Rational and Irrational Numbers - ArundeepSelfhelp.info...1. Rational and Irrational Numbers 1 —...

1. Rational and Irrational Numbers 1 — 22

2. Compound Interest 23 — 45

3. Expansions 46 — 64

4. Factorisation 65 — 79

5. Simultaneous Linear Equations 80 — 102

6. Indices 103 — 121

7. Logarithms 122 — 134

8. Triangles 135 — 161

9. Inequalities 162 — 174

10. Mid-Point Theorem 175 — 190

11. Pythagoras Theorem 191 — 209

12. Rectilinear Figures 210 — 234

13. Theorems on Area 235 — 251

14. Circles 252 — 272

15. Statistics 273 — 283

16. Graphical Representation of Statistical Data 284 — 294

17. Mensuration 295 — 338

18. Surface Area and Volume of Solids 339 — 352

19. Trigonometry 353 — 371

20. Simple 2D Problems in Right Triangle 372 — 385

21. Co-ordinate Geometry 386 — 430

CONTENTS

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PointstoRemember

1. Natural Numbers :The numbers 1, 2, 3, 4, 5, .... are called natural numbers and are denoted

by N then N = {1, 2, 3, 4, 5, ......}. These are infinite in number.

2. Whole Numbers : The numbers 0, 1, 2, 3, 4, .... are called whole numbers and are denoted

by W then W = {0, 1, 2, 3, 4, ....}. These are also infinite in number.

3. Integers : The numbers –3, –2, –1, 0, 1, 2, 3, ..... are called integers and are denoted by I.

Then I = {...... –3, –2, –1, 0, 1, 2, 3, 4, .....}

4. Rational Numbers : The numbers of the type p

q; where p and q are integers and q 0 are

called rationals and are denoted by Q, then Q = {p

q, where p and q are integers and q 0}.

5. Irrational Numbers : The numbers which are not rational numbers are called irrational

numbers and are denoted by ‘Q’.

6. Properties of Irrational Numbers : The irrational numbers are :

(i) Non terminating and non-recurring decimals.

(ii) Square root and cube roots etc. of natural numbers are irrational numbers.

(iii) is an irrational number.

7. Real Numbers : Both rational and irrational numbers make up the set of real numbers and

are denoted by R. The real numbers can be represented on a number line.

8. Facts About Irrational Numbers : (i) Sum of two irrational numbers may be or may not

be irrational.

(ii) The difference between two irrational numbers may or may not be irrational.

(iii) The product of two irrational numbers may or may not be irrational.

(iv) If a and b are rational numbers, then

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

1

Rational and Irrational Numbers

1


FRANK ICSE Mathematics for IX Class2

(c) ab a b (d) a a

b b

9. Surd : If a is a rational number and n is a positive integer such that nth root of a be n a or

1

na is an irrational number, then it is called a surd or radical of order n. So surds are irrational

numbers written in the form of roots of rational numbers. , or n are called the radical sign.

Exercise 1A

1. Insert three rational numbers between

(i) 7 and 8 (ii) 9.9 and 10

(iii) 2 and 2.01

Solution.

(i) Three rational numbers between 7 and 8

(a) 7 8 15

7.52 2

Others two can be 7.6 and 7.7 etc.

(ii) Three rational numbers between 9.9 and 10

9.9 10 19.99.95

2 2

Others two can be 9.96, 9.97

(iii) Three rational numbers between 2 and

2.01.

2 2.01 4.012.005

2 2

Others will be 2.006, 2.007.

2. Insert a rational number between :

(i) 2

3 and

5

7(ii)

4

7 and

9

11

(iii) 5

7 and

6

11

Solution.

(i) Since we know that a rational number

between anda c

b d. Thus, one rational number

between 2

3 and

5

7

2 5 7

3 7 10

.

(ii) One rational numbers between 4

7

and 9

11 is given by

4 7 4 7 11

7 11 7 11 18

.

(iii) One rational number between 5

7 and

6

11is given by

5 6 11

7 11 18

.

3. Insert three rational numbers between :

(i) 5

9 and

8

11(ii)

3

8 and

10

13

(iii) 2

7 and

11

17

Solution.

(i) Three rational numbers between 5

9 and

8

11

First number 5 8 13

9 11 20

Second number between 5

9 and

13

20

5 13 18

9 20 29


3Rational and Irrational Numbers

and third number between 13

20 and

8

11 be

13 8 21

20 11 31

(ii) Three rational numbers between 3

8 and

10

13

First number 3 10 13

8 13 21


8 and

13

21

3 13 16

8 21 29

and third number between 13

21 and

10

13

13 10 23

21 13 34

Three numbers are 13 16 23

, ,21 29 34

.

(iii) Three rational numbers between 2

7 and

11

17

First number 2 11 13

7 17 24


7 and

13

24

2 13 15

7 24 31

Third number between 13

24 and

11

17

13 11 24

24 17 41

Three numbers are 13 15 24

, ,24 31 41

.

4. Without actual division, state which of the

following have a terminating decimal.

(i) 17

125(ii)

19

75 (iii)

41

16

(iv) 37

50(v)

5

11(vi)

23

3125

(vii) 9

14(viii)

18

35(ix)

37

80

(x) 5

12

Solution. We know that if the denominator

of a rational number can be expressed as the

power of 2, 5 or both, then the rational number

can be converted into a terminating decimal.

(i) 17

125 3

17 17

5 5 5 5

It has terminating decimal.

(ii) 19

75

19

5 5 3

Its denominator has 3 as factor.

It has non terminating decimal.

(iii) 41

16 4

41 41

2 2 2 2 2


(iv) 37

50 2

17 17

2 5 5 2 5


(v) 5

11



(vi) 23

3125 5

23 23

5 5 5 5 5 5




(vii) 9

14

9

2 7



(viii) 18

35

18

5 7



(ix) 37

80 4

37 37

2 2 2 2 5 2 5


(x) 5

12

5

2 2 3



5. Convert each of the following fractions to

decimal.

(i) 9

5 (ii)

13

25 (iii)

19

125 (iv)

78

65

Solution.

(i) 9

5

9 2 181.8

5 2 10

(ii) 13

25

13 4 520.52

25 4 100

(iii) 19

125

19 8 1520.152

125 8 1000

(iv) 78

65

78 13

65 13

( HCF of 78 and 65 =

13)

6

5

6 2 121.2

5 2 10

.

6. Express each of the following as a fraction.

(i) 0.7

(ii) 0.39 (iii) 2.3

(iv) 1.42 (v) 0.3105 (vi) 0.213

(vii) 0.285 (viii) 4.321

Solution. (i) 0.7 = 0.777...

Let x = 0.777 ...(i)

then 10x = 7.777.......... ...(ii)

subtracting (i) from (ii); we have

9x = 7; x = 7

9

(ii) 0.39 = 0.393939 ......

Let x = 0.393939.......

then 100x = 39.393939.......

on subtracting; 99x = 39

i.e. x 39 13

99 33

(iii) 2.3 = 2.333.......

Let x = 2.33..... then 10x = 23.333.......

On subtracting; 9x = 21

x 21 7

9 3

(iv) 1.42 = 1.424242.....

Let x = 1.424242......

then 100x = 142.424242......

On subtracting; we have

99x = 142 – 1 = 141

x 141 47

99 33

(v) 0.3105 = 0.3105105105........

Let x = 0.3105105105

Then 10x = 3.105105105... ...(i)

and 10000x = 3125.105,105105........ ...(ii)

Subtracting (i) from (ii); we have

9990x = 3125 – 3 = 3102

x 3102 517

9990 1665 .



(vi) 0.213 = 0.2131313.......

Let x = 0.2131313.......

then 10x = 2.131313......

and 1000x = 213.131313.........


x 211

990

(vii) 0.285 = 0.285285285.........

Let x = 0.285285285........

Then 1000x = 285.285285285.......


999x = 285

x 85 95

999 333

(viii) 4.321 = 4.3212121..........

Let x = 4.3212121

then 10x = 43.212121........ (i)

and 1000x = 4321.212121 ...(ii)


990x = 4321 – 43 = 4278

x 4278 1426 713

990 330 165 .

7. Express 9

as a repeating decimal. Now

represent 2 5

,9 9

and 7

9 as repeating decimal

numbers by deducing from 1

9 without doing

actual calculations.

Solution. (i) 1

9 = 0.1111......... = 0.1

9)1.0000 (

9

10

–9

10

–9

1

2

92

9

= 2 × 0.1111.......

= 0.2222....... = 0.2

5

9

15

9 = 5 × 0.1111........

= 0.5555...... = 0.5

7

9

17

9 = 7 × 0.1111.......

= 0.7777........ = 0.7

1. Fill in the blanks using rational/irrational/

either rational or irrational.

(i) The sum of two rational numbers is

_______.

(ii) The difference of two irrational numbers

is _______.

(iii) The product of a rational and an irrational

number is _______.

Exercise 1B

(iv) The product of two irrational numbers

is _______.

Solution. (i) The sum of two rational

numbers is rational.

(ii) The difference of two irrational numbers

is either rational or irrational.

(iii) The product of a rational and an irrational

number is irrational.

0.1111



(iv) The product of two irrational numbers

is either rational or irrational.

2. State whether true or false in each of the

following :

(i) 4 3 7

(ii) 8 5 2 5 6 5

(iii) 2 72 12

(iv) 75

53

(v) 2 24 5 9

(vi) 9 4 5

(vii) 125 7 5 80 7

Solution.

(i) 4 3 7 is false.

(ii) 8 5 2 5 (8 2) 5 6 5 is true.

(iii) 2 72 2 72 144 12 is true.

(iv) 75

25 53

is true.

(v) 2 24 5 16 25 41 9 is false.

(vi) 9 4 3 2 1 5 is false.

(vii) 125 7 5 80 7

= 5 25 7 5 5 16

5 5 7 5 4 5

= 7 is true.

3. State which of the following are irrational.

(i) 2 5 (ii) 147

3

(iii) 4

5(iv)

2

3

(v) 2

(vi)

33

24

(vii) 18 8 (viii) 7 1

(ix) (3 2) (3 2)

(x) (5 3) (4 2)

Solution.

(i) 2 5 , clearly it is an irrational number..

(ii) 147

3

14749 7

3

it is not irrational.

(iii) 4

5

4 2

5 5 is irrational.

(iv) 2

3

clearly it is not an irrational.

(v) 2

, clearly it is irrational since is

irrational

(vi) 3

32

4

3 3

3

32

4

27

2 2 264

27 27

2 2 264 32

Clearly it is an irrational number.

(vii) 18 8 18 8 144 12

It is rational.

(viii) 7 1 , It is an irrational number..

(ix) (3 2) (3 2) 2 2(3) ( 2)

= 9 – 2 = 7, it is rational.



(x) (5 3) (4 2)

20 5 2 4 3 3 2

20 5 2 4 3 6 It is irrational.

(i), (iii), (v), (vi), (viii) and (x) are

irrational.

4. Represent 10 and 11 on the number line.

Solution. 2 210 9 1 3 1

Draw AB = 3 units and at B, draw a

perpendicular BX at B and cut off BC = 1 unit.

BA3

C10

X

1

Join AC

Then AC = 10 units.

(ii) 11 36 25

2 26 5

(i) Draw AB = 5 units.

BA5

C

6 units

X

(ii) At B, draw a perpendicular BX.

(iii) From A cutoff AC = 6 units and join AC.

Then BC 2 26 5 (6 5)(6 5)

11 1 11 units.

Exercise 1C

Solution. We know that surds are irrational

numbers written in the form of roots of rational

numbers. Therefore

(i) 3 81 = 3 33 3 3 3 3 3

It is a surd.

(ii) 140 2 2 5 7 2 35 It is a surd.

(iii) 4 4250 40 4 4250 40 10000

4 10 10 10 10 10

It is not a surd. It is a rational.

(iv) 33 9 24 3 9 24

3 (3 3 3) (2 2 2)

= 3 × 2 = 6

1. State with reason, which of the following

are surds.

(i) 3 81 (ii) 140

(iii) 4 4250 40 (iv) 33 9 24

(v) 3 32 32 (vi) 3 5

(vii) 1 (viii) 5 243

(ix) 3 2 (x) 3 5

(ix) 9

(xii) 0.04004000400004...

(xiii) 3 6( 7) (xiv) 3

7

5




(v) 3 32 32 3 2 32

3 (2 2 2) (2 2 2)

= 2 × 2 = 4


(vi) 3 5

It is not a surd.

(vii) 1

It is not a surd.

(viii) 5 243

5 3 3 3 3 3 = 3

It is a rational.

(ix) 3 2

It is not a surd.

Since it is not of the form n a where aQ,

nN.

(x) 3 5 1/3 1/2 1/6(5 ) 5 6 5

It is a surd.

(xi) 9 1 3 3 3 1 It is not a surd.

(xii) 0.04004000400004...

It is not a surd.

since 0.04004000400004.... is an irrational

number.

(xiii) 3 6( 7)

6237 7 49

It is rational.

(xiv) 3

7

5

It is a surd.

2. A. Express the following surd in simplest

form.

(i) 75 (ii) 3 16

(iii) 4 80 (iv) 3 54

(v) 4 112

B. Write the simplest rationalising factor of :

(i) 32 (ii) 72

(iii) 3 5 (iv) 5 9

(v) 3 135

Solution.

A. (i) 75 3 5 5 5 3

(ii) 3 16 3 32 2 2 2 2 2

(iii) 4 8044 (2 2 2 2) 5 2 5

(iv) 3 54 33 2 (3 3 3) 3 2

(v) 4 112 4 42 2 2 2 7 2 7

B. The simplest rationalising factor of :

(i) 32 2 2 2 2 2 2 2 2 4 2

Rationalising factor 2 .

(ii) 72 2 2 2 3 3 2 3 2 6 2

Rationalising factor 2 .

(iii) 3 5

Rationalising factor 3 35 5 25 .

(iv) 5 9 5 3 3

Rationalising factor is 5 3 3 3

5 27

(v) 3 135 3 33 3 3 5 3 5

Rationalising factor 3 35 5 25

3. Compare the following surds.

(i) 2 and 3 3 (ii) 3 5 and 4 6

(iii) 5 and 3 10 (iv) 3 4 and 4 6



(v) 4 3 and5 2

Solution.

Compare

(i) 2 and 3 3

1

22 and

1

33

LCM of 2 and 3 = 6

3 31

3 622 2

1 1

6 6(2 2 2) 8

and

1 2 2 1

3 2 6 6 63 3 (3 3) 9

Since 8 < 9

1 1

6 68 9 or 2 < 3 3

(ii) 3 5 and 4 6

i.e.

1

35 and

1

46

LCM of 3, 4 = 12

1 1 4 4 1

43 3 4 12 125 5 5 (5 )

1

12(625)

and

1 31 3 1

34 34 12 126 6 6 (6 )

1 1

12 12(6 6 6) (216)

Since 625 > 216

1 1

12 12(625) (216)

or 3 5 > 4 6

(iii) 5 and 3 10

i.e.

1

2(5) and

1

3(10)

Here, LCM of 2 and 3 = 6

1 3 31

2 3 625 5 5

1 1

6 6(5 5 5) (125)

and

1 1 2 2 1

23 3 2 6 610 10 10 (10 )

1 1

6 6(10 10) (100)

Since 125 > 100

1 1

6 6(125) (100)

or 5 > 3 10

(iv) 3 4 and 4 6

i.e. 1

34 and

1

46

LCM of 3 and 4 = 12

Thus, 1 1 4 4 1

43 3 4 12 124 4 4 (4 )

1 1

12 12(4 4 4 4) (256)

and

1 31 3 1

34 34 12 126 6 6 (6 )

1 1

12 12(6 6 6) (216)

Since 256 > 216

1 1

12 12(256) (216)

3 4 > 4 6

(v) 4 3 and 5 2

Now 4 3 4 4 3

and 5 2 = 5 5 2 = 50

Since 48 < 50

48 < 50 4 3 5 2.



4. Simplify the following :

(i) 75 12 27

(ii) 3 11 8 99 44

(iii) 3 63 2 28 112

(iv) 5 18 7 50 10 8

(v) 6 72 18 8 32

(vi) 7 7 1 63

5 32 8 2 2

(vii) 3 1 27 3

45 2 5 20

(viii) 3

12

40(ix)

3 135

32

(x) 48 12

75 27

(xi)

45 20

245 80

(xii) 8 50

162 98

Solution.

(i) 75 12 27

3 75

5 25

5 5

1

5 5 3 2 2 3 3 3 3

5 3 2 3 3 3

(5 2 3) 3 6 3 .

(ii) 3 11 8 99 44

3 11 8 3 3 11 2 2 11

3 11 8 3 11 2 11

3 11 24 11 2 11

(3 24 2) 11 25 11 .

(iii) 3 63 2 28 112

3 3 3 7 2 2 2 7 4 4 7

3 3 7 2 2 7 4 7

9 7 4 7 4 7

(9 4 4) 7 1 7 7 .

(iv) 5 18 7 50 10 8

5 3 3 2 7 2 5 5 10 2 2 2

5 3 2 7 5 2 10 2 2

15 2 35 2 20 2

(15 35 20) 2 30 2

(v) 6 72 18 8 32

6 2 2 2 3 3 2 3 3

8 (2 2) (2 2) 2

6 2 3 2 3 2 8 2 2 2

36 2 3 2 32 2

= (36 – 3 – 32) 2 1 2 2 .

(vi) 7 7 1 63

5 32 8 2 2

7 7 1 3 3 75 3

2 2 2 2 2 2

7 1 7 1 75 3 3

2 2 2 2 2

7 3 7 3 75

2 2 2 2 2

7 3 3 75 5

2 2 2 2



(vii) 3 1 27 3

45 2 5 20

3 1 3 3 3 34

5 2 5 2 2 5

3 1 3 1 34 3

5 2 5 2 5

3 3 1 3 44 4

5 2 2 5 2

3 3(4 2) 6

5 5 .

(viii) 3

12

40 3 3 3

2 2 3 2 3 3

2 2 2 5 2 5 5

.

(ix)

3 135

32

By prime factorisation, we have

3 135

3 45

3 15

5 5

1

3 3 3 3 5

2 2 2 2 2

33 5

2 2 2

33 5

4 2

33 5

4 2 .

(x) 48 12

75 27

2 2 2 2 3 2 2 3

5 5 3 3 3 3

2 2 3 2 3 4 3 2 3

5 3 3 3 5 3 3 3

2 31

2 3 .

(xi) 45 20

245 80

3 3 5 2 2 5

5 7 7 2 2 2 2 5

3 5 2 5

7 5 2 2 5

3 5 2 5

7 5 4 5

5 1

33 5 .

(xii) 8 50

162 98

2 2 2 2 5 5

2 3 3 3 3 2 7 7

2 2 5 2

9 2 7 2

7 2 7

22 2 .

5. Expand.

(i) 2(2 5) (ii) 2(7 3)

(iii) 2(2 3 5 2) (iv) 2(4 5 3 7)

Solution.

(i) 2(2 5)

= (2)2 + 2 × 2 × 25 ( 5)

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

= 4 + 4 5 + 5 = 9 + 4 5 .

(ii) 2(7 3)

= (7)2 – 2 × 7 × 23 ( 3)



= 49 – 14 3 + 3 = 52 – 14 3

{ (a – b)2 = a2 – 2ab + b2}.

(iii) 2(2 3 5 2)

2 2(2 3) 2 2 3 5 2 (5 2)

= 4 × 3 – 20 6 + 25 × 2

= 12 – 20 6 + 50 = 62 – 20 6 .

(iv) 2(4 5 3 7)

2 2(4 5) 2 4 5 3 7 (3 7)

= 16 × 5 + 24 5 7 9 7

= 80 + 24 35 + 63

= 143 + 24 35 .

6. Find the product.

(i) (3 5) (3 5)

(ii) (5 3 2) (5 3 2)

(iii) (7 3 4) (5 3 1)

(iv) ( 2 3 11) (5 2 2 11)

(v) (3 6 5 2) (4 6 3 2)

(vi) (5 7 3) (4 7)

Solution.

(i) (3 5) (3 5)

2(3) ( 5)

{ (a + b) (a – b) = a2 – b2}

= 9 – 5 = 4.

(ii) (5 3 2) (5 3 2)

2(5) (3 2)

{ (a + b) (a – b) = a2 – b2}

= 25 – 9 × 2 = 25 – 18 = 7.

(iii) (7 3 4) (5 3 1)

7 3(5 3 1) 4(5 3 1)

35 3 7 3 20 3 4

105 7 3 20 3 4

101 13 3 .

(iv) ( 2 3 11) (5 2 2 11)

2(5 2 2 11) 3 11(5 2 2 11)

5 2 2 22 15 22 6 11

10 2 22 15 22 66

56 13 22 .

(v) (3 6 5 2) (4 6 3 2)

3 6(4 6 3 2) 5 2(4 6 3 2)

= 12 × 6 – 9 12 – 20 12 + 15 × 2

= 72 – 29 12 + 30

= 102 – 29 × 2 2 3

= 102 – 29 × 2 3

= 102 – 58 3 .

(vi) (5 7 3) (4 7)

5 7(4 7) 3(4 7)

20 7 5 7 12 3 7

20 7 35 12 3 7

23 17 7 17 7 23 .

7. Rationalise the denominators of :

(i) 5

3 2(ii)

3

4 7



(iii) 2 3

4 3

(iv)

4 5

3 5

(v) 2 5 3

3 2 4

(vi)

3 3 2

2 7 1

(vii) 3

6

24(viii)

5

5

4

50

(ix) 1

2 3 5

Solution.

(i) 5

3 2

5(3 2)

(3 2) (3 2)

{ (a + b) (a – b) = a2 – b2}

2 2

15 5 2 15 5 2

9 2(3) ( 2)

15 5 2

7

.

(ii) 3

4 7

3(4 7)

(4 7) (4 7)

{(a + b) (a – b) = a2 – b2}

12 3 7 12 3 7

16 7 9

4 7

3

(Dividing by 3)

(iii) 2 3

4 3

(2 3) (4 3)

(4 3) (4 3)

{ (a + b) (a – b) = a2 – b2}

2 2

8 2 3 4 3 3

(4) ( 3)

11 6 3

16 3

11 6 3

13

.

(iv) 4 5

3 5

(4 5) (3 5)

(3 5) (3 5)

2 2

12 4 5 3 5 5

(3) ( 5)

17 7 5 17 7 5

9 5 4

.

(v) 2 5 3

3 2 4

(2 5 3)(3 2 4)

(3 2 4)(3 2 4)

2 2

6 10 8 5 9 2 12

(3 2) (4)

{ (a + b) (a – b) = a2 – b2}

6 10 8 5 9 2 12

18 16

6 10 8 5 9 2 12

2

.

(vi) 3 3 2

2 7 1

(3 3 2)(2 7 1)

(2 7 1)(2 7 1)

2

6 21 3 3 4 7 2

4 7 (1)



6 21 3 3 4 7 2

28 1

6 21 3 3 4 7 2

27

.

(vii) 3

6

24

3 3

6 6

2 2 2 3 2 3

2/3

3 1 2

3 3

3 3 3

33 3

2 51

3 3

1 2

3 3

3 3

33

5 5 3 21

3 3 33 3 3

3 323 9 .

(viii) 5

5

4

50

1

5

1

5

(2 2)

(2 5 5)

1 1 3

5 5 5

1 2 3

5 5 5

2 2 5

5 5 5

1 1

5 5

5

5

(2 5 5 5) (2 5 5 5)

55

1

55(250) 250

5 5

(ix) 1

2 3 5

1 ( 2 3) 5)

[( 2 3) 5] [ 2 3 5]

2 2

2 3 5

( 2 3) ( 5)

( a2 – b2 = (a + b) (a – b)

2 3 5

2 3 2 2 3 5

( 2 3 5) 6

2 6 6

12 18 30

2 6

4 3 9 2 30

12

2 3 3 2 30

12

.

8. Insert any two irrational numbers

between the following :

(i) 3 and 6 (ii) 3 5 and 2 3

(iii) 6 and 7.

Solution.

(i) 3 and 6

2( 3) = 3 and 2( 6) = 6

3 < 4, 5 < 6

One irrational number is 5

and second will be 2.010010001....

