Contents - Arundeep Self Help

Contents

1. Rational Numbers 1—41

2. Powers 42—54

3. Squares and Square Roots 55—100

4. Cubes and Cube Roots 101—133

5. Playing With Numbers 134—139

6. Algebraic Expressions and Identities 140—170

7. Factorization 171—190

8. Division of Algebraic Expressions 191—204

9. Linear Equation in One Variable 205—237

10. Direct and Inverse Variations 238—250

11. Time and Work 251—258

12. Percentage 259—266

13. Profit, Loss, Discount and Value Added Tax (VAT) 267—285

14. Compound Interest 286—317

15. Understanding Shapes-I (Polygons) 318—321

16. Understanding Shapes-II (Quadrilaterals) 322—332

17. Understanding Shapes-III (Special Types of Quadrilaterals) 333—353

18. Practical Geometry (Constructions) 354—364

19. Visualising Shapes 365—372

20. Mensurations-I (Area of a Trapezium and a Polygon) 373—392

21. Mensuration-II (Volumes and Surface Areas of a Cuboid

and a Cube) 393—409

22. Mensuration-III (Surface Area and Volume of a Right

Circular Cylinder) 410—429

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23. Data Handling-I (Classification and Tabulation of Data 430—437

24. Data Handling-II (Graphical Representation of Data

as Histograms) 438—443

25. Data Handling-III (Pictorial Representation of Data as

Pie Charts or Circle Graphs) 444—457

26. Data Handling-IV (Probability) 458—465

27. Introduction to Graphs 466—475


1

Points to Remember

1. Rational numbers— A number of the form q

p where p and q are integers and q o, is called

a rational number. In other words, a rational number can be expressed as the quotient of two

integers when divisor is not equal is zero.

2. Lowest form of a rational numbers— A rational number q

p is to be called in the lowest or

simplest form of p and q have no common factor other than 1. q

p is also said to be in standard

form.

3. Equality of rational numbers— Two rational numbers q

p and

s

r are equal if p × s = q × r.

q

p =

mq

mp

where m is a non-zero integer, then q

p =

mq

mp

which is equivalent to q

p.

4. Addition of rational numbers.

(a) If the denominators are same, Then

Let q

p,

q

r,

q

s are the rational number, then

q

p +

q

r +

q

s =

p

rqp

(b) When the denominators are different, then

First convert all the rational numbers of the same denominators by taking LCM.

Then add them as given above in (a).

5. Properties of addition of rational numbers—

(a) Closure property— The sum of any two rational numbers is always a rational number. If b

a,

d

c are the rational numbers the

d

c

b

a is also a rational number..

1RATIONAL NUMBERS


2

(b) Commutative property— The addition of rational number is commutative i.e., if b

a +

d

c are

rational numbers, then b

a +

d

c =

d

c +

b

a.

(c) Associative property— The addition of rational numbers is associative. i.e. If b

a,

d

c,

f

e are

rational numbers, then

b

a +

f

e

d

c –

d

c

b

a +

f

e.

(d) Existence of Additive Identity or zero property— The sum of any rational number and zero

(0) is the rational number itself e.g. If b

a is a rational number, then

b

a + 0 = 0 +

b

a =

b

a.

(e) Existence of negative (Additive Inverse) of a rational number— For every rational number

b

a, there is a rational number

d

c such that

b

a +

d

c = 0 =

d

c +

b

a.

Then the rational numbers b

a and

d

c are called additive inverse or negative of each other..

The additive inverse of b

a is

b

a and of

b

a is

b

a.

b

a +

b

a = 0 or

b

a +

b

a = 0

6. Subtraction of rational numbers— If b

a and

d

c are two rational number, then

b

a –

d

c =

b

a +

d

c.

Properties of subtraction— Only closure property of subtraction holds good, but commutative

property, Associative property do not alway hold good.

Existance of right identity (i.e. zero identity) is true i.e. b

a – 0 =

b

a.

7. Multiplication of rational numbers— Product of two given rational numbers

= rsdenominatotheirofProduct

numenatorstheirofProduct


3

i.e. b

a and

d

c are two rational numbers, then

b

a ×

d

c =

db

ca

.

8. Properties of multiplication—

(i) Closure property— The multiplication of two rational numbers is also a rational number. If

b

a and

d

c are two rational number, then

b

a ×

d

c =

bd

ac is also a rational number..

(ii) Commutative property— The multiplication of two rational is commutative i.e. a × b = b ×

a for any two rational numbers a and b.

(iii) Associative property— The multiplication of rational numbers is associative. If a, b, c are

three rational numbers, then (a × b) × c = a × (b × c).

(iv) Existance of Identity— The rational number 1 is the multiplicative identity

a × 1 = a = 1 × a

(v) Multiplication by zero— Any rational number a we have a × 0 = 0 = 0 × a.

(vi) Existance of multiplicative inverse— For every non zero rational number ‘a’, there exists a

rational number a

1 (reciprocal of a) such that a ×

a

1 = 1 =

a

1 × a.

(vii) Distributive property of multiplication over addition— If x, y, z are any three rational

numbers, then x × (y + z) = x × y + x × z.

9. Division of rational number— If x and y are two rational numbers, y 0, then the result of

dividing x by y is a rational number in multiplying x by the reciprocal or inverse of y.

10. Properties of division of rational numbers—

Properties I— If b

a and

d

c are two rational numbers such that

d

c 0, then

b

a

d

c is

always a rational number.

That is, the set of all non-zero rational numbers is closed under division.

Properties II— For any rational number b

a, we have

b

a 1 =

b

a and

b

a (–1) = –

b

a =

b

a

Properties III— For every non-zero rational number b

a, we have

(i)b

a

b

a = 1 (ii)

b

a �

b

a = –1 (iii)

b

a

b

a = –1

Remark— The division of rational numbers is neither commutative nor associative.


4

11. Representation of rational numbers on a number line—

Any rational number can be represented on a number line. Every positive number lies on the

right-side of 0 and negative number on the left of 0.

O 1 2 3 4 512345 66

12. Rational number between two rational numbers—

Let x and y be the rational numbers than a number between x and y will be 2

yx

Note— We can find as many rational number between two rational numbers.

1. Add the following rational numbers :

(i)7

5 and

7

3(ii)

4

15 and

4

7

(iii)11

8 and

11

4(iv)

13

6 and

13

9

Solution—

(i)7

5 and

7

3 =

7

5 +

7

3 =

7

35 =

7

2

(ii)4

15 and

4

7

= 4

15 +

4

7 =

4

715 =

4

8

= 44

48

= –2

(iii)11

8 and

11

4

= 11

8 +

11

4 =

11

48

= 11

12

EXERCISE 1.1

(iv)13

6 and

13

9

= 13

6 +

13

9 =

13

96

= 13

3

2. Add the following rational numbers :

(i)4

3 and

8

5(ii)

9

5

and

3

7

(iii) –3 and 5

3(iv)

27

7 and

18

11

(v)4

31

and

8

5(vi)

36

5 and

12

7

(vii)16

5 and

24

7(viii)

18

7

and

27

8

Solution—

(i)4

3 and

8

5

LCM of 4, 8 = 8


5

4

3 =

24

23

= 8

6

4

3 +

8

5 =

8

6 +

8

5

= 8

56 =

8

1

(ii)9

5

and

3

7

9

5

and

3

7

LCM of 9, 3 = 9

3

7 =

33

37

= 9

21

Now 9

5 +

9

21 =

9

215 =

9

16

(iii) –3 and 5

3

LCM of 5, 1 = 1

1

3 =

51

53

= 5

15

–3 + 5

3 =

5

15 +

5

3

= 5

315 =

5

12

(iv)27

7 and

18

11

LCM of 27, 18 = 54

27

7 =

227

27

= 54

14

18

11 =

318

311

= 54

33

27

7 +

18

11 =

54

14 +

54

33

= 54

3314 =

54

19

(v)4

31

and

8

5

4

31 and

8

5

LCM of 4, 8, = 8

4

31 =

24

231

= 8

62

4

31 +

8

5 =

8

62 +

8

5

8

562 =

8

67

(vi)36

5 and

12

7

LCM of 36, 12 = 36

12

7 =

312

37

= 36

21

36

5 +

12

7 =

36

5 +

36

21

= 36

215 =

36

16

= 436

416

= 9

4

(vii)16

5 and

24

7

LCM of 16, 24 = 48

16

5 =

316

35

= 48

15

and 24

7 =

224

27

= 48

14


6

Now 16

5 +

24

7 =

48

15 +

48

14

= 48

1415 =

48

1

(viii)18

7

and

27

8

18

7 and

27

8

LCM of 18, 27 = 54

18

7 =

318

37

= 54

21

and 27

8 =

227

28

= 54

16

Now 18

7 +

27

8 =

54

21 +

54

16

= 54

1621 =

54

5

3. Simplify :

(i)9

8 +

6

11(ii)

16

5 +

24

7

(iii)12

1

and

15

2

(iv)

19

8 +

57

4

(v)9

7 +

4

3

(vi)

26

5 +

39

11

(vii)9

16 +

12

5(viii)

8

13 +

36

5

(ix) 0 + 5

3(x) 1 +

5

4

(xi) 3 + 7

5

Solution—

(i)9

8 +

6

11

LCM of 9, 6 = 18

9

8 =

29

28

= 18

16

and 6

11 =

36

311

= 18

33

Now 9

8 +

6

11 =

18

16 +

18

33

= 18

3316 =

18

17

(ii)16

5 +

24

7

LCM of 16, 24 = 48

16

5 =

316

35

= 48

15

and 24

7 =

224

27

= 48

14

Now 16

5 +

24

7 =

48

15 +

48

14

= 48

1415 =

48

1

(iii)12

1

+

15

2

= 12

1 +

15

2

15

2

115

12

15

2and

12

1

112

11

12

1

LCM of 12, 15 = 60

12

1 =

512

51

= 60

5


7

and 15

2 =

415

42

= 60

8

Now = 12

1 +

15

2 =

60

5 +

60

8

= 60

85 =

60

13

(iv)19

8 +

57

4

LCM of 19, 57 = 57

19

8 =

319

38

= 57

24

and 57

4 =

157

14

= 57

4

Now 19

8 +

57

4 =

57

24 +

57

4

= 57

424 =

57

28

(v)9

7 +

4

3

9

7 +

4

3

4

3

14

13

4

3

LCM of 9, 4 = 36

9

7 =

49

47

= 36

28

and 4

3 =

94

93

= 36

27

Now 9

7 +

4

3 =

36

28 +

36

27

= 36

2728 =

36

1

(vi)26

5 +

39

11

26

5 +

39

11

39

11

139

111

39

11

LCM of 26, 39 = 78

26

5 =

326

35

= 78

15

and 39

11 =

239

211

= 78

22

Now 26

5 +

39

11 =

78

15 +

78

22

= 78

2215 =

78

7

(vii)9

16 +

12

5

LCM of 9, 12 = 36

9

16 =

49

416

= 36

64

12

5 =

312

35

= 36

15

Now 9

16 +

12

5 =

36

64 +

36

15

= 36

1564 =

36

79

(viii)8

13 +

36

5

LCM of 8, 36 = 72

8

13 =

98

913

= 72

117


8

36

5 =

236

25

= 72

10

Now 8

13 +

36

5 =

72

117 +

72

10

= 72

10117 =

72

107

(ix) 0 + 5

3 =

5

3

(Zero property of addition)

(x) 1 + 5

4 =

5

5 +

5

4

5

5

51

51

1

1

= 5

45 =

5

1

(xi) 3 + 7

5

= 3 +

7

5

7

5

17

15

7

5

= 1

3 +

7

5

= 7

521 =

7

16

4. Add and express the sum as mixed

fraction :

(i)5

12 and

10

43(ii)

7

24 and

4

11

(iii)6

31 and

8

27(iv)

6

101 and

8

7

Solution—

(i)5

12 and

10

43

LCM of 5, 10, = 10

5

12 =

25

212

= 10

24

5

12 +

10

43 =

10

24 +

10

43

= 10

4324 =

10

19

= 110

9

(ii)7

24 and

4

11

LCM of 7, 4 = 28

7

24 =

47

424

= 28

96

and 4

11 =

74

711

= 28

77

7

24 +

4

11 =

28

96 +

28

77

= 28

7796 =

28

19

(iii)6

31 and

8

27

LCM of 6, 8 = 24

6

31 =

46

431

= 24

124

and 8

27 =

38

327

= 24

81

6

31 +

8

27 =

24

124 +

24

81

= 24

81124 =

24

205


9

= –824

13

(iv)6

101 and

8

7

LCM of 6, 8 = 24

6

101 =

46

4101

= 24

404

and 8

7 =

38

37

= 24

21

6

101 +

8

7 =

24

404 +

24

21

= 24

21404 =

24

425

= 1724

17

1. Verify commutativity of addition of

rational numbers for each of the following

pairs of rational numbers :

(i)5

11 and

7

4(ii)

9

4 and

12

7

(iii)5

3 and

15

2

(iv)7

2

and

35

12

(v) 4 and 5

3(vi) –4 and

7

4

Solution—

(i)5

11 and

7

4

5

11 +

7

4

LCM of 5 and 7 = 35

5

11 =

75

711

= 35

77

and 7

4 =

57

54

= 35

20

5

11 +

7

4 =

35

77 +

35

20

EXERCISE 1.2

= 35

2077 =

35

57

and 7

4 +

5

11 =

35

20 –

35

77

= 35

7720 =

35

57

5

11 +

7

4 =

7

4 +

5

11

(ii)9

4 and

12

7

112

17

= 12

7

Now 9

4 +

12

7

LCM of 9, 12 = 36

9

4 =

49

44

= 36

16

12

7 =

312

37

= 36

21


10

9

4 +

12

7 =

36

16 +

36

21

= 36

2116 =

36

5

and 12

7 +

9

4 =

36

21 +

36

16

= 36

1621 =

36

5

9

4 +

12

7 =

12

7 +

9

4

(iii)5

3 and

15

2

15

2

= 115

12

= 15

2

LCM of 5 and 15 = 15

5

3 =

35

33

= 15

9

Now 5

3 +

15

2 =

15

9 +

15

2

15

29 =

15

7

and 15

2 +

5

3 =

15

2 +

15

9

= 15

92 =

15

7

5

3 +

15

2 =

15

2 +

5

3

(iv)7

2

and

35

12

LCM of 7, 35 = 35

7

2

=

57

52

= 35

10

35

12

=

135

112

= 35

12

Now 7

2 +

35

12 =

35

10 +

35

12

= 35

1210 =

35

22

and 35

12 +

7

2 =

35

12 +

35

10

= 35

1012 =

35

22

7

2 +

35

12 =

35

12 +

7

2

(v) 4 and 5

3

LCM of 1, 5 = 5

1

4 =

51

54

= 5

20

Now 1

4 +

5

3 =

5

20 +

5

3

= 5

320 =

5

17

and 5

3 +

1

4 =

5

3 +

5

20

= 5

203 =

5

17

4 + 5

3 =

5

3 + 4


11

(vi) –4 and 7

4

1

4 and

7

4

7

4

17

14

7

4

LCM of 1, 7 = 7

1

4 =

71

74

= 7

28

and 7

4

–4 + 7

4 =

7

28 +

7

4

= 7

428 =

7

32

and 7

4 + (–4) =

7

4 +

7

28

= 7

284 =

7

32

–4 + 7

4 =

7

4 + (–4)

2. Verify associativity of addition of rational

numbers i.e., (x + y) + z = x + (y + z),

when :

(i) x = 2

1, y =

3

2, z = –

5

1

(ii) x = 5

2, y =

3

4, z =

10

7

(iii) x = 11

7, y =

5

2

, z =

22

3

(iv) x = –2, y = 5

3, z =

3

4

Solution—

(i) x = 2

1, y =

3

2, z = –

5

1

(x + y) + z =

3

2

2

1 +

5

1

= 6

43 +

5

1 =

6

7 +

5

1

= 6

7 –

5

1

= 30

635 =

30

29

and x + (y + z) = 2

1 +

5

1

3

2

= 2

1 +

15

310

= 2

1 +

15

7

= 30

1415 =

30

29

(x + y) + z = x + (y + z)

(ii) x = 5

2, y =

3

4, z =

10

7

(x + y) + z =

3

4

5

2 +

10

7

= 15

206 +

10

7

= 15

14 +

10

7

LCM of 15, 10 = 30

= 30

2128 =

30

7


12

and x + (y + z) = 5

2 +

10

7

3

4

= 5

2 +

30

2140

= 5

2 +

30

19

= 30

1912 =

30

7

(x + y) + z = x + (y + z)

(iii) x = 11

7, y =

5

2

, z =

22

3

x = 11

7, y =

15

12

, z = 22

3

x = 11

7, y =

5

2, z =

22

3

Now (x + y) + z

5

2

11

7 +

22

3

= 55

2235 +

22

3

= 55

57 +

22

3(LCM of 22, 55 = 110)

= 110

15114 =

110

129

and x + (y + z) = 11

7 +

22

3

5

2

= 11

7 +

110

1544

= 11

7 +

110

59

= 110

5970 =

110

129

(x + y) + z = x + (y + z)

(iv) x = –2, y = 5

3, z =

5

4

(x + y) + z =

5

3

1

2 +

5

4

= 5

310 +

5

4

= 5

7 +

5

4

= 5

47 =

5

11

and x + (y + z) = –2 +

5

4

5

3

= –2 + 5

43

= 1

2 +

5

1

= 5

110 =

5

11

(x + y) + z = x + (y + z)

3. Write the additive inverse of each of the

following rational numbers :

(i)17

2(ii)

11

3

(iii)5

17(iv)

25

11

Solution—

(i) Additive inverse of 17

2 is –

17

2 =

17

2


13

(ii) Additive inverse of 11

3

is –

11

3 =

11

3

(iii) Additive inverse of 5

17 is –

5

17 =

5

17

(iv) Additive inverse of 25

11

=

25

11

= –

25

11 =

25

11

4. Write the negative (additive inverse) of

each of the following :

(i)5

2(ii)

9

7

(iii)13

16(iv)

1

5

(v) 0 (vi) 1

(vii) –1

Solution—

(i) Negative of 5

2 is –

5

2 =

5

2

(ii) Negative of 9

7

is –

9

7 =

9

7

(iii) Negative of 13

16 is –

13

16 =

13

16

(iv) Negative inverse of 1

5 is –

1

5 = 5

(v) Negative inverse of 0 is 0

(vi) Negative inverse of 1 is –1

(vii) Negative inverse of –1 is –(–1) = 1

5. Using commutativity and associativity of

addition of rational numbers, express

each of the following as a rational

number :

(i)5

2 +

3

7 +

5

4 +

3

1

(ii)7

3 +

9

4 +

7

11 +

9

7

(iii)5

2 +

3

8 +

15

11 +

5

4 +

3

2

(iv)7

4 + 0 +

9

8 +

7

13 +

21

17

Solution—

(i)5

2 +

3

7 +

5

4 +

3

1

=

5

4

5

2 +

3

1

3

7

=

5

42 +

3

17

= 5

2 +

3

6 =

5

2 +

1

2

= 1

2 –

5

2

= 5

210 =

5

8

(ii)7

3 +

9

4 +

7

11 +

9

7

=

7

11

7

3 +

9

7

9

4

= 7

113 +

9

74

= 7

8 +

9

3


14

= 7

8 +

3

1

3

1

39

33

9

3

= 21

724 =

21

17

(iii)5

2 +

3

8 +

15

11 +

5

4 +

3

2

=

5

4

5

2 +

3

2

3

8 +

15

11

= 5

42 +

3

28 +

15

11

= 5

6 +

3

6 +

15

11

= 15

113018 =

15

1148 =

15

37

(iv)7

4 + 0 +

9

8 +

7

13 +

21

17

=

7

13

7

4 +

21

17

9

8 + 0

= 7

134 +

63

5156

= 7

9 +

63

5

= 63

581 =

63

86

6. Re-arrange suitably and find the sum in

each of the following :

