. 5

1.

r

Si

� 70 km/hr.

= 70 – 30 = 40km/hr. Relative speed � Distance between man & bus in one hr. will be 40 km. Relative speed = diff of speeds

1 1 S �� , T �� T S

p

D = S1T1 � S1T1 = S2T2

a Pu t bl a i

� S1 = T2 S2 T1

2.

D1 S1 = D2 S 2

on

1.

am

2.

If A and B starts walking towards each other. After meeting each other. A covered his remaining distance in t1 time and B covered his remaining distance in t2 time. Then ratio of their speed is A B t1

S L

L S

T = crossing time When train passes a bridge/platform. S LT

LP

distance covered will be equal to length of train + length of bridge/platform. = +

B

t2 A

When train passes a pole or stationary man

T=

D1 T1 = D2 T2

A

= sum of both speeds.

Distance covered will be equal to length of train

Case-III: If speed constant D � T Dl = ST1 D2 = ST2

�

� 30+70 = 100 km/hr.

When train passes

S L

Ch

Ga ga

n

Speed � 2 : 3 Time � 3 : 2 Case-II: If time constant D � S D1 = S1 T D2 = S2 T

70 km/hr�

Relative speed

pi

720 � 8 Hr. 90

�30 km/hr

Relative speed

Pr

D = Constant

ca

B S2 = 90 km/h

720 � 12 Hr. A= 60

When two objects travel in opposite direction:-

ti on

720 km

B=

When the two objects travel in same direction:� 30km/hr.

Case-I:If D = constant

A S1 = 60 km/h

t1 · t 2

Relative speed

1000m

5 1km/hr = 1Hr � 3600sec � m/sec 18

1km/hr =

t=

t1 · t 2

T�

t2

LT � LP S

T= Crossing time

� B= t 1

143

S2 L2

distance covered

= L1+L2 4.

L1 � L 2

T = S �S 1 2

When a train passes another train in same direction. S1 L1

slow

A B

y

Speed increase/decrease If an object increases/decreases its speed from x km/hr to y km/hr. to cover a distance in t2 hours in place of t1 hours then (Here (t2 – t1) will be given). x y

ti on

on

� Product of Speeds � � × (Change in time) � Diff. in Speeds �

Distance = �

2.

pi

n =

xy × (Change in time) x and y �

�Diff. of

or,

total travelled distance total time taken in travelling distance

If an object travels certain distance with the speed of

am

Ga ga =

t2

(t2 – t1)

Distance �

Pr

Average speed If a man travels different distances d1, d2, d3, ..... and so on in different time t1, t2, t3 respectively then, d1, d2, d3, .....

A of its original speed and reaches its B

destination 't' hours before or after, then the time taken by object travelling at original speed is

d1 � d2 � d 3 � ... t1 � t2 � t 3 � ...

A B

't'

If a man travels different distances d1, d2, d3, ..... and so on with diffe re nt spe eds s 1, s 2 , s 3 , respectively then, d1, d2, d3, ..... s 1, s 2, s 3 Average speed

t1

ca

L1 Re lative Speed

Average speed

3.

x

p

S2

t1 , t 2 , t 3

2.

1.

Ch

1.

A

B

� 2xy � � � �x �y�

a Pu t bl a i

T�

where n number of equal parts s1, s2, s3, ......... sn are speeds. n s1, s2, s3, ......... sn If a bus travels from A to B with the speed of x km/h and returns from B to A with the speed of y � 2xy �

S2 L2

When a train passes a person sitting in another moving train S1 L1

n �1 1 1 1� � � � � � � s1 s 2 s 3 s 4 �

km/h, then the average speed will be � x � y � � �

L + L2 T= 1 S1 – S 2

5.

�

Average speed

r

S1 L1

4.

Arithmetic

When a train pases another train in opposite direction

Si

3.

Time = 3.

� d � d2 � d3 � .....� � 1 d1 d2 d3 � � � .... s1 s 2 s 3

If a distance is divided into n equal parts each travelled with different speeds, then, n

� Diff. of

If a man travels at the speed of s1, he reaches his destination t1 late while he reaches t2 before when he travels at s2 speed, then the distance between the two places is s1 t1 t2 Distance �

144

A × time (in hour) A and B �

� s1 � s2 � � � t1 � t2 � s 2 � s1

s2

1.

2.