(ii) 3 5 and 2 3

2(3 5) = 9 × 5 = 45

and 2(2 3) = 4 × 3 = 12

45 > 40, 30, ...... > 12

First irrational number 40

4 10 2 10 and second will be 30 .



(iii) 6 and 7.

(6)2 = 36 and (7)2 = 49

36 < 37, 38, 39, 40,..... and 49

First irrational number 37

and second irrational number 40

9. Write two rational numbers between.

(i) 2 and 3 (ii) 7 and 8

Solution.

(i) 2 and 3

2 > 1.9, 1.8, ..... 3

Two irrational numbers will be 1.9, 1.8.

(ii) 7 and 8

2( 7) = 7 and 8 = 8

Two rational numbers can be 7.29 and

7.84 i.e. 2.7 and 2.8.

10. (i) If x = 7 – 4 3 , find x + 1

x and

x2 + 2

1

x.

(ii) If x = 9 + 4 5 , find x + 1

x and

x2 + 2

1

x.

(iii) If a 7 6

7 6

and b

7 6

7 6

, find

the value of a + b.

(iv) Simplify : 5 3 5 3

5 3 5 3

(v) Simplify : 7 2

3 5 2 3 5 2

(vi) If x = 5 + 5 , find the value of

10x + 7x2 – x3.

Solution.

(i) Given, x = 7 – 4 3

1

x

1 1 (7 4 3)

7 4 3 (7 4 3) (7 4 3)

[Rationalising the denominator]

2 2

7 4 3 7 4 3

49 48(7) (4 3)

7 4 3

1

x + 1

x = 7 – 4 3 + 7 + 4 3 = 14

Squaring both sides, we have

221

(14)xx

x2 + 2

1

x + 2 = (14)2 = 196

x2 + 2

1

x = 196 – 2 = 194.

(ii) Given, x = 9 + 4 5

1

x

1

9 4 5

1 (9 4 5)

(9 4 5) (9 4 5)

(Rationalising the denominator)

2 2

9 4 5

(9) (4 5)

9 4 5

81 80

9 4 5

1

x + 1

x = 9 + 4 5 + 9 – 4 5 = 18



Squaring both sides; we have

21

xx

= (18)2

2

2

1x

x + 2 = 324

2

2

1x

x = 324 – 2 = 322.

(iii) Given, a 7 6

7 6

, b

7 6

7 6

Rationalising the denominators of each

fraction,

a 2

2

7 6 7 6 ( 7 6)

7 6 7 6 ( 7) ( 6)

a 7 6 2 42 13 2 42

7 6 1

and b 7 6

7 6

×

7 6

7 6

2

2

( 7 6)

( 7) ( 6)

7 6 2 42

7 6

13 2 42

1

.

a + b = 13 + 2 42 + 13 – 2 42 = 26.

(iv) 5 3 5 3

5 3 5 3

2( 5 3) ( 5 3)

( 5 3) ( 5 3)

2 2

5 3 2 15 5 3 2 15

( 5) ( 3)

16 168

5 3 2

.

(v) 7 2

3 5 2 3 5 2

7(3 5 2) 2(3 5 2)

(3 5 2) (3 5 2)

2 2

21 5 14 6 5 4

(3 5) (2)

27 5 10

45 4

27 5 10

41

.

(vi) Given x = 5 + 5

x2 = (5 + 5 )2 = 25 + 5 + 2 × 5 5

= 30 + 10 5

Thus, x3 = x2 x = (30 + 10 5 ) (5 + 5 )

= 150 + 30 5 + 50 5 + 50

= 200 + 80 5

Now 10x + 7x2 – x3

= 10(5 + 5 ) + 7(30 + 10 5 ) – (200 + 80 5 )

= 50 + 10 5 + 210 + 70 5 – 200 – 80 5

= 50 + 210 – 200 + 5 (10 + 70 – 80)

= 260 – 200 + 0 5

= 60.



11. Write the following numbers in

ascending order.

(i) 3 42, 5, 7 (ii) 4 3, 2 27, 5 75

Solution.

(i) 3 42, 5, 7

11 1

32 42 , 5 , 7

LCM of 2, 3, 4 = 12

1

22

1 6 6 1

62 6 12 122 2 (2 )

1

12(64)

1

35

1 4 4 1

43 4 12 125 5 (5 )

1

12(625)

1

47

1 3 3 1

34 3 12 127 7 (7 )

1

12(343)

Since, 64 < 343 < 625

1 1 1

12 12 12(64) (343) (620)

Thus, in ascending order

11 1

32 42 7 5 .

(ii) 4 3, 2 27, 5 75

Since 4 3 4 4 3 48

2 27 2 2 27 108

5 75 5 5 75 1875

since 48 < 108 < 1875

48 108 1875

Now in ascending order, we have

4 3 2 27 5 75

12. Write the following numbers in

descending order.

(i) 8

5 2, , 98, 2 182

(ii) 3 45, 3, 8

Solution.

(i) 8

5 2, , 98, 2 182

8 8 2 8 2

4 222 2 2

98 7 7 2 7 2

2 18 2 3 3 2 2 3 2 6 2

Now in descending order, we have

7 2 6 2 5 2 4 2

i.e. 8

98 2 18 5 22

.

(ii) 3 45, 3, 8

i.e.

1 1 1

3 2 4(5) , (3) , (8)

LCM of 3, 2, 4 = 12

1

3(5) =

1 4 4 1 1

43 4 12 12 12(5) 5 (5 ) (625)

1

23 =

1 6 6 1 1

62 6 12 12 123 3 (3 ) (729)

1

48 =

31 3 1 1

334 12 12 128 8 (8 ) (512)

Since 729 > 625 > 512

1 1 1

12 12 12(729) (625) (512)

So required descending order, we have

3 45 8



1. Without actual division, state which of

the following are terminating decimals.

13 4 19 43 5 14 2, , , , , ,

80 9 125 500 11 35 7

Solution.

We know that if the denominator of a

fraction can be expressed as the factors of 2 or

5 or both then the fraction has terminating

decimal.

(i) 13

80

13

2 2 2 2 5

It is terminating.

(ii) 4

9

4

3 3

It is not terminating.

(iii) 19

125

19

5 5 5

It is terminating.

(iv) 43

500

43

2 2 5 5 5

It is terminating.

(v) 5

11


(vi) 14

35

14 7 2

35 7 5

It is terminating.

(vii) 2

7


2. (i) Insert 3 rational numbers between 3

7

and 7

9.

(ii) Insert 3 irrational numbers between 11

and 12.

Solution.

(i) 3

7 and

7

9

First rational number 3 7 0 5

7 9 16 8

Second rational number between3

7 and

5

8

3 5 8

7 8 15

Third rational number between 5

8 and

9

5 7 12

8 9 17

.

Hence required three rational numbers are

7 5 12, ,

9 8 17.

(ii) Three irrational numbers between 11 and

12.

(11)2 = 121 and (12)2 = 144

required three irrational numbers are

125, 130, 140 .

3. Compare.

(i) 3 5 and 3 (ii) 3 4 and 4 5

(iii) 3 and 3 6

Solution.

(i)

1

3 35 5 and

1

23 3

LCM of 3 and 2 = 6

1

35

1 2 1 1

23 2 6 65 (5 ) (25)

Miscellaneous Exercise



1

23

1 3 1 1

32 3 6 63 (3 ) (27)

Since 25 < 27

1 1

6 625 27

Thus, 3 5 3.

(ii) 3 44 and 5

i.e.

1

34 and

1

45

LCM of 3, 4 = 12

1 4 4 1 1

43 4 12 12 124 4 (4 ) (256)

and

1

45

1 3 3 1 1

34 3 12 12 125 5 (5 ) (125)

since 256 > 125

1 1

12 12(256) (125)

i.e. 3 44 > 5

(iii) 3 and 3 6

i.e. 1

23 and

1

36

Since, LCM of 2, 3 = 6

1

23

1 3 3 1 1

32 3 6 6 63 3 (3 ) (27)

and 1

36

1 2 2 1 1

23 2 6 6 66 6 (6 ) (36)

Since 27 < 36

1 1

6 6(27) (36)

Thus, 33 6 .

4. Arrange in ascending order.

, 3 64 3 2 and 5

Solution.

3 64 3, 2 and 5 i.e. 1 11

3 643 , 2 , 5

Since, LCM of 4, 3, 6 = 12

1

43

1 3 3 1 1

34 3 12 12 123 3 (3 ) 27

1

32

1 4 4 1 1

43 4 12 12 122 2 (2 ) 16

1

65

1 2 1 2 1

26 2 12 12 125 5 (5 ) 25

.

Since 16 < 25 < 27

1 1 1

12 12 126 25 27 Thus, required ascending order.

i.e. 3 6 42, 5 3

5. Simplify the following :

(i) 147 27 75 48

(ii) 6

8 3 3 273

(iii) 175 112

28 63

Solution.

(i) 147 27 75 48

7 7 3 3 3 3 3 5 5

2 2 2 2 3

7 3 3 3 5 3 2 2 3

7 3 3 3 5 3 4 3

= (7 – 3 + 5 – 4) 3 5 3 .

(ii) 6

8 3 3 273

6 38 3 3 3 3 3

3 3



6 38 3 3 3 3

3

8 3 9 3 2 3

10 3 9 3 3 .

(iii) 175 112

28 63

5 5 7 4 4 7

2 2 7 3 3 7

5 7 4 7

2 7 3 7

7 1

55 7 .

6. Expand 2( 5 2)

Solution.

Now, 2( 5 2) = 2 2( 5) 2 5 2 ( 2)

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

= 5 + 2 10 + 2

= 7 + 2 10 .

7. Which of the following are irrational.

(i) 80

5(ii) 2( 7 3)

(iii) 2( 50 8) (iv) 288

2

Solution.

(i) 80

5

8016

5 = 4

It is not irrational.

(ii) 2( 7 3)

2 2( 7) ( 3) 2 7 3

= 7 + 3 – 2 21

= 10 – 2 21

It is irrational.

(iii) 2( 50 8)

2 2( 50) (8) 2 50 8

= 50 + 8 + 2 × 400

= 58 + 2 × 20 = 58 + 40

= 98.

It is not an irrational number.

(iv) 288

2

288144

2

= 12

It is not an irrational number.

8. Rationalise the denominator.

(i) 6

2 3(ii)

5 3

5 3

(iii) 1

5 2 1

Solution.

(i) 6

2 3

6(2 3)

(2 3) (2 3)

2

6(2 3) 6(2 3)

4 3(2) ( 3)

[(a + b) (a – b) = a2 – b2]

6(2 3)12 6 3

1

.

(ii) 5 3

5 3

( 5 3) ( 5 3)

( 5 3) ( 5 3)

2

2 2

( 5 3)

( 5) ( 3)

5 3 2 5 3

5 3

.

8 2 154 15

2

.



(iii) 1

5 2 1

1

5 ( 2 1)

1[ 5 ( 2 1)]

[ 5 ( 2 1)] [ 5 2 1]

2 2

5 2 1 5 2 1

5 (2 1 2 2)( 5) ( 2 1)

5 2 1 5 2 1

5 2 1 2 2 2 2 2

( 5 2 1) (2 2 2)

(2 2 2) (2 2 2)

2

2 5 2 2 2 2 10 2 2 2 2

(2) (2 2)

2 5 4 2 2 10 6

4 8

2 10 2 5 4 2 6

4

10 5 2 2 3

2

1(3 2 5 10)

2 .

9. Which of the following are surds.

3 33 47, 0.81, 5 , 2, 3 7 , 676 5 9

Solution.

(i) 3 7 It is a surd.

(ii) 4 0.81

44

4

81 3 3 3 3 3

100 1010 10

It is surd.

(iii) 11 1

3 1234 125 (5 ) 5 5

It is a surd.

(iv) 2

It is not a surd.

Since surd is of the form m a where a Q

and m N.

(v) 3 3 7

It is not a surd.

(vi) 676 2 2 13 13

= 2 × 13 = 26

It is not a surd.

(vii) 5 9 5 3 8

It is a surd.

10. If 3 7

73 7

a b

, find the value of a

and b.

Solution.

Given, 3 7

73 7

a b

3 7

3 7

(3 7) (3 7)

(3 7) (3 7)

[Rationalising the denominator.]

9 7 6 7 16 6 78 3 7

9 7 2

8 + 3 7 = a + b 7

On comparing, we get

a = 8 and b = 3.

11. (i) If x = 3 – 8 , find the value of

x + 1

x,

2 1x

x ,

22

1x

x .

(ii) If x = 3 + 3 , find the value of24x + x2 – x3.



Solution.

(i) x = 3 – 8

1

x

1

3 8

1(3 8)

(3 8) (3 8)

[Rationalising the denominator]

2 2

3 8 3 83 8

9 8(3) ( 8)

Thus, 1x

x 3 8 3 8 6

2

2

1x

x

21

2xx

= (6)2 – 2 = 36 – 2 = 34

Also, 1

xx

3 8 3 8 2 8

22

1 1x x x

x xx

6 ( 2 8) 12 2 2 2

12 2 2 24 2 .

(ii) x 3 3

x2 2 2 2(3 3) 3 ( 3) 2 3 3

= 9 + 3 + 6 3 = 12 + 6 3

x3 = x2 × x = (12 + 6 3 ) (3 + 3 )

= 36 + 12 3 + 18 3 + 6 × 3

= 36 + 30 3 + 18 = 54 + 30 3

Now 24x + x2 – x3

= 24 (3 + 3 ) + (12 + 6 3 )

– (54 + 30 3 )

= 72 + 24 3 + 12 + 6 3 – 54 – 30 3

= 72 + 12 – 54 + (24 + 6 – 30) 3

= 30 + 0 3 = 30 + 0

= 30.

12. State true or false in the following :

(i) 7

8 17

(ii) 75

53

(iii) 49 5 7 5

(iv) 9 7 3 7

(v) 3. 27 9

(vi) ( 11 7) ( 11 7) 4

Solution.

(i) 8 1 7 and

7 7 7 7 77

77 7 7

given statement is true.

(ii) L.H.S. =75

3= 25 5

It is true.

(iii) 49 5 44 7 5

given statement is false.

(iv) 9 7 16 = 4 3 7

It is false.

(v) L.H.S. = 3. 27

= 3 27

= 81 = 9 = R.H.S.

It is true.

(vi) L.H.S. = ( 11 7) ( 11 7)

= 2 2( 11) ( 7)

[ (a – b) (a + b) = a2 – b2]

= 11 – 7 = 4 = R.H.S.

It is true.


PointstoRemember

1. Simple Interest (S.I)

S.I Principal × Rate × Time

=100

P R T=

100

Amount = P + S.I.

2. Compound Interest (C.I).

(i) When CI is asked for fractional years, e.g., Time - 3 years 4 months. First find the amount

for 3 years and then use SI for 4 months 1

year3

on the amount as Principal and finally add.

(ii) While finding the time or rate when you get perfect squares or cubes, use powers and

factorials page of mathematical tables.

3. Compound Interest = Amount – Principal

(i) Amount = P 1100

nr

if compounded yearly..

(ii) Amount 2

P 1200

nr

if compounded half yearly. Observe that time is doubled and rate

is halved.

(iii) Amount 31 2P 1 1 1

100 100 100

rr r

if 3 rates are given for 3 successive years.

(iv) When part of the money is returned, e.g., after every 6 months some money is returned,

find the amount for 6 months and deduct the returned amount. The remaining becomes the principal

for next 6 months and so on.

4. (i) CI – SI

2

P100

r

for 2 years only..

2

Compound Interest

23



(ii) If compounded half-yearly.

CI – SI

2

P200

r

for 1 year only..

5. (i) If amounts of 2 successive years are given (e.g., n, n + 1 years)

Rate % th

Difference in amounts= ×100

Amount for the n years

(ii) If CI of 2 successive years are given,

Rate % th

Difference in CI= ×100

CI for the n years

(iii) If amounts of 2 non-successive years are given, (e.g., n, n + 3 years), rate can be found

by dividing the amounts.

Amount of ( + 3) years

Amount of years

n

n

3

3P 1100

1100

P 1100

n

n

r

r

r

6. If rate of growth = r% every year

(i) Population after n years = Present population 1100

nr

(ii) Present population = Population n years ago 1100

nr

(iii) Population after 3 years 31 2P 1 1 1

100 100 100

rr r

where 3 different years are given.

7. If rate of depreciation = r% of every year.

Value after n years = Present value 1100

nr

Present value = Value n years ago 1100

nr

.


25Compound Interest

Exercise 2A

1. Ramesh invests ` 12,800 for three years

at the rate of 10% per annum compound interest.

Find

(i) The sum due to Ramesh at the end of the

first year.

(ii) The interest he earns for the second year.

(iii) The total amount due to him at the end

of the third year.

Solution.

Investment by Ramesh (P) = ` 12800

Rate (R) = 10% p.a.

Period (T) = 3 years

(i) Interest for the first year PRT

=100

= ̀ 12800 10 1

100

= ` 1280

Amount at the end of first year

= ` 12800 + 1280 = ` 14080.

(ii) Interest for the second year

14080 10 1

100

= ` 1408

(iii) Amount after second year

= ` 14080 + 1408

= ` 15488

Interest for the third year = ` 15488 10 1

100

= ` 1548.80.

Total amount due at the end of third year

= ` 15488 + 1548.80

= ` 17036.80

2. Archana invests ` 5,600 at 14% p.a.

compound interest for 2 years. Calculate.

(i) The interest for the Ist year.

(ii) The amount at the end of the 1st year.

(iii) The interest for the 2nd year, correct to

the nearest ` .

Solution.

Given, investment by Ramesh (P) = ` 5600

Rate (R) = 14% p.a.


(i) Interest for the first year PRT

=100

= ̀5600 14 1

100

= ` 784

(ii) Amount at the end of first year

= ` 5600 + 784 = ` 6384

(iii) Interest for the second year

6384 14 1

100

= ` 893.76 = ` 894 (in rupee)

3. Arpana borrows a sum of ` 2500 for 2

years 3 months at 8% p.a. compounded annually.

Find

(i) the CI for 2 years

(ii) the amount at the end of 2 years 3

months.

Solution.

Sum borrowed by Arpana (P) = ` 2500

Rate (R) = 8% p.a.

Period (P) = 2 years 3 months

Interest for the first year

PRT 2500×8×1= =

100 100

= ̀200


= ` 2500 + 200

= ` 2700



Interest for the second year = ` 2700 8 1

100

= ` 216

Amount after second year = ` 2700 + 216

= ` 2916

(i) C.I for 2 years = ` 200 + 216

= ` 416

Interest for 3 months 1

year4

2916 8 1

100 4

= `

5832

100

= ` 58.32.

(ii) Amount after 2 years 3 months

= ` 2916 + 58.32

= ` 2974.32.

4. How much will ` 20,000 amount to in 2

years at CI if the rates for successive years are

8% and 10% p.a. respectively ?

Solution.

Principal (P) = ` 20000

Rate for the first year (R1) = 8% p.a.

and for the second year (R2) = 10% p.a.

Interest for the first year PRT

=100

20000 8 1

100

= ` 1600

Amount at the end of the first year

= ` 20000 + 1600

= ` 21600.

Principal for the second year = ` 21600

Rate = 10%

Interest for the second year

= ̀ 21600 10 1

100

= ` 2160

Amount at the end of second year

= ` 21600 + 2160

= ` 23760.

5. If the amount after 2 years on a certain

sum is ` 4452 with 6% and 5% p.a. for 2

successive years CI, find the sum.

Solution.

Given, amount (A) = ` 4452

Rate for successive years = 6% and 5%

Period = 2 years

Let the principal (P) = ` 100

then amount after 2 years

= ` 100(100 6) (100 5)

100 100

= ` 100 × 106 105

100 100

11130

100 = ` 111.30

If amount is ` 111.30 then principal = ` 100

and if amount is ` 4452,

then principal 100 4452

111.30

100 100 4452

11130

`

= ` 4000

6. Find the CI on ` 64000 for 11

2 years at

5% p.a., the interest being compounded half-

yearly.

Solution.

Given, principal = ` 64000

Rate (R) = 5% p.a. 5

2 % half yearly

Period (P) = 11

2 years = 3 half years


27Compound Interest

Amount (A) R

P 1+100

n

= ` 64000

35

1+2 100

= ` 64000 ×

341

40

= ` 64000 × 41 41 41

40 40 40

= ` 68921

CI = A – P = ` 68921 – 64000

= ` 4921.

7. Calculate the amount and CI on ` 27,000

for 11

2 years at 6

2

3% p.a., the interest being

compounded semi-annually.

Solution. Given, principal (P) = ` 27,000

Rate (R) = 2 20

6 %3 3 p.a.

10

3 half yearly

Period (n) 1

12

years = 3 half years

Amount (A) R

P 1+100

n

= ` 27000

310

1100 3

= ` 27000 ×

331

30

= ` 27000 × 31 31 31

30 30 30

= ` 29791

Thus, CI = A – P = 29791 – 27000

= ` 2791.

8. The simple interest on a sum of money for

2 years at 4% per annum is ` 340. Find

(i) the sum of money and

(ii) the compound interest on this sum for

one year payable half yearly at the same rate.

Solution.

Simple Interest = ` 340

Rate (R) = 4% p.a.


(i) Principal S.I ×100 340 100

=R × T 4 2

= ` 4250.

(ii) Principal (P) = ` 4250

Rate (R) = 4% p.a. or 2 of half yearly

Period (n) = 1 year = 2 half years

Amount R

P 1100

n

= ` 4250

22

1100

= ` 4250 ×

251

50

= ` 4250 × 51 51

50 50

= ` 44217

10 = 4421.70

CI = A – P = ` 4421.70–4250 = `171.70.

9. Mr. Britto borrowed ` 28,000 for 2 years.

The rate of interest for the two successive years

are 8% and 10% respectively. If he repays ` 5240

at the end of the first year, find the outstanding

amount at the end of the second year.

Solution.

Given, sum borrowed by Mr. Britto (P)

= ` 28000

Rate of successive years (R) = 8% and 10%





=100

= ̀28000 8 1

100

= ` 2240

Amount = ` 28000 + 2240

= ` 30240

Amount paid at the end of the first year

= ` 5240

Remaining amount = ` 30240 – 5240

= ` 25000


= ̀25000 10 1

100

= ` 2500

Amount outstanding at the end of the

second year = ` 25000 + 2500 = ` 27500.

10. Ranbir borrows ` 20,000 at 12% per

annum compound interest. If he repays ` 8400

at the end of the first year and ` 9680 at the end

of the second year, find the amount of loan

outstanding at the beginning of the third year.

Solution.

Given, sum borrowed by Ranbir (P)

= ` 20000

Rate (R) = 12% p.a.

Period (P) = 3 years


=100

= ` 20000 12 1

100

= ` 2400


= ` 20000 + 2400 = ` 22400

Amount repaid = ` 8400

Thus, balance amount = ` 22400 – 8400

= ` 14000


= ̀14000 12 1

100

= ` 1680

Amount at the end of the second year

= ` 14000 + 1680 = ` 15680

Amount paid = ` 9680

Balance amount outstanding at the beginning

of third year or at the end of the second year

= ` 15680 – 9680 = ` 6000.

11. Mr. Dubey borrows ` 1,00,000 from State

Bank of India, at 11% per annum compound

interest. He repays ` 41,000 at the end of first

year and ` 47,700 at the end of the second year.

Find the amount outstanding at the beginning of

the third year.

Solution.

Sum borrowed by Mr. Dubey (P)

= 1,00,000

Rate (R) = 11% p.a.

Time (T) = 3 years


=100

= ̀ 100000 11 1

100

= ` 11000

Amount at the end of the first year

= ` 100000 + 11000

= ` 1,11,000

Amount repaid at the end of first year

= ` 41000


= ` 70,000


= ̀ 70000 11 1

100

= ` 7700


29Compound Interest

Amount at the end of the second year

= ` 700000 + 7700 = ` 77700

Amount repaid at the end of the second year

= ` 47700


= ` 30000

Amount outstanding in the beginning of the

third year = ` 30000.

12. What sum will amount to ` 18522 in 11

2

years at 10% p.a. compounded half yearly ?

Solution.