(i)12

11 +

3

17 +

2

11 +

2

25

(ii)7

6 +

6

5 +

9

4 +

7

15

(iii)5

3 +

3

7 +

5

9 +

15

13 +

3

7

(iv)13

4 +

8

5 +

13

8 +

13

9

(v)3

2 +

5

4 +

3

1 +

5

2

(vi)8

1 +

12

5 +

7

2 +

12

7 +

7

9 +

16

5

Solution—

(i)12

11 +

3

17 +

2

11 +

2

25

=

3

17

12

11 +

2

25

2

11

= 12

6811+

2

2511

= 12

57 +

2

14

= 12

8457 =

12

141

(ii)7

6 +

6

5 +

9

4 +

7

15

=

7

15

7

6 +

9

4

6

5

= 7

156 +

18

815

= 7

21 +

18

23

= 126

161378

LCM of 7, 18 = 126


15

= 126

539 =

7126

7539

= 18

77

(iii)5

3 +

3

7 +

5

9 +

15

13 +

3

7

=

5

9

5

3 +

3

7

3

7 +

15

13

= 5

93 +

3

77 +

15

13 =

5

12 + 0 +

15

13

= 5

12 +

15

13

= 15

1336 =

15

23

(iv)13

4 +

8

5 +

13

8 +

13

9

= 13

4 +

13

8 +

13

9 +

8

5

= 13

984 +

8

5

= 13

5 +

8

5

= 104

6540 (LCM of 13, 8 = 104)

= 104

25

(v)3

2 +

5

4 +

3

1 +

5

2

=

3

1

3

2 +

5

2

5

4

= 3

12 +

5

24

= 3

3 +

5

2

= 1 + 5

2 =

5

25 =

5

3

(vi)8

1 +

12

5 +

7

2 +

12

7 +

7

9 +

16

5

=

16

5

8

1 +

12

7

12

5 +

7

9

7

2

= 16

52 +

12

75 +

7

92

= 16

3 +

12

12 +

7

11

= 16

3 + 1 +

7

11

= 112

17611221 =

112

28821

= 112

267

1. Subtract the first rational number from the second in each of the following :

(i)8

3,

8

5(ii)

9

7,

9

4(iii)

11

2,

11

9(iv)

13

11,

13

4

EXERCISE 1.3


16

(v)4

1,

8

3(vi)

3

2,

6

5

(vii)7

6,

14

13(viii)

33

8,

22

7

Solution—

(i)8

3 from

8

5 =

8

5 –

8

3 =

8

35

= 8

2 =

28

22

= 4

1

(ii)9

7 from

9

4 =

9

4–

9

7

= 9

4 +

9

7 =

9

74

= 9

11

(iii)11

2 from

11

9 =

11

9–

11

2

11

9 +

11

2 =

11

29

= 11

7

(iv)13

11 from

13

4 =

13

4 –

13

11

= 13

114 =

13

15

(v)4

1 from

8

3 =

8

3–

4

1

= 8

3 –

4

1

= 8

23(LCM 8, 4 = 8)

= 8

5

(vi)3

2 from

6

5=

6

5–

3

2

= 6

5 +

3

2

= 6

45 =

6

9 =

36

39

= 2

3

(vii)7

6 from

14

13 =

14

13–

7

6

= 14

13 +

7

6

= 14

1213 =

14

1

(viii)33

8 from

22

7 =

22

7–

33

8

= 66

1621 =

66

5

2. Evaluate each of the following :

(i)3

2 –

5

3(ii)

7

4–

3

2

(iii)7

4–

7

5

(iv) –2 –9

5

(v)8

3

– 7

2(vi)

13

4 –

26

5

(vii)14

5 –

7

2(viii)

15

13 –

25

12

(ix)13

6 –

13

7(x)

24

7 –

36

19

(xi)63

5 –

21

8


17

Solution—

(i)3

2 –

5

3

= 15

910 (LCM of 3, 5 = 15)

= 15

1

(ii)7

4 –

3

2

=

7

4 +

3

2

= 21

1412 (LCM of 7, 3 = 21)

= 21

2

(iii)7

4 –

7

5

= 7

4 –

7

5

= 7

54 =

7

1

(iv)1

2 –

9

5

= 9

518 =

9

23

(v)8

3

– 7

2 =

8

3 +

7

2

= 56

1621(LCM of 8, 7 = 56)

= 56

37

(vi)13

4 –

26

5 =

13

4 +

26

5

= 26

58(LCM of 13, 26 = 26)

= 26

3

(vii)14

5 –

7

2 =

14

5 +

7

2

= 14

45(LCM of 14, 7 = 14)

= 14

1

(viii)15

13 –

25

12

75

3665 (LCM of 15, 25 = 75)

= 75

29

(ix)13

6 –

13

7 =

13

6 +

13

7 =

13

76

= 13

1

(x)24

7–

36

19

LCM of 24, 36 = 72

= 24

7 –

36

19

= 72

3821 =

72

17

(xi)63

5 –

21

8 =

63

5 +

21

8

= 63

245(LCM of 63, 21 = 63)

= 63

29


18

3. The sum of two numbers is 9

5. If one of

the numbers is 3

1, find the other..

Solution—

Sum of two numbers = 9

5

One number = 3

1

Second number = 9

5 –

3

1

= 9

35(LCM of 9, 3 = 9)

= 9

2


1. If one

of the numbers is 3

12, find the other..

Solution—


1

One number = 3

12

Second number = 3

1 –

3

12

= 3

1 +

3

12

= 3

121 =

3

11


4. If one

of the number is –5, find the other.

Solution—


4

One number = –5

Second number = 3

4 – (–5)

= 3

4 +

1

5

= 3

154 =

3

11

6. The sum of two rational numbers is –8.

If one of the numbers is 7

15, find the

other.

Solution—

Sum of two rational numbers = –8

One number = 7

15

Second number = –8 –

7

15

= 1

8 +

7

15

= 7

1556 =

7

41

7. What should be added to 8

7 so as to get

9

5 ?


19

Solution—

The required number = 9

5 –

8

7

= 9

5 +

8

7

= 72

6340 (LCM of 9, 8 = 72)

= 72

103

8. What number should be added to 11

5 so

as to get 33

26 ?

Solution—


26 –

11

5

= 33

26 +

11

5(LCM of 33, 11 = 33)

= 33

1526 =

33

41

9. What number should be added to 7

5 to

get 3

2 ?

Solution—


2 –

7

5

= 3

2 +

7

5

= 21

1514 (LCM of 3, 7 = 21)

= 21

1

10. What number should be subtracted from

3

5 to get

6

5 ?

Solution—


5 –

6

5

= 6

510(LCM of 3, 6 = 6)

= 6

15 =

36

315

= 2

5

11. What number should be subtracted from

7

3 to get

4

5 ?

Solution—


3 –

4

5

= 7

3 –

4

5

= 28

3512(LCM of 7, 4 = 28)

= 28

23

12. What should be added to

5

3

3

2 to get

15

2 ?

Solution—


2 –

5

3

3

2


20

= 15

2 –

15

19 =

15

192

= 15

21=

315

321

= 5

7

13. What should be added to

5

1

3

1

2

1

to get 3 ?

Solution—

The required number = 3 –

5

1

3

1

2

1

= 1

3 –

30

61015

(LCM of 2, 3, 5, = 30)

= 1

3 –

30

31

= 30

3190 =

30

59

14. What should be subtracted from

3

2

4

3

to get 6

1 ?

Solution—

The required number =

3

2

4

3 –

6

1

= 4

3 –

3

2 +

6

1

= 12

289 (LCM of 4, 3, 6 = 12)

= 12

811 =

12

3

= 312

33

= 4

1

15. Fill in the blanks :

(i)13

4 –

26

3 = ............

(ii)14

9 + ............ = –1

(iii)9

7 + ............ = 3

(iv) ............ + 23

15 = 4

Solution—

(i)13

4 –

26

3 = ............

Required number

= 13

4 –

26

3 =

13

4 +

26

3

= 26

38 =

26

5

13

4 –

26

3 =

26

5

(ii)14

9 + ............ = –1

Required number = –1 –

14

9

= –1 + 14

9

= 14

914 =

14

5

(iii)9

7 + ............ = 3

Required number = 3 –

9

7


21

= 1

3 +

9

7

= 9

727

= 9

34

(iv) ............ + 23

15 = 4

Required number = 1

4 –

23

15

= 23

1592 =

23

77

1. Simplify each of the following and write

as a rational number of the form q

p :

(i)4

3 +

6

5 +

8

7(ii)

3

2 +

6

5 +

9

7

(iii)2

11 +

6

7 +

8

5

(iv)5

4 +

10

7 +

15

8

(v)10

9 +

15

22 +

20

13

(vi)3

5 +

2

3

+

3

7 + 3

Solution—

(i)4

3 +

6

5 +

8

7

LCM of 4, 6, 8 = 24

4

3 +

6

5 +

8

7

24

212018 =

24

2138 =

24

17

EXERCISE 1.4

(ii)3

2+

6

5 +

9

7

LCM of 3, 6, 9 = 18

3

2+

6

5 +

9

7

18

141512 =

18

2912 =

18

17

(iii)2

11 +

6

7 +

8

5

LCM of 2, 6, 8 = 24

2

11 +

6

7 +

8

5

24

1528132 =

24

28147

= 24

119

(iv)5

4 +

10

7 +

15

8

LCM of 5, 10, 15 = 30

5

4 +

10

7 +

15

8


22

30

162124 =

30

61

(v)10

9 +

15

22 +

20

13

= 10

9 +

15

22 +

20

13

LCM of 10, 15, 20 = 60

10

9 +

15

22 +

20

13

= 60

398854 =

60

8893=

60

5=

12

1

(vi)3

5 +

2

3

+

3

7 + 3

= 3

5 +

2

3 +

3

7 +

1

3

LCM of 3, 2, 3 = 6

3

5 +

2

3 +

3

7 +

1

3

6

1814910 =

6

2328 =

6

5

2. Express each of the following as a rational

number of the form q

p :

(i)3

8 +

4

1 +

6

11 +

8

3 – 3

(ii)7

6 + 1 +

9

7 +

21

19 +

7

12

(iii)2

15 +

8

9 +

3

11 + 6 +

6

7

(iv)4

7 + 0 +

5

9 +

10

19 +

14

11

(v)4

7 +

3

5 +

2

1 +

6

5 + 2

Solution—

(i)3

8 +

4

1 +

6

11 +

8

3 – 3

LCM of 3, 4, 6, 8 = 24 2 3, 4, 6, 82 3, 2, 3, 43 3, 1, 3, 2

1, 1, 1, 2

LCM = 2 × 2 × 3 × 2 = 24

3

8 +

4

1 +

6

11 +

8

3 –

1

3

24

72944664

= 24

9186 =

24

177 =

324

3177

= 8

59

(ii)7

6 + 1 +

9

7 +

21

19 +

7

12

LCM of 7, 9, 21, 7, = 63 3 7, 9, 21, 77 7, 3, 7, 73 1, 3, 1, 1

LCM = 3 × 7 × 3 = 63

7

6 +

1

1 +

9

7 +

21

19 +

7

12

63

10857496354

= 63

157174 =

63

17

(iii)2

15 +

8

9 +

3

11 + 6 +

6

7

LCM of 2, 8, 3, 6 = 24 2 2, 8, 3, 63 1, 4, 3, 34 1, 4, 1, 1


23

LCM = 2 × 3 × 4 = 24

2

15 +

8

9 +

3

11 +

1

6 +

6

7

= 24

281448827180

= 24

116351 =

24

235

(iv)4

7 + 0 +

5

9 +

10

19 +

14

11

LCM of 4, 5, 10, 14 = 140

2 4, 5, 10, 145 2, 5, 5, 72 2, 1, 1, 7

LCM = 2 × 5 × 2 × 7 = 140

4

7 + 0 +

5

9 +

10

19 +

14

11

140

1102662520245

= 140

497376 =

140

121

(v)4

7 +

3

5 +

2

1 +

6

5 + 2

LCM of 4, 3, 2, 6 = 12

2 4, 3, 2, 63 2, 3, 1, 32 2, 1, 1, 1

LCM = 2 × 3 × 2 = 12

4

7 +

3

5 +

2

1 +

6

5 +

1

2

12

241062021

= 12

3744 =

12

7

3. Simplify :

(i)2

3 +

4

5 –

4

7(ii)

3

5 –

6

7 +

3

2

(iii)4

5 –

6

7 –

3

2(iv)

5

2 –

10

3 –

7

4

(v)6

5 +

5

2–

15

2(vi)

8

3 –

9

2 +

36

5

Solution—

(i)2

3 +

4

5–

4

7

= 4

756 (LCM of 2, 4 = 4)

= 4

8 = –2

(ii)3

5 –

6

7 +

3

2

= 6

4710 (LCM of 3, 6 = 6)

= 6

1

(iii)4

5 –

6

7 –

3

2

= 12

81415 (LCM of 4, 6, 3, = 12)

= 12

9=

312

39

= 4

3

(iv)5

2 –

10

3 –

7

4

= 5

2 +

10

3 +

7

4

= 70

402128 (LCM of 5, 10, 7 = 70)


24

= 70

2861 =

70

33

(v)6

5 +

5

2 –

15

2

= 30

41225 (LCM of 6, 5, 15, = 30)

= 30

1229 =

30

17

(vi)8

3 –

9

2 +

36

5

= 72

101627 (LCM of 8, 9, 36 = 72)

= 72

1043 =

72

33

= 372

333

= 24

11

1. Multiply :

(i)11

7 by

4

5(ii)

7

5 by

4

3

(iii)9

2 by

11

5(iv)

17

3 by

4

5

(v)7

9

by

11

36

(vi)

13

11 by

7

21

(vii) –5

3 by –

7

4(viii) –

11

15 by 7

Solution—

(i)11

7 by

4

5 =

411

57

= 44

35

(ii)7

5 by

4

3 =

47

35

= 28

15

(iii)9

2 by

11

5 =

119

52

= 99

10

(iv)17

3 by

4

5

= 417

53

= 68

15

= 168

115

= 68

15

EXERCISE 1.5

(v)7

9

by

11

36

= 117

369

= 77

324

(vi)13

11 by

7

21 =

713

2111

= 91

231 =

791

7231

= 13

33

(vii) –5

3 by –

7

4 =

75

43

= 35

12

(viii)–11

15 by 7 =

11

715 =

11

105

2. Multiply :

(i)17

5 by

60

51

(ii)

11

6 by

36

55

(iii)25

8 by

16

5(iv)

7

6 by

36

49

(v)9

8

by

16

7

(vi)9

8 by

64

3

Solution—

(i)17

5 by

60

51


25

17

5 ×

60

51

60

51

160

151

60

51

=

6017

515

= 1020

255 =

2551020

255255

= 4

1

(ii)11

6 by

36

55 =

11

6 ×

36

55

=

3611

556

=

61

51

= 6

5

(iii)25

8 by

16

5 =

1625

58

=

25

11

= 10

1

(iv)7

6 by

36

49 =

7

6 ×

36

49

=

367

496

=

61

71

= 6

7

(v)9

8

by

16

7

= 9

8

×

16

7

=

169

78

=

29

71

= 18

7

(vi)9

8 by

64

3

= 649

38

= 83

11

= 24

1

3. Simplify each of the following and express

the result as a rational number in

standard form :

(i)21

16 ×

5

14(ii)

6

7 ×

28

3

(iii)36

19 × 16 (iv)

9

13 ×

26

27

(v)16

9 ×

27

64

(vi)7

50 ×

3

14

(vii)9

11 ×

88

81

(viii)9

5 ×

25

72

Solution—

(i)21

16 ×

5

14 =

521

1416

= 53

216

= 15

32

(ii)6

7 ×

28

3 =

286

37

=

42

11

= 8

1

(iii)36

19 × 16 =

36

1619 =

9

419 =

9

76

(iv)9

13 ×

26

27

= 269

2713

= 21

31

= 2

3

= 2

3

(v)16

9 ×

27

64

=

2716

649

= 31

41

= 3

4

=

13

14

= 3

4


26

(vi)7

50 ×

3

14 =

37

1450

= 31

250

= 3

100

(vii)9

11 ×

88

81

=

889

8111

=

81

91

= 8

9

=

18

19

= 8

9

(viii)9

5 ×

25

72

= 259

725

= 51

81

= 5

8

= 5

8

4. Simplify :

(i)

5

2

8

25 –

9

10

5

3

(ii)

4

1

2

1 +

62

1

(iii)

15

25 –

9

26

(iv)

3

5

4

9+

6

5

2

13

(v)

5

12

3

4+

15

21

7

3

(vi)

3

8

5

13 –

3

11

2

5

(vii)

26

11

7

13 –

6

5

3

4

(viii)

2

3

5

8 +

16

11

10

3

Solution—

(i)

5

2

8

25 –

9

10

5

3

= 58

225

–

95

103

= 14

15

–

31

21

= 4

5 –

3

2

= 12

815 =

12

23

(ii)

4

1

2

1 +

62

1

= 42

11

+ 2

1 ×

1

6 =

8

1 +

1

3

= 8

241 =

8

25

(iii)

15

25 –

9

26

= 15

25 –

9

26

= 3

21 –

3

22 =

3

2 –

3

4

= 3

42 =

3

2

(iv)

3

5

4

9 +

6

5

2

13


27

= 34

59

+ 62

513

= 14

53

+ 12

65

= 4

15 +

12

65

= 12

6545 =

12

20 =

412

420

= 3

5

(v)

5

12

3

4 +

15

21

7

3

= 53

124

+ 157

213

= 51

44

+ 51

31

= 5

16

+ 5

3

= 5

16 +

5

3 =

5

316 =

5

19

(vi)

3

8

5

13 –

3

11

2

5

= 35

813

– 32

115

= 15

104 –

6

55

= 30

275208 (LCM of 15, 6 = 30)

= 30

483 =

10

161

(vii)

26

11

7

13 –

6

5

3

4

= 267

1113

– 63

54

= 27

111

– 33

52

= 14

11 –

9

10

= 126

14099(LCM of 14, 9 = 126)

= 126

239

(viii)

2

3

5

8 +

16

11

10

3

=

25

38

+ 1610

113

=

15

34

+ 1610

113

= 5

12 +

160

33

= 160

33384 =

160

417

5. Simplify :

(i)

6

1

2

3 +

2

7

3

5 –

3

4

8

13

(ii)

7

2

4

1 –

3

2

14

5 +

2

9

7

3

(iii)

2

15

9

13 +

5

8

3

7 +

2

1

5

3

(iv)

6

5

11

3 –

3

4

12

9 +

15

6

13

5

Solution—

(i)

6

1

2

3 +

2

7

3

5 –

3

4

8

13


28

= 62

13

+ 23

75

– 38

413

= 22

11

+ 23

75

– 32

113

= 4

1 +

6

35 –

6

13

= 12

26703 (LCM of 4, 6 = 12)

= 12

2673 =

12

47

(ii)

7

2

4

1 –

3

2

14

5 +

2

9

7

3

= 74

21

–

314

25

+ 27

93

= 72

11

–

37

15

+ 27

93

= 14

1 –

21

5 +

14

27

= 42

81103 (LCM of 14, 21 = 42)

= 42

94 =

242

294

= 21

47

(iii)

2

15

9

13 +

5

8

3

7 +

2

1

5

3

=

29

1513

+ 53

87

+ 25

13

= 23

513

+ 53

87

+ 25

13

= 6

65 +

15

56 +

10

3

= 30

33256565

(_ LCM of 6, 15, 10 = 30)

= 30

9112325 =

30

204 =

15

102

(iv)

6

5

11

3 –

3

4

12

9 +

15

6

13

5

= 611

53

– 312

49

+ 1513

65

= 211

51

– 3

13 +

113

21

= 22

5 – 1 +

13

2

= 286

4428665 (LCM of 22, 13 = 286)

= 286

286109 =

286

177

EXERCISE 1.6

1. Verify the property : x × y = y × x by taking :

(i) x = –3

1, y =

7

2(ii) x =

5

3, y =

13

11(iii) x = 2, y =

8

7

(iv) x = 0, y =

8

15


29

Solution—

(i) x × y = y × x

x = –3

1, y =

7

2

L.H.S. = x × y = –3

1 ×

7

2 =

73

21

= 21

2

R.H.S. = y × x = 7

2 ×

3

1 =

37

12

= 21

2

L.H.S. = R.H.S.