Some important points Formula to calculate the no. of rounds. Circular Distance = (circumference) × No of rounds, × D=2 r×n If any one overtakes or follows another, then time taken to catch

'a' 'b'

�a�b� �� � seconds �x �y�

If a train crosses a standing man/a pole in 't1' sec time and crosses 'P' meter long platform in 't2' sec P � t1 time, then length of the train � t � t � 2 1�

r

distance between them Relative speed

or, meeting time

� Speed of 1st traveller � � time � Diff. of speeds �

Total travelled distance to catch the thief

cross each other �

Formula to calculate the no. of poles,

ca

Pr

�

8.

on

1 part of Journey at u km/h, x

n

If a man covers

v

1 z

Ch

pi

on, then his average speed for the whole journey will be u

P � t1

� t2 � t1 �

2 � Product of time Diff. of time

t2

2 � Product of time Diff. of time

If two trains of same length are coming from opposite directions and cross a man in t1 seconds and t2 seconds then time taken by both trains to

1 y

t1

2 � Product of time Sum of time

t2 �

w

9.

1 1 1 1 � � � .... xu yv zw

5.

t1

cross each other �

am

Ga ga

1 1 part at v km/h and part at w km/hr and so y z

1 x

�

p

� Product of speeds � � time � Diff. of speeds �

Distance = (n – 1)x where n = No. of poles. x = distance between consecutive two poles. x 4.

't2'

If two trains of (same lengths) are coming from same direction and cross a man in t1 and t 2 seconds, then time taken by both the trains to

ti on

3.

7.

a Pu t bl a i

�

't1'

'P'

Si

�

'y' m/

s

6.

�

Arithmetic

'x' m/s

Let 'a' metre long train is travelling with the speed 'x' m/s and 'b' metre long train is travelling with the speed 'y' m/s in the opposite direction on parallel path. Then, time taken by the trains to cross each other

145

2 � Product of time Sum of time

If a train of length l m passes a bridge/ platform of 'x' m in t1 sec, then the time taken by the same train to cross another bridge/platform of length 'y' m is, �l � y � Time taken � � � t1 �l � x �

l

'x'

t1 'y' �l �y � �� � t1 �l � x �

10. From stations A and B, two trains start travelling towards each other at speeds a and b, respectively. When they meet each other, it was found that one train covers distance d more than that of another train. The distance between stations A and B is given as A B a b d

11. Excluding stoppage, the average speed of a train is u and with stoppage its average speed is v. Then, the stoppage time per hour u

on pi

am

Ch

Ga ga

n

Pr

With u > v and u, v � 0

ca

ti on

a Pu t bl a i

u� v u

B

(D)

Where 't' : change in the time taken 't'

p

v

�

A

r

�a � b� � � ��d �a � b�

Diff. between their average speed Speed without stoppage

t2

� t t � � S' � D � S � 1 2 � or � � t1t 2 � t1 � t 2 � � t ' �

B

�

S

Si

A

Arithmetic

12. A train covers a distance between stations A and B in time t1. If the speed is changed by S then the time taken to cover the same distance is t2. Then the distance (D) between A and B is given by A B t1

146

Arithmetic

Boat & Stream

by =

(a � b)(a � b) km/hr a

r

Si

p

't' d d � � t, x �y x �y

b d (a � b)(a � b) a

If a man or a boat covers x km distance in t1 hours along the direction of stream (downstream) and covers the same distance in t2 hours against the stream i.e. upstream, then t1

d=

t( x 2 � y 2 ) 2x

t(x 2 � y 2 ) 2x

2dx

t = x 2 – y2 8.

x

If a boat travels in downstream and upstream. then,

Speed of boat =

Sum of distances d1 � d 2 = 2 � time 2 � time

Speed of stream =

t2 speed of man/boat

Let the speed of stream be y km/h and speed of boat be x km/h. A boat travels equal distance (d) upstream as well as downstream in 't' hours, then y x

d is the fixed distance or, d =

a

4.

7.

If the speed of a boat or swimmer in still water is a km/hr and river is flowing with a speed of b km/hr. then average speed in going to a certain place and coming back to starting point is given

Ga ga

3.

Speed of swimmer t1 � t2 � Speed of stream t1 � t2

on

d2 d1 � �t x �y x �y

n

't'

ti on

D>U Let the speed of boat is x km/h and speed of stream is y km/h. To travel d1 km downstream and d2 km upstream, the time is 't' hours, then x y d2

A swimmer or boat travels a certain distance upstream in t1 hours, while it takes t2 hours to travel same distance downstream. then, t1 t2

ca

D�U = Speed of stream. 2

d1

Speed of swimmer or boat d1 � d 2 � Speed of stream d1 � d2

6.