Let P be the sum

Amount (A) = ` 18522

Rate (R) = 10% p.a. or 5% half yearly

Period (n) = 11

2 years or 3 half yearly

we know that

A R

P 1100

n

18522

35

P 1100

321

P20

P = 18522 ×

320

21

= 18522 × 20 20 20

21 21 21

= ` 16000

13. What sum of money will amount to

` 11,025 in 2 years at 5% p.a. compounded

annually ?

Solution. Let P be the required sum

Given, amount (A) = ` 11025

Rate (R) = 52 p.a.

Period (n) = 2 years

We know that

A R

P 1100

n

11025

2 25 21

P 1 P100 20

P = 11025 ×

220

21

= 11025 × 20

21 ×

20

21

= ` 10000

Sum = ` 10,000

14. A sum of ` 16,000 earns ` 1640 as

interest in 2 years when compounded annually.

Find the rate of interest.

Solution.

Given, sum (P) = ` 16000

Interest = ` 1640


Let R% be the required rate of interest per

annum.

Amount = 16000 + 1640 = `17640

A R

1P 100

n

217640 R

116000 100

2 221 R

120 100

On comparing; we have

R 211

100 20

R 21 11

100 20 20

R = 100 × 1

20 = 5

Hence, required rate = 5% p.a.

15. At what rate per cent per annum

compound interest, will ` 4000 amount to ` 5324

in 3 years ?

Solution.

Principal (P) = ` 4000

Amount (A) = ` 5324




we know that,

A R

1P 100

n

35324 R

14000 100

3 311 R

110 100

On comparing, we get

R 11 R 11 1

1 1100 10 10 10

R 1

100 1010

Thus, required Rate = 10% p.a.

16. The CI earned by a sum of money in 2

years at 8% p.a. is ` 832. Find the sum.

Solution.

Given, C.I. on a sum = ` 832

Rate (R) = 8% p.a.


Let sum = ` 100

Then A

2R 8

P 1+ 100 1100 100

n

227 27 27

100 10025 25 25

= ̀2916

25

CI = A – P 2916 2916 2500

10025 25

= ̀416

25

If CI is 416

25 then sum = ` 100

and if C.I is ` 832 then sum

= ̀100 25 832

410

= ` 5000

thus required sum = ` 5000.

17. Arun borrows ` 24,000 from Bryan at SI

for 2 years at 5% p.a. and immediately lends out

this money to Chand for 2 years at 5% p.a.

compounded annually. What is Arun’s gain in this

transaction ?

Solution.

Sum borrowed by Arun (P) = ` 24000

Rate (R) = 5% p.a.


we know that

Simple Interest PRT 24000×5× 2

= =100 100

= ` 2400

Rate (R) = 5% p.a. on compound interest


Amount (A) R

P 1100

n

25

24000 1100

= ` 2400 × 21

20 ×

21

20

= ` 26460

CI = A – P = ` 26460 – 24000 = ` 2460

Required gain = CI – SI

= ` 2460 – 2400 = ` 60.

18. The difference between CI and SI for 2

years on the same sum at 5% p.a. is ` 240. Find

the sum.

Solution.

Difference between CI and SI on a sum

= ` 240


31Compound Interest

Rate (R) = 5%

Period = 2 years

Let sum of money = ` 100

Then SI PRT 100×5× 2

= =100 100

= ` 10

and in case of compound interest, we have

A = PR 5

1 100 1100 100

n

= 100 × 21 21

20 20 = `

441

4

SI = A – P = ` 441

4 – 100 = `

41

4

Difference between CI and SI = ` 41

4 – 10

= ` 41 40

4

= `

4

If difference is ` 4

then principal = ` 100

if difference is ` 240 then principal

= ` 100 4 240

1

= ` 96000.

19. The difference between CI and SI on the

same sum of ` 8000 for 2 years is ` 20. Find

the rate of interest.

Solution. Sum (P) = ` 8000



Let required rate of interest be R%

SI PRT 8000× 2× R

= =100 100

= 160R

and CI R

P 1+ P100

n

2R

P 1+ 1100

2R

8000 1+ 1100

2R 2R8000 1+ 1

10000 100

2R 2R8000

10000 100

24R= +160R

5

Difference between CI and SI

2 24R 4R= +160R 160R

5 5

according to given condition, we have

24R

5 = 20 R2

0 5

4

= 25 = (5)2

R = 5

required rate be 5% p.a.

20. On a certain sum of money, the difference

between the CI for a year, payable half-yearly and

the SI for a year is ` 120. Find the sum lent out,

if the rate of interest in both cases is 10% p.a.

Solution.

Difference in CI and SI = ` 120

Let sum (P) = ` 100

Rate (R) = 10% p.a.

Period (P) = 1 year

SI PRT 100×10×1

= =100 100

= ` 10

In case of C.I

Rate (R) = 10% p.a. or 5% half yearly

Period (n) = 1 year = 2 half years



We know that

A

2R 5

P 1+ 100 1100 100

n

221 21 21

100 10020 20 20

= `

441

4

CI – A.P = ` 441

1004

=441 400

4

= `

41

4

Now difference in CI ans SI = ` 41

4 – 10

= ` 41 40 1

4 4


4 sum = ` 100

and if difference is ` 120 then sum

= ̀00 4

1201

= ` 48000

Hence required sum = ` 48000.

21. On what sum of money will the difference

between the compound interest and simple

interest for 2 years be equal to ` 25 if the rate of

interest charged for both is 5% p.a. ?

Solution.

Given difference between CI and SI = ` 25

Rate (R) = 5% p.a.


Let sum (P) = ` 100

then SI PRT 100×5× 2

= =100 100

= ` 10

and in case of CI

A

2R 5

P 1+ 100 1100 100

n

A = 100 ×

21

20

= 100 × 1

20

×

1

20

= ` 441

4

Thus, CI = A – P = ` 441

4 – 100

= ` 441 400

4

= `

41

4


104

= ` 41 40

4

= `

1

4


4 then sum = ` 100

and if difference is ` 25

then sum = ` 100 4

251

= ` 10000

Thus, required sum = ` 10000

22. Asha invests ` 8000 at a certain rate for

3 years compounded annually. She finds that at

the end of one year it amounts to ` 9200.

Calculate :

(i) the rate of interest

(ii) the interest accrued in the second year

(iii) the amount at the end of third year.

Solution.

Sum (P) = ` 8000

Let Rate = R% p.a.


Amount at the end of the one year = ` 9200

Interest = A – P = ` 9200 – 8000 = ` 1200

(i) Rate SI ×100

=P × T

1206 100

8000 1

= 15%


9200 15 1

100

= ` 1380


33Compound Interest

Amount after 2nd year = ` 9200 + 1380

= ` 10580

(iii) Interest for the third year

10580 15 1

100

= ` 1587

Amount at the end of the third year

= ` 10580 + 1587 = ` 12167.

23. Ranjan invests ` 15,360 for 2 years at a

certain rate compounded annually. At the end of

the year, it amounts to ` 16,320. Calculate.


(ii) the amount at the end of the second year.

Solution.

Sum (P) = ` 15360

Period (A) = 2 years

At the end of the first year, amount = ` 16320

Interest for one year = ` 16320 – 15360

= ̀960

(i) Rate (R) SI ×100 960 100

=P ×T 15360 1

25%

4

16

4 of p.a.


16320 25 1

100 4

= ` 1020

(ii) Amount after second year

= ` 16320 + 1020

= ` 17340

24. Seema invests ` 5000 compounded

yearly. At the end of the one year, it amounts to

` 5200. Calculate :


(ii) the CI for the second year.

Solution.

Sum (P) = ` 5000

Amount at the end of the one year = ` 5200

Interest for the first year = ` 5200 – 5000

(i) Rate SI ×100 200 100

=P ×T 5000 1

= 4% p.a.

(ii) CI for the second year 5200 4 1

100

= ` 208.

Exercise 2B

1. A certain sum amounts to ` 17,640 in 2

years and to ` 18,522 in 3 years at compound

interest. Find the rate and the sum.

Solution. Amount in 2 years = ` 17640

and amount in 3 years = ` 18522

On subtracting, we get

Interest for 1 year on ` 17640 = ` 882

Rate SI×100 882×100

= =P× T 17640×1

= 5

Rate = 5% p.a.

Now A R

P 1+100

n

` 17640

25

P 1+100

` 17640 = P

221

20

P = ` 17640 ×

220

21

= ` 17640 × 20

21 ×

20

21

= ` 16000

Principal = ` 16000 and rate = 5% p.a.



2. The amount at compound interest which

is calculated yearly on a certain sum of money

is ` 4840 in 2 years and ` 5324 in 3 years.

Calculate the rate and the sum.

Solution.

Amount in 2 years = ` 4840


Subtracting we get interest for 1 year on

` 4840 = ` 484

Rate SI ×100

=P × T

484 100

4840 1

= 10%

Now A

2R

= P 1+100

4840

210

= P 1+100

4840

211

= P10

P = ` 4840 ×

210

11

P = ` 4840 × 10

11 ×

10

11 = ` 4000

Sum = ` 4000 and rate = 10% p.a.

3. A sum ammounts to ` 8820 in 2 years and

` 9261 in 3 years compounded annually. Find the

rate and the sum.

Solution.

Amounts in 2 years = ` 8820


Subtracting we get interest for 1 year on

` 8820 = ` 441

Rate Interest × 100 441×100

= =P × T 8820×1

= 5%

Now A R

P 1+100

n

8820

25

P 1+100

8820 = P ×

221

20

P = 8820 ×

220

21

= 8820 × 20 20

21 21

= ` 8000

Thus, required sum = ` 8000 and rate 5%

p.a.

4. A sum amounts to ` 6600 in 1 year and

` 7986 in 3 years at compound interest. Find the

rate and the sum.

Solution.

Given amount in 1 year = ` 6600


Since A R

P 1

n

Now A1

1R

P 1

A3

3R

P 1

3

1

A

A

3

2

1

RP 1+

R1001+

100RP 1+

100

27986 R

16600 100


35Compound Interest

2 2R 121 11

1100 100 10

R 11

1100 10

R 11

1100 10

11 10 1

10 10

R 1

10010

= 10

Now A = P

110

1100

6600 = P × 11

10

P = 6600 10

11

= 6000

Required sum = `6000 and rate = 10% p.a.

5. A sum of money if invested at compound

interest for 2 years amounts to ` 57,600 and `

65,536 in 4 years. Find the rate and the sum.

Solution.


and amounts in 4 years = ` 65536

we know that

A R

P 1

n

given, A4 4

RP 1

and A2 2

RP 1

On dividing, we get

4

2

A

A

4

2

RP 1+

100

RP 1+

100

65536

57600

2R

1100

2R

1100

2256 16

225 15

R 16

1100 15

16 16 15 1

1100 15 15 15

R

R 1 20 2

100 615 3 3

thus required rate 2

63

%.

Now A = PR

1100

n

220

P 13 100

21

P 13 5

57600

216

P15

P = 57600×

215

16

P = 57600 × 15

16 ×

15

16

P = 225 × 225 = 50625

Sum = ` 50625 and Rate = 62

3%.

6. Amar invests some money at compound

interest and finds that it will amount to ` 21,600

in 2 years and ` 31,104 in 4 years. Find the rate

and the sum.

Solution.

Given, amount in 2 years = ` 21600


Now A

RP 1

n



A4

4R

P 1

A2

2R

P 1

On dividing; we get

4

42

2

RP 1+

A 100

A RP 1+

100

31104

21600

4

2

2

RP 1+

R1001

100RP 1+

100

2 2R 31104 36 6

1100 21600 25 5

R 6 R 6 6 5 1

1 1100 5 100 5 5 5

R 1

1005

= 20

Thus, A2

2 220 6

= P 1+ P100 5

P = ` 21600 ×

2

6

= ` 21600 × 6

×

6

= ` 15000

Sum = 15000 and rate = 20%.

7. On a sum of money, the compound

interest calculated yearly is 880 for the second

year and ` 968 for the third year. Calculate the

rate of interest and the sum of money.

Solution.

CI for the second year = ` 880

and CI for the third year = ` 968

Subtracting, we get interest for 1 year on R

800 = ` 88

Rate SI ×100 88 100

=P ×T 880 1

= 10%

Let principal (P) = ` 100

CI for the first year = ` 100 10 1

100

= ` 10

Amount = ` 100 + 10 = ` 110

CI for the second year 110 10 1

100

= ` 111

If CI for 2nd year is ` 11 then principal

= ` 100

and if CI for 2nd year is ` 880 then principal

= ` 100 880

11

= ` 8000

8. When Arpita invests a certain sum of

money, she observes that the CI for the second

year is ` 1380 and for the third year it is ` 1587,

when it is compounded yearly. Calculate the rate

of interest and the sum of money invested.

Solution.

CI for second year = ` 1380

and CI for third year = ` 1587

Subtracting we get interest on ` 1380 for 1

year = ` 207

Rate = SI × 100 207 100

P × T 138 100

= 15%

Let principal = ` 100

then interest for the third year PRT

=100

100 15 1

100

= ` 15

Amount = ` 100 + 15 = ` 115


37Compound Interest

CI interest on second year = ` 115 15 1

100

1725

100

If CI for 2nd year is ` 1380 then principal

= ` 100 100

13801725

= ` 8000


9. Ayaan invests a sum of money for 3 years

compounded annually. He finds that CI for the

second year is ` 1260 and ` 1323 is the CI for

the third year. Calculate the rate of interest and

the sum invested by him.

Solution.

CI for the second year = ` 1260

and CI for the third year = ` 1323

Subtracting, we get the interest on ` 1260 for

1 year = ` 63

required rate RSI ×100 63 100

=P × T 1260 1

= 5%

Let principal (P) = ` 100

Now CI for the first year = ` 100 5 1

100

= ` 5

Amount = ` 100 + 5 = ` 105

CI on the second year = ` 105 8 1

100

= ̀21

4

If CI on second year is ` 21

4 then for

principal = ` 100

If CI on second year is ` 1260 then principal

= ` 100 4

126021

= ` 24000


10. The SI on a certain sum for 2 years is

` 1200 and the CI on the same sum at the same

rate for 2 years is ` 1272. Find the rate of interest

and the sum invested.

Solution.

SI on a certain sum for 2 years = ` 1200

and CI = ` 1272

SI for the first year = ` 1200

2 = ` 600

CI interest for the first year = ` 600

and for second year = 1272 – 600 = ` 672

Difference = ` 672 – 600 = ` 72

` 72 is interest on ` 600 for 1 year

Rate Interest ×100 72×100

= =P × T 600×1

= 12%

Now principal SI ×100 1200 100

=R × T 12 2

= ` 5000


11. The SI on a certain sum is ` 400 for 2

years and the CI on the same sum at the same

rate for 2 years is ` 410. Find the rate of interest

and the sum borrowed.

Solution.

SI on a sum for 2 years = ` 400

and CI for 2 years = ` 410

SI for the first year = ` 400

2 = ` 200

then CI for the first year = ` 200



and CI for the second year = ` 410 – 200

= ` 210

Difference = ` 210 – 200 = ` 10

` 10 is SI on ` 200 for, year

Rate SI × 100 10×100

= =P × T 200×1

= 5%

Now sum SI × 100 400×100

= =R × T 5× 2

= ` 4000

Sum = ` 4000 and rate = 5% p.a.

Exercise 2C

I. Growth

1. There were 7,20,000 people in a town. If

the population of this town increase at the rate

of 5% p.a., find the population at the end of 3

years.

Solution.

Given, people in a town (P) = 720000

Rate of increase (r) = 5%


Population of the town after 3 years

RP 1

n

= 720000

35

1100

= 720000 ×

321

20

= 720000 × 21

20 ×

21

20 ×

21

20

= 833490.

2. The present population of a town is

7,98,600. If it increased at the rate of 10% every

year. What was its population 3 yeas ago ?

Solution.

Present population of a town (A) = 7,98,600

Rate of increase (r) = 10%


we know that

A R

P 1

n

798600

3 310 11

P 1 P100 10

P = 798600 ×

310

11

= 798600 × 10

11 ×

10

11 ×

10

11

= 6,00,000

Thus, required population of the town 3 years ago

= 6,00,000.

3. The production of toys in a factory was

40,000 two years ago. The production decreased

by 3% in the first year but increased by 5% in

the second year. Find the present day production.

Solution.

2 years ago, in a factory, Production of toys

= 40000

Rate decreased by 3% in first year and

increase of 5% in second year

Present production 1 2P 1 1100 100

r r

3 540000 1 1

100 100


39Compound Interest

97 2140000

100 20

= 40740.

4. Two years ago, a plant was 192 cm tall.

At present its height is 243 cm. Find the rate of

its growth.

Solution.

2 years ago, height of plant = 192cm

Present height = 243 cm

Let rate of increase = r%

A

1P 100

nr

2 2243

1 1192 100 100

r r

2

1100

r

29

8

9

1100 8

r

9 9 8 11

100 8 8 8

r

r 1 25

100 % 12.58 2

required rate of increase = 12.5%.

5. Three years ago, the price of a plot of land

was ` 1,20,000. If the price of land increased at

10% p.a. what is its present price ?

Solution.

3 years ago, price of plot was = ` 120000

Rate of increase =10% of p.a.

Present price R

P 1

n

= ` 120000

310

1100

= ` 120000 ×

311

10

= ` 120000 × 11

10 ×

11

10 ×

11

10

= ` 159720.

6. The population of a town increases by 10%

every year. If the present population is 60,500, find

its population (i) after 2 years (ii) 2 years ago.

Solution.

Given, increase of population = 10% p.a.

Present population = 60500

(i) Population after two years

2

P 1100

r

= 60500

210

1100

= 60500 ×

211

10

= 60500 × 11

10 ×

11

10

= 73205

(ii) Further population 2 years ago, we have

A R

P 1

n

60500

210

= P 1+100

211

= P10

P = 60500 ×

210

11

P = 60500 × 10

11 ×

10

11 = 50000

Thus required population 2 years ago=50000.

II. Depreciation

7. The value of machine depreciates every

year by 5% p.a. If the present value is ` 4,11,540,

what was its value 3 years ago ?

Solution.

Present value of machine (P) = ` 411540



Rate of decrease = 5% p.a.

3 years ago its value was given by

A R

P 1

n

` 411540

35

P 1100

` 411540

319

P20

P = ` 411540 ×

320

19

= ` 411540 × 20

19 ×

20

19 ×

20

19

= 4,80,000

Thus, 3 years ago, value of machine was

= ` 480,000.

8. A waching machine was bought at

`15,625 three years ago. If its value depreciates

at 16% p.a. calculate.

(i) its present value

(ii) its value after one years from now

Solution.

3 years ago, the value of washing machine

= ` 15625

depreciation in value = 16% p.a.

Present value P

P 1

n

= ` 15625

316

1100

= ` 15625 ×

321

25

= ` 15625 × 21

25 ×

21

25 ×

21

25

= ` 9261

(ii) Value after 1 year now

= 9261 × 16

1100

= ` 9261 × 21

25 = `

194481

25

= ` 7775.24

9. The present value of some equipment in a

firm is ` 1,09,350. If the rate of depreciation is

1% p.a., how many years ago was its value

` 1,50,000.

Solution.

Given, present value of some equipment

= ` 1,09,350

and rate of depreciation = 10%

Some years ago, its value was ` 150000

Since, A R

P 1

n

A

P

101

100

n

3109350 9 9

150000 10 10

n n

n = 3

required period = 3 years.

10. A piece of machinery was purchased by

a factory. It depreciates at 10% every year. At the

end of second year its value is ` 1,29,600. What

was its cost price ?

Solution.

Rate of depreciation = 10% p.a.

Value of machinery at the end of the second

year = ` 129600

we know that

A P 1

nr

129600

2 210 9

P 1 P100 10


41Compound Interest

P = ` 129600 ×

210

9

= ` 129600 × 10

9 ×

10

9

= ` 160000

Required cost price = ` 160000.

11. A scooter is bought for ` 31,250. If its

value after 2 years is ` 24,200, find its rate of

depreciation.

Solution.

Cost price of a scooter (P) = ` 31250

Value after 2 years (A) = ` 24200

We know that

A

P 1

100

nr

224200

131250 100

r

484

625

2

1100

r

2

1100

r

222

25

1100

r

22

25

100

r

22 25 22 31

25 25 25

r 3

100 1225

required rate of depreciation = 12% p.a.

12. The machinery of a factory was valued

at ` 18,400 at the end of 2000. If it depreciated

at 8% of the value at the beginning of the year,

calculate its value at the end of 1999 and 200.

Solution. At the end of 2000,

The value of machinery of a factory

= ` 18400

Rate of depreciation = 8%

(i) Value of machinery at the end of 1999

18400

1100

nr

1

18400

81

100

18400

23

25

= ` 18400 × 25

23 = ` 20000

(ii) Value at the end of 2000

= ` 18400

1

1100

r

= 18400

81

100

= ` 18400 × 23

25 = ` 16928.


1. Rahul borrows ` 15,000 for 2 years at CI

annually from a bank. After one year, it amounts

to ` 16,200. Calculate (i) the rate of interest

(ii) the interest for the second year.

Solution.

Sum borrowed by Rahul (P) = ` 15000


Amount after one year (A) = ` 16200

Interest for the first year = A – P

= ` 16200 – 15000 = ` 1200

(i) Rate SI ×100 200 100

=P ×T 15000 1

= 8%


16200 8 1

100

= ` 1296.

2. A sum of ` 20000 yields ` 3328 as

compound interest in 2 years. Find the rate of

interest.



Solution.

Principal (P) = ` 20000, Interest = ` 3328

Amount = P + S.I. = 20000 + 3328

= ` 23328


We know that

A

P

R1

100

n

223328 R

120000 100

2 2R 729 27

1100 625 25

R 27

1100 25

R 27 27 25 2

1100 25 25 25

R 2

10025

= 8

required rate of interest be 8% p.a.

3. Amrit Raj borrows ` 50,000 from a bank

at 9% p.a. compound interest. He repays

` 24,500 at the end of first year and ` 12,700 at

the end of the second year. Find the amount

outstanding at the beginning of the third year.

Solution.

Sum borrowed by Amrit Raj = ` 50,000

Rate of interest = 9% p.a.


=100

= ` 50000 9 1

100

= ` 4500

Total amount after first year

= ` 50000 + 4500

= ` 54500

Amount returned = ` 24500

Balance = ` 54500 – 24500 = 30000


= ̀30000 9 7

100

= ` 2700

Amount after second year = ` 30000 + 2700

= ` 32700

Amount repaid = ` 12700

Balance in the end of the second year

= ` 32700 – 12700 = ` 20000

Balance in the beginning of third year

= ` 20000.

4. A certain sum amounts of ` 26460 in 2

years and ` 27,783 in 3 years when compounded

annually. Calculate the rate of interest and the sum

borrowed.

Solution.




S.I. on ` 26460 = ` 27783 – 26460 = ` 1323

` 1323 is interest on ` 26460 for 1 year

Rate Interest ×100

=P × T

1323 100

26460 1

= 5%

Now A R

P 1+100

n

26460

25

P 1+100

= P

221

20

P = 26460 ×

220

21

= 26460 × 20

21 ×

20

21

= ` 24000



43Compound Interest

5. Khushnaz borrows a sum of ` 8000 for 2

years at SI and immediately lends out this money

to another person for the same period at the same

rate. If she makes a profit of ` 96.80 on this

transaction at the end of two years, calculate the

rate of interest.

Solution.

Sum borrowed by Khushnaz = ` 8000


Difference in CI and SI = ` 96.80

Let r be the rate % p.a.

Then SI PRT 8000 2

100 100

r = 160r

and CI P 1 P100

nr

2

P 1 1100

r

2

8000 1 110000 50

r r

2

800010000 50

r r

24

1605

rr

Now CI – SI 24

1605

rr – 160r

29680 4

100 5

r r2

9680 5

100 4 = 121

r2 = (11)2

r = 11

required rate of interest = 11% p.a.

6. On a sum of money, the difference

between CI for a year payable half yearly and the

SI for a year is ` 80. If the rate of interest is

5%, find the sum.

Solution.


Period (n) = 1 year

Rate = 5% p.a.

Let the required sum be ` 100

then SI PRT 100×5×1

= =100 100

= ` 5

and in case of CI

Rate = 5% p.a. or 5

2% half yearly

Period (n) = 1 year or 2 half years.