(ii) x × y = y × x

x = 5

3, y =

13

11

L.H.S. = x × y = 5

3 ×

13

11

=

135

113

= 65

33

R.H.S. = y × x = 13

11 ×

5

3

=

513

311

= 65

33

L.H.S. = R.H.S.

(iii) x × y = y × x

x = 2, y = 8

7

=

18

17

= 8

7

L.H.S. = x × y = 2 × 8

7 =

8

72

= 4

7

R.H.S. = y × x = 8

7 × 2 =

4

7 × 1 =

4

7

L.H.S. = R.H.S.

(iv) x × y = y × x

x = 0, y = 8

15

L.H.S. = x × y = 0 × 8

15 = 0

R.H.S. = y × x = 8

15 × 0 = 0

L.H.S. = R.H.S.

2. Ver ify the property : x × (y × z) = (x × y) ×

z by taking :

(i) x = 3

7, y =

5

12, z =

9

4

(ii) x = 0, y = 5

3, z =

4

9

(iii) x = 2

1, y =

4

5

, z =

5

7

(iv) x = 7

5, y =

13

12, z =

18

7

Solution—

(i) x × (y × z) = (x × y) × z

x = 3

7, y =

5

12, z =

9

4

L.H.S. = x × (y × z)

= 3

7 ×

9

4

5

12 =

3

7 ×

35

44

= 3

7 ×

15

16 =

153

167

= 45

112

R.H.S. = (x × y) × z =

5

12

3

7 ×

9

4

= 51

47

× 9

4 =

5

28 ×

9

4


30

= 95

428

= 45

112

L.H.S. = R.H.S.

(ii) x × (y × z) = (x × y) × z

x = 0, y = 5

3, z =

4

9

L.H.S. = x × (y × z) = 0 ×

4

9

5

3

= 0 ×

45

93

= 0 × 20

27 = 0

R.H.S. = (x × y) × z =

5

30 ×

4

9

= 0 × 4

9 = 0

L.H.S. = R.H.S.

(iii) x × (y × z) = (x × y) × z

x = 2

1, y =

4

5

, z =

5

7

L.H.S. = x × (y × z) = 2

1 ×

5

7

4

5

= 2

1 ×

54

75

= 2

1 ×

4

7 =

2

1 ×

4

7 =

42

71

= 8

7

R.H.S. = (x × y) × z

=

4

5

2

1 ×

5

7

= 42

51

× 5

7 =

8

5

×

5

7

=

58

75

=

18

71

= 8

7

= 8

7

L.H.S. = R.H.S.

(iv) x × (y × z) = (x × y) × z

x = 7

5, y =

13

12, z =

18

7

L.H.S. = x × (y × z) = 7

5 ×

18

7

13

12

= 7

5 ×

1813

712 =

7

5 ×

313

72

= 7

5 ×

39

14 =

397

145

= 391

25

= 39

10

R.H.S. = (x × y) × z =

13

12

7

5 ×

18

7

=

137

125

× 18

7 =

91

60 ×

18

7

= 1891

760

= 313

110

= 39

10

L.H.S. = R.H.S.

3. Verify the property : x × (y + z) = x × y +

x × z by taking :

(i) x = 7

3, y =

13

12, z =

6

5

(ii) x = 5

12, y =

4

15, z =

3

8

(iii) x = 3

8, y =

6

5, z =

12

13

(iv) x = 4

3, y =

2

5, z =

6

7

Solution—

(i) x × (y + z) = x × y + x × z


31

x = 7

3, y =

13

12, z =

6

5

L.H.S. = x × (y + z) = 7

3 ×

6

5

13

12

= 7

3 ×

78

6572

(LCM of 13, 6 = 78)

= 7

3 ×

78

7 =

787

73

= 261

11

= 26

1

R.H.S. = x × y + x × z

= 7

3 ×

13

12 +

7

3 ×

6

5

= 137

123

+ 67

53

= 91

36 +

27

51

= 91

36 +

14

5

= 182

6572 =

182

7

= 26

1

L.H.S. = R.H.S.

(ii) x × (y + z) = x × y + x × z

x = 5

12, y =

4

15, z =

3

8

L.H.S. = x × (y + z) = 5

12 ×

3

8

4

15

= 5

12 ×

12

3245 =

5

12 ×

12

13

= 125

1312

=

15

131

= 5

13

R.H.S. = x × y + x × z =

4

15

5

12 +

3

8

5

12

= [–3 × (–3)] +

5

84 = 9 +

5

32

= 1

9 +

5

32

= 5

3245 =

5

13

L.H.S. = R.H.S.

(iii) x × (y + z) = x × y + x × z

x = 3

8, y =

6

5, z =

12

13

L.H.S. = x × (y + z) = 3

8 ×

12

13

6

5

= 3

8 ×

12

1310 =

3

8 ×

12

3

= 31

12

= 3

2

R.H.S. = x × y + x × z

=

6

5

3

8 +

12

13

3

8

= 33

54

+ 33

132

= 9

20 +

9

26 =

9

2620


32

= 9

6 =

39

36

= 3

2

L.H.S. = R.H.S.

(iv) x × (y + z) = x × y + x × z

x = 4

3, y =

2

5, z =

6

7

L.H.S. = x × (y + z) = 4

3 ×

6

7

2

5

= 4

3 ×

6

715 =

4

3 ×

6

8

=

64

83

=

21

21

= 2

2 = 1

R.H.S. = x × y + x × z

=

2

5

4

3 +

6

7

4

3

=

24

53

+ 64

73

= 8

15 +

24

71

= 8

15 +

8

7 =

8

715 =

8

8 = 1

L.H.S. = R.H.S.

4. Use the distributivity of multiplication of

rational numbers over their addition to

simplify :

(i)5

3 ×

1

10

24

35(ii)

4

5 ×

5

16

5

8

(iii)7

2 ×

4

21

16

7(iv)

4

3 ×

409

8

Solution—

(i)5

3 ×

1

10

24

35

= 5

3 ×

24

35 +

5

3×

1

10 =

245

353

+ 15

10

= 81

71

+ 11

23

= 8

7 + 6 =

8

7 +

1

48

= 8

487 =

8

55

(ii)4

5 ×

5

16

5

8 =

4

5 ×

5

8 +

4

5 ×

5

16

= 54

85

+ 54

165

= 11

21

+ 11

41

= –2 – 4 = –6

(iii)7

2 ×

4

21

16

7 =

7

2 ×

16

7 –

7

2 ×

4

21

= 167

72

– 47

212

= 81

11

– 21

31

= 8

1 –

2

3

= 8

121 =

8

11

(iv)4

3 ×

409

8 =

4

3 ×

9

8 –

4

3 ×

1

40

= 94

83

– 14

403

= 31

21

– 11

103

= 3

2 –

1

30

= 3

902 =

3

88

5. Find the multiplicative inverse

(reciprocal) of each of the following

rational numbers :

(i) 9 (ii) –7

(iii)5

12(iv)

9

7


33

(v)5

3

(vi)3

2 ×

4

9

(vii)8

5 ×

15

16(viii) –2 ×

5

3

(ix) –1 (x)3

0

(xi) 1

Solution—

(i) Multiplicative inverse of 9 = 9

1

(ii) Multiplicative inverse of –7 = 7

1

(iii) Multiplicative inverse of 5

12 =

12

5

(iv) Multiplicative inverse of 9

7 =

7

9

(v) Multiplicative inverse of 5

3

= 3

5

= 3

5

(vi) Multiplicative inverse of 3

2 ×

4

9 =

2

3×

9

4

= 92

43

= 31

21

= 3

2

(vii) Multiplicative inverse of 8

5 ×

15

16

= 5

8

×

16

15 =

165

158

= 21

31

= 2

3

= 12

13

= 2

3

(viii)Multiplicative inverse of –2 × 5

3

= 2

1

×

3

5

= 32

51

= 6

5

(ix) Multiplicative inverse of –1 = –1

(x) Multiplicative inverse of 0 = Does not exist

as division by 0 is not admissible.

(xi) Multiplicative inverse of 1 = 1

6. Name the property of multiplication of

rational numbers illustrated by the

following statements :

(i)16

5 ×

15

8 =

15

8 ×

16

5

(ii)5

17 × 9 = 9 ×

5

17

(iii)4

7×

12

13

3

8=

4

7×

3

8 +

4

7 ×

12

13

(iv)9

5 ×

8

9

15

4 =

15

4

9

5 ×

8

9

(v)17

13

× 1 =

17

13

= 1 ×

17

13

(vi)16

11 ×

11

16

= 1

(vii)13

2 × 0 = 0 = 0 ×

13

2

(viii)2

3 ×

4

5 +

2

3 ×

6

7 =

2

3 ×

6

7

4

5

Solution—

(i)16

5 ×

15

8 =

15

8 ×

16

5

It is commutative property

(ii)5

17 × 9 = 9 ×

5

17

It is commutative property


34

(iii)4

7 ×

12

13

3

8 =

4

7 ×

3

8 +

4

7 ×

12

13

It is distributive property over addition

(iv)9

5 ×

8

9

15

4 =

15

4

9

5 ×

8

9

It is associative property

(v)17

13

× 1 =

17

13

= 1 ×

17

13

It is multiplicative identity.

(vi)16

11 ×

11

16

= 1

It is existance of multiplicative inverse.

(vii)13

2 × 0 = 0 = 0 ×

13

2

It is zero property of multiplication.

(viii)2

3 ×

4

5 +

2

3 ×

6

7 =

2

3 ×

6

7

4

5

It is distributive law of multiplication over

addition.


(i) The product of two positive rational

numbers is always ________.

(ii) The product of a positive rational number

and a negative rational number is always

________.

(iii) The product of two negative rational

numbers is always ________.

(iv) The reciprocal of a positive rational

number is ________.

(v) The reciprocal of a negative rational

number is ________.

(vi) Zero has ________ reciprocal.

(vii) The product of a rational number and its

reciprocal is ________.

(viii) The numbers ________ and ________ are

their own reciprocals.

(ix) If a is reciprocal of b, then the reciprocal

of b is ________.

(x) The number 0 is ________ the reciprocal

of any number.

(xi) Reciprocal of a

1, a 0 is ________.

(xii) (17 × 12)–1 = 17–1 × ________.

Solution—

(i) The product of two positive rational numbers

is always positive.

(ii) The product of a positive rational number

and a negative rational number is always

negative.

(iii) The product of two negative rational numbers

is always positive.

(iv) The reciprocal of a positive rational number

is positive.

(v) The reciprocal of a negative rational number

is negative.

(vi) Zero has no reciprocal.

(vii) The product of a rational number and its

reciprocal is 1.

(viii)The numbers 1 and –1 are their own

reciprocals.

(ix) If a is reciprocal of b, then the reciprocal of

b is a.

(x) The number 0 is not the reciprocal of any

number.

(xi) Reciprocal of a

1, a 0 is a.

(xii) (17 × 12)–1 = 17–1 × 12–1.


(i) – 4 × 9

7 =

9

7 × ________.

(ii)11

5 ×

8

3 =

8

3 × ________


35

(iii)2

1 ×

12

5

4

3 =

2

1 × _____________ +

________ × 12

5

(iv)5

4 ×

9

8

7

5 =

___

5

4 ×

9

8

Solution—

(i) – 4 × 9

7 =

9

7 × – 4

(ii)11

5 ×

8

3 =

8

3 ×

11

5

(iii)2

1 ×

12

5

4

3 =

2

1 ×

4

3 +

2

1 ×

12

5

(iv)5

4 ×

9

8

7

5 =

7

5

5

4 ×

9

8

1. Divide :

(i) 1 by 2

1(ii) 5 by

7

5

(iii)4

3 by

16

9

(iv)

8

7 by

16

21

(v)4

7

by

64

63(vi) 0 by

5

7

(vii)4

3 by –6 (viii)

3

2 by

12

7

(ix) – 4 by 5

3(x)

13

3 by

65

4

Solution—

(i) 1 by 2

1 = 1

2

1 = 1 ×

1

2 = 2

(ii) 5 by 7

5 = 5

7

5 = 5 ×

5

7

= 15

17

=5

7 –

15

57

=5

35 = –7

EXERCISE 1.7

(iii)4

3 by

16

9

=

4

3

16

9

= 4

3 ×

9

16 =

94

163

= 31

41

= 3

4

(iv)8

7 by

16

21 =

8

7

16

21

= 8

7 ×

21

16

= 31

21

= 3

2

= 3

2

(v)4

7

by

64

63 =

4

7

64

63 =

4

7

×

63

64

= 91

161

= 9

16

=

19

116

= 9

16


36

(vi) 0 by 5

7= 0

5

7 = 0 ×

7

5

= 0

(vii)4

3 by –6 =

4

3

1

6 =

4

3 ×

6

1

= 24

11

= 8

1

= 8

1

(viii)3

2 by

12

7 =

3

2

12

7 =

3

2 ×

7

12

= 71

42

= 7

8

=

17

18

= 7

8

(ix) – 4 by 5

3 = –4

5

3 = –4 ×

3

5

= 3

54

= 3

20

= 3

20

(x)13

3 by

65

4 =

13

3

65

4 =

13

3 ×

4

65

= 41

53

= 4

15

= 4

15

2. Find the value and express as a rational

number in standard form :

(i)5

2

15

26(ii)

3

10

12

35

(iii) – 6

17

8(iv)

99

40 (–20)

(v)27

22

18

110(vi)

125

36

75

3

Solution—

(i)5

2

15

26 =

5

2 ×

26

15 =

265

152

= 131

31

= 13

3

(ii)3

10

12

35 =

3

10 ×

35

12

= 353

1210

= 71

42

= 7

8

=

17

18

= 7

8

(iii) – 6 17

8 = – 6 ×

8

17

=

4

173

= 4

51

= 4

51

(iv)99

40 (–20) =

99

40 ×

20

1

= 199

12

= 99

2

= 99

2

(v)27

22

18

110 =

27

22 ×

110

18

= 11027

1822

= 53

21

= 15

2

= 15

2

(vi)125

36

75

3 =

125

36 ×

3

75

= 3125

7536

= 15

312

= 5

36

= 5

36

3. The product of two rational numbers is

15. If one of the numbers is –10, find the

other.

Solution—

Product of two numbers = 15

One number = –10

Second number = 15 (–10) = 15 × 10

1

= 2

13

= 2

3

=

12

13

= 2

3


37

4. The product of two rational numbers is

9

8. If one of the numbers is

15

4, find

the other.

Solution—


8

One number = 15

4

Second number = 9

8

15

4

= 9

8 ×

4

15

= 13

52

= 3

10

= 3

10

5. By what number should we multiply 6

1

so that the product may be 9

23 ?

Solution—

Product = 9

23

and given number = 6

1

Required number = 9

23

6

1

= 9

23 ×

1

6

= 13

223

= 3

46

= 3

46

6. By what number should we multiply

28

15, so that the product may be

7

5 ?

Solution—


5

One number = 28

15

Required number = 7

5

28

15

= 7

5 ×

15

28

= 31

41

= 3

4

= 3

4

7. By what number should we multiply 13

8

so that the product may be 24 ?

Solution—


One number = 13

8

Required number = 24 13

8

= 24 × 8

13

=

1

133

= 1

39

= 11

139

= 1

39 = –39

8. By what number should 4

3 be

multiplied in order to produce 3

2 ?

Solution—

Product = 3

2


38

Let x be multiplied to 4

3 to get

3

2

x + 4

3 =

3

2 x =

3

2

4

3

x = 3

2 ×

3

4

= 33

42

= 9

8

= 19

18

= 9

8

Required number = 9

8

9. Find (x + y) (x – y), if

(i) x = 3

2, y =

2

3(ii) x =

5

2, y =

2

1

(iii) x = 4

5, y =

3

1(iv) x =

7

2, y =

3

4

(v) x = 4

1, y =

2

3

Solution—

(i) x = 3

2, y =

2

3

x + y = 3

2 +

2

3

= 6

94 =

6

13

and x – y = 3

2 –

2

3

= 6

94 =

6

5

Now (x + y) (x – y) = 6

13

6

5

= 6

13 ×

5

6

= 5

13

=

15

113

= 5

13

(ii) x = 5

2, y =

2

1

x + y = 5

2 +

2

1

= 10

54 =

10

9

x – y = 5

2 –

2

1

= 10

54 =

10

1

(x + y) (x – y) = 10

9

10

1 =

10

9 ×

1

10

= 1

9

= –9

(iii) x = 4

5, y =

3

1

x + y = 4

5 +

3

1

= 12

415 =

12

11

x – y = 4

5 –

3

1

= 12

415 =

12

19

(x + y) (x – y) = 12

11

12

19=

12

11×

19

12=

19

11

(iv) x = 7

2, y =

3

4


39

x + y = 7

2 +

3

4

= 21

286 =

21

34

x – y = 7

2 –

3

4

21

286 =

21

22

(x + y) (x – y) = 21

34

21

22

= 21

34 ×

22

21

= 111

117

= 11

17

=

111

117

= 11

17

(v) x = 4

1, y =

2

3

x + y = 4

1 +

2

3 =

4

61 =

4

7

x – y = 4

1 –

2

3

= 4

61 =

4

5

(x + y)(x – y) =4

7

4

5=

4

7×

5

4

=

5

7

= 15

17

= 5

7

10. The cost of 73

2 metres of rope is Rs 12

4

3.

Find its cost per metre.

Solution—

Cost of 73

2 m of rope = Rs. 12

4

3

or cost of 3

23 m of rope = Rs.

4

51

Cost of 1 m of rope = Rs. 4

51

3

23

= Rs. 4

51 ×

23

3 = Rs.

92

153 = Rs. 1

92

61

11. The cost of 23

1 metres of cloth is Rs.

754

1. Find the cost of cloth per metre.

Solution—

Cost of 23

1 m or

3

7 m of cloths = Rs. 75

4

1

= Rs. 4

301

Cost of 1 m cloth = Rs. 4

301

3

7

= Rs. 4

301 ×

7

3 = Rs.

14

343

= Rs. 4

129 = Rs. 32

4

1 = Rs. 32.25

12. By what number should 16

33 be divided

to get 4

11 ?

Solution—

Let x be divided, then

16

33 x =

4

11

16

33 × x

1 =

4

11

x = 16

33 ×

11

4

=

14

13

= 4

3

Required number = 4

3


40

13. Divide the sum of 5

13 and

7

12 by the

product of 7

31 and

2

1.

Solution—

Sum of 5

13 and

7

12 =

5

13 +

7

12

= 35

6091 =

35

31

Product of 7

31 and

2

1 =

7

31 ×

2

1

= 27

131

= 14

31

Required number = 35

31

14

31

= 35

31 ×

31

14 =

15

21

= 5

2

14. Divide the sum of 12

65 and

7

12 by their

difference.

Solution—

Sum of 12

65 and

7

12 =

12

65 +

7

12

= 84

144455 =

84

599

and difference of 12

65 and

7

12 =

12

65 –

7

12

= 84

144455 =

84

311


599

84

311

= 84

599 ×

311

84 =

311

599

15. If 24 trousers of equal size can be

prepared in 54 metres of cloth, what

length of cloth is required for each

trouser ?

Solution—

Cloth required for 24 trousers = 54 m

Cloth required for 1 trouser = (54 24) m

= 24

54 =

4

9 = 2

4

1m

1. Find a rational number between –3 and

1.

Solution—

We know that a number between two rational

numbers a, b = 2

ba

Rational number between –3 and 1 = 2

13

= 2

2 = –1

2. Find any five rational number less than 1.

Solution—

_ 1 = 5

5 and number 0, 1, 2, 3, 4 are less than 5

Five numbers less than 1 can be

0, 5

1,

5

2,

5

3,

5

4

3. Find four rational numbers between 9

2

and 9

5.

EXERCISE 1.8


41

Solution—

_ –1, 0, 1, 2, 3, 4 lie between –2 and 5

Four rational numbers between 9

2 and

9

5

can be 9

1, 0,

9

1,

9

2 ..........

4. Find two rational numbers between 5

1 and

2

1.

Solution—

5

1,

2

1(LCM of 5, 2 = 10)

5

1 =

25

21

= 10

2 and

2

=

52

5

= 10

5

_ 3 and 4 lie between 2 and 5

Number between 10

2 and

10

5,

can be 10

3,

10

4

5. Find ten rational numbers between 4

1 and

2

1.