D�U = Speed of boat. 2

If a swimmer takes same time to travel d1 km downstream and d2 km upstream, then, d1 d2

a Pu t bl a i

x>y x= y=

2.

5.

1�

pi

�

x �1

= 2 � t � t � km / hr 2 � � 1

speed of stream

=D=x+y Speed of boat in opposite direction of stream = upstream = U = x – y =U=x– y

Pr

�

x �1 1 � � � � km / hr 2 � t1 t 2 �

am

�

Speed of boat in still water = xkm/hr xkm/hr Speed of current/stream = ykm/hr ykm/hr Spe ed of boat in same direction of stream = downstream = D = x + y

Ch

1.

=

147

Difference of distances d1 � d2 = 2 � time 2 � time

Arithmetic

Race Ex: A gives B a start of 200 m and still beats him by 5 sec in a race of 1 km. Find the speed of B it speed of A is 10 m/sec. 200 1 A, B

x � A beats B by d metres. Race:- ��� x �A B d ���

A

x x–d

d

B

Time is same Hence, D � S � distance & speed

A

5

Speed of B B

ratio same

Si � 7

5.

13 m/sec 21

Pr

SA 40 � S � 27 B

If in a race of length L, the time taken by A and B be tA and tB (tB > tA), then the distance (d) by which A beats B given by, L A B tA tB

on pi

Ga ga

�L� d � � � (t – t ) � tB � B A

A B

am

(d)

n

(tB > tA)

P

Circumference

= S � S = Re lative Speed = A B

Ex: A and B walk around a circle of circumference 132 m with speeds 18 m/sec and 7 m/sec respectively. If they start simultaneously from the same point, the time after which they will be together again for the first time? B 18 7 A 132

Ch

or, d = B’s speed × (tB – tA) If in a race of length L, A can give B a start of ‘b’ and C a start of ‘c’ then the start that B can give C. L A B ‘b’

ca

ti on

DA 400 40 � � DB 270 27

‘c’

B C

�c�b�

= L� � �L �b� 4.

If A gives B a start of distance ‘d’ and still beats him by time ‘t’ in a race of length ‘L’. then B’s speed is A B ‘d’ ‘L’ ‘t’ SB �

800 1000 � 200 = 1000 105 �5 10

A and B walk around a circle of circumference ‘p’ with speeds SA and SB respectively. If they start simultaneously from the same point, the time after which they will be together again for the first time B SA SB ‘p’ A

a Pu t bl a i

80

C

=

p

Example:400m race — A give a start of 50m to B and still beat him by 80m 400 A B 50

3.

A

10

DA SA x = = DB SB x -d

2.

B

r

1.

B

L �d Dis tan ce cov ered by B � L Total time taken by B �t SA

Where, SA : A’S speed

SA : A

148

132 132 � = 12 sec 18 � 7 11

Permutation & Combinations

Fundamental Principle of Counting

(vi)

n–1

Pr+r�n–1Pr–1= nPr

n

�

If an event can occur in m different ways, following

(vii)

which another event can occur in n different ways,

n

Pr =n–r+1 Pr �1

Factorial notation The notation (n!) represents the product of first n natural number.

then the total no. of occurrence of the events in the given order is m × n.

n!

m

r

n

n! = 1 × 2 × 3 × .............. × (n-1) × n

Si

m×n

1! = 1

Ex. Mohan has 3 pants and 2 shirts. How many

2! = 2 × 1 = 2

up with?

3! = 3 × 2 × 1 = 6

a Pu t bl a ic p a

different pairs of a pant and a shirt, can be dress 3

2

4! = 4 × 3! = 4 × 3 × 2! = 4 × 3 × 2 × 1 = 24 5! = 5 × 4! = 5 × 4 × 3! = 5 × 4 × 3 × 2! = 5 × 4 × 3 × 2 × 1 = 120

of shirt.

Note

Permutation

Pr

�3×2=6

0! = 1

ti o

n

n

For every choice of a pant, there are two choices

Therefore � 3 × 2 = 6 pairs of pant and shirt.

�

n!

Pr = � n � r � ! , o < r < n (Repetition not allowed)

Note:-

A permutation is an arrangement in a definite

n

Pn = n!

n

Po = 1

n

on

pi

r(0