We know that

CI P 1 P100

nr

= ` 100

25

1 1002 100

= ` 100 ×

241

40

– 100

= 41 41

100 10040 40

= ` 1681

16 – 100

1681 1600

16

= `

81

16

Difference in CI and SI

81 85 80 1

16 1 16 16


16 the principal = ` 100

and if difference ` 80 then principal

= ̀100 16 80

1

= ` 128000

Principal = ` 128000.



7. The price of a car is ` 3,50,000. The value

of the car depreciates by 20% in the first year

and after that it depreciates by 25% every year.

What will be the car’s value after 3 years ?

Solution.

Given, Price of a car (P) = ` 350000

Rate of depreciation = 20% p.a. in the first

and 25% p.a. for the next year

Total period (n) = 3 years

Amount after 3 years

21 2P 1 1

100 100

r r

= ̀

220 25

350000 1 1100 100

24 3

3500005 4

`

= ` 350000 × 4 3 3

5 4 4

= ` 1,57,500.

8. The value of a machine depreciates every

year at the rate of 10%. If the value after 3 years

after it was purchased is ` 10,935 at what price

was it purchased.

Solution.

Given, rate of depreciation (r) = 10%

Value of machine after 3 years (A) = ` 10935

We know that

A P 1100

nr

10935

310

P 1100

10935

39

P10

P = 10935×

310 10 10 10

1099 9 9 9

= ` 15000

Thus, required cost price of the machine

= ` 15000.

9. At what rate of p.a. will a sum of

` 4000 yield ` 1324 as compound interest in 3

years.

Solution. Given, principal (P) = ` 4000

CI = ` 1324

Amount = ` 4000 + 1324 = ` 5324


We know that

A1

P 100

nr

3324

14000 100

r

3

1100

r

3331 1

1000 10

11

1100 10

r

11

1100 10

r

11 10 1

10 10

r 1

10010

= 10

required rate of interest = 10% p.a.

10. The compound interest calculated yearly,

on a certain sum of money for the second year

is ` 1320 and for the third year is ` 1452.

Calculate the rate of interest and the original sum

of money.

Solution.

Given, CI for 2nd year = ` 1320

CI in 3rd year = ` 1452


45Compound Interest

Subtracting, ` 132 is interest on ` 1320 for1 year.

Rate S.I.×100 132×100

= =P× T 1320×1

= 10%.

Let P = ` 10

then interest for the first year 100 10 1

100

= ` 10

and amount after 1 year = 100 + 10 = ` 110

C.I. for the second year = 110 10 1

100

= ` 11

If C.I. is ` 11 then principal = ` 100

and if interest is ` 1320 then principal

1001320

11

= ` 12000

11. The CI earned by a sum of money is 2

years at 5% p.a. is ` 4305. Find the sum.

Solution.

Given, CI = ` 4305

Rate (R) = 5%


Let sum (P) = ` 100

CI P 1 P100

nr

4305

25

P 1 P100

221 441

P 1 P 120 400

441 400 41P P

400 400

P 4305 400

41

= ` 42000

required sum = ` 42000.

12. The machinery of a factory was valued

at ` 18,400 at the end of 2000. If it depreciated

at 8% of the value at the beginning of the year,

calculate its value at the end of 1999 and 2001.

Solution.

Given value of machinery at end of year 2000

= ` 18400

Also given rate depreciated at 8% of the value

of beginning of year.

Let, value of machinery at the end of year

1999 (or beginning of year 2000) = x

So,

x – 8% of x = 18400,

8

18400100

xx

2

1840025

xx

23

1840025

x

x = 25

1840023

x = 800 × 25

x = 20000

So,

Value of machinery at the end of year 1999

= ` 20,000

And

Value of machinery at the end of year 2001

= 18400 – 8% of 18400

= 18400 – 18400 8

100

= 18400 – 184 × 8

= 18400 – 1472 = ` 16,928


PointstoRemember

Formula used are given below :

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

2. (a – b)2 = a2 – 2ab + b2

3. (a + b) (a – b) = a2 – b2

4. (a + b)2 – (a – b)2 = 4ab

5.

22

2

1 12a a

a a

6.

22

2

1 12a a

a a

7.2

2

1 1 1a a a

a a a

8.

2 21 1

4a aa a

9. (i) (x + a) (x + b) = x2 + (a + b) x + ab

(ii) (x + a) (x – b) = x2 + (a – b) x – ab

(iii) (x – a) (x + b) = x2 – (a – b)x – ab

(iv) (x – a) (x – b) = x2 – (a + b)x + ab

10. (a + b)3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab (a + b)

11. (a – b)3 = a3 – 3a2b + 3ab2 – b3 = a3 – b3 – 3ab (a – b)

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

13. If a + b + c = 0, then a3 + b3 + c3 = 3abc.

3

Expansions

46


47Expansions

Exercise 3A

1. Expand the following :

(i) (3a + 5b)2 (ii) (4a – 7b)2

(iii)

22 5

5 2

a b

b a

(iv) (2ab + 3cd)2

(v) (2a3 + 5b2)2 (vi) (–2a – 5b)2

Solution.

(i) (3a + 5b)2

= (3a)2 + 2 × 3a × 5b + (5b)2

= 9a2 + 30ab + 25b2.

(ii) (4a – 7b)2

= (4a)2 – 2 × 4a × 7b + (7b)2

= 16a2 – 56ab + 49b2

(iii)

22 5

5 2

a b

b a

2 22 5 5

25 5 2 2

a a b b

b b a a

[ (a – b)2 = a2 – 2ab + b2]

2 2

2 2

4 252

25 4

a b

b a .

(iv) (2ab + 3cd)2

= (2ab)2 + 2 × 2ab × 3cd + (3cd)2

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

= 4a2b2 + 12abcd + 9c2d2.

(v) (2a3 + 5b2)2

= (2a3)2 + 2 × 2a3 × 5b2 + (5b2)2

= 4a6 + 20a3b2 + 25b4.

(vi) (–2a – 5b)2

= (–2a)2 + 2(–2a) (–5b) + (–5b)2

= 4a2 + 20ab + 25b2.

2. Evaluate using algebraic formula.

(i) (104)2 (ii) (99)2

(iii) (10.2)2 (iv) (9.7)2

(v) 992 + 2(99) + 1

(vi) 4.82 + 2(4.8) (0.2) + (0.2)2

Solution.

(i) (104)2 = (100 + 4)2

= (100)2 + 2 × 100 × 4 + (4)2

= 10000 + 800 + 16 = 10816

(ii) (99)2 = (100 – 1)2

= (100)2 – 2 × 100 × 1 + (1)2

= 10000 – 200 + 1 = 10001 – 200

= 9801.

(iii) (10.2)2 = (10 + 0.2)2

= (10)2 + 2 × 10 × (0.2) + (0.2)2

= 100 + 20 × 0.2 + 0.04

= 100 + 4 + 0.04 = 104.04

(iv) (9.7)2 = (10 – 0.3)2

= (10)2 – 2 × 10 × 0.3 + (0.3)2

[ (a – b)2 = a2 – 2ab + b2]

= 100 – 6 + 0.009

= 94.09.

(v) 992 + 2(99) + 1

= (99)2 + 2 × 99 × 1 + (1)2

= (99 + 1)2 = (100)2 = 10000.

(vi) 4.82 + 2(4.8) (0.2) + (0.2)2

= (4.8)2 + 2 × 4.8 × 0.2 × (0.2)2

= (4.8 + 0.2)2 = (5.0)2 = 25.

3. If a + b = 9 and ab = 18, find (a – b).

Solution. Given, a + b = 9, ab = 18

(a – b)2 = (a + b)2 – 4ab

= (9)2 – 4 × 18

= 81 – 72 = 9 = (±3)2

a – b = ±3.

4. a – b = 5 and ab = 6, find a + b.

Solution.

Given, a – b = 5, ab = 6

(a + b)2 = (a – b)2 + 4ab

= (5)2 + 4 × 6

= 25 + 24 = 49 = (±7)2

a + b = ±7.



5. If 1

4xx

, find 2

2

1x

x .

Solution.

Given,1

4xx


21

xx

= (4)2

2

2

12x

x = 16

2

2

1x

x = 16 – 2 = 14

Hence 2

2

1x

x = 14.

6. If 1

8xx

, find 2

2

1x

x .

Solution.

Given,1

8xx

Squaring both sides

21

xx

= (8)2,

2

2

12x

x = 64

2

2

1x

x = 64 + 2

i.e. 2

2

1x

x = 66

7. If x + y 5

2 and xy = 1, find x – y.

Solution.

Given, x + y 5

2


(x + y)2

25

2

x2 + y2 + 2xy 5

4

x2 + y2 + 2 × 1 5

4

x2 + y2 + 25

4

x2 + y2 5 7

24 4

Now (x – y)2 = x2 + y2 – 2xy

17

2 14

217 9 3

24 4 2

Thus, x – y = ±3

2.

8. If 3x + 4y = 13 and xy = 1, find 3x – 4y.

Solution.

Given, 3x + 4y = 13, xy = 1

(3x – 4y)2 = (3x + 4y)2 – 4 × 3x × 4y

[ (a – b)2 = (a + b)2 – 4ab]

= (3x + 4y)2 – 48xy

= (13)2 – 48 × 1 = 169 – 48

= 121 = (±11)2.

3x – 4y = ± 11

9. If x2 – 4x = 1, find x2 + 2

1

x.

Solution.

x2 – 4x = 1,

Dividing both sides by x; we have

2 4x x

x x x

x – 4

1

x

1

xx

= 4


49Expansions

Squaring both sides; we get

21

xx

= (4)2 2

2

12x

x = 16

2

2

1x

x = 16 + 2 = 18.

10. If x2 – 8x + 1 = 0, find 2

2

1x

x .

Solution.

x2 – 8x + 1 = 0

Dividing both sides by x, we have

2 8 1x x

x x x = 0 x – 8

1

x = 0

1

xx

= 8


21

xx

= (8)2 2

2

12x

x = 64

2

2

1x

x = 64 – 2 = 62.

11. If a2 + b2 = 65 and ab = 8, find the value

of a2 – b2.

Solution.

Given, a2 + b2 = 65, ab = 8

(a2 – b2)2 = (a2 + b2)2 – 4a2b2

= (65)2 – 4(8)2 = 4225 – 4 × 64

= 4225 – 256 = 3969

= (±63)2

a2 – b2 = ±63.

12. If 3x + 4y = 16, and xy = 4, find the value

of 9x2 + 16y2.

Solution.

Given, 3x + 4y = 16, xy = 4

On squaring both sides we have

(3x + 4y)2 = (16)2

9x2 + 16y2 + 2 × 3x × 4y = 256

9x2 + 16y2 + 24xy = 256

9x2 + 16y2 + 24 × 4 = 256

9x2 + 16y2 + 96 = 256

9x2 + 16y2 = 256 – 96 = 160

9x2 + 16y2 = 160.

13. If 2

2

151x

x , find

1x

x .

Solution. Now,

22

2

1 12x x

x x

= 51 – 2 = 49 = (±7)2

1

xx

= ±7.

14. If 2

2

1 24 14

39x

x , find

12

3x

x

.

Solution.

Given, 22

1 24 14

39x

x

22

2

1 1 12 (2 ) 2 2

3 3(3 )x x x

x xx

22

1 44

39x

x

44 4 4816

3 3 3

= (±4)2

Thus, 1

23

xx

= ±4

15. If 4

4

1119x

x , find

(i) 2

2x

x

(ii)

1x

x .



Solution.

Given, 4

4

1119x

x

(i)

22 4

2 4

1 12x x

x x

= 119 + 2 = 121 = (11)2

2

2

1x

x = 11.

22

10x

x

(ii)

22

2

1 12x x

x x

= 11 – 2 = 9 = (±3)2

1

xx

= ±3.

16. If x – y = 5, xy = 84, find x + y.

Solution.

Given, x – y = 5, xy = 84

Now, (x + y)2 = (x – y)2 + 4xy

= (5)2 + 4 × 84 = 25 + 336

= 361 = (±19)2.

Thus, x + y = ±19.

17. 2 1

5x

x

find

2xx

.

Solution.

Given, 2 1x

x

= 5

1x

x = 5


21

xx

= (5)2

2

2

12x

x = 25

2

2

1x

x = 25 – 2 = 23

2

2

1x

x = 23.

18. If 2 1

7x

x

find

22

1x

x .

Solution.

Given, 2 1

7x

x

1

xx

= 7


21

xx

= (7)2

2

2

12x

x = 49

2

2

1x

x = 49 + 2 = 51

2

2

1x

x = 51.

19. Find the missing terms in the following

expressions to make each a perfect square.

(i) 18 9x

(ii) 4x2 – ? + 49

(iii) 25x2 + 40xy + ?

Solution. (i) 18 9x

+ 2 × 3x × 3 + (3)2

First term will be (3x)2 = 9x2

9x2 + 18x + 9 = (3x + 3)2

(ii) 4x2 – + 49

(2x)2 – + (7)2

Middle term = 2 × 2x × 7 = 28x

4x2 – 28x + 49 = (2x – 7)2

(iii) 25x2 + 40xy +

(5x)2 + 2 × 5x × 4y +

Last term = (4y)2 = 16y2

25x2 + 4xy + 16y2 = (5x + 4y)2


51Expansions


(i) (2a + 3b)3 (ii) (5a + 4b)3

(iii)

31

2aa

(iv) (5a – 3b)3

(v) (2a – 7b)3 (vi)

31

3aa

(vii) (5ab + 2)3 (viii) (3 – 4ab)3

Solution.

(i) (2a+3b)3=(2a)3+(3b)3+3×2a×3b (2a+3b)

[ (a + b)3 = a3 + b3 + 3ab (a + b)]

= 8a3 + 27b3 + 18ab (2a + 3b)

= 8a3 + 27b3 + 36a2b + 54ab2

(ii) (5a + 4b)2 = (5a)3 + 3(5a)2 (4b)

+ 3(5a) (4b)2 + (4b)3

= 125a3 + 3 × 25a2 × 4b + 3 × 5a × 16b2

+ 64b3

= 125a3 + 300a2b + 240ab2 + 64b3.

(iii)

31

2aa

= (2a)3 + 3 × (2a)2 × a

+ 3 × (2a)

2 31

a a

.

= 8a3 + 3 × 4a2 × 1

a + 3 × 2a × 2

1 1

a a

= 8a3 + 12a + 3

6 1

a a .

(iv) (5a – 3b)3 = (5a)3 – 3(5a)2 (3b)

+ 3(5a) (3b)2 – (3b)3

= 125a3 – 3 × 25a2 × 3b + 3 × 5a × 9b2

– 27b3

= 125a3 – 225a2b + 135ab2 – 27b3.

(v) (2a – 7b)3 = (2a)3 – 3(2a)2 (7b)

+ 3(2a) (7b)2 – (7b)3

= 8a3–3×4a2×7b + 3 × 2a × 49b2 – 343b3

= 8a3 – 84a2b + 294ab2 – 343b3.

(vi)

31

3aa

= (3a)3–3×(3a)23

2

1 1 13(3 )a

a aa

.

= 27a3 – 3 × 9a2 × 1

a + 3 × 3a × 2 3

1 1

a a

= 27a3 – 27a + 3

9 1

a a .

(vii) (5ab + 2)3 = (5ab)3 + 3 × (5ab)2 (2)

+ 3(5ab) (2)2 + (2)3.

= 125a3b3 + 3 × 25a2b2×2+3 × 5ab × 4 + 8

= 125a3b3 + 150a2b2 + 60ab + 8

(viii) (3 – 4ab)3 = (3)3 – 3(3)2 (4ab)

+ 3(3) (4ab)2 – (4ab)3.

= 27 – 3 × 9 × 4ab + 3 × 3×16a2b2–64a3b3

= 27 – 108ab + 144a2b2 – 64a3b3.

2. (i) If a + b = 6, ab = 8, find a3 + b3.

(ii) If a + b = 8, ab = 15, find a3 + b3.

Solution.

(i) Given a + b = 6, ab = 8

a + b = 6

Cubing both sides; we have

(a + b)3 = a3 + b3 + 3ab (a + b)

(6)3 = a3 + b3 + 3 × 8 × 6

216 = a3 + b3 + 144

a3 + b3 = 216 – 144 = 72.

(ii) Given a + b = 8, ab = 15

since, a + b = 8

Exercise 3B



Cubing both sides, we have

(a + b)3 = (8)3

a3 + b3 + 3ab (a + b) = 512

a3 + b3 + 3 × 15 × 8 = 512

a3 + b3 + 360 = 512

a3 + b3 = 512 – 360 = 152

a3 + b3 = 152.

3. (i) If a – b = 7, ab = 30, find a3 – b3.

(ii) If a – b = 9, ab = 10, find a3 – b3.

Solution.

(i) a – b = 7, ab = 30

a – b = 7


(a – b)3 = (7)3

a3 – b3 – 3ab (a – b) = 343

a3 – b3 – 3 × 30 × 7 = 343

a3 – b3 – 630 = 343

a3 – b3 = 343 + 630 = 973

a3 – b3 = 973.

(ii) a – b = 9, ab = 10

(a – b) = 9


(a – b)3 = (9)3

a3 – b3 – 3ab (a – b) = 729

a3 – b3 – 3 × 10 × 9 = 729

a3 – b3 – 270 = 729

a3 – b3 = 729 + 270 = 999

a3 – b3 = 999.

4. If 1

2xx

, find 3

3

1x

x .

Solution. Given, 1

2xx


31

xx

= (2)3

3

3

1 13x x

xx

= 8

3

3

1x

x + 3 × 2 = 8

3

3

1x

x + 6 = 8

3

3

1x

x = 8 – 6 = 2

3

3

1x

x = 2.

5. If 1

5xx

, find 3

3

1x

x .

Solution.

Given, 1

5xx


3

1x

x

= (5)3

3

3

13x x

xx

= 125

3

315x

x

= 125

3

3x

x

= 125 + 15 = 140

3

3x

x

= 140.

6. If x2 – 4x + 1 = 0, find 33

xx

.

Solution.

Given, x2 – 4x + 1 = 0


53Expansions

Dividing by x

2 4 1x x

x x x = 0 x – 4 +

1

x = 0

1

xx

= 4


31

xx

= (4)3

3

3

1 13x x

xx

= 64

3

3

13x

x × 4 = 64

3

3

1x

x + 12 = 64

3

3

1x

x = 64 – 12 = 52

3

3

1x

x = 52.

7. If x2 – 6x – 1 = 0, find 3

3

1x

x .

Solution.

Given, x2 – 6x – 1 = 0

Dividing by x, we have

2 6 1x x

x x x = 0

x – 6 – 1

x = 0

x – 1

x = 6


31

xx

= (6)3

3

3

1x

x – 3 × 6 = 216

3

3

1x

x – 18 = 216

3

3

1x

x = 216 + 18 = 234

3

3

1x

x = 234.

8. If 3

4xx

, find 3

3

27x

x

Solution.

Given, 3

4xx


3

xx

= (4)3

3

3 3 3 3( ) 3x x x

x x x

= 64

3

3

279 4x

x = 64

3

3

2736x

x = 64

3

3

27x

x = 64 + 36 = 100

3

3

27x

x = 100.

9. If 22

19 13

4x

x , find 27x3 + 3

1

8x.

Solution.

Given, 22

19 13

4x

x



22(3 )

2x

x

= 13

22 1

(3 ) 2 32 2

x xx x

= 13 + 2 × 3x × 1

2x

21

32

xx

= 13 + 3 = 16 = (4)2

1

32

xx

= 4


31

32

xx

= (4)3

33 1 1 1

(3 ) 3 3 32 2 2

x x xx x x

= 64.

3

3

1 927 4

28x

x = 64

3

3

127 18

8x

x = 64

3

3

127

8x

x = 64 – 18 = 46

3

3

127

8x

x = 46.

10. If 2

225 20

4x

x

, find

33

1125

8x

x .

Solution.

Given, 22

25 204

xx

2

2 1 1(5 ) 2 5

2 2x x

x x

= 20 + 2 × 5x × 1

2x

2

15

2x

x

= 20 + 5 = 25 = (5)2

2 2 2[ 2 ( ) ]a ab b a b

1

52

xx

= 5 ...(i)


31

52

xx

= (5)3

3

3 1 1 1(5 ) 3 5 5

2 2 2x x x

x x x

= 125

3

3

1 15125 5

28x

x = 125

[using eqn. (i)]

3

3

1 75125

28x

x = 125

3

3

1125

8x

x = 125 –

75

2

250 75 175

2 2

= 87.5

Thus, 3

3

1125

8x

x = 87.5.

11. If 2 1

7x

x

, find

33

1x

x .

Solution.

Given 2 1x

x

= 7

2 1x

x x = 7

1x

x = 7 ...(i)


3

3

1 13x x

xx

= (7)3 = 343


55Expansions

3

3

1x

x + 3 × 7 = 343

[ using eqn. (i)]

3

3

1x

x + 21 = 343

3

3

1x

x = 343 – 21 = 322

3

3

1x

x = 322.

12. If 2 1

8x

x

, find

33

18.x

x

Solution.

Given,2 1

8x

x

,

2 1x

x x = 8

1x

x = 8 ...(i)


31

xx

= (8)3

3

3

13x x

xx

= 512

[ using eqn. (i)]

3

3

13x

x × 8 = 512

3

3

124x

x = 512

3

3

1x

x = 512 + 24 = 536

3

3

1x

x = 536.

13. (i) If a + b = 5 and a3 + b3 = 35, find

a2 + b2.

(ii) If a + b = 8 and a3 + b3 = 152, find

a2 + b2.

Solution.

(i) a + b = 5, a3 + b3 = 35 ...(i)

(a + b)3 = a3 + b3 + 3ab(a + b) [using(i)]

(5)3 = 35 + 3ab × 5

125 = 35 + 15ab 15ab = 125 – 35 = 90

ab 90

15 = 6

Now a2 + b2 = (a + b)2 – 2ab

= (5)2 – 2 × 6 = 25 – 12 = 13

(ii) Given a + b = 8, a3 + b3 = 152

(a + b)3 = a3 + b3 + 3ab (a + b)

(8)3 = 152 + 3ab × 8

512 – 152 = 24ab

24ab = 360 ab 360

24 = 15

Now a2 + b2 = (a + b)2 – 2ab

= (8)2 – 2 × 15 = 64 – 30 = 34

a2 + b2 = 34.

14. (i) If a – b = 3 and a3 – b3 = 63, find

a2 + b2.

(ii) If a – b = 4 and a3 – b3 = 124, find

a2 + b2.

Solution.

(i) a – b = 3, a3 – b3 = 63 ...(i)

(a – b)3 = a3 – b3 – 3ab (a – b)

(3)3 = 63 – 3ab × 3 [using eqn. (i)]

27 = 63 – 9ab 9ab = 63 – 27 = 36

ab 36

9 = 4.

Now a2 + b2 = (a – b)2 + 2ab

= (3)2 + 2 × 4 = 9 + 8 = 17.

(ii) Given, a – b = 4, a3 – b3 = 124.

(a – b)3 = a3 – b3 – 3ab (a – b)

(4)3 = 124 – 3ab × 4

64 = 124 – 12ab

12ab = 124 – 64 = 60



ab 60

12 = 5

Now a2 + b2 = (a – b)2 + 2ab

= (4)2 + 2 × 5

= 16 + 10 = 26.

15. Evaluate using the formula.

(i) 793 + 3(79)2 + 3(79) + 1

(ii) 483 + 6(48)2 + 12(48) + 8.

Solution.

(i) 793 + 3(79)2 + 3(79) + 1

= (79)3 + 3(79)2 × 1 + 3(79) (1)2 + (1)3

= (79 + 1)3 = (80)3 = 512000.

[ a3 + 3a2b + 3ab2 + b3 = (a + b)3]

(ii) 483 + 6(48)2 + 12(48) + 8

= (48)2 + 3(48)2 × 2 + 3(48) × 22 + 23

= (48 + 2)2 = (50)3 = 125000.

16. Without calculating the cubes, find the

value of :

(i) (–35)3 + (18)3 + (17)3

(ii) (25)3 + (–17)3 + (–8)3.

Solution.

(i) (–35)3 + (18)3 + (17)3

Let a = –35, b = 18 and c = 17

then a + b + c = –35 + 18 + 17 = 0

Thus, a3 + b3 + c3 = 3abc

(–35)3 + (18)3 + (17)3 = 3 × (–35) (18) (17)

= –32130.