Solution—

LCM of 4, 2 = 4

4

1 =

4

1 and

2

1 =

22

21

= 4

2

_ We need 10 rational numbers

We extend them as 80

20 and

80

40

Rational numbers between 80

20 and

80

40

Can be 80

21,

80

22,

80

23,

80

24, .........

80

39


2

and 2

1.

Solution—

5

2 =

45

42

= 20

8

and 2

1 =

102

101

= 20

10

Now numbers lying between –8, 10 will be

–7, –6, –5, –4, –3, –2, –1, 0, 1, 2, 3, 4,.....9

The rational numbers will be

20

7,

20

6,

20

5,

20

4, .....,

20

8,

20

9


3 and

4

3.

Solution—

5

3 and

4

3

LCM of 5, 4 = 20

But we need 10 numbers.

5

3 =

165

163

= 80

48 and

4

3 =

204

203

= 80

60

The numbers lie between 48 and 60 will be

49, 50, 51, 52, 53, 54, ......... 60

Rational numbers are

80

49,

80

50,

80

51,

80

52,

80

53 .........

80

59


42

Points to Remember

1. If ‘a’ is multiplied with itself for n-time

then it is a × a × a × a ...... × a (n-times) =

an and is read as ‘a’ raised to the power n.

2. Laws fo exponents—

(i) am × an = am + n

(ii) am an = am – n (m > n)

(iii) (am)n = amn = (an)m

(iv) (ab)n = anbn

(v)

n

b

a

=

n

n

b

a

(vi) a1 = a and a0 = 1

Here a and b are non-zero rational numbers.

3. Negative Integral Exponents—

We know that :

100 = 1

101 = 10

102 = 10 × 10 = 100

103 = 10 × 10 × 10 = 1000

Therefore

10–1 = 10

1

2POWERS

10–2 = 100

1

10–3 = 1000

1 and so on

a–n = na

1 and an = n

a

1

4. Use of exponents to express small

numbers in standard form—

Method :

(i) Obtain the number and see whether it is

between 1 and 10 or it is less than 1.

(ii) If the number is between 1 and 10, then

write it as the product of the number itself

and 100.

(iii) If the number is less than one, then move

the decimal point to the right so that there

is just one digit on the left side of the

decimal point. Write the given number as

the product of the number so obtained and

10–n, when n is the number of places the

decimal point has been moved to the right.

The number so obtained is the standard

form of the given number.

EXERCISE 2.1

1. Express each of the following as a rational number of the form q

p, where p and q are

integers and q 0 :


43

(i) 2–3 (ii) (–4)–2

(iii) 23

1 (iv)

5

2

1

(v)

2

3

2

Solution—

(i) 2–3 = 32

1 =

m

m

aa

1

= 222

1

=

8

1

(ii) (–4)–2 = 24

1

= 44

1

= 16

1

(iii) 23

1 = 32

m

ma

a

1

= 3 × 3 = 9

(iv)

5

2

1

=

5

1

2

= 2 × 2 × 2 × 2 × 2 = 32

(v)

2

3

2

=

2

2

3

=

2

3 ×

2

3 =

4

9

2. Find the values of each of the following :

(i) 3–1 + 4–1 (ii) (30 + 4–1) × 22

(iii) (3–1 + 4–1 + 5–1)0

(iv)

111

4

1

3

1

Solution—

(i) 3–1 + 4–1 = 3

1 +

4

1

= 12

34 =

12

7

(ii) (3º + 4–1) × 22 =

4

11 × 4

mama

m 1,10

= 4

14 × 4 =

4

5 × 4 = 5

(iii) (3–1 + 4–1 + 5–1)0

=

0

5

1

4

1

3

1

= 1 (_ a0 = 1)

(iv)

111

4

1

3

1

= {(3)1 – (4)1}–1 = (3 – 4)–1 = (–1)–1

= 11

1

= 1

1

= –1

3. Find the values of each of the following :

(i)

1

2

1

+

1

3

1

+

1

4

1

(ii)

2

2

1

+

2

3

1

+

2

4

1

(iii) (2–1 × 4–1) + 2–2

(iv) (5–1 × 2–1) + 6–1

Solution—

(i)

1

2

1

+

1

3

1

+

1

4

1

= (2)1 + (3)1 + (4)1 = 2 + 3 + 4 = 9

(ii)

2

2

1

+

2

3

1

+

2

4

1

= (2)2 + (3)2 + (4)2

= 4 + 9 + 16 = 29

(iii) (2–1 × 4–1) 2–2

=

4

1

2

1 22

1


44

= 8

1

4

1 =

8

1 ×

1

4 =

2

1

(iv) (5–1 × 2–1) 6–1 =

2

1

5

1

6

1

= 10

1 ×

1

6 =

10

6 =

210

26

= 5

3

4. Simplify :

(i) (4–1 × 3–1)2 (ii) (5–1 6–1)3

(iii) (2–1 + 3–1)–1 (iv) (3–1 × 4–1)–1 × 5–1

Solution—

(i) (4–1 × 3–1)2 =

2

3

1

4

1

=

2

12

1

= 12

1 ×

12

1 =

144

1

(ii) (5–1 6–1)3 =

3

6

1

5

1

=

3

1

6

5

1

=

3

5

6

=

555

666

= 125

216

(iii) (2–1 + 3–1)–1 =

1

3

1

2

1

=

1

6

23

=

1

6

5

=

5

6

(iv) (3–1 × 4–1)–1 × 5–1 =

1

4

1

3

1

× 5–1

=

1

12

1

×

5

1

= (12)1 × 5

1 =

5

12

5. Simplify :

(i) (32 + 22) ×

3

2

1

(ii) (32 – 22) ×

3

3

2

(iii)

33

2

1

3

1

3

4

1

(iv) (22 + 32 – 42)

2

2

3

Solution—

(i) (32 + 22) ×

3

2

1

= (9 + 4) ×

8

1 = 13 ×

8

1

= 8

13

(ii) (32 – 22) ×

3

3

2

= (9 – 4) ×

3

2

3

= 5 × 8

27 =

8

135

(iii)

33

2

1

3

1

3

4

1

= [(3)3 – (2)3] (4)3

= (27 – 8) 64 = 19 64 = 64

19

(iv) (22 + 32 – 42)

2

2

3

= (4 + 9 – 16) 4

9 = –3 ×

9

4

= 3

41 =

3

4

6. By what number should 5–1 be multiplied

so that the product may be equal to

(–7)–1 ?

Solution—

Let x be multiplied to 5–1, then

x × 5–1 = (–7)–1

x × 5

1 = 17

1

5

x =

7

1


45

x = 5 × 7

1

=

7

5

Required number = 7

5

7. By what number should

1

2

1

be

multiplied so that the product many be

equal to

1

7

4

?

Solution—

Let x be the required number, then

x ×

1

2

1

=

1

7

4

x (2)1 =

1

4

7

2x =

4

7

x = 24

7

= 8

7

Required number = 8

7

8. By what number should (–15)–1 be divided

so that the quotient may be equal to

(–5)–1 ?

Solution—

Let the number should be divided by = x

their (–15)–1 x = (–5)–1 15

1

x

= 15

1

15

1 ×

x

1 =

5

1

x15

1 =

5

1

–15x = –5

x = 15

5

= 3

1

Required number = 3

1

1. Write each of the following in exponential

form :

(i)

1

2

3

×

1

2

3

×

1

2

3

×

1

2

3

(ii)

2

5

2

×

2

5

2

×

2

5

2

Solution—

(i)

1

2

3

×

1

2

3

×

1

2

3

×

1

2

3

1111

2

3

=

4

2

3

EXERCISE 2.2

(ii)

2

5

2

×

2

5

2

×

2

5

2

222

5

2

=

6

5

2

2. Evaluate :

(i) 5–2 (ii) (–3)–2

(iii)

4

3

1

(iv)

1

2

1

Solution—

(i) 5–2 = 25

1 =

55

1

=

25

1


46

(ii) (–3)–2 = 23

1

= 33

1

= 9

1

(iii)

4

3

1

=

4

1

3

= 3 × 3 × 3 × 3 = 81

(iv)

1

2

1

= (–2)1 = –2

3. Express each of the following as a rational

number in the form q

p :

(i) 6–1 (ii) (–7)–1

(iii)

1

4

1

(iv) (–4)–1 ×

1

2

3

(v)

1

5

3

×

1

2

5

Solution—

(i) 6–1 = 6

1

(ii) (–7)–1 = 7

1

(iii)

1

4

1

=

1

4 = 4

(iv) (–4)–1 ×

1

2

3

=

1

4

1

×

1

3

2

= 4

1 ×

3

2 =

32

11

= 6

1

(v)

1

5

3

×

1

2

5

=

1

3

5

×

1

5

2

= 3

5 ×

5

2 =

3

2

4. Simplify :

(i) {4–1 × 3–1}2 (ii) {5–1 6–1}3

(iii) (2–1 + 3–1)–1 (iv) {3–1 × 4–1}–1 × 5–1

(v) (4–1 – 5–1)–1 3–1

Solution—

(i) { 4–1 × 3–1}2 =

2

3

1

4

1

=

2

12

1

= 12

1 ×

12

1 =

144

1

(ii) {5–1 6–1}3 =

3

6

1

5

1

=

3

1

6

5

1

=

3

5

6

=

555

666

= 125

216

(iii) (2–1 + 3–1)–1 =

1

3

1

2

1

=

1

6

23

=

1

6

5

=

5

6

(iv) {3–1 × 4–1}–1 × 5–1 =

1

4

1

3

1

×

5

1

=

1

12

1

×

5

1 =

1

12 ×

5

1 =

5

12

(v) (4–1 – 5–1)–1 3–1 =

1

5

1

4

1

3

1

=

1

20

45

×

1

3 =

1

20

1

×

1

3

= 1

20 ×

1

3 = 60

5. Express each of the following rational

numbers with a negative exponent :

(i)

3

4

1

(ii) 35 (iii)

4

5

3

(iv)

34

2

3

(v)

34

3

7


47

Solution—

(i)

3

4

1

=

3

1

4

= (4)–3

(ii) 35 =

5

3

1

(iii)

4

5

3

=

4

3

5

(iv)

34

2

3

=

34

2

3

=

12

2

3

(v)

34

3

7

=

34

3

7

=

12

3

7

6. Express each of the following rational

numbers with a positive exponent :

(i)

2

4

3

(ii)

8

4

5

(iii) 43 × 4–9

(iv)

43

3

4

(v)

24

2

3

Solution—

(i)

2

4

3

=

2

3

4

(ii)

5

4

=

5

5

4

(iii) 43 × 4–9 = 43 – 9 = 4– 6 =

6

4

1

(iv)

43

3

4

=

43

3

4

=

12

3

4

(v)

24

2

3

=

24

2

3

=

8

2

3

=

8

3

2

7. Simplify :

(i)

33

2

1

3

1

3

4

1

(ii) (32 – 22) ×

3

3

2

(iii) 1

11

42

1

(iv)

12

2

4

1

(v)

32

3

2

×

4

3

1

× 3–1 × 6–1

Solution—

(i)

33

2

1

3

1

3

4

1

= (33 – 23) (4)3

= (27 – 8) 64 = 64

19

(ii) (32 – 22) ×

3

3

2

= (32 – 22) ×

3

2

3

= (9 – 4) ×

8

27

= 5 × 8

27 =

8

135

(iii) 1

11

42

1

= 1

11

4

12

=

1

4

12


48

=

1

2

1

= (–2)1 = –2

(iv)

12

2

4

1

=

222

4

1

=

4

4

1

=

4

1 ×

4

1 ×

4

1 ×

4

1 =

256

1

(v)

32

3

2

×

4

3

1

× 3–1 × 6–1

=

32

3

2

×

4

1

3

× 13

1 × 16

1

=

6

3

2

× (3)4 ×

3

1 ×

6

1

= 333333

222222

× 3 × 3 × 3 × 3 × 3

1

× 6

1

= 6333

64

=

281

64

=

81

32

8. By what number should 5–1 be multiplied

so that the product may be equal to (–7)–1 ?

Solution—

Let x be multiplied, the

x × 5–1 = (–7)–1

x = 1

1

5

7

=

1

5

7

=

1

7

5

= 7

5

Required number = 7

5


1

2

1

be

multiplied so that the product may be

equal to

1

7

4

?

Solution—

Let x be multiplied, then

x ×

1

2

1

=

1

7

4

x =

1

7

4

1

2

1

=

1

4

7

1

1

2

=

4

7

2

= 4

7

×

2

1 =

8

7

=

8

7

Required number = 8

7

10. By what number should (–15)–1 be divided

so that the quotient may be equal to

(–5)–1 ?

Solution—

Let (–15)–1 be divided by x

(–15)–1 x = (–5)–1

15

1 x =

1

5

1

15

1

×

x

1 =

5

1

x15

1

=

5

1

x = 15

5

= 3

1

Required number = 3

1


49


2

3

5

be

multiplied so that the product may be

1

3

7

?

Solution—

Let x be multiplied, then

x ×

2

3

5

=

1

3

7

x =

1

3

7

2

3

5

x =

1

7

3

2

5

3

=

7

3

25

9 =

7

3 ×

9

25

= 21

25


25

12. Find x, if

(i)

4

4

1

×

8

4

1

=

x4

4

1

(ii)

19

2

1

8

2

1

=

12

2

1

x

(iii)

3

2

3

×

5

2

3

=

12

2

3

x

(iv)

3

5

2

×

15

5

2

=

x32

5

2

(v)

x

4

5

4

4

5

=

5

4

5

(vi)

12

3

8

x

×

5

3

8

=

2

3

8

x

Solution—

(i)

4

4

1

×

8

4

1

=

x4

4

1

84

4

1

=

x4

4

1

12

4

1

=

x4

4

1

Comparing, we get :

–4x = –12 x = 4

12

= 3

x = 3

(ii)

19

2

1

8

2

1

=

12

2

1

x

819

2

1

=

12

2

1

x

27

2

1

=

12

2

1

x

Comparing, we get :

–2x + 1 = –27

–2x = –27 – 1 = –28

x = 2

28

= 14

x = 14

(iii)

3

2

3

×

5

2

3

=

12

2

3

x

53

2

3

=

12

2

3

x

2

2

3

=

12

2

3

x

Comparing, we get :

2x + 1 = 2

2x = 2 – 1 = 1

x = 2

1


50

(iv)

3

5

2

×

15

5

2

=

x32

5

2

153

5

2

=

x32

5

2

12

5

2

=

x32

5

2

Comparing, we get :

2 + 3x = 12

3x = 12 – 2 = 10

x = 3

10

(v)

x

4

5

4

4

5

=

5

4

5

4

4

5

x

=

5

4

5

Comparing, we get :

–x + 4 = 5 –x = 5 – 4 = 1

x = –1

(vi)

12

3

8

x

×

5

3

8

=

2

3

8

x

512

3

8

x

=

2

3

8

x

Comparing, we get :

2x + 1 + 5 = x + 2

2x – x = 2 – 1 – 5 x = –4

x = –4

13. (i) If x =

2

2

3

×

4

3

2

, find the value

of x–2.

(ii) If x =

2

5

4

2

4

1

, find the value of x–1.

Solution—

(i) x =

2

2

3

×

4

3

2

2

2

3

×

4

3

2

=

42

2

3

=

6

2

3

x–2 =

26

2

3

=

12

2

3

=

12

3

2

(ii) x =

2

5

4

2

4

1

=

2

4

5

2

4

1

=

2

4

1

4

5

=

2

1

4

4

5

= (5)2 = 25

x–1 = (25)–1 = 25

1

14. Find the value of x for which 52x 5–3 = 55.

Solution—

52x 5–3 = 55

52x 55 = 5–3

52x = 55–3 = 52

Comparing, we get :

2x = 2 x = 2

2 = 1

x = 1


51

1. Express the following numbers in

standard form :

(i) 6020000000000000

(ii) 0.00000000000942

(iii) 0.00000000085 (iv) 846 × 107

(v) 3759 × 10–4 (vi) 0.00072984

(vii) 0.000437 × 104 (viii) 4 100000

Solution—

(i) 6020000000000000

= 6.02 × 1000000000000000

= 6.02 × 1015

(ii) 0.00000000000942

= 9.42 × 0001000000000

1 = 9.42 × 1210

1

= 9.42 × 10–12

(iii) 0.00000000085

= 8.5 × 01000000000

1 = 8.5 × 1010

1

= 8.5 × 10–10

(iv) 846 × 107 = 8.46 ×100 × 107

= 8.46 × 102 × 107

= 8.46 × 109

(v) 3759 × 10–4 = 3.759 × 1000 × 10–4

= 3.759 × 103 × 10–4

= 3.759 × 10–1

(vi) 0.00072984 = 10000

2984.7

= 410

2984.7 = 7.2984 × 10–4

(vii) 0.000437 × 104 = 10000

1037.4 4 = 4

4

10

1037.4

= 4.37 × 104 × 10–4

= 4.37 × 100

= 4.37 × 1 = 4.37

(viii)4 100000

= 100000

4 = 510

4

= 4 × 10–5

2. Write the following numbers in the usual

form :

(i) 4.83 × 107 (ii) 3.02 × 10–6

(iii) 4.5 × 104 (iv) 3 × 10–8

(v) 1.0001 × 109 (vi) 5.8 × 102

(vii) 3.61492 × 106 (viii) 3.25 × 10–7

Solution—

(i) 4.83 × 107

= 4.83 × 10000000

= 48300000

(ii) 3.02 × 10–6

= 610

02.3 =

1000000

02.3

= 0.00000302

(iii) 4.5 × 104 = 4.5 × 10000

= 45000

(iv) 3 × 10–8 = 810

3

= 100000000

3

= 0.00000003

(v) 1.0001 × 109

= 1.0001 × 1000000000

= 1000100000

(vi) 5.8 × 102

= 5.8 × 100 = 580

(vii) 3.61492 × 106

= 3.61492 × 1000000

= 3614920

(viii)3.25 × 10–7

= 710

25.3 =

10000000

25.3

= 0.000000325

EXERCISE 2.3


52

Choose the correct alternative in each of

the following :

1. Square of

3

2 is

(a) –3

2(b)

3

2

(c) –9

4(d)

9

4

Solution—

(d)

9

4

3

2

3

2

3

22

2. Cube of 2

1 is

(a)8

1(b)

16

1

(c) –8

1(d)

16

1

Solution—

(c)

8

1

2

1

2

1

2

1

2

13

3. Which of the following is not equal to

4

5

3

?

(a)

4

4

5

3(b) 4

4

5

3

(c) – 4

4

5

3(d)

5

3 ×

5

3 ×

5

3 ×

5

3

Solution—

(c) _

4

5

3

=

4

4

5

3 or

4

4

5

3

or

5

3 ×

5

3

× 5

3 ×

5

3

4. Which of the following in not reciprocal

of

4

3

2

?

(a)

4

2

3

(b)

4

3

2

(c)

4

2

3

(d)

4

4

2

3

Solution—

(c)

44

2

3ofreciprocalis

2

3

5. Which of the following numbers is not

equal to 27

8 ?

(a)

3

3

2

(b)

3

3

2

(c)

3

3

2

(d)

3

2 ×

3

2 ×

3

2

Solution—

(a)

8

27

2

3

3

233

6.

5

3

2

is equal to

(a)

5

3

2

(b)

5

2

3

(c)3

52 x(d)

53

2

Exercise (MCQs)


53

Solution—

(b)

55

2

3

3

2

7.

5

2

1

×

3

2

1

is equal to

(a)

8

2

1

(b)

8

2

1

(c)

8

4

1

(d)

15

2

1

Solution—

(a)

83535

2

1

2

1

2

1

2

1

8.

3

5

1

8

5

1

is equal to

(a)

5

5

1

(b)

11

5

1

(c) (–5)5 (d)

5

5

1

Solution—

(c)

558383

55

1

5

1

5

1

5

1

9.

7

5

2

5

5

2

is equal to

(a)25

4(b)

25

4

(c)

12

5

2

(d)

4

25

Solution—

(a)

25

4

5

2

5

2

5

2

5

2

5

2

5

2

2

5757

10.

42

3

1

is equal to

(a)

6

3

1

(b)

8

3

1

(c)

24

3

1

(d)

16

3

1

Solution—

(b)

42

3

1

=

42

3

1

=

8

3

1

11.

0

5

1

is equal to

(a) 0 (b)5

1

(c) 1 (d) 5

Solution—

(c)

)1(1

5

1 0

0

a

12.

1

2

3

is equal to

(a)3

2(b) –

3

2

(c)2

3(d) none of these


54

Solution—

(b)

3

2

3

2

2

311

13.

5

3

2

×

5

7

5

is equal to

(a)

10

7

5

3

2

(b)

5

7

5

3

2

(c)

25

7

5

3

2

(d)

25

7

5

3

2

Solution—

(b)

555

7

5

3

2

7

5

3

2

{_ am.bm = (ab)m}

14.

5

4

3

5

3

5

is equal to

(a)

5

3

5

4

3

(b)

1

3

5

4

3

(c)

0

3

5

4

3

(d)

10

3

5

4

3

Solution—

(a)

555

3

5

4

3

3

5

4

3

m

mm

b

a

b

a

15. For any two non-zero rational numbers a

and b, a4 b4 is equal to

(a) (a b)1 (b) (a b)0

(c) (a b)4 (d) (a b)8

Solution—

(c) {_ a4 b4 = (a b)4}

16. For any two rational numbers a and b, a5

× b5 is equal to

(a) (a × b)0 (b) (a × b)10

(c) (a × b)5 (d) (a × b)25

Solution—

(c) {_ a5 × b5 = (a × b)5}

17. For a non-zero rational number a, a7 a12 is equal to

(a) a5 (b) a–19

(c) a–5 (d) a19

Solution—

(c) {a7 a12 = a7 – 12 = a–5}

18. For a non-zero rational number a, (a3)–2

is equal to

(a) a6 (b) a–6

(c) a–9 (d) a1

Solution—

(b) {(a3)–2 = a3 × (–2) = a–6}


55

Points to Remember

1. Square— Square of a number is that

number raised to the power i.e. a × a = a2,

a2 is the square of a.

2. A perfect square— A natural number ‘n’

is called a perfect square or a square

number, if there exist a natural number m,

such that n = m2.

3. Check of a perfect square :

(i) Write the natural number.

(ii) Write the number as a product of prime

numbers.

(iii) Group the factors in pair such a way that

both the factors in each pair are equal.

(iv) See whether the some factors are left over

or not. If no factor is left over, their it is a

perfect square. Otherwise, it is not a

perfect square.

(v) Take one factor out of each pair and

multiply them. It is the square root of the

given number.

4. Properties of the square numbers—

(i) A number ending 2, 3, 7 or 8 is never a

perfect square.

(ii) The number of zeros at the end of a perfect

square is never odd. So, a number ending

one zero is not a perfect square.

(iii) Square of even numbers is always even

number.

(iv) Square of odd numbers is always an odd

number.

(v) Square of a natural number n = 1 + 3 + 5 +

....... n.

(vi) Square of a natural number other than 1, is

either a multiple of 3 or exceeds a multiple

of 3 by 1.

(vii) Square of a natural number other than 1, is

either a multiple of 4 or exceeds a multiple

of 4 by 1.

(viii)For any natural number n greater than 1.

(2n, n2 – 1, n2 + 1) is a pythagorean triplets

or A triplet (m, n, p) of three natural

numbers, m, n, p are called Phythagorean

triplet it m2 + n2 = p2

(ix) There is no natural numbers p and q such

that p2 = 2q2

(x) (n + 1)2 – n2 = (n + 1) + n.

(xi) The square root of a given natural number

n is that natural number which when

multiplied itself given n, as the product and

we denote the square root of n by n or

2

1

n.

If n or 2

1

n = m, then m2 = n.

(xii) Square root can be formed by

(a) By prime factorization method.

(b) By division method.

(c) By the given table of square roots.

(xiii) Some formulas are also used

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

and (a – b)2 = a2 – 2ab + b2

(xv) Number ending 5

n (n + 1)

(n5)2 = write n (n + 1) before 25

(xvi) Number of the form 5a (two digit number)

(5a)2 = (25 + a) × 100 + a2

(xvii) Number of the form 5ab

(5ab)2 = (250 + ab) × 1000 + (ab)2

3SQUARES AND

SQUARE ROOTS


56

1. Which of the following numbers are

perfect squares ?

(i) 484 (ii) 625

(iii) 576 (iv) 941

(v) 961 (vi) 2500

Solution—

(i) 484 = 2 × 2 × 11 × 11

2 4842 242

11 12111 11

1

Grouping the factors in pairs, we have left

no factor unpaired

484 is a perfect square of 22

(ii) 625 = 5 × 5 × 5 × 5

5 6255 1255 255 5

1

Grouping the factors in pairs, we have left

no factor unpaired

625 is a perfect square of 25.

(iii) 576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

2 5762 2882 1442 722 362 183 93 3

1

Grouping the factors in pairs, we see that no

factor is left unpaired


(iv) 941 has no prime factors

941 is not a perfect square.

(v) 961 = 31 × 31

Grouping the factors in pairs,

we see that no factor is left unpaired


(vi) 2500 = 2 × 2 × 5 × 5 × 5 × 5

2 25002 12505 6255 1255 255 5

1




2. Show that each of the following numbers

is a perfect square. Also find the number

whose square is the given number in each

case :

(i) 1156 (ii) 2025

(iii) 14641 (iv) 4761

Solution—

(i) 1156 = 2 × 2 × 17 × 17

2 11562 578

17 28917



1156 is a perfect square of 2 × 17 = 34

(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5

3 20253 6753 2253 755 255 5

1

EXERCISE 3.1

31 96131 31

1


57



2025 is a perfect square of 3 × 3 × 5 = 45

(iii) 14641 = 11 × 11 × 11 × 11

11 1464111 133111 12111 11

1




(iv) 4761 = 3 × 3 × 23 × 23

3 47613 1587

23 52923 23

1




3. Find the smallest number by which the

given number must be multiplied so that

the product is a perfect square.

(i) 23805 (ii) 12150

(iii) 7688

Solution—

(i) 23805 = 3 × 3 × 5 × 23 × 23

3 238053 79355 2645

23 52923 23

1

Grouping the factors in pairs of equal factors,

we see that 5 is left unpaird

In order to complete the pairs, we have to

multiply 23805 by 5, then the product will

be the perfect square.

Requid smallest number = 5

(ii) 12150 = 2 × 3 × 3 × 3 × 3 × 3 × 5 × 5

2 121503 60753 20253 6753 2253 755 255 5

1


we see that factors 2 and 3 are left unpaired

In order to complete the pairs, we have to

multiply 12150 by 2 × 3 = 6 i.e., then the

product will be the complete square.

Required smallest number = 6

(iii) 7688 = 2 × 2 × 2 × 31 × 31

2 76882 38442 1922

31 96131 31

1


we see that factor 2 is left unpaired

In order to complete the pairs we have to

multiply 7688 by 2, then the product will be

the complete square

Required smallest number = 2

4. Find the smallest number by which the

given number must be divided so that the

resulting number is a perfect square.

(i) 14283 (ii) 1800

(iii) 2904

Solution—

(i) 14283 = 3 × 3 × 3 × 23 × 23

3 142833 47613 1587

23 52923 23

1


58

Grouping the factrors in pairs of equal

factors, we see that 3 is left unpaired

Dividing by 3, the quotient will be the perfect

square.

(ii) 1800 = 2 × 2 × 2 × 3 × 3 × 5 × 5

2 18002 9002 4503 2253 755 255 5

1

Grouping the factors in pair of equal factors,

we see that 2 is left unpaired.

Dividing by 2, the quotient will be the perfect

square.

(iii) 2904 = 2 × 2 × 2 × 3 × 11 × 11

2 29042 14522 7263 363

11 12111 11

1


we see that 2 × 3 we left unpaired

Dividing by 2 × 3 = 6, the quotient will be

the perfect square.


perfect squares ?

11, 12, 16, 32, 36, 50, 64, 79, 81, 111, 121

Solution—

11 is not a perfect square as 11 = 1 × 11

12 is not a perfect square as 12 = 2 × 2 × 3

16 is a perfect square as 16 = 2 × 2 × 2 × 2


× 2 × 2




× 2 × 2




121 is a perfect square as 121 = 11 × 11

Hence 16, 36, 64, 81 and 121 are perfect

squares.

6. Using prime factorization method, find

which of the following numbers are

perfect squares ?

189, 225, 2048, 343, 441, 2916, 11025, 3549

Solution—

(i) 189 = 3 × 3 × 3 × 7 3 1893 633 217 7

1

Grouping the factors in pairs, we see that

are 3 and 7 are left unpaired

189 is not a perfect square

(ii) 225 = 3 × 3 × 5 × 5

3 2253 755 255 5

1

Grouping the factors in pairs, we see no

factor left unpaired

225 is a perfect square

(iii) 2048 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×

2 × 2

2 20482 10242 5122 2562 1282 642 322 162 82 42 2

1


59


one 2 is left unpaired


(iv) 343 = 7 × 7 × 7

7 3437 497 7

1


one 7 is left unpaired


(v) 441 = 3 × 3 × 7 × 7

3 4413 1477 497 7

1



441 is a perfect square.

(vi) 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

2 29162 14583 7293 2433 813 273 93 3

1




(vii) 11025 = 3 × 3 × 5 × 5 × 7 × 7

3 110253 36755 12255 2457 497 7

1




(viii)3549 = 3 × 7 × 13 × 13 3 35497 1183

13 16913 13

1

Grouping the factors in pairs, we see that 3,

7 are left unpaired


7. By what number should each of the

followng numbers be multiplied to get a

perfect square in each case ? Also, find

the number whose square is the new

number.

(i) 8820 (ii) 3675

(iii) 605 (iv) 2880

(v) 4056 (vi) 3468

(vii) 7776

Solution—

(i) 8820 = 2 × 2 × 3 × 3 × 5 × 7 × 7

2 88202 44103 22053 7355 2457 497 7

1

Grouping the factors in pairs, we see that 5

is left unpaired

By multiplying 8820 by 5, we get the perfect

square and square root of product will be

= 2 × 3 × 5 × 7 = 210

(ii) 3675 = 3 × 5 × 5 × 7 × 7

3 36755 12255 2457 497 7

1


60


is left unpaired

Multiplying 3675 by 3, we get a perfect

square and square of the product will be = 3

× 5 × 7 = 105

(iii) 605 = 5 × 11 × 11

5 60511 12111 11

1


is left unpaired

Multiplying 605 by 5, we get a perfect square

and square root of the product will be = 5 ×

11 = 55

(iv) 2880 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5

2 28802 14402 7202 3602 1802 903 453 155 5

1


is left unpaired

Multiplying 2880 by 5, we get the perfect

square.

Square root of product will be = 2 × 2 × 2 ×

3 × 5 = 120

(v) 4056 = 2 × 2 × 2 × 3 × 13 × 13

2 40562 20282 10143 507

13 16913 13

1


and 3 are left unpaired

Multiplying 4056 by 2 × 3 i.e., 6, we get the

perfect square.

and square root of the product will be = 2 ×

2 × 3 × 13 = 156

(vi) 3468 = 2 × 2 × 3 × 17 × 17

2 34682 17343 867

17 28917 17

1


is left unpaired

Multiplying 3468 by 3 we get a perfect square.

and square root of the product will be 2 × 3

× 17 = 102

(vii) 7776 = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3

2 77762 38882 19442 9722 4863 2433 813 273 93 3

1



Multiplying 7776 by 2 × 3 or 6

We get a perfect square

and square root of the product will be

= 2 × 2 × 2 × 3 × 3 × 3 = 216

8. By what numbers should each of the

following be divided to get a perfect

square in each case ? Also find the

number whose square is the new number.

(i) 16562 (ii) 3698

(iii) 5103 (iv) 3174

(v) 1575


61

Solution—

(i) 16562 = 2 × 7 × 7 × 13 × 13 2 165627 82817 1183

13 16913 13

1


is left unpaired

Dividing by 2, we get the perfect square

and square root of the quotient will be

7 × 13 = 91

(ii) 3698 = 2 × 43 × 43 2 369843 184943 43

1


is left unpaired,

Dividing 3698 by 2, the quotient is a perfect

square

and square of quotient will be = 43

(iii) 5103 = 3 × 3 × 3 × 3 × 3 × 3 × 7 3 51033 17013 5673 1893 633 217 7

1


is left unpaired

Dividing 5103 by 7, we get the quotient a

perfect square.

and square root of the quotient will be 3 × 3

× 3 = 27

(iv) 3174 = 2 × 3 × 23 × 23 2 31743 1587

23 52923 23

1



Dividing 3174 by 2 × 3 i.e. 6, the quotient

will be a perfect square and square root of

the quotient will be = 23

(v) 1575 = 3 × 3 × 5 × 5 × 7 3 15753 5255 1755 357 7

1

Grouping the factors in pairs, we find that 7

is left unpaired

Dividing 1575 by 7, the quotient is a perfect

square

and square root of the quotient will be = 3 ×

5 = 15

9. Find the greatest number of two digits

which is a perfect square.

Solution—

The greatest two digit number = 99

We know, 92 = 81 and 102 = 100

But 99 is in between 81 and 100

81 is the greatest two digit number which is

a perfect square.

10. Find the least number of three digits

which is perfect square.

Solution—

The smallest three digit number = 100

We know that 92 = 81, 102 = 100, 112 = 121

We see that 100 is the least three digit number


11. Find the smallest number by which 4851

must be multiplied so that the product

becomes a perfect square.

Solution—

By factorization :

4851 = 3 × 3 × 7 × 7 × 11 3 48513 16177 5397 77

11 111


62


is left unpaired

The least number is 11 by which multiplying

4851, we get a perfect square.


must be divided so that the quotient

becomes a perfect square.

Solution—

By factorization,

28812 = 2 × 2 × 3 × 7 × 7 × 7 × 7

2 288122 144063 72037 24017 3437 497 7

1


is left unpaired

Dividing 28812 by 3, the quotient will be a

perfect square.


must be divided so that it becomes a

perfect square. Also find the number

whose square is the resulting number.

Solution—

By factorization,

1152 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

2 11522 5762 2882 1442 722 362 183 93 3

1


one 2 is left unpaired.

Dividing 1152 by 2, we get the perfect square

and square root of the resulting number 576,

will be 2 × 2 × 2 × 3 = 24

1. The following numbers are not perfect

squares. Give reason :

(i) 1547 (ii) 45743

(iii) 8948 (iv) 333333

Solution—

We know that if the units digit is 2, 3, 7 or 8

of a number, then the number is not a perfect

square.

(i) _ 1547 has 7 as units digit.

It is not a perfect square.

(ii) 45743 has 3 as units digit


(iii) _ 8948 has 8 as units digit


(iv) _ 333333 has 3 as units digits


2. Show that the following numbers are not

perfect squares :

(i) 9327 (ii) 4058

(iii) 22453 (iv) 743522

Solution—

(i) 9327

_ The units digit of 9327 is 7

This number can’t be a perfect square.

(ii) 4058



(iii) 22453



(iv) 743522


EXERCISE 3.2


63


3. The square of which of the following

numbers would be an odd number ?

(i) 731 (ii) 3456

(iii) 5559 (iv) 42008

Solution—

We know that the square of an odd number

is odd and of even number is even. Therefore

(i) Square of 731 would be odd as it is an odd

number.

(ii) Square of 3456 should be even as it is an

even number.

(iii) Square of 5559 would be odd as it is an odd

number.

(iv) The square of 42008 would be an even

number as it is an even number.

Therefore suqares of (i) 731 and (iii) 5559

will be odd numbers.

4. What will be the units digit of the squares

of the following numbers ?

(i) 52 (ii) 977

(iii) 4583 (iv) 78367

(v) 52698 (vi) 99880

(vii) 12796 (viii) 55555

(ix) 53924

Solution—

(i) Square of 52 will be 2704 or (2)2 = 4

Its units digit is 4.

(ii) Square of 977 will be 954529 or (7)2 = 49

Its units digit is 9

(iii) Square of 4583 will be 21003889 or (3)2 = 9


(iv) Is 78367, square of 7 = 72 = 49


(v) In 52698, square of 8 = (8)2 = 64


(vi) In 99880, square of 0 = 02 = 0


(vii) In 12796, square of 6 = 62 = 36


(viii)In In 55555, square of 5 = 52 = 25


(ix) In 53924, square of 4 = 42 = 16


5. Observe the following pattern

1 + 3 = 22

1 + 3 + 5 = 32

1 + 3 + 5 + 7 = 42

and write the value of 1 + 3 + 5 + 7 + 9 +

....... upto n terms.

Solution—

The given pattern is

1 + 3 = 22

1 + 3 + 5 = 32

1 + 3 + 5 + 7 = 42

1 + 3 + 5 + 7 + 9 + ....... upto n terms

(number of terms)2 = n2

6. Observe the following pattern :

22 – 12 = 2 + 1

32 – 22 = 3 + 2

42 – 32 = 4 + 3

52 – 42 = 5 + 4

Find the value of

(i) 1002 – 992 (ii) 1112 – 1092

(iii) 992 – 962

Solution—

From the given pattern,

22 – 12 = 2 + 1

32 – 22 = 3 + 2

42 – 32 = 4 + 3

52 – 42 = 5 + 4

Therefore

(i) 1002 – 990 = 100 + 99 = 199

(ii )1112 – 1092 = 1112 – 1102 – 1092

= (1112 – 1102) + (1102 – 1092)

= (111 + 110) + (110 + 109)

= 221 + 219 = 440

(iii) 992 – 962 = 992 – 982 + 982 – 972 + 972 – 962

= (992 – 982) + (982 – 972) + (972 – 962)

= (99 + 98) + (98 + 97) + (97 + 96)

= 197 + 195 + 193 = 585

7. Which of the following triplets are

pythagorean ?


64

(i) (8, 15, 17) (ii) (18, 80, 82)

(iii) (14, 48, 51) (iv) (10, 24, 26)

(v) (16, 63, 65) (vi) (12, 35, 38)

Solution—

A pythagorean triplet is possible if

(greatest number)2 = (sum of the two smaller

numbers)

(i) 8, 15, 17

Here, greatest number = 17

(17)2 = 289

and (8)2 + (15)2 = 64 + 225 = 289

_ 82 + 152 = 172

8, 15, 17 is a pythagorean triplet

(ii) 18, 80, 82

Greatest number = 82

(82)2 = 6724

and 182 + 802 = 324 + 6400 = 6724

182 + 802 = 822


(iii) 14, 48, 51


(51)2 = 2601

and 142 + 482 = 196 + 2304 = 2500

_ 512 142 + 482

14, 48, 51 is not a pythagorean triplet

(iv) 10, 24, 26

Greatest number is 26

262 = 676

and 102 + 242 = 100 + 576 = 676

262 = 102 + 242


(v) 16, 63, 65


652 = 4225

and 162 + 632 = 256 + 3969 = 4225

_ 652 = 162 + 632


(vi) 12, 35, 38


382 = 1444

and 122 + 352 = 144 + 1225 = 1369

_ 382 122 + 352

12, 35, 38 is not a pythagorean triplet.