(ii) (25)3 + (–17)3 + (–8)3

Let a = 25, b = –17, c = –8

then a + b + c = 25 – 17 – 8 = 0

Thus, a3 + b3 + c3 = 3abc

(25)3 + (–17)3 + (–8)3 = 3 × 25 × –17 × (–8)

= 10200.

17. (i) If 4xx

, find

33

27x

x

(ii) If 1

2 1xx

, find 3

3

18x

x .

Solution.

(i) Given, 4xx


33

xx

= (4)3

3

3 3 3 3( ) 3x x x

x x x

= 64

3

3

27x

x + 9 × 4 = 64

3

3

27x

x + 36 = 64

3

3

27x

x = 64 – 36 = 28

3

3

27x

x = 28.

(ii) Given 1

2 1xx


31

2xx

= (1)3

(2x)3 –

3

x

– 3 × 2x × 1 1

2 1xx x

3

3

18 3 2 1x

x = 1

3

3

18x

x – 6 = 1

3

3

18x

x = 1 + 6 = 7

3

3

18x

x = 7.


57Expansions

18. If x + y + z = 0, prove that

2 2 2( ) ( ) ( )3

x y y z z x

xy yz zx

Solution. Given x + y + z = 0

x3 + y3 + z3 = 3xyz ...(i)

Now x + y + z = 0

x + y = –z (x + y)2 = z2

Similarly (y + z)2 = x2 and (z + x)2 = y2

Now 2 2 2 2 2 2( ) ( ) ( )x y y z z x z x y

xy yz zx xy yz zx

.

3 3 3z x y

xyz

3xyz

xyz = 3 = RHS.

19. If x = 43, y = –61, z = 18, find the value

of

2 2 2) ( ) ( )x y y z z x

xy yz zx

.

Solution. Given, x = 43, y = –61, z = 18

x + y + z = 43 – 61 + 18 = 0

Thus, x3 + y3 + z3 = 3xyz

Now x + y + z = 0

x + y = –z

(x + y)2 = (–z)2 = z2

Similarly (y + z)2 = (–x)2 = x2

(z + x)2 = (–y)2 = y2

Now

2 2 2) ( ) ( )x y y z z x

xy yz zx

2 2z x y

xy yz zx

3 3 3 3 3 3 33

z x y x y z xyz

xyz xyz xyz

.

20. 2 1 1

33

x

x

, find

(i) 1

xx

(ii) 3

1x

x

Solution.

Given 2 1 1

33

x

x

10

3

1 10

3x

x

(i)

2 21 1

4x xx x

210 100 4

43 9 1

2100 36 64 8

9 9 3

1

xx

8 2

23 3

.

(ii) Given, 1 8

3x

x


3 31 8

3x

x

3

3

1 1 5123

27x x

xx

3

3

1 8 5123

3 27x

x

3

3

1 5128

27x

x



3

3

1 5128

27x

x 512 216 28

27 27

3

3

1x

x

728 2626

27 27 .

21. If 44

1527x

x , find

(i) 2

2

1x

x (ii)

1x

x

(iii) 3

3

1x

x

Solution. Given, 44

1x

x = 527

(i) Now,

22

2

1x

x

44

12x

x

= 527 + 2 = 529 = (23)2

2

2

1x

x = 23.

(ii)

22

2

1 12x x

x x

= 23 + 2 = 25 = (± 5)2

1

xx

= ±5.

(iii) Since, 1

xx

= 5


31

xx

= (5)3

3

3

1 13x x

xx

= 125

3

3

13 5x

x = 125

3

3

115x

x = 125

3

3

1x

x = 125 – 15 = 110

3

3

1x

x = 110.

22. If a + b = 8, a2 + b2 = 34, find a3 + b3.

Solution.

Given, a + b = 8, a2 + b2 = 34


(a + b)2 = (8)2

a2 + b2 + 2ab = 64

34 + 2ab = 64

2ab = 64 – 34 = 30

ab 30

152

Now a + b = 8


(a + b)3 = (8)3

a3 + b3 + 3ab(a + b) = 512

a3 + b3 + 3 × 15 × 8 = 512

a3 + b3 + 360 = 512

a3 + b3 = 512 – 360 = 152

a3 + b3 = 152.

23. Evaluate using algebraic formula.

(i) 1023 (ii) 993

(iii) 10.13 (iv) 9.83

(v) 993 + 3(99)2 + 3(99) + 1

Solution.

(i) (102)3 = (100 + 2)3

= (100)3 + 3 × 100 × 2(100 + 2) + (2)3


59Expansions

= 1000000 + 600 × 102 + 8

= 1000000 + 61200 + 8 = 1061208.

(ii) (99)3 = (100 – 1)3

= (100)3 – (1)3 – 3 × 100 × 1 (100 – 1)

= 1000000 – 1 – 300 × 99

= 1000000 – 1 – 29700

= 1000000 – 29701

= 970299.

(iii) (10.1)3 = (10 + 0.1)3

= (10)3 + (0.1)3 + 3 × 10 × 0.1 (10 + 0.1)

= 1000 + 0.001 + 3 × 10.1

= 1000 + 0.001 + 30.3

= 1030.301

(iv) (9.8)3 = (10 – 0.2)3

= (10)3 – (0.2)3 – 3 × 10 × 0.2(9.8)

= 1000 – 0.008 – 58.8

= 1000 – 58.808

= 941.192.

(v) 993 + 3(99)2 + 3(99) + 1

= (99)3 + 3(99)2 × 1 + 3(99) (1)2 + (1)3

[ (a + b)3 = a3 + b3 + 3ab (a + b)]

= (99 + 1)3 = (100)3

= 1000000.

Exercise 3C

Expand using the formula.

1. (x + 8) (x + 3).

Solution. (x + 8) (x + 3)

= x2 + (8 + 3)x + 8 × 3

= x2 + 11x + 24.

2. (x – 5) (x + 6).

Solution. (x – 5) (x + 6)

= x2 + (–5 + 6) x + (–5) × 6

= x2 + x – 30.

3. (a – 4) (a – 3).

Solution. (a – 4) (a – 3)

= a2 + (–4 – 3) x + (–4) (–3)

= a2 – 7x + 12.

4. (a + 7) (a – 2)

Solution. (a + 7) (a – 2)

= a2 + (7 – 2)a + (7) (–2)

= a2 + 5a – 14.

5. (3x + 5y + 2) (3x + 5y – 2).

Solution. (3x + 5y + 2) (3x + 5y – 2)

= [(3x + 5y) + 2] [(3x + 5y) – 2]

= (3x + 5y)2 – (2)2

= (3x)2 + (5y)2 + 2 × 3x × 5y – 4

= 9x2 + 25y2 + 30xy – 4.

6. (4x + 2y + 3) (4x + 2y – 3).

Solution. (4x + 2y + 3) (4x + 2y – 3)

= [(4x + 2y) + 3] [(4x + 2y) – 3]

= (4x + 2y)2 – (3)2

= (4x)2 + (2y)2 + 2 × 4x × 2y – 9

= 16x2 + 4y2 + 16xy – 9.

7. (2a + 3b + 5) (2a – 3b + 5).

Solution.

(2a + 3b + 5) (2a – 3b + 5)

= [(2a + 5) + 3b] [(2a + 5) – 3b]

= (2a + 5)2 – (3b)2

= (2a)2 + (5)2 + 2 × 2a × 5 – 9b2

= 4a2 + 25 + 20a – 9b2

= 4a2 – 9b2 + 20a + 25.

8. (4a + 5b – 7) (4a + 5b + 2).

Solution.

(4a + 5b – 7) (4a + 5b + 2)

= [(4a + 5b – 7)] [(4a + 5b) + 2]

= (4a + 5b)2 + (–7 + 2) (4a + 5b) + (–7) (2)

= (4a)2 + (5b)2 + 2 × 4a × 5b – 5(4a + 5b) – 14

= 16a2 + 25b2 + 40ab – 20a – 25b – 14.

9. Expand :

(i) (2a – 3b + 5c)2 (ii) (4a – 5b – 6)2



(iii)

2

2 42

ab c

(iv)

2

5 42

ba c

Solution.

(i) (2a – 3b + 5c)2

= (2a)2 + (–3b)2 + (5c)2 + 2 × 2a × (–3b)

+ 2(–3b) (5c) +2(5c) × 2a

= 4a2 + 9b2 + 25c2 – 12ab – 30bc + 20ca.

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

(ii) (4a – 5b – 6)2

= (4a)2 + (–5b)2 + (–6)2 + 2(4a) (–5b)

+ 2(–5b) (–6) + 2(–6) (4a)

= 16a2 + 25b2 + 36 – 40ab + 60b – 48a

= 16a2 + 25b2 – 40ab + 60b – 48a + 36.

(iii)

2

2 42

ab c

22 2(2 ) ( 4 ) 2 2

2 2

a ab c b

2 2b ( 4 )c 2 ( 4 )2

ac

22 24 16 2 16 4

4

ab c ab bc ca .

(iv)

2

5 42

ba c

= (5a)2 +

2

2

b

+ (4c)2 + 2 × 5a × 2

b

+ 2 × 2

b

× 4c + 2 × 4c × 5a

= 25a2 + 2

4

b + 16c2 – 5ab – 4bc + 40ca.

10. (i) If a + b + c = 7, a2 + b2 + c2 = 45,

find the value of ab + bc + ca.

(ii) If a + b + c = 9, ab + bc + ca = 14, find

the value of a2 + b2 + c2.

(iii) If ab + bc + ca = 27, a2 + b2 + c2 = 90,

find the value of a + b + c.

(iv) If a2 + b2 + c2 = 29 and

ab + bc + ca = 26, find the value of a + b + c.

Solution.

(i) Given a + b + c = 7, a2 + b2 + c2 = 45

(a + b + c)2 = (7)2

a2 + b2 + c2 + 2(ab + bc + ca) = 49

45 + 2(ab + bc + ca) = 49

2(ab + bc + ca) = 49 – 45 = 4

ab + bc + ca 4

2 = 2

Hence ab + bc + ca = 2.

(ii) Given, a + b + c = 9, ab + bc + ca = 14

(a + b + c)2 = (9)2

a2 + b2 + c2 + 2(ab + bc + ca) = 81

a2 + b2 + c2 + 2 × 14 = 81

a2 + b2 + c2 + 28 = 81

a2 + b2 + c2 = 81 – 28 = 53

a2 + b2 + c2 = 53.

(iii) ab + bc + ca = 27, a2 + b2 + c2 = 90

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

= 90 + 2 × 27 = 90 + 54

= 144 = (±12)2.

a + b + c = ±12

(iv) a2 + b2 + c2 = 29, ab + bc + ca = 26

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

= 29 + 2 × 26

= 29 + 52 = 81 = (±9)2

a + b + c = ±9.


61Expansions


(i) (5ab – 4cd)2 (ii) (3x3 – 8y2)2

(iii) (–6x – 7y)2 (iv) (5x – 4y)3

(v)

31

33

xx

(vi) (2a – 5b – 6c)2

Solution.

(i) (5ab – 4cd)2

= (5ab)2 + (4cd)2 – 2 × 5ab × 4cd

[ (a – b)2 = a2 – 2ab + b2]

= 25a2b2 + 16c2d2 – 40abcd.

(ii) (3x3 – 8y2)2

= (3x3)2 + (8y)2 – 2 × 3x3 × 8y2

= 9x6 + 64y2 – 48x3y2.

(iii) (–6x – 7y)2

= (–6x)2 + (–7y)2 + 2(–6x) (–7y)

= 36x2 + 49y2 + 84xy.

(iv) (5x – 4y)3

= (5x)3 – 3(5x)2 (4y) + 3(5x) (4y)2 – (4y)3

= 125x3 – 3 × 25x2 × 4y + 3 × 5x

× 16y2 – 64y3.

= 125x3 – 300x2y + 240xy2 – 64y3

(v)

31

33

xx

= (3x)3 + 3 × (3x)2 1

3x

+ 3 (3x)

2 31 1

3 3x x

= 27x3 + 3 × 9x2 ×1

3x+ 3 × 3x

2 3

1 1

9 27x x

= 27x3 + 9x + 3

1 1

27x x .

(vi) (2a – 5b – 6c)2

= (2a)2 + (–5b)2 + (–6c)2 + 2 × 2a (–5b)

+ 2 (–5b) (–6c) + 2(–6c) (2a)

= 4a2 + 25b2 + 36c2 – 20ab + 60bc – 24ca.

2. Evaluate using algebraic formula

983 + 6(98)2 + 12(98) + 23.

Solution.

983 + 6(98)2 + 12(98) + 23

= 983 + 3(98)2 × 2 + 3 × (98) × 22 + 23

= (98 + 2)3 = (100)3 = 1000000.

3. (i) If x2 – 9x + 1 = 0, find the value of

1x

x and

33

1x

x .

(ii) Given 3

5xx

, find the value of

x3 – 3

27

x.

(iii) If 9x2 + 2

1

4x = 13, find the value of

3x + 1

2x and 27x3 + 3

1

8x.

(iv) If x4 + 4

1

x = 2 find the value of

x2 + 2

1 1, x

xx and

33

1x

x .

Solution.

(i) Given, x2 – 9x + 1 = 0

Dividing by x, we get

2 9 1x x

x x x = 0

x – 9 + 1

x = 0 x +

1

x = 9




1

xx

= 9


31

xx

= (9)3

3

3

1 13x x

xx

= 729

3

3

13 9x

x = 729

3

3

127x

x = 729

3

3

1729 27 702x

x

Thus, 3

3

1x

x = 702

(ii) Given, 3

xx

= 5


33

xx

= (5)3

3

3 3 3 3( ) 3x x x

x x x

= 125

3

3

273 3 5x

x = 125

x3 – 3

27

x – 45 = 125

x3 – 3

27

x = 125 + 45 = 170

x3 – 3

27

x = 170

(iii) Given 9x2 + 2

1

4x = 13 ...(i)

21

32

xx

22 1 1

(3 ) 2 32 2

x xx x

22

9 34

xx

= 13 + 3 = 16 = (4)2

[using eqn. (i)]

1

32

xx

= 4


31

32

xx

= (4)3

3

3 1 1 1(3 ) 3 3 3

2 2 2x x x

x x x

= 64

27x3 + 3

1 9

28x × 4 = 64

27x3 + 3

1

8x + 18 = 64

27x3 + 3

1

8x = 64 – 18 = 46

Thus, 27x3 + 3

1

8x = 46

(iv) Given, 44

12x

x

22 2 2

2 2 2

1( ) 2

( )x x

x x

4

4

12x

x

= 2 + 2 = 4 = (2)2


63Expansions

2

2

12x

x 2

2

10x

x

22

2

1 12x x

x x

= 2 + 2 = 4 = (2)2

1

xx

= 2.

Thus,

31

xx

3

3

1 13x x

xx

(2)3 3

3

13 2x

x

3

3

16 8x

x

3

3

18 6 2x

x

Thus, 3

3

12x

x .

4. Find the product of

(5x + 4y + 2) (5x + 4y – 3).

Solution.

(5x + 4y + 2) (5x + 4y – 3)

Let 5x + 4y = a, then

given exp. = (a + 2) (a–3)

=a2+(2–3)a+2× (–3) = a2 – a – 6

= (5x + 4y)2 – (5x + 4y) – 6

= 25x2 + 16y2 + 2 × 5x × 4y – 5x – 4y – 6

= 25x2 + 16y2 + 40xy – 5x – 4y – 6.

5. If a – b = 2, ab = 15 find (i) a + b

(ii) a3 – b3.

Solution.

(i) Given, a – b = 2, ab = 15

Now, (a + b)2 = (a–b)2+4ab = (2)2 + 4 × 15

= 4 + 60 = 64

a + b = ±8

(ii) a – b = 2


(a – b)3 = (2)3

a3 – b3 – 3ab (a – b) = 8

a3 – b3 – 3 × 15 × (2) = 8

a3 – b3 – 90 = 8

a3 – b3 = 8 + 90 = 98

a3 – b3 = 98.

6. Without calculating the cubes, find the

value of (25)3 + (–12)3 + (–13)3.

Solution.

(25)3 + (–12)3 + (–13)3

Let a = 25, b = –12, c = –13

Now a + b + c = 25 – 12 – 13 = 0

a3 + b3 + c3 = 3abc

(25)3 + (–12)3 + (–13)3

= 3 × 25 × (–12) (–13)

= 11700

7. If x2 + y2 = 34 and xy = 101

2, find the

value of 2(x + y)2 + (x – y)2.

Solution.

Given, x2 + y2 = 34, xy = 101

2

21

2

Now, (x+y)2 = x2 + y2 + 2xy = 34 + 2 × 21

2

= 34 + 21 = 55

and (x – y)2 = x2 + y2 – 2xy = 34 – 2 × 21

2

= 34 – 21 = 13

Now 2(x + y)2 + (x – y)2 = 2 × 55 + 13

= 110 + 13 = 123.

8. If 2a + 5b = 11, ab = 3, find the value of

8x3 + 125b3.

Soluton.

Given, 2a + 5b = 11, ab = 3



2a + 5b = 11


(2a + 5b)3 = (11)3

(2a)3 + (5b)3 + 3 × 2a × 5b (2a + 5b)

= 1331

8a3 + 125b3 + 30 × 3 ×11 = 1331

8a3 + 125b3 + 990 = 1331

8a3 + 125b3 = 1331 – 990 = 341

8a3 + 125b3 = 341.

9. Evaluate

(x – 4)3 + 6(x – 4)2 + 12(x – 4) + 8 when

x = 3.2.

Solution.

(x – 4)3 + 6(x – 4)2 + 12(x – 4) + 8

= (x – 4)3 + 3(x– 4)2× 2+ 3(x– 4) ×22 + 23

[ a3 + 3a2b + 3ab2 + b3 = (a + b)3]

= (x – 4 + 2)3 = (x – 2)3

= (3.2 – 2)3 = (1.2)3 = 1.728.

10. If a2 + b2 + c2 = 26 and a + b + c = 8,

find the value of ab + bc + ca.

Solution.

Given, a2 + b2 + c2 = 26, a + b + c = 8

Since, (a + b + c) = 8

Squaring both sides, we have

(a + b + c)2 = (8)2

a2 + b2 + c2 + 2 (ab + bc + ca) = 64

26 + 2 (ab + bc + ca) = 64

2(ab + bc + ca) = 64 – 26 = 38

ab + bc + ca 38

2 = 19

Thus, ab + bc + ca = 19.


PointstoRemember

We can factorise the given algebraic expression by the following methods :

1. Taking out common factors

2. Grouping

3. By spliting the middle term of the type ax2 + bx + c.

4. Difference of two squares :

(a2 – b2) = (a + b) (a – b)

5. Sum or difference of cubes.

a3 + b3 = (a + b) (a2 – ab + b2)

a3 – b3 = (a – b) (a2 + ab + b2)

Exercise 4A

4

Factorisation

65

Factorise the following :

1. 12a2b – 16ab2 – 28a2b2

Solution. HCF of 12, 16, 28 = 4

12a2b – 16ab2 – 28a2b2 = 4ab (3a – 4b – 7ab).

2. 7ax – 7x + ab – b.

Solution.

7ax – 7x + ab – b

By grouping, we have

= 7x (a – 1) + b(a – 1)

= (a – 1) (7x + b).

3. cd – c – 4d + 4

Solution.

cd – c – 4d + 4

By grouping; we have

= c(d – 1) – 4(d – 1) = (d – 1) (c – 4).

4. ax – ay – 5x + 5y

Solution.

ax – ay – 5x + 5y


= a(x – y) – 5(x – y) = (x – y) (a – 5).

5. 2a3 – a2 – 6a + 3

Solution.

2a3 – a2 – 6a + 3


= a2(2a–1)–3(2a – 1) = (2a – 1) (a2 – 3).



6. 5a – b + x(b – 5a)

Solution.

5a – b + x(b – 5a)

= (5a – b) – x(5a – b)

= (5a – b) (1 – x).

7. (2c – d) – a(d – 2c)

Solution.

(2c – d) – a(d – 2c)

= (2c – d) + a(2c – d)

= (2c – d) (1 + a).

8. b(x – y) – 3(y – x) + a(y – x).

Solution.

b(x – y) – 3(y – x) + a(y – x)

= b(x – y) + 3(x – y) – a(x – y)

= (x – y) (b + 3 – a).

9. 3a2 – 6a – 12a + 2b + am – 2m.

Solution.

3a2 – 6a – 12a + 2b + am – 2m

= 3a(a – 2) – k(a – 2) + m(a – 2)

= (a – 2) (3a – k + m).

10. a2 – ab (1 – b) – b3.

Solution.

a2 – ab (1 – b) – b3

= a2 – ab + ab2 – b3

= a(a – b) + b2(a – b)

= (a – b) (a + b2).

Exercise 4B

I. Factorise the following :

1. 72x2 – 50y2

Solution.

72x2 – 50y2 = 2(36x2 – 25y2)

= 2[(6x)2 – (5y)2]

= 2(6x + 5y) (6x – 5y).

[a2 – b2 = (a – b) (a + b)]

2. a4 – 16b4

Solution.

a4 – 16b4 = (a2)2 – (4b2)2

= (a2 + 4b2) (a2 – 4b2)

= (a2 + 4b2) [(a)2 – (2b)2]

= (a2 + 4b2) (a + 2b) (a – 2b).

3. m4n – 81n5.

Solution.

m4n – 81n5 = n(m4 – 81n4)

= n[(m2)2 – (9n2)2]

= n[(m2 + 9n2) (m2 – 9n2)]

= n(m2 + 9n2) [(m)2 – (3n)2]

= n(m2 + 9n2) (m + 3n) (m – 3n).

4. 80y3 – 5y.

Solution.

Now 80y3 – 5y = 5y (16y2 – 1)

= 5y [(4y)2 – (1)2]

= 5y(4y + 1) (4y – 1).

5. (a + 8)2 – (b + 5)2

Solution.

(a + 8)2 – (b + 5)2

= (a + 8 + b + 5) (a + 8 – b – 5)

= (a + b + 13) (a – b + 3).

6. (x + 3) (x – 3) – 40

Solution.

(x + 3) (x – 3) – 40

= (x2 – 9) – 40 = x2 – 9 – 40

= x2 – 49 = (x)2 – (7)2

= (x + 7) (x – 7).

7. (x + 1) (x – 1) – 5

4

Solution.

(x + 1) (x – 1) – 5

4

= x2 – 1 – 5

4 = x2 –

9

4


67Factorisation

= (x)2 –

23

2

3 3

2 2x x

.

8. 9x2 – 4(y + 2x)2.

Solution.

9x2 – 4(y + 2x)2

= (3x)2 – [2(y + 2x)]2

= (3x)2 – (2y + 4x)2

= (3x + 2y + 4x) (3x – 2y – 4x)

= (2y + 7x) (–x – 2y).

9.

2 249

3 9

y yx

.

Solution.

2 249

3 9

y yx

2 27

3 3

yx y

7 7

3 3 3 3

y yx y x y

8( 2 )

3x y x y

.

10. x2 + 4y2 + 4xy – 4.

Solution.

x2 + 4y2 + 4xy – 4

= (x)2 + (2y)2 + 2 × x × 2y – (2)2

= (x + 2y)2 – (2)2

= (x + 2y + 2) (x + 2y – 2).

11. 18ab3 – 8a3b.

Solution.

18ab3 – 8a3b

= 2ab (9b2 – 4a2)

= 2ab [(3b)2 – (2a)2]

= 2ab (3b + 2a) (3b – 2a).

12. 32ab4 – 162a

Solution.

32ab4 – 162a = 2a (16b4 – 81)

= 2a[(4b2)2–(9)2]=2a(4b2 + 9)(4b2–9)

= 2a(4b2 + 9) [(2b)2 – (3)2]

= 2a(4b2 + 9) (2b + 3) (2b – 3).

13. 9x2 – (x2 – 4)2.

Solution.