(1 × 2) + (2 × 3) = 3

432

(1 × 2) + (2 × 3) + (3 × 4) = 3

543

(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5)

= 3

654

and find the value of

(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 × 6)

Solution—

From the given pattern

(1 × 2) + (2 × 3) = 3

432

(1 × 2) + (2 × 3) + (3 × 4) = 3

543

(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) =

3

654

(1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + (5 ×

6) = 3

765 = 70


1 =2

1 {1 × (1 + 1)}

1 + 2 =2

1 {2 × (2 + 1)}

1 + 2 + 3 =2

1 {3 × (3 + 1)}

1 + 2 + 3 + 4 =2

1 {4 × ( 4 + 1)}

and find the values of each of the

following :

(i) 1 + 2 + 3 + 4 + 5 + ..... + 50


65

(ii) 31 + 32 + ..... + 50

Solution—


1 =2

1 {1 × (1 + 1)}

1 + 2 =2

1 {2 × (2 + 1)}

1 + 2 + 3 =2

1 {3 × (3 + 1)}

1 + 2 + 3 + 4 =2

1 {4 × ( 4 + 1)}

Therefore

(i) 1 + 2 + 3 + 4 + 5 + ..... + 50 = 2

1 {50 × (50

+ 1)}

= 2

1 × 50 × 51 = 1275

(ii) 31 + 32 + ..... + 50

= (1 + 2 + 3 + 4 + ..... + 50) – (1 + 2 + 3 +

4 + ..... + 30)

= 2

1 {50 × (50 + 1)} –

2

1{30 × (30 + 1)}

= 2

1 × 50 × 51 –

2

1 × 30 × 31

= 1275 – 465 = 810


12 = 6

1 [1 × (1 + 1) × (2 × 1 + 1)]

12 + 22 = 6

1 [2 × (2 + 1) × (2 × 2 + 1)]

12 + 22 + 32 = 6

1 [3 × (3 + 1) × (2 × 3) + 1)]

12 + 2 2 + 32 + 42 = 6

1 [4 × (4 + 1) × (2 × 4

+ 1)]

and find the values of each of the

following :

(i) 12 + 22 + 32 + 42 + ..... + 102

(ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122

Solution—


12 = 6

1 [1 × (1 + 1) × (2 × 1 + 1)]

12 + 22 = 6

1 [2 × (2 + 1) × (2 × 2 + 1)]

12 + 22 + 32 = 6

1 [3 × (3 + 1) × (2 × 3) + 1)]

12 + 22 + 32 + 42 = 6

1 [4 × (4 + 1) × (2 × 4 +

1)]

Therefore :

(i) 12 + 22 + 32 + 42 + .... + 102

= 6

1 {10 × (10 + 1) × (2 × 10 + 1)]

= 6

1 [10 × 11 × 21] =

6

211110

= 6

2310 = 385

(ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122

= [12 + 22 + 32 + 42 + ..... + 122] – [12 + 22 +

32 + 42]

= 6

1 [12 × (12 + 1) × (2 × 12 + 1)] –

6

1 [4

× (4 + 1) × (2 × 4 + 1)]

= 6

1 × (12 × 13 × 25) –

6

1 (4 × 5 × 9)

= 6

251312 –

6

954 = 650 – 30 = 620


squares of even numbers ?

121, 225, 256, 324, 1296, 6561, 5476, 4489,

373758


66

Solution—

We know that squares of even numbers is

also are even number. Therefore numbers

256, 324, 1296, 5476 and 373758 have their

units digit an even number.

These are the squares of even numbers.

12. By just examining the units digits, can

you tell which of the following cannot be

whole squares ?

(i) 1026 (ii) 1028

(iii) 1024 (iv) 1022

(v) 1023 (vi) 1027

Solution—

We know that a perfect square cannot ends

with the digit 2, 3, 7, or 8

By examining the given number, we can say

that 1028, 1022, 1023, 1027 can not be

perfect squares.

13. Write five numbers for which you cannot

decide whether they are squares.

Solution—

A number which ends with 1, 4, 5, 6, 9 or 0

can’t be a perfect square

2036, 4225, 4881, 5764, 3349, 6400

14. Write five numbers which you cannot

decide whether they are square just by

looking at the unit’s digit.

Solution—

A number which does not end with 2, 3, 7

or 8 can be a perfect square

The five numbers can be 2024, 3036, 4069,

3021, 4900

15. Write true (T) or false (F) for the

following statements.

(i) The number of digits in a square number

is even.

(ii) The square of a prime number is prime.

(iii) The sum of two square numbers is a

square number.

(iv) The difference of two square numbers is

a square number.

(v) The product of two square numbers is a

square number.

(vi) No square number is negative.

(vii) There is not square number between 50

and 60.

(viii) There are fourteen square number upto

200.

Solution—

(i) False : In a square number, there is no

condition of even or odd digits.

(ii) False : A square of a prime is not a prime.

(iii) False : It is not necessarily.

(iv) False : It is not necessarily.

(v) True.

(vi) True : A square is always positive.

(vii) True : As 72 = 49, and 82 = 64.

(viii)True : As squares upto 200 are 1, 4, 9, 16,

25, 36, 49, 64, 81, 100, 121, 144, 169, 196

which are fourteen in numbers.

1. Find the squares of the following numbers

using column method. Verify the result

by finding the square using the usual

multiplication :

(i) 25 (ii) 37

(iii) 54 (iv) 71

(v) 96

Solution—

(i) (25)2

EXERCISE 3.3

Here a = 2, b = 5

a2 2ab b2

(2)2 2 × 2 × 5 (5)2

= 4 = 20 = 25

6

2

22

2

(25)2 = 25 × 25 = 625

(25)2 = 625


67

(ii) (37)2

Here a = 3, b = 7

a2 2ab b2

(3)2 2 × 3 × 7 (7)2

= 9 = 42 = 49

+ 4 + 4

13 46

(37)2 = 37 × 37 = 1369

(37)2 = 1369

(iii) (54)2

a2 2ab b2

(5)2 2 × 5 × 4 (4)2

= 25 = 40 = 16

+ 4 + 1

29 41

(54)2 = 54 × 54 = 2916

(54)2 = 2916

(iv) (71)2

Here a = 7, b = 1

a2 2ab b2

(7)2 2 × 7 × 1 (1)2

= 49 = 14 = 1

+ 1

50

(71)2 = 71 × 71 = 5041

(71)2 = 5041

(v) (96)2

Here a = 9, b = 6

a2 2ab b2

(9)2 2 × 9 × 6 (6)2

= 81 = 108 = 36

+ 11 + 3

92 111

(96)2 = 96 × 96 = 9216

(96)2 = 9216


using diagonal method :

(i) 98 (ii) 273

(iii) 348 (iv) 295

(v) 171

Solution—

(i) (98)2

(98)2 = 9604

(ii) (273)2

(273)2 = 74529

(iii) (348)2

(348)2 = 121104

(iv) (295)2

(295)2 = 87025

8

1

7

2

72

64

9 8

9

8

10 4

15+116

8+1

9

2 006 1 9

1 00

4 214 9 1

4 4 6

2 7 3

2

7

3

9

0

225

6+1

7

12+214

3 624 2 4

1 20

1 312 6 2

9 2 4

3 4 8

3

4

8

410

11+112

0+1

1

20+121

9+211

4 210 5 5

1 10

8 418 1 5

4 8 0

2 9 5

2

9

5

0

9+110

26+127 512

6+28


68

(v) (171)2

(171)2 = 29241

3. Find the squares of the following

numbers :

(i) 127 (ii) 503

(iii) 451 (iv) 862

(v) 265

Solution—

(i) (127)2 = (120 + 7)2

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

= (120)2 + 2 × 120 × 7 + (7)2

= 14400 + 1680 + 49 = 16129

(ii) (503)2 = (500 + 3)2

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

= (500)2 + 2 × 500 × 3 + (3)2

= 250000 + 3000 + 9 = 253009

(iii) (451)2 = (400 + 51)2

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

= (400)2 + 2 × 400 × 51 + (51)2

= 160000 + 40800 + 2601 = 203401

(iv) (862)2 = (800 + 62)2

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

= (800)2 + 2 × 800 × 62 + (62)2

= 640000 + 99200 + 3844 = 743044

(v) (265)2

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

(200 + 65)2 = (200)2 + 2 × 200 × 65 + (65)2

= 40000 + 26000 + 4225 = 70225


(i) 425 (ii) 575

(iii) 405 (iv) 205

(v) 95 (vi) 745

(vii) 512 (viii) 995

Solution—

(i) (425)2

Here n = 42

n (n + 1) = 42 (42 + 1) = 42 × 43 = 1806

(425)2 = 180625

(ii) (575)2

Here n = 57

n (n + 1) = 57 (57 + 1) = 57 × 58 = 3306

(575)2 = 330625

(iii) (405)2

Here n = 40

n (n + 1) = 40 (40 + 1) = 40 × 41 = 1640

(405)2 = 164025

(iv) (205)2

Here n = 20

n (n + 1) = 20 (20 + 1) = 20 × 21 = 420

(205)2 = 42025

(v) (95)2

Here n = 9

n (n + 1) = 9 (9 + 1) = 9 × 10 = 90

(95)2 = 9025

(vi) (745)2

Here n = 74

n (n + 1) = 74 (74 + 1) = 74 × 75 = 5550

(745)2 = 555025

(vii) (512)2

Here a = 1, b = 2

(5ab)2 = (250 + ab) × 1000 + (ab)2

(512)2 = (250 + 12) × 1000 + (12)2

= 262 × 1000 + 144

= 262000 + 144 = 262144

(viii)(995)2

Here n = 99

n (n + 1) = 99 (99 + 1) = 99 × 100 = 9900

(995)2 = 990025


using the identity (a + b)2 = a2 + 2ab + b2

(i) 405 (ii) 510

(iii) 1001 (iv) 209

(v) 605

Solution—

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

(i) (405)2 = (400 + 5)2

= (400)2 + 2 × 400 × 5 + (5)2

0 001 7 1

0 00

4 007 9 7

1 7 1

1 7 1

1

7

1

11411+112

0

1+1

2

18+119


69

= 160000 + 4000 + 25

= 164025

(ii) (510)2 = (500 + 10)2

= (500)2 + 2 × 500 × 10 × (10)2

= 250000 + 10000 + 100

= 260100

(iii) (1001)2 = (1000 + 1)2

= (1000)2 + 2 × 1000 × 1 + (1)2

= 1000000 + 2000 + 1

= 1002001

(iv) (209)2 = (200 + 9)2

= (200)2 + 2 × 200 × 9 × (9)2

= 40000 + 3600 + 81

= 43681

(v) (605)2 = (600 + 5)2

= (600)2 + 2 × 600 × 5 + (5)2

= 360000 + 6000 + 25

= 366025


using the identity (a – b)2 = a2 – 2ab + b2 :

(i) 395 (ii) 995 (iii) 495

(iv) 498 (v) 99 (vi) 999

(vii) 599

Solution—

(a – b)2 = a2 – 2ab + b2

(i) (395)2 = (400 – 5)2

= (400)2 – 2 × 400 × 5 + (5)2

= 160000 – 4000 + 25

= 160025 – 4000

= 156025

(ii) (995)2 = (1000 – 5)2

= (1000)2 – 2 × 1000 × 5 + (5)2

= 1000000 – 10000 + 25

= 1000025 – 10000

= 990025

(iii) (495)2 = (500 – 5)2

= (500)2 – 2 × 500 × 5 + (5)2

= 250000 – 5000 + 25

= 250025 – 5000 = 245025

(iv) (498)2 = (500 – 2)2

= (500)2 – 2 × 500 × 2 + (2)2

= 250000 – 2000 + 4

= 250004 – 2000 = 248004

(v) (99)2 = (100 – 1)2

= (100)2 – 2 × 100 × 1 + (1)2

= 10000 – 200 + 1 = 10001 – 200

= 9801

(vi) (999)2 = (1000 – 1)2

= (1000)2 – 2 × 1000 × 1 + (1)2

= 1000000 – 2000 + 1

= 1000001 – 2000 = 998001

(vii) (599)2 = (600 – 1)2

= (600)2 – 2 × 600 × 1 + (1)2

= 360000 – 1200 + 1

= 360001 – 1200 = 358801


by visual method :

(i) 52 (ii) 95

(iii) 505 (iv) 702

(v) 99

Solution—

(a + b)2 = a2 – ab + ab + b2

(i) (52)2 = (50 + 2)2

= 2500 + 100 + 100 + 4

= 2704

4

2500

2

100

100

50

250

(ii) (95)2 = (90 + 5)2

= 8100 + 450 + 450 + 25

= 9025

25

8100

5

45

0

450

90

590

(iii) (505)2 = (500 + 5)2

= 250000 + 2500 + 2500 + 25

= 255025


70

25

250000

5

2500

2500

500

5500

(iv) (702)2 = (700 + 2)2

= 490000 + 1400 + 1400 + 4

= 492804

4

490000

2

140

0

1400

700

2700

(v) (99)2 = (90 + 9)2

= 8100 + 810 + 810 + 81

= 9801

819

810

810

90

990

1. Write the possible unit’s digits of the

square root of the following numbers.

Which of these numbers are odd square

roots ?

(i) 9801 (ii) 99856

(iii) 998001 (iv) 657666025

Solution—

(i) In 9801 _ the units digits is 1, therefore,

the units digit of the square root can be 1 or

9

(ii) In 99356 _ the units digit is 6

The units digit of the square root can be 4 or 6

(iii) In 998001 _ the units digit is 1

The units digit of the square root can be 1

or 9

(iv) In 657666025

_ The unit digit is 5

The units digit of the square root can be 5

2. Find the square root of each of the

following by prime factorization.

(i) 441 (ii) 196

(iii) 529 (iv) 1764

EXERCISE 3.4

(v) 1156 (vi) 4096

(vii) 7056 (viii) 8281

(ix) 11664 (x) 47089

(xi) 24336 (xii) 190969

(xiii)586756 (xiv) 27225

(xv) 3013696

Solution—

(i) 441 = 7733 3 4413 1477 497 7

1

= 3 × 7 = 21

(ii) 196 = 7722 2 1962 987 497 7

1

= 2 × 7 = 14

(iii) 529 = 2323

= 23

23 52923 23

1


71

(iv) 1764 = 773322

2 17642 8823 4413 1477 497 7

1

= 2 × 3 × 7 = 42

(v) 1156 = 171722

= 2 × 17 = 34

(vi) 4096 = 22222

2222222

2 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2

1

= 2 × 2 × 2 × 2 × 2 × 2 = 64

(vii) 7056 = 77332222

= 2 × 2 × 3 × 7 = 84

(viii) 8281 = 131377 7 82817 1183

13 16913 13

1

= 7 × 13 = 91

(ix) 11664 = 3333

332222

2 116642 58322 29162 14583 7293 2433 813 273 93 3

1

= 2 × 2 × 3 × 3 × 3 = 108

(x) 47089 = 313177 7 470897 6727

31 96131 31

1

= 7 × 31 = 217

(xi) 24336 = 1313332222

2 243362 121682 60842 30423 15213 507

13 16913 13

1

= 2 × 2 × 3 × 13 = 156

(xii) 190969 = 23231919

2 11562 578

17 28917 17

1

2 70562 35282 17642 8823 4413 1477 497 7

1


72

19 19096919 1005123 52923 23

1

= 19 × 23 = 437

(xiii) 586756 = 38338322

2 5867562 293378

383 146689383 383

1

= 2 × 383 = 766

(xiv) 27225 = 11115533

3 272253 90755 30255 605

11 12111 11

1

= 3 × 5 × 11 = 165

(xv) 3013969 = 313177

222222

2 30136962 15068482 7534242 3767122 1883562 941787 470897 6727

31 96131 31

1

= 2 × 2 × 2 × 7 × 31 = 1736


must be multiplied so that it becomes a

perfect square. Also, find the square root

of the perfect square so obtained.

Solution—

Factorising 180,

2 1802 903 453 155 5

1

180 = 2 × 2 × 3 × 3 × 5

Grouping the factors in pairs

we see that factor 5 is left unpaired.

Multiply 180 by 5, we get the product 180 ×

5 = 900

Which is a perfect square

and square root of 900 = 2 × 3 × 5 = 30


must be multiplied so that it becomes a


of the number so obtained.

Solution—

Factorising 147,

3 1477 497 7

1

147 = 3 × 7 × 7

Grouping the factors in pairs of the equal

factors, we see that one factor 3 is left

unpaired

Multiplying 147 by 3, we get the product

147 × 3 = 441

Which is a perfect square

and its square root = 3 × 7 = 21




of the resulting number.


73

Solution—

Factorising 3645

3 36453 12153 4053 1353 453 155 5

1

3645 = 3 × 3 × 3 × 3 × 3 × 3 × 5

Grouping the factors in pair of the equal

factors, we see that one factor 5 is left

unpaired

Dividing 3645 by 5, the quotient 729 will be

the perfect square and square root of 729

= 27




of the number so obtained.

Solution—

Factorsing 1152,

2 11522 5762 2882 1442 722 362 183 93 3

1

1152 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

Grouping the factors in pairs of the equal

factors, we see that factor 2 is left unpaired.

Dividing by 2, the quotient 576 is a perfect

square

Square root of 576, it is 24

7. The product of two numbers is 1296. If

one number is 16 times the others find

the numbers.

Solution—


Let one number = x

Second number = 16x

16x × x = 1296

16x2 = 1296

x2 = 16

1296 = 81 = (9)2

x = 9

First number = 9

and second number = 16 × 9 = 144

8. A welfare association collected Rs. 202500

as donation from the residents. If each

paid as many rupees as there were

residents find the number of residents.

Solution—

Total donation collected = Rs. 202500

Let number of residents = x

Then donation given by each resident = Rs.

x

Total collection = Rs. x × x

x2 = 202500

x = 202500

2 2025002 1012503 506253 168753 56253 18755 6255 1255 255 5

1

= 5555333322

= 2 × 3 × 3 × 5 × 5 = 450

Number of residents = 450

9. A society collected Rs. 92.16. Each

member collected as many paise as there


74

were members. How many members were

there and how much did each contribute?

Solution—

Total amount collected = Rs. 92.16

= 9216 paise

Let the number of members = x

Then amount collected by each member = x

paise

x + x = 9216 x2 = 9216

x = 9216

2 92162 46082 23042 11522 5762 2882 1442 722 362 183 93 3

1

= 332222222222

= 2 × 2 × 2 × 2 × 2 × 3 = 96

Number of members = 96

and each member collected = 96 paise

10. A school collected Rs. 2304 as fees from

its students. If each student paid as many

paise as there were students in the school,

how many students were there in the

school ?

Solution—

Total fee collected = Rs. 2304

Let number of students = x

Then fee paid by each student = Rs. x

x × x = 2304 x2 = 2304

x = 2304

2 23042 11522 5762 2882 1442 722 362 183 93 3

1

= 3322222222

= 2 × 2 × 2 × 2 × 3 = 48

Number of students = 48

11. The area of a square field is 5184 m2. A

rectangular field, whose length is twice

its breadth has its perimeter equal to the

perimeter of the square field. Find the

area of the rectangular field.

Solution—

The area of a square field = 5184 m2

Let side of the square = x

x × x = 5184 x2 = 5184

x = 5184

2 51842 25922 12962 6482 3242 1623 813 273 93 3

1

= 3333222222

= 2 × 2 × 2 × 3 × 3 = 72

Side of square = 72 m

Perimeter of square field = 72 × 4 m

= 288 m


75

Perimeter of rectangle = 288 m

Let breadth of rectangular field (b) = x

Then length (l) = 2x

Perimeter = 2 (l + b)

= 2 (2x + x) = 2 × 3x

= 6x

6x = 288 x = 6

288 = 48

Length of rectangular field = 2x = 2 × 48

= 96 m

and breadth = 48 m

and area = l × b = 96 × 48 m2

= 4608 m2

12. Find the least square number, exactly

divisible by each one of the numbers :

(i) 6, 9, 15 and 20 (ii) 8, 12, 15 and 20

Solution—

(i) 6, 9, 15, 20

2 6, 9, 15, 203 3, 9, 15, 105 1, 3, 5, 10

1, 3, 1, 2

LCM of 6, 9, 15, 20 = 2 × 3 × 5 × 3 × 2

= 180

= 2 × 2 × 3 × 3 × 5

We see that after grouping the factors in pairs,

5 is left unpaired

Least perfect square = 180 × 5 = 900

(ii) 8, 12, 15, 20

LCM of 8, 12, 15, 20

2 8, 12, 15, 202 4, 6, 15, 103 2, 3, 15, 55 2, 1, 5, 5

2, 1, 1, 1

= 2 × 2 × 2 × 3 × 5 = 120

We see that after grouping the factors,

factors 2, 3, 5 are left unpaired

Perfect square = 120 × 2 × 3 × 5

= 120 × 30 = 3600

13. Find the square roots of 121 and 169 by

the method of repeated subtraction.

Solution—

(i) 121

121 – 1 = 120 85 – 13 = 72

120 – 3 = 117 72 – 15 = 57

117 – 5 = 112 57 – 17 = 40

112 – 7 = 105 40 – 19 = 21

105 – 9 = 96 21 – 21 = 0

96 – 11 = 85

Square root = 11

(ii) 169

169 – 1 = 168 120 – 15 = 105

168 – 3 = 165 105 – 17 = 88

165 – 5 = 160 88 – 19 = 69

160 – 7 = 153 69 – 21 = 48

153 – 9 = 144 48 – 23 = 25

144 – 11 = 133 25 – 25 = 0

133 – 13 = 120

Square root = 13

14. Write the prime factorization of the

following numbers and hence find their

square roots.

(i) 7744 (ii) 9604

(iii) 5929 (iv) 7056

Solution—

Factorization, we get :

(i) 7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11

2 77442 38722 19362 9682 4842 242

11 12111 11

1

Square root of 7744

= 7744 = 1111222222



76

7744 = 2 × 2 × 2 × 11 = 88

Square root of 7744 = 88

(ii) 9604 = 2 × 2 × 2 × 2 × 7 × 7

2 96042 48027 24017 3437 497 7

1

Square root of 9604

9604 = 777722

= 2 × 7 × 7 = 98

(iii) 5929 = 7 × 7 × 11 × 11

7 59297 847

11 12111 11

1

Square root of 5929

= 5929 = 111177

= 7 × 11 = 77

(iv) 7056 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7

2 70562 35282 17642 8823 4413 1477 497 7

1

Square root of 7056

= 7056 = 77332222

= 2 × 2 × 3 × 7 = 84

15. The students of class VIII of a school

donated Rs. 2401 for PM’s National Relief

Fund. Each student donated as many

rupees as the number of students in the

class. Find the number of students in the

class.

Solution—

Total amount of donation = 2401

Let number of students in VIII = x

Amount donoted by each student = Rs. x

x × x = 2401 x2 = 2401

x = 2401 = 7777

7 24017 3437 497 7

1

= 7 × 7 = 49

Number of students of class VIII = 49

16. A PT teacher wants to arrange maximum

possible number of 6000 students in a field

such that the number of rows is equal to

the number of columns. Find the number

of rows if 71 were left out after

arrangement.

Solution—

Number of students = 6000

Students left out = 71

Students arranged in a field = 6000 – 71

= 5929

Let number of rows = x

Then number of students in each row = x

x × x = = 5929 x2 = 5929

7 59297 847

11 12111 11

1

x = 5929 = 111177

= 7 × 11 = 77

Number of rows = 77


77


following by long division method.

(i) 12544 (ii) 97344

(iii) 286225 (iv) 390625

(v) 363609 (vi) 974169

(vii) 120409 (viii) 1471369

(ix) 291600 (x) 9653449

(xi) 1745041 (xii) 4008004

(xiii)20657025 (xiv) 152547201

(xv) 20421361 (xvi) 62504836

(xvii) 82264900 (xviii) 3226694416

(xix) 6407522209 (xx) 3915380329

Solution—

(i) Square root of 12544

= 12544 = 112

(ii) Square root of 97344

3123 9 73 44

961 73

61622 1244

1244 ×

= 97344

= 312

(iii) Square root of 286225

= 286225

= 535

(iv) Square root of 390625

6256 39 06 25

36122 306

2441245 6225

6225 ×

= 390625

= 625

(v) Square root of 363609

6036 36 36 09

361203 3609

3609 ×

= 363609

= 603

(vi) Square root of 974169

= 974169

= 987

(vii) Square root of 120409

= 120409

= 347

EXERCISE 3.5

1122 1 25 44

121 25

21222 444

444 ×

5355 28 62 25

25103 362

3091065 5325

5325 ×

9879 97 41 69

81188 1641

15041967 13769

13769 ×

3473 12 04 09

964 304

256687 4809

4809 ×


78

(viii)Square root of 1471369

= 1471369

= 1213

(ix) Square root of 291600

= 291600

= 540

(x) Square root of 9653449

= 9653449

= 3107

(xi) Square root of 1745041

= 1745041

= 1321

(xii) Square root of 4008004

= 4008004

= 2002

(xiii) Square root of 20657025

45454 20 65 70 25

1685 465

425904 4070

36169085 45425

45425 ×

= 20657025

= 4545

(xiv) Square root of 152557201

123511 1 52 54 72 01

122 52

44243 854

7292465 12572

1232524701 24701

24701 ×

= 152547201

= 12351

(xv) Square root of 20421361

= 20421361

= 4519

12131 1 47 13 69

122 47

44241 313

2412423 7269

7269 ×

31073 9 65 34 49

961 65

616207 43449

43449 ×

13211 1 74 50 41

123 74

69262 550

5242641 2641

2641 ×

5405 29 16 00

25104 416

4161080 00

00 00

20022 4 00 80 04

44002 008004

8004 ×

45194 20 42 13 61

1685 442

425901 1713

9019029 81261

81261 ×


79

(xvi) Square root of 62504836

= 62504836

= 7906

(xvii) Square root of 82264900

90709 82 26 49 00

811807 12649

1264919140 00

00 ×

= 82264900

= 9070

(xviii) Square root of 3226694416

568045 32 26 69 44 16

25106 726

6361128 9069

9024113604 454416

454416 ×

= 3226694416

= 56804

(xix) Square root of 6407522209

800478 64 07 52 22 09

6416004 075222

64016160087 1120609

1120609 ×

= 6407522209

= 80047

(xx) Square root of 3915380329

= 3915380329

= 62573

2. Find the least number which must be

subtracted from the following numbers

to make them a perfect square :

(i) 2361 (ii) 194491 (iii) 26535

(iv) 16160 (v) 4401624

Solution—

(i) 2361

Finding the square root of 2361

484 23 61

1688 761

704 57

We get 48 as quotient and remainder = 57

To make it a perfect square, we have to

subtract 57 from 2361

Least number to be subtracted = 57

(ii) 194491


4414 19 44 91

1684 344

336881 891

881 10

79067 62 50 48 36

49149 1350

134115806 94836

94836 ×

625736 39 15 38 03 29

36122 315

2441245 7138

622512507 91303

87549125143 375429

375429 ×


80

We get 441 as quotient and remainder = 10




(iii) 26535


1621 2 65 35

126 165

156322 935

644 291

We get 162 as quotient and 291 as remainder




(iv) 16160


1271 1 61 60

122 61

44247 1760

1729 31





(v) 4401624

Find the square root of 4401624

20982 4 40 16 24

4409 4016

36814188 33524

33504 20





3. Find the least number which must be

added to the following numbers to make

them a perfect square :

(i) 5607 (ii) 4931

(iii) 4515600 (iv) 37460

(v) 506900

Solution—

(i) 5607

747 56 07

49144 707

576 131

757 56 07

49145 707

725

Finding the square root of 5607, we see that

742 = 5607 – 131 = 5476

and 752 = 5625

_ 5476 < 5607 < 5625

5625 – 5607 = 18 is to be added to get a

perfect square

Least number to be added = 18

(ii) 4931

707 49 31

49140 31

0 31

717 49 31

49141 31 141

Finding the square root of 4931, we see that

702 = 4900

712 = 5041

4900 < 4931 < 5041


perfect square.



81

(iii) 4515600

21242 4 51 56 00

441 51

41422 1056

8444244 21200

16976 4224

21252 4 51 56 00

441 51

41422 1056

8444245 21200

21225

Finding the square root of 4515600, we see

that 21242 = 4511376

and 21252 = 4515625

4511376 < 4515600 < 4515625

4515625 – 4515600 = 25 is to be added to

get a perfect square.


(iv) 37460

1931 3 74 60

129 274

261383 1360

1149 211

1941 3 74 60

129 274

261384 1360

1536


that 1932 = 37249, 1942 = 37636

37249 < 37460 < 37636


perfect square.


(v) 506900

7117 50 69 00

49141 169

1411421 2800

1421 1379

7127 50 69 00

49141 169

1411422 2800

2844


that

7112 = 505521, 7122 = 506944

505521 < 506900 < 506944

506944 – 506900 = 44 is to be added to get

a perfect square.


4. Find the greatest number of 5 digits which

is a perfect square.

Solution—

Greatest number of 5-digits = 99999

Finding square root, we see that 143 is left

as remainder

3163 9 99 99

961 99

61626 3899

3756 143

Perfect square = 99999 – 143 = 99856

If we add 1 to 99999, it will because a

number of 6 digits

Greatest square 5-digits perfect square

= 99856

5. Find the least number of four digits which


Solution—

Least number of 4-digits = 10000

Finding square root of 1000

We see that if we subtract 39

323 10 00

962 100

124

313 10 00

961 100

61 39

From 1000, we get three digit number

We shall add 124 – 100 = 24 to 1000 to get a

perfect square of 4-digit number


82

1000 + 24 = 1024

Least number of 4-digits which is a perfect

square = 1024

6. Find the least number of six-digits which


Solution—

Least number of 6-digits = 100000

3173 10 00 00

961 100

61627 3900

4389

3163 10 00 00

961 100

61626 3900

3756 544


that if we subtract 544, we get a perfect

square of 5-digits.

So we shall add

4389 – 3900 = 489

to 100000 to get a perfect square

Past perfect square of six digits= 100000 +

489 = 100489

7. Find the greatest number of 4-digits


Solution—

Greatest number of 4-digits = 9999

999 99 99

81189 1899

1701 198

Finding the square root, we see that 198 has

been left as remainder

4-digit greatest perfect square

= 9999 – 198 = 9801

8. A General arranges his soldiers in rows

to form a perfect square. He finds that in

doing so, 60 soldiers are left out. If the

total number of soldiers be 8160, find the

number of soldiers in each row.

Solution—

Total number of soldiers = 8160

Soldiers left after arranging them in a square

= 60

Number of soldiers which are standing in a

square = 8160 – 60 = 8100

909 81 00

81180 00

00 ×

Number of soldiers in each row = 8100

= 90

9. The area of a square field is 60025 m2. A

man cycle along its boundry at 18 km/hr.

In how much time will be return at the

starting point.

Solution—

Area of a square field = 60025 m2

2452 6 00 25

444 200

176485 2425

2425 ×

Side of the square field (a) = 60025 m

= 245 m

and perimeter = 4a

= 4 × 245 m

= 980 m

Speed of the man = 18 km/hr = 18000 m/hr

Time taken = 18000

1980 hr

= 18000

601980 minutes

= 315

4 minutes = 3 minutes 16 seconds


83

10. The cost of levelling and turfing a square

lawn at Rs. 2.50 per m2 is Rs. 13322.50.

Find the cost of fencing it at Rs. 5 per

metre ?

Solution—

Cost of levelling a square field

= Rs. 13322.50

Rate of levelling = Rs. 2.50 per m2

Area of square field = 250

1332250 = 5329 m2

Length of each side (a) = Area

= 5329 m = 73 m

and perimeter = 4a = 4 × 73 = 292 m

Rate of fencing the field = Rs. 5 per m

Total cost of fencing = Rs. 5 × 292

= Rs. 1460

11. Find the greatest number of three digits


Solution—

3-digits greatest number = 999

313 9 99

961 99

61 38

Finding the square root, we see that 38 has

been left

Perfect square = 999 – 38 = 961

Greatest 3-digit perfect square = 961

12. Find the smallest number which must be

added to 2300 so that it becomes a perfect

square.

Solution—


484 23 00

1688 700

704

484 23 00

1687 700

609 91

We see that we have to add 704 – 700 = 4 to

2300 in order to get a perfect square

Smallest number to be added = 4

1. Find the square root of :

(i)961

441(ii)

841

324

(iii) 449

29(iv) 2

25

14

(v) 2196

137(vi) 23

121

26

(vii) 25729

544(viii) 75

49

46

(ix) 32209

942(x) 3

3025

334

EXERCISE 3.6

(xi) 213364

2797(xii) 38

25

11

(xiii)23729

394(xiv) 21

169

51

(xv) 10225

151

Solution—

(i)961

441 =

961

441

= 31

21

212 4 41

441 41

41 ×

313 9 61

961 61

61 ×


84

(ii)841

324 =

841

324

181 3 24

128 224

224 ×

292 8 41

449 441

441 ×

= 29

18

(iii)49

294 =

49

29494

= 49

29196 =

49

225

151 2 25

125 125

125 ×

= 7

15 = 2

7

1

(iv)25

142 =

25

14252

= 25

64 =

25

64 =

5

8 = 1

5

3

(v)196

1372 =

196

1371962

= 196

137392 =

196

529

= 196

529 232 5 29

443 129

129 ×

141 1 96

124 96

96 ×

= 14

23 = 1

14

9

(vi)121

2623 =

121

2612123 =

121

262783

= 121

2809 535 28 09

25103 309

309 ×

= 11

53 = 4

11

9

(vii)729

54425 =

729

54472925

= 729

54418225 =

729

18769

272 7 29

447 329

329 ×

1371 1 87 69

123 87

69267 1869

1869 ×

= 27

137 = 5

27

2

(viii)49

4675 =

49

464975

= 49

463675 =

49

3721

= 7

61 = 8

7

5

616 37 21

36121 121

121 ×


85

(ix)2209

9423 =

2209

94222093

= 2209

9426627 =

2209

7569

878 75 69

64167 1169

1169 ×

474 22 09

1687 609

609 ×

= 47

87 = 1

47

40

(x)3025

3343 =

3025

33430253

= 3025

3349075 =

3025

9409

979 94 09

81187 1309

1309 ×

555 30 25

25105 525

525 ×

= 55

97 = 1

55

42

(xi)3364

279721 =

3364

2797336421

= 3364

279770644 =

3364

73441

585 33 64

25108 864

864 ×

2712 7 34 41

447 334

329541 541

541 ×

= 58

271 = 4

58

39

(xii)25

1138 =

25

112538

= 25

11950 =

25

961

313 9 61

961 61

61 ×

= 5

31 = 6

5

1

(xiii)729

39423 =

729

39472923

= 729

39416767 =

729

17161

1311 1 71 61

123 71

69261 261

261 ×

272 7 29

447 329

329 ×

= 27

131 = 4

27

23

(xiv)169

5121 =

169

5116921

= 169

513549 =

169

3600

= 13

60 = 4

13

8


86

(xv)225

15110 =

225

15122510

= 225

1512250 =

225

2401

494 24 01

1689 801

801 ×

151 2 25

125 125

125 ×

= 15

49 = 3

15

4

2. Find the value of :

(i)405

80(ii)

625

441

(iii)1728

1587(iv) 72 × 338

(v) 45 × 20

Solution—

(i)405

80 =

405

80 =

81

16(Dividing by 5)

= 81

16 =

9

4

(ii)625

441 =

625

441 =

25

21

(iii)1728

1587 =

1728

1587 =

31728

31587

=

576

529

232 5 29

443 129

129 ×

242 5 76

444 176

176 ×

576

529 =

24

23

(iv) 72 × 338 = 33872 = 24336

= 156

(v) 45 × 20 = 2045 = 900 = 30

3. The area of a square field is 80729

244 square

metres. Find the length of each side of

the field.

Solution—

Area of a square field = 80729

244

= 729

24472980 m2

= 729

24458320 =

729

58564 m2

Side of square field = Area

= 729

58564 =

729

58564

272 7 29

447 329

329 ×

2422 5 85 64

444 185

176482 964

964 ×

= 27

242 = 8

27

26

1561 2 43 36

125 143

125306 1836

1836 ×


87

4. The area of a square field is 304

1 m2.

Calculate the length of the side of the

squre.

Solution—

Area of the square field = 304

1 =

4

121m2

Side = Area = 4

121m

= 22

1111

=

2

11 = 5.5 m

5. Find the length of a side of a square

playground whose area is equal to the

area of a rectangular field of dimensions

72 m and 338 m.

Solution—

Length of rectangular field (l) = 338 m

and breadth (b) = 72 m

Area = l × b = 338 × 72 m2

Area of square = 338 × 72 m2

= 24336 m2

and length of the side of the square

= Area = 24336

1561 2 43 36

125 143

125306 1836

1836 ×

= 156 m

Find the square root of the following

numbers in decimal form :

1. 84.8241

Solution—

8241.84 = 10000

848241

9.219 84 82 41

81182 382

3641841 1841

1841 ×

= 10000

848241 =

100

921 = 9.21

EXERCISE 3.7

2. 0.7225

Solution—

7225.0 = 10000

7225

= 10000

7225 =

100

85 = 0.85

3. 0.813604

Solution—

813604.0 = 1000000

813604

858 72 25

64165 825

825 ×


88

9029 81 36 04

811802 3604

3604 ×

= 1000000

813604 =

1000

902 = 0.902

4. 0.00002025

Solution—

00002025.0 = 100000000

2025

454 20 25

1685 425

425 ×

= 100000000

2025 =

10000

45 = 0.0045

5. 150.0625

Solution—

0625.150 = 10000

1500625

= 10000

1500625

= 100

1225 = 12.25

6. 225.6004

Solution—

6004.225

= 10000

2256004 =

10000

2256004

15021 2 25 60 04

125 125

1253002 6004

6004 ×

= 100

1502 = 15.02

7. 3600.720036

Solution—

720036.3600

= 1000000

3600720036 =

1000000

3600720036

600066 36 00 72 00 36

36120006 00720036

720036 ×

= 1000

60006 = 60.006

8. 236.144689

Solution—

144689.236 = 1000000

236144689

= 1000000

236144689

12251 1 50 06 25

122 50

44242 606

4842445 12225

12225 ×


89

153671 2 36 14 46 89

125 136

125303 1114

9093066 20546

1839630727 215089

215089 ×

= 1000

15367 = 15.367

9. 0.00059049

Solution—

00059049.0

= 100000000

59049 =

100000000

59049

2432 5 90 49

444 190

176483 1449

1449 ×

= 10000

243 = 0.0243

10. 176.252176

Solution—

252176.176

= 1000000

176252176 =

1000000

176252176

132761 1 76 25 21 76

123 76

69262 725

5242647 20121

1852926546 159276

159276 ×

= 1000

13276 = 13.276

11. 9998.0001

Solution—

0001.9998 = 10000

99980001

= 10000

99980001

99999 99 98 00 01

81189 1898

17011989 19700

1790119989 179901

179901 ×

= 100

9999 = 99.99

12. 0.00038809

Solution—

00038809.0 = 100000000

38809


90

= 100000000

38809

1971 3 88 09

129 288

261387 2709

2709 ×

= 10000

197 = 0.0197

13. What is that fraction which when

multiplied by itself gives 227.798649 ?

Solution—

Let the fraction = x

Then x2 = 227.798649

x = 798649.227

= 1000000

227798649 =

1000000

227798649

150931 2 27 79 86 49

125 127

1253009 27986

2708130183 90549

90549 ×

= 1000

15093 = 15.093

Required fraction = 15.093

14. The area of a square playground is

256.6404 square metres. Find the length

of one side of the playground.

Solution—

Area of square playground = 256.6404 sq. m

Side of the square = Area

= 6404.256 = 10000

2566404

16021 2 56 64 04

126 156

1563202 6404

6404 ×

= 10000

2566404 =

100

1602 = 16.02 m

15. What is the fraction which when

multiplied by itself gives 0.00053361 ?

Solution—

Let the given fraction = x

x × x = 0.00053361

x2 = 0.00053361

x = 00053361.0 = 100000000

53361

2312 5 33 61

4431 133

129461 461

461 ×

= 100000000

53361 =

10000

231 = 0.0231

Required fraction = 0.0231

16. Simplify :

(i)2952959

2952959

..

..


91

(ii)0.17640.2304

0.17640.2304

Solution—

(i)5.2959.29

5.2959.29

7.77 59.29

49147 1029

1029 ×

2.32 5.29

443 129

129 ×

= 3.27.7

3.27.7

= 10

4.5 = 0.54

(ii)0.17640.2304

0.17640.2304

0.484 0.23 04

1688 704

704 ×

0.424 0.17 64

1682 164

164 ×

= 42.048.0

42.048.0

= 06.0

90.0 =

6

90 = 15

17. Evaluate 50625 and hence find the

value of 506.25 + 5.0625 .