9x2 – (x2 – 4)2

= (3x)2 – (x2 – 4)2

= (3x + x2 – 4) (3x – x2 + 4)

= (x2 + 3x – 4) (–x2 + 3x + 4)

= {x2 + 4x – x – 4} {–x2 + 4x – x + 4}

= {x(x + 4) –1 (x + 4)} {–x(x – 4) – 1(x – 4)}

= (x + 4) (x – 1) (– x – 1) (x – 4)

= (x + 4) (x – 1) (4 – x) (x + 1).

14. (x2 + y2 – z2)2 – 4x2y2.

Solution.

(x2 + y2 – z2)2 – 4x2y2

= (x2 + y2 – z2)2 – (2xy)2

= (x2 + y2 – z2 + 2xy) (x2 + y2 – z2 – 2xy)

= (x2 + y2 + 2xy – z2) (x2 + y2 – 2xy – z2)

= {(x + y)2 – (z)2} {(x – y)2 – (z)2}

= (x + y + z) (x + y – z) (x – y + z) (x – y – z).

15. 4(a + b)2 – a2 + 2ab – b2

Solution.

4(a + b)2 – a2 + 2ab – b2

= 4(a + b)2 – (a2 + b2 – 2ab)

= 4(a + b)2 – (a – b)2

= {2(a + b)}2 – (a – b)2

= (2a + 2b + a – b) (2a + 2b – a + b)

= (3a + b) (a + 3b).

16. 25x2 – y2 + 5x + y.

Solution.

25x2 – y2 + 5x + y



= {(5x)2 – (y)2} + 1(5x + y)

= (5x + y) (5x – y) + 1(5x + y)

= (5x + y) (5x – y + 1).

17. 36x2 – 49y2 + 6x – 7y.

Solution.

36x2 – 49y2 + 6x – 7y

= (6x)2 – (7y)2 + 1(6x – 7y)

= (6x + 7y) (6x – 7y) + 1(6x – 7y)

= (6x – 7y) (6x + 7y + 1).

18. a(a – 1) – b(b – 1).

Solution.

a(a – 1) – b(b – 1)

= a2 – a – b2 + b = a2 – b2 – a + b

= (a + b) (a – b) –1(a – b)

= (a – b) (a + b – 1)

19. a4 + b4 – 7a2b2.

Solution.

a4 + b4 – 7a2b2

= (a2)2 + (b2)2 + 2a2b2 – 9a2b2

= (a2 + b2)2 – (3ab)2

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

= (a2 + b2 + 3ab) (a2 + b2 – 3ab)

[ a2 – b2 = (a – b) (a + b)]

20. a4 + 9b4 + 2a2b2.

Solution.

a4 + 9b4 + 2a2b2

= (a2)2 + (3b2)2 + 6a2b2 + 2a2b2 – 6a2b2

= (a2 + 3b2)2 – (2ab)2

= (a2 + 3b2 + 2ab) (a2 + 3b2 – 2ab).

21. a4 + 25b4 + a2b2.

Solution.

a4 + 25b4 + a2b2

= (a2)2 + (5b2)2 + 10a2b2 – 10a2b2 + a2b2

= (a2 + 5b2)2 – 9a2b2

= (a2 + 5b2)2 – (3ab)2

= (a2 + 5b2 + 3ab) (a2 + 5b2 – 3ab)

[ a2 – b2 = (a + b) (a – b)]

22. a2 – a + 1

4 – ab +

2

b.

Solution.

a2 – a + 1

4 – ab +

2

b

= (a)2 – 2 × a × 1 1 1

2 2 2b a

21 1

2 2a b a

1 1

2 2a a b

1 1

2 2a a b

II. Find the value using algebraic formula.

(i) 2

99813 99815 1

(99814)

(ii) 99982 – 9999 × 9997

(iii) 8.952 – 1.052

Solution.

(i) 2

99813 99815 1

(99814)

2

(99814 1) (99814 1) 1

(99814)

2 2

2 2

[(99814) 1] 1 (99814) 1 1

(99814) (99814)

2

2

(99814)1

(99814) .

(ii) 99982 – 9999 × 9997

= (9998)2 – (9998 + 1) (9998 – 1)

= (9998)2 – (99982 – 1)

= (9998)2 – (9998)2 + 1 = 1.

(iii) 8.952 – 1.052

= (10 – 1.05)2 – (1.05)2

= 102 – 2 × 10 × 1.05 + (1.05)2 – (1.05)2

= 102 – 21 = 100 – 21 = 79.


69Factorisation

2. Simplify :

8 2 4 22 4 2

1 1 1x x x x

x x x

Solution.

8 2 4 22 4 2

1 1 1x x x x

x x x

8 2 2 42 2 4

1 1 1x x x x

x x x

8 4 44 4

1 1x x x

x x

8 88

1x x

x

8 88

1x x

x

8

1

x .

III. Factorise :

1. (1 – x2) (1 – y2) + 4xy

Solution.

(1 – x2) (1 – y2) + 4xy

= 1 – y2 – x2 + x2y2 + 4xy

= 1 + x2y2 + 2xy – (x2 + y2 – 2xy)

= [(1)2 + (xy)2 + 2xy] – [(x)2 + (y)2 – 2xy]

= (1 + xy)2 – (x – y)2

= (1 + xy + x – y) (1 + xy – x + y).

2. a4 – 9b2 + 6b – 1.

Solution.

a4 – 9b2 + 6b – 1

= a4 – (9b2 – 6b + 1)

= (a2)2 – [(3b)2 – 2 × 3b × 1 + (1)2]

= (a2)2 – (3b – 1)2

= (a2 + 3b – 1) (a2 – 3b + 1).

3. (a2 + b2 – c2)2 – 4a2b2

Solution.

(a2 + b2 – c2)2 – 4a2b2

= (a2 + b2 – c2)2 – (2ab)2

= (a2 + b2 – c2 + 2ab) (a2 + b2 – c2 – 2ab)

= (a2 + b2 + 2ab – c2) (a2 + b2 – 2ab – c2)

= [(a + b)2 – (c)2] [(a – b)2 – (c)2]

= (a + b + c) (a + b – c) (a – b + c) (a – b – c).

4. (a2 + 4b2 – c2)2 – 16a2b2.

Solution.

(a2 + 4b2 + c2)2 – 16a2b2

= (a2 + 4b2 – c2)2 – (4ab)2

= (a2 + 4b2 – c2 + 4ab) (a2 + 4b2 – c2 – 4ab)

= (a2 + 4b2 + 4ab – c2) (a2 + 4b2 – 4ab – c2)

= [(a)2 + (2b)2 + 4ab – (c)2] [(a)2 + (2b)2

– 4ab – (c)2]

= [(a + 2b)2 – (c)2] [(a – 2b)2 – (c)2]

= (a + 2b + c)(a + 2b – c)(a – 2b + c)(a–2b–c).

5. a2 – ab + 4b – 16.

Solution.

a2 – ab + 4b – 16

= a2 – 16 – ab + 4b

= a2 – (4)2 – b(a – 4)

= (a + 4) (a – 4) – b(a – 4)

= (a – 4) (a + 4 – b) = (a – 4) (a – b + 4).

6. a2 – 5b + ab – 25

Solution.

a2 – 5b + ab – 25

= a2 – 25 + ab – 5b

= (a)2 – (5)2 + b(a – 5)

= (a + 5) (a – 5) + b(a – 5)

= (a – 5) (a + 5 + b) = (a – 5) (a + b + 5).

IV. Factorize :

1.2

2

1 32 3x x

xx



Solution.

22

1 32 3x x

xx

21 1

3x xx x

1 13x x

x x

2. 2

2

1 59 2 15

39x x

xx

Solution.

22

1 59 2 15

39x x

xx

22

1 1(3 ) 2 15

3(3 )x x

xx

21 1

3 5 33 3

x xx x

1 13 3 5

3 3x x

x x

Exercise 4C

I. Factorize the following :

1. a2 – 3a – 54

Solution.

a2 – 3a – 54Since, 54 9 6

3 9 6

= a2 – 9a + 6a – 54

= a(a – 9) + 6(a – 9)

= (a – 9) (a + 6).

2. a2 – 25a + 84.

Solution.

a2 – 25a + 8484 21 ( 4)

25 21 4

= a2 – 4a – 21a + 84

= a(a – 4) – 21(a – 4)

= (a – 4) (a – 21).

3. 1 – 18a – 63a2.

Solution.

1 – 18a – 63a2

= 1 – 21a + 3a – 63a2 63 21 3

18 21 3

= 1(1 – 21a) + 3a(1 – 21a)

= (1 – 21a) (1 + 3a).

4. x2 – 10x – 24

Solution.

x2 – 10x – 2424 12 2

10 12 2

= x2 – 12x + 2x – 24

= x(x – 12) + 2(x – 12)

= (x – 12) (x + 2).

5. x2 – x – 6

Solution.

x2 – x – 66 3 2

1 3 2

= x2 – 3x + 2x – 6

= x(x – 3) + 2(x – 3)

= (x – 3) (x + 2).

6. x2 – 5x + 6.

Solution.

x2 – 5x + 66 3 ( 2)

5 3 2

= x2 – 3x – 2x + 6

= x(x – 3) – 2(x – 3)

= (x – 3) (x – 2).


71Factorisation

7. 3x2 – 2x – 16.

Solution.

3x2 – 2x – 16

16 3 48

48 8 6

2 8 6

= 3x2 – 8x + 6x – 16

= x(3x – 8) + 2(3x – 8)

= (3x – 8) (x + 2).

8. 5x2 – 13x – 6.

Solution.

5x2 – 13x – 6

5 ( 6) 30

30 15 2

13 15 2

= 5x2 – 15x + 2x – 6

= 5x(x – 3) + 2(x – 3)

= (x – 3) (5x + 2).

9. 63x2 – 18x – 1.

Solution.

63x2 – 18x – 1

63 ( 1) 63

63 21 3

18 21 3

= 63x2 + 3x – 21x – 1

= 3x (21x + 1) – 1(21x + 1)

= (21x + 1) (3x – 1).

10. 3x2 – 5x – 12.

Solution.

3x2 – 5x – 12

3 ( 12) 36

36 9 4

5 9 4

= 3x2 – 9x + 4x – 12

= 3x(x – 3) + 4(x – 3)

= (x – 3) (3x + 4).

11. 5x2 – 17x – 12.

Solution.

5x2 – 17x – 12

[ –12×5=60=–20×3 and –20 +3 = –17]

= 5x2 – 20x + 3x – 12

= 5x(x – 4) + 3(x – 4)

= (x – 4) (5x + 3).

12. 2x2 – 7x – 39.

Solution.

2x2 – 7x – 39

2 ( 39) 78

78 13 6

7 13 6

= 2x2 – 13x + 6x – 39

= x(2x – 13) + 3(2x – 13)

= (2x – 13) (x + 3).

13. 6x2 – x – 12.

Solution.

6x2 – x – 12

12 6 72

72 9 8

1 9 8

= 6x2 – 9x + 8x – 12

= 3x (2x – 3) + 4(2x – 3)

= (2x – 3) (3x + 4).

14. 2x2 – 3x – 65.

Solution.

2x2 – 3x – 65

65 2 130

130 13 10

3 13 10

= 2x2 – 13x + 10x – 65

= x(2x – 13) + 5(2x – 13)

= (2x – 13) (x + 5).

15. 4x2 – 8x – 21.

Solution.

4x2 – 8x – 21

21 4 84

84 14 6

8 14 6



= 4x2 – 14x + 6x – 21

= 2x(2x – 7) + 3(2x – 7)

= (2x – 7) (2x + 3).

16. 5a2 – 10a – 15.

Solution.

5a2 – 10a – 15

= 5(a2 – 2a – 3)

= 5(a2 – 3a + a – 3)

= 5{a(a – 3) + 1(a – 3)}

= 5(a – 3) (a + 1).

17. 4a2 + 12a – 216.

Solution.

4a2 + 12a – 216

= 4(a2 + 3a – 54)

= 4{a2 + 9a – 6a – 54}

[ –54 = 9 × (–6) and 9 + (–6) = 3]

= 4{a(a + 9) – 6(a + 9)}

= 4(a + 9) (a – 6).

II. Factorize by substituting terms.

1. 12(a + b)2 – 5(a + b) – 3.

Solution.

12(a + b)2 – 5(a + b) – 3

12 ( 3) 36

36 9 4

5 9 4

Let a + b = x, then given expression

12x2 – 5x – 3

= 12x2 – 9x + 4x – 3

= 3x (4x – 3) + 1(4x – 3)

= (4x – 3) (3x + 1)

= [4(a + b) – 3][3(a + b)+1] ( x = a + b)

= (4a + 4b – 3) (3a + 3b + 1).

2. 3(x – y)2 – 4x + 4y – 4.

Solution.

3(x – y)2 – 4x + 4y – 4

= 3(x – y)2 – 4(x – y) – 4

Let x – y = a, then

= 3a2 – 4a – 4

= 3a2 – 6a + 2a – 4

= 3a (a – 2) + 2(a – 2)

= (a – 2) (3a + 2)

= (x – y – 2) [3 (x – y) + 2] ( a = x – y)

= (x – y – 2) (3x – 3y + 2).

3. (2x – y)2 – 14x + 7y – 18.

Solution.

(2x – y)2 – 14x + 7y – 18

= (2x – y)2 – 7 (2x – y) – 18

Let 2x – y = a, then

= a2 – 7a – 18

= a2 – 7a + 2a – 1818 9 2

7 9 2

= a(a – 9) + 2(a – 9)

= (a – 9) (a + 2)

= (2x – y – 9) (2x – y + 2).

4. 6(x + 2)2 – 5(x + 2) – 4.

Solution.

6(x + 2)2 – 5(x + 2) – 4

Let x + 2 = a, then

6a2 – 5a – 4

= 6a2 – 8a + 3a – 4

= 2a(3a – 4) + 1(3a – 4)

= (3a – 4) (2a + 1)

= [3(x + 2) – 4] [2(x + 2) + 1]

= (3x + 6 – 4) (2x + 4 + 1)

= (3x + 2) (2x + 5).

5. 5(x + y)2 – 6x – 6y – 8

Solution.

5(x + y)2 – 6x – 6y – 8

= 5(x + y)2 – 6(x + y) – 8

= 5a2 – 6a – 8; where x + y = a

6 ( 4) 24

24 8 3

5 8 3

3 ( 4) 12

12 6 2

4 6 2


73Factorisation

= 5a2 – 10a + 4a – 8

= 5a(a – 2) + 4(a – 2)

= (a – 2) (5a + 4)

= (x + y – 2) (5x + 5y + 4).

6. (a2 – 2a)2 – 18(a2 – 2a) + 45.

Solution.

(a2 – 2a)2 – 18(a2 – 2a) + 45

Let a2 – 2a = x, then

given expression x2 – 18x + 45

= x2 – 3x – 15x + 45

= x(x – 3) – 15(x – 3)

= (x – 3) (x – 15)

= (a2 – 2a – 3) (a2 – 2a – 15)

= {a2 – 3a + a – 3}{a2 – 5a + 3a – 15}

= {a(a – 3)+1(a–3)}{a(a – 5) + 3 (a – 5)}

= (a – 3) (a + 1) (a – 5) (a + 3).

III. Write the following in a product of

factors.

1. x(3x – 11) + 6.

Solution.

x(3x – 11) + 6

= 3x2 – 11x + 6

= 3x2 – 9x – 2x + 6

3 6 18

18 9 ( 2)

and 9 2 11

= 3x (x – 3) – 2(x – 3)

= (x – 3) (3x – 2).

2. x(2x + 5) – 75.

Solution.

x(2x + 5) – 75

= 2x2 + 5x – 25

= 2x2 + 10x – 5x – 25

= 2x (x + 5) – 5 (x + 5)

= (x + 5) (2x – 5).

3. x (2x + 1) – 6.

Solution.

x (2x + 1) – 6

6 2 12

12 4 ( 3)

1 4 3

= 2x2 + x – 6

= 2x2 + 4x – 3x – 6

= 2x (x + 2) – 3(x + 2)

= (x + 2) (2x – 3).

4. x(2x + 5) – 3.

Solution.

x(2x + 5) – 3

3 2 6

6 6 ( 1)

5 6 1

= 2x2 + 5x – 3

= 2x2 + 6x – x – 3

= 2x(x + 3) – 1 (x + 3)

= (x + 3) (2x – 1).

IV. Factorize the spliting the last term.

1. 2

118x

x

.

Solution.

2

118x

x

2

12 16x

x

221

(4)xx

1 14 4x x

x x



2. 22

1xx

.

Solution.

Now, 2

21x

x

22

2 1xx

22

1 12 1x x

xx

22(1)x

x

11 1x x

x x

3. 22

19 23

9x

x

Solution.

Now, 2

2

19 23

9x

x 2

2

19 2 25

9x

x

22(3 ) 2 3 25

3 3x x

x x

223 (5)

3x

x

13 5 3 5

3 3x x

x x

.

Exercise 4D

I. Factorize :

1. 8x3 + 343y3.

Solution.

8x3 + 343y3 = (2x)3 + (7y)3

= (2x + 7y) {(2x)2 – 2x × 7y + (7y)2}

[ a3 + b3 = (a + b) (a2 – ab + b2)]

= (2x + 7y) (4x2 – 14xy + 49y2).

2. 9a2 + 1125y3.

Solution.

9a2 + 1125y3

= 9[x3 + 125y3]

= 9[(x)3 + (5y)3]

= 9(x + 5y) [x2 – x × 5y + (5y)2]

= 9(x + 5y) (x2 – 5xy + 25y2).

3. 3

364343

ba .

Solution.

3364

343

ba

33(4 )

7

ba

224 (4 ) 4

7 7 7

b b ba a a

[ a3 – b3 = (a – b) (a2 + ab + b2)]

22 4

4 167 7 49

b ba a ab

.

4. 3

3

1

27a

a .

Solution.

33

1

27a

a

33 1

( )3

aa

221 1 1

3 3 3a a a

a a a

22

1 1 1

3 3 9a a

a a

.

5. a2 – 8

a.

Solution.

a2 – 8

a

31( 8)a

a

3 31( ) (2)a

a


75Factorisation

21( 2)( 2 4)a a a

a

21( 2)( 2 4)a a a

a .

6.

3

3

8

27

a

a .

Solution.

3

3

8

27

a

a

3 32

3

a

a

2

2

2 2 4

3 9 3

a a

a a

2

2

2 2 4

3 9 3

a a

a a

.

7. 2a4b – 432ab4.

Solution.

2a4b – 432ab4

= 2ab (a3 – 216b3)

= 2ab[(a)3 – (6b)3]

= 2ab(a – 6b) {a2 + 6ab + (6b)2}

= 2ab(a – 6b) (a2 + 6ab + 36b2).

8. 3 1

39

x .

Solution.

3 13

9x

3 13

27x

33 1

3 ( )3

x

221 1 1

33 3 3

x x x

21 1 13

3 3 9x x x

.

9. x2 + x5.

Solution.

x2 + x5 = x2 (1 + x3)

= x2[(1)3 + (x)3]

= x2[(1 + x) (1 – x + x2)]

10. 24a4 + 81a.

Solution.

24a4 + 81a = 3a[8a3 + 27]

= 3a[(2a)3 + (3)3]

= 3a (2a + 3) [(2a)2 – 2a × 3 + 32]

= 3a(2a + 3) (4a2 – 6a + 9).

11. 2a7 – 128a.

Solution.

2a7 – 128a = 2a[a6 – 64]

= 2a[(a3)2 – (8)2]

= 2a(a3 + 8) (a3 – 8)

= 2a[a3 + 23] [a3 – 23]

= 2a(a + 2) (a2–2a +4)(a–2)(a2 + 2a + 4).

12. 64a6 – 729b6.

Solution.

64a6 – 729b6

= (8a3)2 – (27b3)2

= (8a3 + 27b3) (8a3 – 27b3)

= {(2a)3 + (3b)3} {(2a)3 – (3b)3}

= (2a + 3b) {(2a)2 – 2a × 3b + (3b)2}

{2a – 3b} {(2a)3 + 2a × 3b + (3b)2}

= (2a + 3b) (4a2 – 6ab + 9b2) (2a – 3b)

(4a2 + 6ab + 9b2).

13. 15625a6 – 64b6.

Solution.

15625a6 – 64b6

= (125a3)2 – (8b3)2

= (125a3 + 8b3) (125a3 – 8b3)

= {(5a)3 + (2b)3}{(5a)3 – (2b)3}



= (5a + 2b) [(5a)2 – 5a × 2b + (2b)2]

(5a – 2b) [(5a)2 + 5a × 2b + (2b)2]

= (5a + 2b) (25a2 – 10ab + 4b2) (5a – 2b)

(25a2 + 10ab + 4b2).

14. 7

64

aa .

Solution.

7

64

aa

6 1

64a a

23 2 1

( )8

a a

3 31 1

8 8a a a

3 33 31 1

( ) ( )2 2

a a a

2 21 1 1 1

2 2 4 2 2 4

a aa a a a a

.

15. x4 + x3 + x + 1.

Solution.

x4 + x3 + x + 1

= x3 (x + 1) + 1(x + 1)

= (x + 1) (x3 + 1)

= (x + 1) {(x)3 + (1)3}

= (x + 1) (x + 1) (x2 – x + 1).

16. x4 + x3 + 8x + 8.

Solution.

x4 + x3 + 8x + 8

= x3(x + 1) + 8(x + 1)

= (x + 1) (x3 + 8)

= (x + 1) {(x)3 + (2)3}

= (x + 1) {(x + 2) (x2 – 2x + 22)}

= (x + 1) (x + 2) (x2 – 2x + 4).

17. 3a7 – 192ab6.

Solution.

3a7 – 192ab6

= 3a[a6 – 64b6]

= 3a[(a3)2 – (8b3)2]

= 3a(a3 + 8b3) (a3 – 8b3)

= 3a[(a)3 + (2b)3] [(a)3 – (2b)3]

= 3a(a + 2b) (a2 – 2ab + 4b2) (a – 2b)

(a2 + 2ab + 4b2).

II. Evaluate using algebraic formula.

1. 3 3

2 2

(0.68) (0.13)

(0.68) (0.68 0.13) (0.13)

.

Solution.

3 3

2 2

(0.68) (0.13)

(0.68) (0.68 0.13) (0.13)

Let 0.68 = a, and 0.13 = b.

3 3 2 2

2 2 2 2

( )( )

( )

a b a b a ab ba b

a ab b a ab b

= 0.68 – 0.13 = 0.55.

2. 3 3

2 2

(0.75) (0.22)

(0.78) (0.78) (0.22) (0.22)

.

Solution.

3 3

2 2

(0.75) (0.22)

(0.78) (0.78) (0.22) (0.22)

Let 0.78 = a and 0.22 = b

3 3 2 2

2 2 2 2

( )( )

( )

a b a b a ab ba b

a ab b a ab b

= 0.78 + 0.22 = 1.00 = 1.

III. Factorize :

1. a6 + 28a3 + 27.

Solution.

a6 + 28a3 + 27

= a6 + a3 + 27a3 + 27


77Factorisation

= a3(a3 + 1) + 27(a3 + 1)

= (a3 + 1) (a3 + 27)

= [(a)3 + (1)3] [(a)3 + (3)3]

= (a + 1) (a2 – a + 1) (a + 3) (a2 – 3a + 9).

2. a6 – 35a3 + 216.

Solution.

a6 – 35a3 + 216

= a6 – 8a3 – 27a3 + 216

= a3 (a3 – 8) – 27(a3 – 8)

= (a3 – 8) (a3 – 27)

= [(a)3 – (2)3] [(a)3 – (3)3]

= (a – 2) (a2 + 2a + 4) (a – 3) (a2 + 3a + 9).

3. x3 + 15x2y + 75xy2 + 126y3.

Solution.

x3 + 15x2y + 75xy2 + 126y3

= x3 + 15x2y + 75xy2 + 125y3 + y3

= (x)3+ 3 ×x2× 5y+3× x×25y2 + (5y)3+(y)3

[ a3 + 3a2b + 3ab2 + b3 = (a + b)3]

= (x + 5y)3 + (y)3

= [x + 5y + y] [(x + 5y)2 – y(x + 5y) + y2]

= (x + 6y) (x2 +25y2 + 10xy – xy – 5y2 + y2)

= (x + 6y) (x2 + 9xy + 21y2).

4. 8x3 + 36x2y + 54xy2 + 26y3.

Solution.

8x3 + 36x2y + 54xy2 + 26y3

= 8x3 + 36x2y + 54xy2 × 27y3 – y3

= (2x)3 + 3 × (2x)2 × 3y + 3 × (2x) × (3y)2

+ (3y)3 – (y)3

= (2x + 3y)3 – (y)3

= (2x + 3y – y) [(2x + 3y)2 + y (2x +3y)+y2]

= (2x + 2y) (4x2 + 9y2 + 12xy+2xy + 3y2+y2)

= 2(x + y) (4x2 + 13y2 + 14xy)

= 2(x + y) (4x2 + 14xy + 13y2).


1. Factorize.

1. 25(a + b)2 – 36(a – b)2.

Solution.

25(a + b)2 – 36(a – b)2

= [5(a + b)]2 – [6(a – b)]2

= (5a + 5b)2 – (6a – 6b)2

= (5a + 5b + 6a – 6b) (5a + 5b – 6a + 6b)

= (11a – b) (–a + 11b)

= (11a – b) (11b – a).

2. (y2 + 3)2 – 16y2.

Solution.

(y2 + 3)2 – 16y2

= (y2 + 3)2 – (4y)2

= (y2 + 3 + 4y) (y2 + 3 – 4y)

= (y2 + 4y + 3) (y2 – 4y + 3)

= [y2 + y + 3y + 3] [y2 – y – 3y + 3]

= [y (y + 1) + 3(y + 1)] [y(y – 1) – 3(y–1)]

= (y + 1) (y + 3) (y – 1) (y – 3).

3. 5x4 – 80y4.

Solution.

5x4 – 80y4 = 5(x4 – 16y4)

= 5[(x2)2 – (4y2)2]

= 5(x2 + 4y2) (x2 – 4y2)

= 5(x2 + 4y2) [(x)2 – (2y)2]

= 5(x2 + 4y2) (x + 2y) (x – 2y).

4. 27a2b – 75b3.

Solution.

27a2b – 75b3 = 3b(9a2 – 25b2)

= 3b [(3a)2 – (5b)2]

= 3b(3a + 5b) (3a – 5b).

II. Factorize the following :

1. 5x2 – 5x – 30.

Solution.

5x2 – 5x – 30

= 5[x2 – x – 6]

= 5[x2 – 3x + 2x – 6]

= 5[x(x – 3) + 2(x – 3)]



= 5(x – 3) (x + 2).

2. x2 – 15xy – 54y2

Solution.

x2 – 15xy – 54y2

= x2 – 18xy + 3xy – 54y2

= x(x – 18y) + 3y(x – 18y)

= (x – 18y) (x + 3y).

3. 6x2 – 15x – 9

Solution.

6x2 – 15x – 9

= 3(2x2 – 5x – 3)

= 3(2x2 – 6x + x – 3)

= 3[2x(x – 3) + 1 (x – 3)]

= 3(x – 3) (2x + 1).

4. 3(x – 2y)2 + 4(x – 2y) – 15

Solution.

3(x – 2y)2 + 4(x – 2y) – 15

Let x – 2y = a, then

given exp. 3a2 + 4a – 15

= 3a2 + 9a – 5a – 15

= 3a (a + 3) – 5(a + 3)

= (a + 3) (3a – 5)

= (x – 2y + 3) [3(x – 2y) – 5]

= (x – 2y + 3) (3x – 6y – 5).

5. a4 + 4b4 – 5a2b2.

Solution. a4 + 4b4 – 5a2b2

= a4 + 4b4 – 4a2b2 – a2b2

= (a2)2 – 4a2b2 + (2b2)2 – (ab)2

= (a2 – 2b2)2 – (ab)2

= (a2 – 2b2 + ab) (a2 – 2b2 – ab)

= (a2 + ab – 2b2) (a2 – ab – 2b2)

= {a2 + 2ab – ab – 2b2} {a2 – 2ab + ab – 2b2}

= [a(a +2b) – b (a +2b)] [a(a – 2b) + b(a – 2b)]

= (a + 2b) (a – b) (a – 2b) (a + b).

6. x4 – 13x2 + 36

Solution. x4 – 13x2 + 36

= x4 – 12x2 + 36 – x2

= (x2)2 – 2 × x2 × 6 + (6)2 – (x)2

= (x2 – 6)2 – (x)2

= (x2 – 6 + x) (x2 – 6 – x)

= (x2 + x – 6) (x2 – x – 6)

= (x2 + 3x – 2x – 6) (x2 – 3x + 2x – 6)

= [x(x + 3) – 2(x + 3)] [x(x – 3) + 2(x – 3)]

= (x + 3) (x – 2) (x – 3) (x + 2)

Aliter:

x4 – 13x2 + 36

= x4 – 4x2 – 9x2 + 36

= x2 (x2 – 4) – 9(x2 – 4)

= [(x)2 – (2)2] [(x)2 – (3)2]

= (x + 2) (x – 2) (x + 3) (x – 3).

III. Factorise :

1. 8x3 – 3

343

y

Solution.

8x3 – 3

343

y = (2x)3 –

3

7

y

222 (2 ) 2

7 7 7

y y yx x x

22 2

2 47 7 49

y yx x xy

2. 54x3y + 250y4.

Solution.

54x3y + 250y4

= 2y (27x3 + 125y3)

= 2y [(3x)3 + (5y)3]

= 2y (3x + 5y) [(3x)2 – 3x × 5y + (5y)2]

= 2y (3x + 5y) (9x2 – 15xy + 25y2).

3. 2a6 – 128 b6

Solution.

2a6 – 128 b6 = 2(a6 – 64b6)

= 2[(a3)2 – (8b3)2]


79Factorisation

= 2(a3 + 8b3) (a3 – 8b3)

= 2[(a)3 + (2b)3] [(a)3 – (2b)3]

= 2(a + 2b) {a2 – 2ab + (2b)2}

(a – 2b){a2 + 2ab + (2b)2}

= 2(a + 2b) (a2 – 2ab + 4b2) (a – 2b)

(a2 + 2ab + 4b2).

4. 729a6 – b6.

Solution.

729a6 – b6

= (27a3)2 – (b3)2 = (27a3 + b3) (27a3 – b3)

= [(3a)3 + (b)3] [(3a)3 – (b)3]

= (3a + b) [(3a)2 – 3a × b + b2]

(3a – b) {(3a)2 + 3a × b × b2}

= (3a + b) (9a2 – 3ab + b2) (3a – b)

(9a2 + 3ab + b2).

IV. Find the value of using algebraic

formulae :

3 3

2 2

(743) (543)

(743) (743) (543) (543)

Solution.

3 3

2 2

(743) (543)

(743) (743) (543) (543)

2 2

2 2

(743 543) [(743) 743 543 (543) ]

(743) 743 543 (543)

= 743 – 543 = 200.

V. The area of rectangle is (14x2–29xy–15y2)

sq. units. Find its sides and the perimeter of the

rectangle.

Solution.

Area of rectangle = 14x2 – 29xy – 15y2

= 14x2 – 35xy + 6xy – 15y2

= 7x (2x – 5y) + 3y (2x – 5y)

= (2x – 5y) (7x + 3y)

Sides of rectangle are 2x – 5y and 7x + 3y

Now perimeter of the rectangle

= 2(Sum of sides)

= 2(2x – 5y + 7x + 3y)

= 2(9x – 2y) = 18x – 4y units.


PointstoRemember

1. System of Linear Equations

Two or more linear equations having same variables taken together form a system of linear

equations in two variables e.g. a1x + b1y + 4 = 0 and a2x + b2y + c2 = 0.

2. Methods of Solving the System of Two equations

(i) Method of substitutions

(ii) Method of eliminations

(iii) Method of cross multiplication

When we solve the simultaneous or system of linear equation, then three will be have possibilities.

(i) If 1 1

2 2

a b

a b , then we get a unique solution and the lines will intersect each other at one point.

(ii) If 1 1 1

2 2 2

,a b c

a b c , then we get no solution and the lines will be parallel.

(iii) If 1 1 1

2 2 2

,a b c

a b c then we get infinitely many solutions and the lines will coincide.

Exercise 5A

5

Simultaneous Linear Equations

80

Solve the following pairs of simultaneous

equations :

1. (i) 3x + y = 2 (ii) 9x – 5y = 52

2x + 3y = 20 4x – 3y = 27

Solution.

(i) Given, 3x + y = 2 ...(i)

2x + 3y = 20 ...(ii)

From (i); y = 2 – 3x

Substituting the value of y in (ii); we have

2x + 3(2 – 3x) = 20

2x + 6 – 9x = 20

–7x = 20 – 6 = 14

x 14

7

= –2

From (i); y = 2 – 3 × (–2) = 2 + 6 = 8

x = –2, y = 8.


81Simultaneous Linear Equations

(ii) Given, 9x – 5y = 52 ...(i)

4x – 3y = 27 ...(ii)

Multiply (i) by 3 and (ii) by 5

27x – 15y = 156

20x – 15y = 135

– + –


x21

7 = 3

From (i); we have

9 × 3 – 5y = 52

27 – 5y = 52 –5y = 52 – 27 = 25

y 25

5

= –5

x = 3, y = –5.

2. (i) 3x – 2y = 9

5

3 6 6

x y

(ii) 2 3

73 4

x y

5x = 3y + 18

Solution.

(i) Given, 3x – 2y = 9 ...(i)

5

3 6 6

x y

2x – y = 5 ...(ii)

(LCM of 3, 6 = 6)

From (ii); y = 2x – 5

Substituting the value of y in (i); we have

3x – 2 (2x – 5) = 9

3x – 4x + 10 = 9

–x = 9 – 10 = –1

x = 1

and y = 2x – 5 = 2 × 1 – 5 = 2 – 5 = –3

x = 1, y = –3.

(ii) Given, 2 3

73 4

x y 8x + 9y = 84 ...(i)

and 5x = 3y + 18

5x – 3y = 18 ...(ii)

Multiply (i) by 1 and (ii) by 3; we have

8x + 9y = 84

15x – 9y = 54

On adding; 23x = 138

x 138

23 = 6

and from (ii) 5 × 6 – 3y = 18

30 – 3y = 18

3y = 30 – 18 = 12

y 12

3 = 4.

Hence x = 6, y = 4.

3. (i) 92 3

x y (ii) 3x + 2.6y = 16

15 4

x y x + 5.2y = 27

Solution.

(i) 92 3

x y ...(i)

15 4

x y ...(ii)

Multiply eqn. (i) by1

4and (ii) by

1

3; we have

9

8 12 4

x y

1

15 12 3

x y



On adding; 9 1

8 15 4 3

x x

15 8

120

x x 27 4

12

3 23

120 12

x

x 23 120

1012 23

From (i); we have

109

2 3

y

53

y = 9

3

y = 9 – 5 = 4

y = 4 × 3 = 12

Hence x = 10, y = 12.

(ii) 3x + 2.6y = 16 ...(i)

x + 5.2y = 27 ...(ii)


6x + 5.2y = 32

x + 5.2y = 27

– – –

On subtracting; 5x = 5 x 5

5 = 1

From (i); 3 × 1 + 2.6y = 16

2.6y = 16 – 3 = 13

y 13

2.6 = 5

Hence x = 1, y = 5.

4. (i) 73x + 27y = 19

27x + 73y = –119

(ii) 31x + 37y = 25

37x + 31y = 43

(iii) 43x + 31y = 98

31x + 43y = 50

(iv) 65x + 49y = 293

49x + 65y = 277

Solution.

(i) Given, 73x + 27y = 19 ...(i)

27x + 73y = –119 ...(ii)

On adding, we get

100x + 100y = –100

x + y = –1 ...(iii)

Subtracting, we get

46x – 46y = 138

x – y 138

46 = 3 ...(iv)

Adding (iii) and (iv); we have

2x = 2 x 2

2 = 1

and on subtracting; we have

2y = –4 y 4

2

= –2

Hence x = 1, y = –2.

(ii) Given, 31x + 37y = 25 ...(i)

37x + 31y = 43 ...(ii)

On adding we get

68x + 68y = 68

x + y = 1 ...(iii)


–6x + 6y = –18

x – y = 3 ...(iv)


2x = 4 x 4

2 = 2

On subtracting, we have

2y = –2 y 2

2

= –1

Hence x = 2, y = –1.

(iii) Given, 43x + 31y = 96 ...(i)

31x + 43y = 50 ...(ii)



On adding, we get

74x + 74y = 148

x + y = 2 ...(iii)


12x – 12y = 48

x – y = 4 ...(iv)


2x = 6 x 2

= 3


2y = –2 y = –1

x = 3, y = –1.

(iv) Given, 65x + 49y = 293 ...(i)

49x + 65y = 277 ...(ii)

Adding (i) and (ii); we have

114x + 114y = 570

x + y = 5 ...(iii)


16x – 16y = 16

x – y = 1 ...(iv)


2x = 6 x 6

2 = 3


2y = 4 y 4

2 = 2

Hence x = 3, y = 2.

5. (i) 3

4 18xy

(ii) 5

3 7yx

2

3 13xy

3

2 4yx

Solution.

(i) Given, 3

4 18xy

...(i)

and2

3 13xy

...(ii)


8x + 6

y = 36

9x + 6

y = 39

– – –

On subtracting; –x = –3 x = 3

From (i); 4 × 3 + 3

y = 18 12 +

3

y = 18

3

y = 18 – 12 = 6 y

3 1

6 2

Hence x = 3, y 1

2 .

(ii) Given, 53 7y

x ...(i)

32 4y

x ...(ii)


15

9yx = 21

1510y

x = 20

– + –

On subtracting, y = 1

From (i); 5

x – 3 × 1 = 7

5

x – 3 = 7

5

x = 7 + 3 = 10

5 1

10 2x

Thus, x 1

2 , y = 1.



6. (i) 7 6

71x y (ii)

8 53

x y

5 8

23x y

6 252

x y

(iii) 15 4

7x y (iv)

2 110

x y

9 8

1x y

3 21

x y

Solution.

(i) Given, 7 6

71x y ...(i)

5 8

23x y ...(ii)


28 24284

x y

15 2469

x y

On adding ; 43

215x

x 43 1

215 5

From (i); 7 6

711

5

y

7 5 671

1 y

35 + 6

y = 71

6

y = 71 – 35 = 36

y 6

36 6

Hence x 1

5 , y

6

(ii) Given, 8 5

3x y ...(i)

6 252

x y ...(ii)


40 25

15x y

6 252

x y

– – –

On subtracting ;34

x = –17 x

34

17

= –2

From (i) ;8 5

2 y

= –3 –4

5

y = –3

5

y = –3 + 4 = 1 y

5

1 = 5

Hence x = –2, y = 5.

(iii) Given, 15 4

7x y ...(i)

9 8

1x y (ii)


30 8

x y = 14

9 8

x y = –1

On adding ; 39

x = 13 x

39

13 = 3

From (i) ; 15 4

3 y = 7 5

4

y = 7



4

y = 7 – 5 = 2 y

4

2 = 2

Hence x = 3, y = 2.

(iv) Given, 2 1

10x y ...(i)

3 2

1x y ...(ii)


4 2

x y = 20

3 2

x y = 1

On adding ; 7

x = 21 x

7 1

21 3

From (i); we have

2 110

1

3

y

2 3 110

1 y

1

6y

= 10 1

y = 10 – 6 = 4

y 1

4

Hence x 1

3 , y

1

4 .

7. (i) 9 5

8x y x y

12 7

11x y x y

(ii) 7 10

9x y x y

8 15

11x y x y

(iii) 5 18

7x y x y

20 9

1x y x y

Solution.

(i) Given, 9 5

8x y x y

12 7

11x y x y

Let x + y = a and x – y = b, then given eqns.

becomes;

9 5

8a b ...(i)

12 711

a b ...(ii)

Multiply (i) by 7, and (ii) by 5; we have

63 3556

a b

60 3555

a b

– – –

On subtracting; 3

1a a

3

1 = 3

From (i); 9 5

83 b

53 8

b

5

b = 8 – 3 = 5 b

5

5 = 1

Now x + y = 3

x – y = 1

On adding, 2x = 4 x 4

2 = 2

On subtracting, 2y = 2 y 2

2 = 1

Hence x = 2, y = 1.



(ii) Given, 7 10

9x y x y

8 1511

x y x y

Let x + y = a and x – y = b, then given eqns.

becomes;

7 10

9a b ...(i)

8 1511

2a b ...(ii)


21 3027

a b

16 3022

a b

– – –

on subtracting; 5

a = 5 a

5

5 = 1

and from (i); 7 10

91 b

10

b= 9 – 7 = 2

b 10

52

Now a = 1, b = 5

x + y = 1

x – y = 5

On adding, 2x = 6 x 6

2 = 3


2y = –4 y 4

22

Hence x = 3, y = – 2

(iii) Given, 5 18

7x y x y

20 9

1x y x y

Let x + y = a, x – y = b, then, given eqns.

becomes;

5 187

a b ...(i)

20 91

a b ...(ii)


5 18

7a b

40 182

a b

On adding; 5 45

9 59

aa

From (i); we have

5 18

75 b

181 7

b

18

b = 7 – 1 = 6

b = 18

6 = 3

a = 5, b = 3

x + y = 5 ...(iii)

x – y = 3 ...(iv)

On adding; 2x = 8 x 8

2 = 4

On subtracting; 2y = 2 y 2

2 = 1

Hence x = 4, y = 1.

8. Solve the folllowing pair of equations by

cross multiplication method :

(i) 4x – 5y + 7 = 0

3x – 4y + 6 = 0



(ii) 5x – 2y + 9 = 0

4x + 3y = 2

(iii) 7x – y = 23

8x + 3y = 18

(iv) 9x + 5y = 7

6x – y = 22

Solution.

(i) Given, 4x – 5y + 7 = 0

3x – 4y + 6 = 0

1

30 28 21 24 16 15

x y

1

2 3 1

x y

Now 1

2 1

x

x 2

21

and 1 3

33 1 1

yy

Hence x = 2, y = 3.

(ii) Given eqns. are 5x – 2y + 9 = 0,

4x + 3y – 2 = 0

By cross-multiplication method, we have

1

4 27 36 10 15 8

x y

1

23 46 23

x y

Now 1

23 23

x

x

23

23

= –1

and 1

46 23

y y

46

23 = 2

Hence x = –1, y = 2.

(iii) Given eqns. are

7x – y – 23 = 0

and 8x + 3y – 18 = 0

1

18 69 184 126 21 8

x y

1

87 58 29

x y

Now 1

87 29

x x

87

29 = 3

and 1

58 29

y

y

58

29

= –2

Hence x = 3, y = –2.

(iv) Given eqns. are

9x + 5y – 7 = 0 ...(i)

6x – y – 22 = 0 ...(ii)

By cross-multiplication method, we have

1

110 7 42 198 9 30

x y

1

117 156 39

x y

117

339

x

and156

439

y

3, 4x y



1. The difference between two numbers is 4

and six times the smaller is equal to four times

the greater. Find the numbers.

Solution.

Let first number = x

and second (smaller) number = y

According to the given conditions

x – y = 4 ...(i)

6y = 4x ...(ii)

from (ii); x 6 3

4 2y y

from (i); 3 3 2

4 42 2

y yy y

y = 8 and x 8 122

required numbers are 12, 8.

2. The sum of two numbers is 48 and fve

times the smaller number is equal to three times

number. Find the numbers.

Solution. Let first (greater) number = x

and second number = y

According to the given condition, we have

x + y = 48 ...(i)

and 5y = 3x ...(ii)

x 5

3y

From (i); we have

5

3y y = 48

5 3

3

y y = 48

8y = 144 y 144

8 = 18

from (i); x = 48 – 18 = 30

Required numbers are 30, 18.

3. A certain number of two-rupee coins and

a certain number of five-rupee coins in a piggy

bank of a child amount to ` 47. If the number of

each kind are interchanged, they would amount

to ` 3 less than before. Find the number of coins

o each kind.

Solution. Total amount = ` 47

Let number of 2-rupee coins = x

and number of 5-rupee coins = y


2x + 5y = 47 ...(i)

5x + 2y = 47 – 3 = 44 ...(ii)

On adding; we get

7x + 7y = 91

x + y 91

137

...(iii)


–3x + 3y = 3

x – y = –1 ...(iv)

On adding (iii) and (iv); we have

2x = 12 x = 6


2y = 14 y = 7

Number of 2 rupee coins = 6

and 5 rupee coins = 7.

4. Two numbers are in the ratio 4 : 7. If 4 is

subtracted from each of the numbers, then the

ratio becomes 7 : 13. Find the numbers.

Solution. Let first number be = x


According to the conditions,

x : y = 4 : 7

7x = 4y ...(i)

and4 7

4 13

x

y

Exercise 5B



13x – 52 = 7y – 28

13x – 7y = 52 – 28 = 24 ...(ii)

From (i); x 4

7y

Substituting the value of x in (ii); we have

13 × 4

7y – 7y = 24

52 7

7 1y y = 24

52y – 49y = 168

3y = 168 y = 168

3 = 56

x 4

56 327

required numbers are 32 and 56.

5. Two numbers are in the ratio 3 : 5. If 5 is

added to each, the ratio becomes 2 : 3. Find the

numbers.

Solution.

Let first number = x


According to the conditions; we have

3

5

x

y

x 3

5y ...(i)

and 5 2

5 3

x

y

3x + 15 = 2y + 10

3x – 2y = 10 – 15 = –5 ...(ii)

using eqn. (i) in eqn. (ii); we have

3 × 3

5y – 2y = –5

9

5y – 2y = –5

9y – 10y = –25

–y = –25

y = 25

and x = 3

5y =

3

5 × 25 = 15

required numbers are 15, 25.

6. The sum of the present ages of Rani and

her father is 43 years. Six years hence, father will

be 4 times as old as his daughter. Find their

present ages.

Solution. Let Rani’s age = x years

and her father’s age = y years

According to given conditions; we have

x + y = 43 years ...(i)

6 years hence age of Rani = x + 6 years

and father’s age = y + 6 years

y + 6 = 4(x + 6)

y + 6 = 4x + 24

y = 4x + 24 – 6 = 4x + 18 ...(ii)

From (i); we have

x + 4x + 18 = 43

5x = 43 – 18 = 25

x 25

5 = 5

from (ii); y = 43 – 5 = 38

Hence Rani’s age = 5 years and father’s age

= 38 years

7. In 5 years, Jagriti will be two-thirds as old

as her aunt. Three years ago, she was half as old

as her aunt is now. How old are they now ?

Solution.

Let present age of Jagriti = x years

and age of her aunt = y years

5 years hence, age of Jagriti = x + 5

and age of her aunt = y + 5

x + 5 2

3 (y + 5)

3x + 15 = 2y + 10



3x – 2y = 10 – 15 = –5 ...(i)

and 3 years ago,

age of Jagriti = x – 3

and age of her aunt = y – 3

x – 3 1

2 (y)

2x – 6 = y ...(ii)

and substituting the value of y in (i); we have

3x – 2 (2x – 6) = –5

3x – 4x + 12 = –5

–x = – 5 – 12 = –17

i.e. x = 17

and y = 2x – 6 = 2 × 17 – 6 = 34 – 6 = 28

Hence age of Jagriti = 17 years

and age of her aunt = 28 years.

8. The sum of the ages of Arunima and

Kareena is 64 years. Arunima is 6 years older than

Kareena. Find their present ages.

Solution. Let age of Arunima = x years

and age of Kareena = y years

x + y = 64 ...(i)

x = y + 6 ...(ii)

From (i); y + 6 + y = 64

2y = 64 – 6 = 58 y 58

2 = 29

x = 29 + 6 = 35

Present age of Arunima = 35 years

and age of Kareena = 29 years

9. Five years ago, A’s age was 5 years less

than twice B’s age. Three years from now, one

third of B’s age will be 12 years less than A’s age.

Find their present ages.

Solution.

Let A’s present age = x years

and B’s age = y years


5 years ago,

A’s age = (x – 5) years

B’s age = (y – 5) years

x – 5 = 2(y – 5)

x = 2y – 10 = 2y – 10 ...(i)

After 3 years;

A’s age = x + 3 years

and B’s age = y + 3 years

1

3(y + 3) = x – 3 – 12

y + 3 = 3x + 9 – 36

y = 3x + 9 – 36 – 3

y = 3x – 30 ...(ii)

From (i); we have

y = 3(2y – 10) – 30 = 6y – 30 – 30

6y – y = 60 5y = 60 y 60

5 = 12

x = 2y – 10 = 12 × 2 – 10 = 24 – 20 = 14

Hence A’s age = 14 years

and B’s age = 12 years.

10. A two-digit number is such that its ten’s

place digit is 2 more than twice the unit’s place

digit. When the digits are interchanged, the

reversed number is 5 more than thrice the sum

of the digits. Find the number.

Solution.

Let unit digit = x

and tens digit = y

required number = x + 10 y

By interchanging the digit

Unit digit = y

and tens digit = x

Number = y + 10x


y = 2x + 2 ...(i)

and y + 10x = 3(x + y) + 5

y + 10x = 3x + 3y + 5

10x + y – 3x – 3y = 5

7x – 2y = 5 ...(ii)



From (ii); we have

7x – 2 (2x + 2) = 5

7x – 4x – 4 = 5

3x = 5 + 4 = 9

x 9

3 = 3

and y = 2x + 2 = 2 × 3 + 2 = 8

required number = x + 10y

= 3 + 10 × 8 = 3 + 80 = 83

11. When a two-digit number is divided by

the sum of its digits, the quotient is 4. If the digits

are interchanged, the reversed number is 6 less

than twice the original number. Find the number.

Solution.

Let unit digit = x

and tens digit = y


By interchanging the digits,

Unit digit = y

and tens digit = x

Number = y + 10x

According to the given conditions, we have

10x y

x y

= 4 x + 10y = 4x + 4y

x + 10y – 4x – 4y = 0

–3x + 6y = 0 6y = 3x

y 3 1

6 2x x ...(i)

and y + 10x = 2(x + 10y) – 6

y + 10x = 2x + 20y – 6

10x – 2x + y – 20y = –6

8x – 19y = –6 ...(ii)

8x – 19 1

2x

= –6 [using (i)]

8x – 19

2x = –6

3

2x = –6 x

6 2

3

= 4

y 1 4

22 2

x


= 4 + 10 × 2 = 4 + 20 = 24.

12. The sum of the digits of a two-digit

number is 7. When the digits are interchanged,

the reversed number is 5 times the ten’s digit of

the original number. Find the original number.

Solution.

Let unit digit = x

and tens digit = y



Unit digit = y

tens digit = x

Number = y + 10x


x + y = 7 ...(i)

and y + 10x = 5y

10x = 5y – y = 4y

x 4

10y ...(ii)

From (i);4

710

y y

(4 + 10)y = 70

14y = 70 y 70

514

x 4 4

5 210 10

y

and required number = x + 10y

= 2 + 10 × 5

= 2 + 50 = 52.

13. The sum of two-digit number and the

number obtained by interchanging the digits is

154. If the ten’s digit number is 2 more than

unit’s digit number, find the original number.

Solution.

Let unit digit = x



and tens digit = y

Number = x + 10y

By interchanging the digits

Unit digit = y

and tens digit = x

Number = y + 10x

According to the conditions; we have

x + 10y + y + 10x = 154

11x + 11y = 154

x + y = 14 ...(i)

and y = x + 2 ...(ii)

From (i); we have

x + x + 2 = 14

2x = 14 – 2 = 12 x 12

62

and y = x + 2 = 6 + 2 = 8

required number

= x + 10y = 6 + 10 × 8 = 86.

14. The sum of digits of two-digit number

is 8. If the digits are interchanged, new number

is 10 more than double the original number. Find

the original number.

Solution.

Let unit digit = x

and tens digit = y



Unit digit = y

and tens digit = x

Number = y + 10x

According to the given condition,

x + y = 8 ...(i)

and y + 10x = 2(x + 10y) + 10

y + 10x = 2x + 20y + 10

10x + y – 2x – 20y = 10

8x – 19y = 10 ...(ii)

From (i); x = 8 – y

From (ii); we have

8(8 – y) – 19y = 10

64 – 8y – 19y = 10

–27y = 10 – 64 = –54

y 54

27

= 2

x = 8 – y = 8 – 2 = 6

required number = x+10y= 6+10×2 = 26.

15. A two-digit number becomes 5

6 of the

reversed number obtained when the digits are

interchanged. The difference between the digits

is 1. Find the number.

Solution.

Let unit digit = x

ten’s digit = y


By interchanging the digits

unit digit = y

ten’s digit = x

Number = y + 10x

According to the conditions, we have

x + 10y 5

6 (y + 10x)

6x + 60y = 5y + 50x

6x + 60y – 50x – 5y = 0

55y + 44x = 0 55y = 44x ...(i)

and x – y = 1

x = 1 + y

from (i); we have

55y = 44(1 + y) = 44 + 44y

55y – 44y = 44

11y = 44 y 44

411

x = 1 + y = 1 + 4 = 5

required number = x+10y=5+10×4 = 45.



16. ABC is a equivalent triangle. If AB 2 1,x

BC = y + 7. AC = 2y + 3, find x and y.

Solution.

ABC is an equilateral triangle.

AB = BC and BC = AC

2x – 1 = y + 7

2x – y = 7 + 1 2x – y = 8 ...(i)

and y + 7 = 2y + 3 2y – y = 7 – 3

y = 4

from (i); 2x – 4 = 8 2x = 8 + 4 = 12

x 12

2 = 6

x = 6, y = 4.

17. If the length of a rectangle is increased

by 12cm and the width is decreased by 8cm, the

area is unchanged. If the original length is

increased by 5cm and the original width is

decreased by 4cm, also the are remains the same.

Find the original dimensions.

Solution.

Let length of a rectangle = x m

and breadth of rectangle = y m

Area of rectangle = xy cm2

A B

CD

x

y


(x + 12) (y – 8) = xy

xy – 8x + 12y – 96 = xy

–8x + 12y = 96

8x – 12y = –96

2x – 3y = –24 ...(i)

and (x + 5) (y – 4) = xy

xy – 4x + 5y – 20 = xy

–4x + 5y = +20 ...(ii)

Multiply (i) by 2 and (ii) by 1, we have

4x – 6y = –48

–4x + 5y = +20

on adding; –y = –28

y = 28

and 2x –3y = –24

2x – 3 × 28 = –24

2x – 84 = –24

2x = –24 + 84 = 60

x 60

302

Length of rectangle = 30 cm

and breadth of rectangle = 28 cm.

18. In a right-angled triangle, the sides

containing the right angle are x cm and (2x + y)

cm. The hypotenuse is (3x – y) cm and it is 1cm

longer than the greater of the two sides. Write

down two equations in x and y and solve them.

Also find the sides of the triangle.

Solution.

In right angled triangle,

Sides containing the right angle are x cm and

(2x + y) cm

A

CB x

2x + y3x – y

Hypotenuse = (3x – y) cm

and 3x – y = 2x + y + 1

3x – y – 2x – y = 1

x – 2y = 1



x = 1 + 2y ...(i)

(3x – y)2 = (2x + y)2 + x2

(Pythagoras Theorem)

9x2 + y2 – 6xy = 4x2 + y2 + 4xy + x2

9x2 – 4x2 – x2 = 4xy + 6xy

4x2 = 10xy

2x = 5y ...(ii)

(Dividing by 2x)

2(1 + 2y) = 5y 2 + 4y = 5y

5y – 4y = 2 y = 2

and x = 1 + 2 × 2 = 1 + 4 = 5

x = 5cm, y = 2cm.

Now sides are

2x + y = 2 × 5 + 2 = 10 + 2 = 12 cm

3x – y = 3 × 5 – 2 = 15 – 2 = 13 cm

and x = 5cm.

19. The length of a rectangle is greater than

4 times its breadth by 5cm. If its length is

reduced by 2cm and breadth is incrased by 2cm,

then the area of the rectangle increases by 36cm2.

Find the length and breadth of the original

rectangle.

Solution.

Let length of rectangle = x cm

and breadth of rectangle = y cm

Area of rectangle = xy cm2

According to the given conditions; we have

x = 4y + 5 ...(i)

and (x – 2) (y + 2) = xy + 36

xy + 2x – 2y – 4 = xy + 36

2x – 2y = 36 + 4 = 40

x – y = 20 ...(ii)

From (ii); we have

4y + 5 – y = 20 [using (i)]

3y = 20 – 5

y 15

3 = 5

and x = 4y + 5 = 4 × 5 + 5 = 20 + 5 = 25

Length = 25 cm and breadth = 5cm.

20. The area of a rectangle decreases by

10cm2 if the length is decreased by 5cm and the

breadth is increased by 3cm. If the length is

increased by 5cm and the breadth is increased by

2cm, then the area increases by 80cm2. Find the

perimeter of the rectangle.

Solution.

Let length of the rectangle = x cm

and breadth of rectangle = y cm

Area = xy cm2


(x – 5) (y + 3) = xy – 10

xy + 3x – 5y – 15 = xy – 10

3x – 5y = 15 – 10 = 5 ...(i)

and (x + 5) (y + 2) = xy + 80

xy + 2x + 5y + 10 = xy + 80

2x + 5y = 80 – 10 = 70 ...(ii)

On adding (i) and (ii); we have

5x = 75 x 75

155

and from (i); 3 × 15 – 5y = 5

45 – 5y = 5 –5y = 5 – 45 = – 40

y 40

85

Length of rectangle = 15cm

and breadth of rectangle = 8cm

Now perimeter of rectangle

= 2(l + b) = 2(15 + 8) cm

= 2 × 23 = 46cm.

21. Find the fraction which becomes 1

2

when the denominator is increased by 4 and is

equal to 1

8 when numerator is diminished by 5.

Solution.

Let numerator of a fraction = x

and denominator = y



required fraction x

y

According to the given condition; we have

1

4 2

x

y

and 5 1

8

x

y

2x = y + 4 y = 2x – 4 ...(i)

y = 8x – 40 ...(ii)

From (i) and (ii); we have

2x – 4 = 8x – 40

8x – 2x = 40 – 4 = 36

6x = 36 x 36

6 = 6

and y = 2x – 4 = 2 × 6 – 4 = 8

required fraction 6

8 .

22. When the numerator of a fraction is

increased by 2 and the denominator by 1, the

fraction becomes equal to 5

8 and if the numerator

and denominator are each diminished by 1, the

fraction becomes 1

2. Find the fraction.

Solution.


and denominator of a fraction = y

required fraction x

y


2 5

1 8

x

y

8x + 16 = 5y + 5

8x – 5y = 5 – 16 = –11 ...(i)

and 1 1

1 2

x

y

2x – 2 = y – 1

2x – y = 2 – 1 = 1 ...(ii)

From (ii), y = 2x – 1

Thus from (i); we have

8x – 5 (2x – 1) = –11

8x – 10x + 5 = –11

–2x = –11 – 5 = –16

x 16

82

Thus, y = 2x – 1 = 2 × 8 – 1 = 16 – 1 = 15

required fraction 8

15 .

23. A fraction’s value is 4

5. When its

numerator is increased by 9, the new fraction

equals the reciprocal of the value of the original

fraction. Find the original fraction.

Solution.

Value of fraction 4

5

Let the required fraction be 4

5

x

x.


4 9

5

x

x

5

4

x

x 25x2 = 16x2 + 36x

25x – 16x2 = 36x 9x2 = 36x

9x2 – 36x = 0 9x (x – 4) = 0

x(x – 4) = 0

Either x is 0 which is not possible

or x – 4 = 0 then x = 4

required fraction 4 16

5 20

x

x .

24. A boat takes 9 hours to travel 30 km

upstream and 40km downstream but it takes 12

hours to travel 42km upstream and 50 km

downstream. Find the speed of the boat in still

water and the speed of the stream.

Solution.

Let speed of the boat = x km/h



and speed of stream water = y km/h

upstream spread = (x – y) km/h

and downstream speed = (x + y)km/h


30 40

x y x y

= 9

42 50

x y x y

= 12

Let x – y = a and x + y = b

Thus given eqns becomes;

30 40

a b = 9 ...(i)

42 50

a b = 12 ...(ii)


150 200

a b = 45

168 200

a b = 48

– – –

on subtracting; 18

3 6aa

and from (i); 30 40

6 b = 9 5 +

40

b = 9

40

b = 9 – 5 = 4

b 40

4 = 10

Now x – y = 6 ...(iii)

x + y = 10 ...(iv)

on adding, 2x = 16 x 16

2 = 8

and from (iv); y = 10 – x = 10 – 8 = 2

Speed of boat = 8 km/h

and speed of stream = 2 km/h.

25. A plane can fly 1120 km in 1 hour 20

minutes with the wind. Flying against the same

wind, the plane travels the same distance in 1

hour 24 minutes. Find the speed of the plane and

the speed of the wind.

Solution.

Let speed of the plane = x km/h

and speed of wind = y km/h

Speed with wind = (x + y) km/h

and speed against wind = (x – y) km/h


1120 20 1 41 1

60 3 3x y

and1120 24 2 7

1 160 5 5x y

4x + 4y = 3360

x + y = 840 ...(i)

and 7x – 7y = 5600 ...(ii)


7x + 7y = 5880

7x – 7y = 5600

On adding we get

14x = 11480

x 11480

14 = 820

and from (i); x + y = 840

820 + y = 840

y = 840 – 820 = 20

Speed of plane = 820 km/h

and speed of wind = 20 km/h

26. If Laisha walks for 1 hour and cycles for

2 hours she can travel 33 km. But if she walks

for 2 hours and cycles for 1 hour she can over

24 km. What are ther walking and cylcing

speeds ?

Solution.

Let speed of walking = x km/h

and speed of cycling = y km/h




x + 2y = 33 ...(i)

2x + y = 24 ...(ii)

Adding (i) and (ii); we have

3x + 3y = 57 x + y 57

3 = 19 ...(iii)


–x + y = 9 x – y = –9 ...(iv)


2x = 10 x 10

2 = 5


2y = 28

y 28

2 = 14

Speed of walking = 5km/h

and speed of cycling = 14km/h.

27. An aeroplane takes 3 hours to fly 1200

km against the wind. The return trip takes 2

hours. Find the speed of the plane in still air and

the wind speed.

Solution.

Let speed of an aeroplane = x km/h

and speed of wind = y km/h

speed of plane against the wind

= (x – y) km/h

and speed with the wind = x + y km/h

Distance covered = 1200 km


1200

x y = 3 x – y =1200

3 = 400 ...(i)

and 1200

x y = 2 x + y = 1200

2 = 600

...(ii)

on adding we get

2x = 1000 x 1000

2 = 500

and on subtracting; –2y = –200

y 200

2

= 100

Speed of plane = 500 km/h

and speed of wind = 100 km/h.

28. A train covers a certain distance at a

uniform speed. If the train had been 20km/h

faster, it would have taken 4 hours less than the

scheduled time. If the train were slower by 10km/h,

it would have taken 4 hours more than the

scheduled time. Find the distance covered.

Solution.

Let speed of train = x km/h

and distance = y km

Time D

S

y

x


420

y y

x x

4

20

y y

x x

xy – xy – 20y = –4x (x + 20) = –4x2 – 80x

–4x2 – 80x + 20y = 0 5y = x2 + 20x ...(i)

and 410

y y

x x

xy – xy + 10y = 4x (x – 10) = 4x2 – 40x

10y = 4x2 – 40x

5y = 2x2 – 20x ...(ii)

From (i) and (ii); we have

x2 + 20x = 2x2 – 20x 2x2 – x2 – 40x = 0

x2 – 40x = 0 x (x – 40) = 0

Either x = 0 which is not possible

x – 40 = 0 then x = 40

Thus, 5y = x2 + 20x = (40)2 + 20 40

= 1600 + 800 = 2400

y 2400

5 = 480

required distance considered bytrain = 480km.



29. Maheep travels 600 km partly by train

and partly by car. He takes 8 hours to cover the

whole distance if he travels 120 km by train and

rest by car. But he takes 20 minutes longer if he

travels 200 km by train and the rest by car. Find

the speed of train and car.

Solution.

given total distance covered = 600 km

Let speed of trian = x km/h

and speed of car = y km/h


120 480

8x y

1 4 8 1

120 15x y

and200 400 1 25

83 3x y

2 25 1

3 200 24x y

1 4 1

15x y ...(i)

and 1 4 1

24x y ...(ii)


2

y

1 1

15 24 =

8 5 3 1

120 120 40

y = 2 × 40 = 80

and 1 4 1

80 15x

1 1 4 1 1

15 80 15 20x

1

x

4 3 1

60 60

x = 60

Speed of train = 60 km/h

and speed of car = 80 km/h.

Solve each pair of equations given below.

(i) 43 2

x y , 4

2 4

x y

Solution.

Given eqns. are; 43 2

x y

2x + 3y = 24 ...(i)

and 42 4

x y

2x + y = 16 ...(ii)

On subtraction; 2y = 8

y 8

2 = 4

Since, 2x + 4 = 16 2x = 16 – 4 = 12

x 12

2 = 6

x = 6, y = 4.

2. 97x – 78y = 310

78x – 97y = 215

Solution.

Given equations are;

97x – 78y = 310 ...(i)

78x – 97y = 215 ...(ii)

On adding; 175x – 175y = 525

x – y 525

3175

x – y = 3 ...(iii)

on subtracting; we have

19x + 19y = 95

x + y 95

19 = 5 ...(iv)


2 = 4


2 = 1

x = 4, y = 1.




3. 14 3 21 8

5, 5x y x y

Solution.

Given eqns. are ;

14 35

x y ...(i)

21 85

x y ...(ii)


42 915

x y

42 1610

x y


2525

y 25y = 25

y = 1

From (i); we have

14 3

51x

14

x= 5 – 3 = 2

2x = 14 x 14

2 = 7

x = 7, y = 1.

4. 12 4

1,x y x y

28 45

x y x y

Solution.

Given equations are;

12 4

1,x y x y

28 45

x y x y

Let x + y = a, x – y = b, then given eqns

becomes;

12 41

a b ...(i)

28 45

a b ...(ii)

On subtracting (ii) from (i) ; we have

16

a

= –4

16

a = 4 a

16

4 = 4

from (i); we have

12 4

a b = 1 3 –

4

b = 1

3 – 1 4

b

4

b = 2 b

4

2 = 2

a = 4, b = 2.

Thus,

x + y = 4 ...(iii)

x – y = 2 ...(iv)


2 = 3


2 = 1

x = 3, y = 1.

5. Amrita came first in x races and second

in y races. A score of 5 points is given for coming

first and 3 points for the second place in a race.

She scored 34 points but if the number of games

in which she came first and second were

interchaged she would have scored 4 points less,

Find x and y.

Solution.

Given Amrita first in x races and second in y

races


5x + 3y = 34 ...(i)

and 3x + 5y = 34 – 4 = 30 ...(ii)


8x + 8y = 64

x + y = 8 ...(iii)




2x – 2y = 4

x – y = 2 ...(iv)

on adding (iii) and (iv); we have

2x = 10 x 10

2 = 5

on subtracting; we have

2y = 6 y 6

2 = 3

x = 5, y = 3.

6. For a school concert 210 tickets are sold.

Of the tickets some were for adults and others

for students. The total money collected is ` 4300.

If adult ticket costs ` 25 each and student ticket

costs ` 15 each, find the number of tickets sold

of each kind.

Solution.

Total number of ticket sold = 210

Let number of adult tickets = x

and number of other students ticket = y

Cost of one adult ticket = ` 25

and for students = ` 15

Total collection = ` 4300


x + y = 210 ...(i)

and 25x + 15y = 4300

5x + 3y = 860 ...(ii)

From (i); x = 210 – y

and from (ii); we have

5(210 – y) + 3y = 860

1050 – 5y + 3y = 860

–2y = 860 – 1050 = –190

y 190

2

= 95

and x + 95 = 210 x = 210 – 95 = 115

Number of adult tickets = 115

and students tickets = 95.

7. An instructor scored a student’s test of 50

questions subtracting two times the number of

wrong answers from the number of right

answers. If the score is 26. Find out the number

of right answers.

Solution.

Number of questions = 50

Let number of right questions = x

Then wrong questions = 50 – x

Let marks for each right question = 1

According to the question; we have

x × 1 – (50 – x) × 2 = 26

x – 100 + 2x = 26

3x = 26 + 100 = 126

x 126

3 = 42

required no. of right questions = 42.

8. When a tank is filled half by water, its

weight is 20 kg. When it is filled completely, its

weight is 38 kg. Find the weight of the empty

tank.

Solution.

Let weight of empty tank = x kg

and weight of water in full tank = y kg

2y x

= 20 ...(i)

and y + x = 38 ...(ii)


y + 2x = 40

y + x = 38

On subtracting; x = 2

required weight of empty tank = 2 kg.

9. For a club anniversary dinner, some

members and some of their guests attended the

dinner. The guests were 1

3 of the number of

members. Each guest paid ` 400 and each

member paid ` 300. The total sum collected for

the event was ` 1,04,000. Find the number of

members who attended the event.



Solution.

Let number of members = x

Then number of guests 3

x

Each guest pays = ` 400

and each member pays = ` 300


x × 300 + 3

x × 400 = 104000

900 400

3

x x = 104000

1300

3

x = 104000

x 104000 3

1300

= 240

required numbers of members = 240.

10. Pramod is 25 kg heavier than his wife

Pranita. After diet Pramod loses 20 kg and Pranita

loses 15 kg and their total weight is 140 kg. Find

their original weight.

Solution.

Let weight of Pranita = x kg

and weight of Pramod = y kg


y – x = 25 ...(i)

and (y – 20) + (x – 15) = 140

y – 20 + x – 15 = 140

y + x = 140 + 20 + 15 = 175 ...(ii)


y = 200 y 200

2 = 100

and y – x = 25

100 – x = 25

x = 100 – 25 = 75

Hence, required weight of pramod be 100 kg

and weight of his wife be 75 kg.

11. In a farmyard, there were some goats and

some chickens. When their number was counted,

it was found that there were 60 heads and 148

legs. How many goats and how many chickens

were in the farmyard ?

Solution.

In a farmyard

Let number of goats = x

and number of chickens = y

Number of heads = 60

and number of legs = 148


and x + y = 60 ...(i)

4x + 2y = 148

2x + y = 74 ...(ii)


x = 14

and y = 60 – x = 60 – 14 = 46

required number of goats = 14

and number of chickens = 46.

12. The sum of the numerator and

denominator of a fraction is 2 more than twice

the numerator. If the numerator and the

denominator are reduced by 3, they are in the ratio

3 : 4. Find the fraction.

Solution.


and denominator = y

Frequired fraction x

y


x + y = 2x + 2

y = 2x – x + 2 = x + 2 ...(i)

and 3 3

3 4

x

y

4x – 12 = 3y – 9 ...(ii)

4x – 12 = 3 (x + 2) – 9 [using (i)]

4x – 12 = 3x + 6 – 9

x = 6 – 9 + 12 = 9

and y = x + 2 = 9 + 2 = 11

required fraction 9

11

x

y .



13. 2 men and 7 boys complete a certain

piece of work in 8 days. 4 men and 4 boys can

do the same in only 6 days. Find the number of

days required to complete the work by 1 man.

Solution.

Let one man’s 1 day’s work 1

x

and boy’s 1 day’s work 1

y


2 7 1

8x y ...(i)

and4 4 1

6x y ...(ii)


8 28 4 1

8 2x y

28 28 7

6x y

On subtracting; 20 1 7 3 7 4

2 6 6 6x

x 20 6

304

one man can do the work in 30 days.

14. A boat goes 24 km downstream and

returns in 5 hours. It takes 8 hours to go 36km

downstream and 40km upstream. Find the speed

of the boat in still water.

Solution.

Let speed of boat = x km/h

and speed of stream = y km/h

Speed of downstream = (x + y) km/h

and speed of upstream = (x – y) km/h


24 245

x y x y

and 40 36

8x y x y

Let x + y = a and x – y = b, then given eqns

become;

4 24

a b

= 5 ...(i)

and36 40

a b = 8 ...(ii)


72 215

a b

72 8016

a b


8

b = –1

8

b = 1 b = 8

and 24 24

8a = 5

24

3 5a

24

5 3 2a

a 24

2 = 12

Now a = 12, b = 8

Now x + y = 12

x – y = 8


2 = 10


2 = 2

required speed of boat = 10 km/h

and speed of stream = 2 km/h.


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