Solution—

50625 = 225

2252 5 06 25

442 106

84445 2225

2225 ×

Now 25.506 + 0625.5

= 100

50625 +

10000

50625

= 10

225 +

100

225 = 22.50 + 2.25

= 24.75

18. Find the value of 103.0225 and hence

find the value of

(i) 10302.25 (ii) 1.030225

Solution—

0225.103 = 10000

1030225 =

10000

1030225

(i) 25.10302 = 100

1030225

= 10

1015 = 101.5

(ii) 030225.1 = 1000000

1030225

= 1000

1015 = 1.015

10151 1 03 02 25

1201 0302

2012025 10125

10125 ×


92


following correct to three places of

decimal :

(i) 5 (ii) 7

(iii) 17 (iv) 20

(v) 66 (vi) 427

(vii) 1.7 (viii) 23.1

(ix) 2.5 (x) 237.615

(xi) 15.3215 (xii) 0.9

(xiii)0.1 (xiv) 0.016

(xv) 0.00064 (xvi) 0.019

(xvii) 8

7(xviii)

12

5

(xix) 22

1(xx) 287

8

5

Solution—

(i) 5 = 000000.5

= 2.236

(ii) 7 = 00000000.7

= 2.6457

= 2.646

(iii) 17 = 000000.17

4.1234 17.00 00 00

1681 100

81822 1900

16448243 25600

24729 871

= 4.123

(iv) 20 = 000000.20

4.4724 20.00 00 00

1684 400

336887 6400

62098942 19100

17884 1216

= 4.472

(v) 66 = 000000.66

8.1248 66.00 00 00

64161 200

1611622 3900

324416244 65600

64976 624

= 8.124

EXERCISE 3.8

2.2362 5.00 00 00

442 100

84443 1600

13294466 27100

26796 304

2.64572 7.00 00 00 00

446 300

276524 2400

20965285 30400

2642552907 397500

370349 27151


93

(vi) 427 = 000000.427

= 20.664

(vii) 7.1 = 70000000.1

1.30371 1.70 00 00 00

123 70

692603 10000

780926068 219100

208544 10556

= 1.3038 = 1.304

(viii) 1.23 = 100000.23

= 4.806

(ix) 5.2 = 500000.2

1.5811 2.50 00 00

125 150

125308 2500

24643161 3600

3161 639

= 1.581

(x) 615.237 = 615000.237

15.41471 2 37.61 50 00 00

125 137

125304 1261

12163081 4550

308130824 146900

123296308287 2360400

2158009 202391

= 15.4147 = 15.415

(xi) 3215.15 = 321500.15

3.91423 15.32 15 00 00

969 632

621781 1115

7817824 33400

3129678282 210400

156564 53836

= 3.9142 = 3.914

(xii) 9.0 = 900000.0

= 0.9486 = 0.949

20.6642 4 27.00 00 00

4406 2700

24364126 26400

2475641324 164400

165296

4.8064 23.10 00 00

1688 710

7049606 60000

57636 2364

0.94869 0.90 00 00 00

81184 900

7361888 16400

1510418966 129600

113796 15804


94

(xiii) 1.0 = 100000.0

= 0.316

(xiv) 016.0 = 016000.0

0.12641 0.01 60 00 00

122 60

44246 1600

14762524 12400

10096 2304

= 0.1264 = 0.126

(xv) 00064.0 = 000640.0

0.02522 0.00 06 40 00

445 240

225502 1500

1004 496

= 0.0252 = 0.025

(xvi) 019.0 = 019000.0

= 0.1378 = 0.138

(xvii) 8

7 = 875.0

0.93549 0.87 50 00 00

81183 650

5491865 10100

932518704 77500

74816 2684

) 7.000 64 60 56 40 40 ×

8

0.875

= 0.9354 = 0.935

(xviii) 12

5 = 416666.0

0.64546 0.41 66 66 66

36124 566

4961285 7066

642512904 64166

51616 12550

) 5.0000 48 20 12 80 72 80 72 80

12

0.41666

= 0.6454 = 0.645

(xix)2

12 =

2

5 = 5.2

= 1.581

0.3163 0.10 00 00

961 100

61626 3900

3756 144

0.13781 0.01 90 00 00

123 90

69267 2100

18692748 23100

21984 1116

1.5811 2.50 00 00

125 150

125308 2500

24643161 3600

3161 439


95

(xx)8

5287 =

8

2301 = 625.287

16.9591 2 87.62 50 00

126 187

156329 3162

29613385 20150

1692533909 322500

305181 12550

) 5.000 48 20 16 40 40 ×

8625

= 16.959

2. Find the square root of 12.0068 correct

to four decimal places.

Solution—

0068.12

= 3.46508 = 3.4651

3. Find the square root of 11 correct to five

decimal places.

Solution—

11 = 3.31662

4. Given that 2 = 1.414, 3 = 1.732, 5

= 2.236 and 7 = 2.646. Evaluate each

of the following :

(i)7

144(ii)

3

2500

Solution—

Given = 2 = 1.414, 3 = 1.732,

5 = 2.236 and 7 2.646

(i)7

144 =

7

144 =

646.2

12

) 12000.000 ( 10584 14160 13230 9300 7938 13620 13230 3900 2646 1251

2646 4.5351

= 2646

100012 =

2646

12000

= 4.5351

= 4.535

OR

7

144 =

7

144 =

77

7144

=

7

712

= 7

646.212 =

7

752.31

= 4.536

(ii)3

2500 =

3

2500 =

33

32500

= 3

350 =

3

732.150

3.465083 12.00 68 00 00 00

964 300

256686 4468

41166925 35200

34625693008 5750000

5544064 205936

3.316623 11.00 00 00 00 00

963 200

189661 1100

6616626 43900

3975666326 414400

397956663322 1644400

1326644 317756


96

= 3

6.86 = 28.8666

= 28.867

5. Given that 2 = 1.414, 3 = 1.732, 5

= 2.236 and 7 = 2.646, find the square

roots of the following :

(i)75

196(ii)

63

400

(iii)7

150(iv)

5

256

(v)50

27

Solution—

We are given : 2 = 1.414, 3 = 1.732,

5 = 2.236 and 7 = 2.646

(i)75

196 =

325

196

=

325

196

=

35

196

= 335

314

=

35

314

=

15

314

= 15

732.114 =

15

248.24

= 1.6165 = 1.617

(ii)63

400 =

79

400

=

73

20

= 773

720

=

73

720

=

21

720

= 21

646.220 =

21

92.52

= 2.52

(iii)7

150 =

7

150 =

77

7150

= 7

7150

= 7

73225

= 7

7325

=

7

646.2732.1414.15

= 7

4801.65 =

7

4009.32

= 4.6287 = 4.629

(iv)5

256 =

5

256 =

5

16

= 55

516

=

5

236.216

= 5

776.35 = 7.155

(v)50

27 =

50

27 =

225

39

= 25

33 =

225

233

= 25

233

=

10

233

=

10

414.1732.13 =

10

449048.23

= 10

3471.7

= 0.7347 = 0.735


97

Using square root table, find the square

roots of the following :

1. 7

Solution—

Let x = 7

Then x = 7

From the table, in the column of x

We find that x = 7 = 2.646

2. 15

Solution—

Let x = 15

Then x = 15

From the table, in column of x , we find

that

x = 15 = 3.873

3. 74

Solution—

Let x = 74

x = 74

From the table, in the column of x , we

find that

x = 74 = 8.602

4. 82

Solution—

Let x = 82

x = 82

From the table in the column of x , we

find that x = 82 = 9.055

5. 198

Solution—

198 = 229

= 3 22

From the table we see from the x ,

22 = 4.690

198 = 3 × 22 = 3 × 4.690

= 14.070

6. 540

Solution—

540 = 1536

6 5406 90

15

= 6 15

From the table, we see in the column x ,

that 15 = 3.873

540 = 6 × 15 = 6 × 3.873

= 23.238 = 23.24

7. 8700

Solution—

8700 = 87100

= 10 87


that 87 = 9.327

8700 = 10 87 = 10 × 9.327

= 93.27

8. 3509

Solution—

3509 = 29121

= 11 × 29

EXERCISE 3.9


98

From the table, we see in the column x

that 29 = 5.385

11 350911 319

29

3509 = 11 × 29 = 11 × 5.385

= 59.235

9. 6929

Solution—

6929 = 41169 = 13 41


that 41 = 6.403

13 692913 533

41

6929 = 13 41

= 13 × 6.403 = 83.239

10. 25725

Solution—

25725 = 217755

5 257255 51457 10297 147

21

= 5 × 7 21

= 35 21

From the table we see the column of x ,

that 21 = 4.583

25725 = 35 21 = 35 × 4.583

= 160.40

11. 1312

Solution—

1312 = 8244

4 13124 328

82

= 4 × 82


that 82 = 9.055

1312 = 4 × 82 = 4 × 9.055

= 36.220 = 36.22

12. 4192

Solution—

4192 = 26216

2 41922 20962 10482 524

262

= 4 262

= 4 × 16.186

= 64.7456 = 64.75

13. 4955

Solution—

4955 = 10055.49

= 10 55.49

55.49 lies between 49 and 50 or 7

and 7.071

Diff. between 49 and 50 = 1 and between 7

and 7.071 = 0.071

Diff. between 0.55 = 0.55 × 0.71 = 0.039

10 55.49 = 10 × 7.039 = 70.39

14.144

99


99

Solution—

144

99 =

1212

1133

=

1212

1133

= 12

311 =

4

111

From the table,

11 = 3.317

144

99 =

4

1 × 11 =

4

1 × 3.317

= 4

317.3 = 0.829

15.169

57

Solution—

169

57 =

169

57 =

13

157

From the table, 57 = 7.550

169

57 =

13

157 =

13

550.7 = 0.581

16.169

101

Solution—

169

101 =

169

101 =

13

1101 =

13

1 × 10.05

= 13

05.10 = 0.773

17. 13.21

Solution—

21.13 is between 13 and 14

13 = 3.606 and 14 = 3.742

Difference between 14 and 13

= 3.742 – 3.606 = 0.136

and difference 13 and 14 = 1

Difference in 0.21 = 0.136 × 0.21

= 0.02856

21.13 = 3.606 + 0.029 = 3.635

18. 21.97

Solution—

97.21

97.21 is between 21 and 22

21 = 4.583 and 22 = 4.690

Difference between 22 and 21

= 4.690 – 4.583 = 0.107

Difference between 21 and 22 = 1

and difference in 0.97 = 0.97 × 0.107

= 0.1038

97.21 = 4.583 + 0.104 = 4.687

19. 110

Solution—

110 = 1011 = 11 × 10

= 3.317 × 3.162

= 10.488

20. 1110

Solution—

1110 = 3037

= 6.083 × 5.477 (from the table)

= 33.3165 = 33.317

21. 11.11

Solution—

11.11 lies between 11 and 12

11 = 3.317 and 12 = 3.464

Difference between 11 and 12 = 1

and difference between 3.464 and 3.317

= 3.464 – 3.317 = 0.147

Difference in 0.11 = 0.147 × 0.11 = 0.01617

11.11 = 3.317 + 0.0162

= 3.3332 = 3.333


100

22. The area of a square field is 325 m2. Find

the approximate length of one side of the

field.

Solution—

Area of the square field = 325 m2

Length of one side = Area = 325 m

= 325 = 1325

= 25 × 13 = 5 × 13

= 5 × 3.606 m (from the table)

= 18.030 = 18.030 m

23. Find the length of a side of a square, whose

area is equal to the area of the rectangle

with sides 240 m and 70 m.

Solution—

Lenght of rectangle (l) = 240 m

and breadth (b) = 70 m

Area = l × b = 240 × 70 m2

= 16800 m2

Area of square field = 16800 m2

and side = Area = 16800 m

= 42400

= 400 × 42

= 20 × 6.481

= 129.620

= 129.62 m


101

Points to Remember

1. Cube— The cube of a number is that

number raised to the power 3 e.g. x3, 53, 23

etc.

2. Perfect cube— An integer is said to be a

perfect cube if it is the cube of some integer.

3. Properties of cubes—

The cubes of natural numbers have the

following interesting properties :

Property 1 : Cubes of all even natural

numbers are even.

Property 2 : Cubes of all odd natural

numbers are odd.

Property 3 : The sum of the cubes of first

n natural numbers is equal to the square of

their sum. That is,

13 + 23 + 33 + ..... + n3 = (1 + 2 + 3 + .....

+ n)2

Property 4 : Cubes of the numbers ending

in digits 1, 4, 5, 6 and 9 are the numbers

ending in the same digit. Cubes of numbers

ending in digit 2 ends in digit 8 and the cube

of numbers ending in digit 8 ends in digit

2. The cubes of the numbers ending in

digits 3 and 7 ends in digit 7 and 3

respectively.

4. Methods of finding the cube of two-digit

number—

(i) Using formula— (a + b)3 = a3 + 3a2b

+ 3ab2 + b3

5. Cube of negative numbers— If m is a

positive integer, (–m)3 = –m × –m × –m = –

m3

–m3 is the cube of (–m).

6. Cube of rational numbers—

3

n

m =

3

3

n

m, where

n

m is a rational number where

m and n are non-zero integers.

7. Cube root— A number m is the cube root

of a number n if n = m3 and it is denoted as

3 n or 3

1

n where 3 is called radical or

3

1 is called index.

8. Computation of cube root through a

pattern—

The pattern is

13 = 1,

23 = 1 + 7

33 = 1 + 7 + 19 (4)3 = 1 + 7 + 19 + 37

(5)2 = 1 + 7 + 19 + 37 + 61

(6)3 = 1 + 7 + 19 + 37 + 61 + 91

and so on.

The procedure for finding the cube root of

the given number will be possible if the

number is a small number.

The sequence can be written as

1, 1 + 2

12 × 6 = 7 1 +

2

67 × 6 = 127

1 + 2

23 × 6 = 19 1 +

2

78 × 6 = 169

1 + 2

34 × 6 = 37 1 +

2

89 × 6 = 217

4CUBES AND CUBE-ROOTS


102

1 + 2

45 × 6 = 61 1 +

2

56 × 6 = 91

and so on.

If the numbers which are to be subtracted

in succession from the given number.

9. Cube root by factors—

(i) Resolve the numbers into factors and then

group them in triplets of equal of a factors.

(ii) Take one factor from each triplet and find

the product of these factors.

This product is the cube root of the given

number.

10. Cube root of a.b or b

a

(i) Cube root of a.b = 3 ab = 3 a × 3 b

and cube root of b

a = 3

b

a = 3

3

b

a, b 0

11. Cube root tables— We can find cube root

of a given number by using the given cube

root tables also.

12. Cube root of negative numbers— Cube

root of –n = 3 n = – 3 n

1. Find the cubes of the following numbers:

(i) 7 (ii) 12

(iii) 16 (iv) 21

(v) 40 (vi) 55

(vii) 100 (viii) 302

(ix) 301

Solution—

(i) (7)3 = 7 × 7 × 7 = 343

(ii) (12)3 = 12 × 12 × 12 = 1728

(iii) (16)3 = 16 × 16 × 16 = 4096

(iv) (21)3 = 21 × 21 × 21

= 441 × 21 = 9261

(v) (40)3 = 40 × 40 × 40 = 64000

(vi) (55)3 = 55 × 55 × 55

= 3025 × 55 = 166375

(vii) (100)3 = 100 × 100 × 100

= 1000000

(viii)(302)3 = 302 × 302 × 302

= 91204 × 302 = 27543608

(ix) (301)3 = 301 × 301 × 301

= 90601 × 301 = 27270901

2. Write the cubes of all natural numbers

between 1 and 10 and verify the following

statements :

EXERCISE 4.1

(i) Cubes of all odd natural numbers are odd.

(ii) Cubes of all even natural numbers are

even.

Solution—

Cubes of first 10 natural numbers :

(1)3 = 1 × 1 × 1 = 1

(2)3 = 2 × 2 × 2 = 8

(3)3 = 3 × 3 × 3 = 27

(4)3 = 4 × 4 × 4 = 64

(5)3 = 5 × 5 × 5 = 125

(6)3 = 6 × 6 × 6 = 216

(7)3 = 7 × 7 × 7 = 343

(8)3 = 8 × 8 × 8 = 512

(9)3 = 9 × 9 × 9 = 729

(10)3 = 10 × 10 × 10 = 1000

We see that the cubes of odd numbers is also

odd and cubes of even numbers is also even.

3. Observe the following pattern :

13 = 1

13 + 23 = (1 + 2)2

13 + 23 + 33 = (1 + 2 + 3)2

Write the next three rows and calculate

the value of 13 + 23 + 33 + ..... + 93 + 103 by

the above pattern.


103

Solution—

We see the pattern

13 = 1

13 + 23 = (1 + 2)2

13 + 23 + 33 = (1 + 2 + 3)2

13 + 23 + 33 + 43 = (1 + 2 + 3 + 4)2

13 + 23 + 33 + 43 + 53 = (1 + 2 + 3 + 4 + 5)2

and 13 + 23 + 33 + 43 + 53 + 63 = (1 + 2 + 3 + 4 + 5 + 6)2

and 13 + 23 + 33 + .... + 93 + 103 = (1 + 2 + 3 + .... + 9 + 10)2

4. Write the cubes of 5 natural numbers

which are multiples of 3 and verify the

followings :

‘The cube of a natural number which is a

mutiple of 3 is a multiple of 27’

Solution—

5 natural numbers which are multiples of 3

3, 6, 9, 12, 15

(3)3 = 3 × 3 × 3 = 27

Which is multiple of 27

(6)3 = 6 × 6 × 6 = 216 27 = 8


(9)3 = 9 × 9 × 9 = 729 27 = 27


(12)3 = 12 × 12 × 12 = 1728 27 = 64


(15)3 = 15 × 15 × 15 = 3375 27 = 125


Hence, cube of multiple of 3 is a multiple of

27

5. Write the cubes of 5 natural numbers

which are of the form 3n + 1 (e.g. 4, 7,

10, .....) and verify the following :

‘The cube of a natural number of the

form 3n + 1 is a natural number of the

same form i.e. when divided by 3 it leaves

the remainder 1’.

Solution—

3n + 1

Let n = 1, 2, 3, 4, 5, then

If n = 1, then 3n + 1 = 3 × 1 + 1 = 3 + 1 = 4

If n = 2, then 3n + 1 = 3 × 2 + 1 = 6 + 1 = 7

If n = 3, then 3n + 1 = 3 × 3 + 1 = 9 + 1 = 10

If n = 4, then 3n + 1 = 3 × 4 + 1 = 12 + 1

= 13

If n = 5, then 3n + 1 = 3 × 5 + 1 = 15 + 1

= 16

Now

(4)3 = 4 × 4 × 4 = 64

Which is 3

64 = 21, Remainder = 1

(7)3 = 7 × 7 × 7 = 343

Which is 3

343 = 114, Remainder = 1

(10)3 = 10 × 10 × 10 = 1000 3 = 333,

Remainder = 1

(13)3 = 13 × 13 × 13 = 2197 3 = 732,

Remainder = 1

(16)3 = 16 × 16 × 16 = 4096 3 = 1365,

Remainder = 1

Hence cube of natural number of the form,

3n + 1, is a natural of the form 3n + 1

6. Write the cubes of 5 natural numbers of

the form 3n + 2 (i.e. 5, 8, 11, .....) and

verify the following :

‘The cube of a natural number of the

form 3n + 2 is a natural number of the

same form i.e. when it is dividend by 3

the remainder is 2’.

Solution—

Natural numbers of the form 3n + 2, when n

is a natural number i.e. 1, 2, 3, 4, 5, .......

If n = 1, then 3n + 2 = 3 × 1 + 2 = 3 + 2 = 5


104

If n = 2, then 3n + 2 = 3 × 2 + 2 = 6 + 2 = 8

If n = 3, then 3n + 2 = 3 × 3 + 2 = 9 + 2 = 11

If n = 4, then 3n + 2 = 3 × 4 + 2 = 12 + 2 =

14

and if n = 5, then 3n + 2 = 3 × 5 + 2 = 15 +

2 = 17

Now(5)3 = 5 × 5 × 5 = 125

125 3 = 41, Remainder = 2

(8)2 = 8 × 8 × 8 = 512

512 3 = 170, Remainder = 2

(11)3 = 11 × 11 × 11 = 1331

1331 3 = 443, Remainder = 2

(14)3 = 14 × 14 × 14 = 2744

2744 3 = 914, Remainder = 2

(17)3 = 17 × 17 × 17 = 4913

4913 3 = 1637, Remainder = 2

We see the cube of the natural number of the

form 3n + 2 is also a natural number of the

form 3n + 2.

7. Write the cubes of 5 natural numbers of

which are multiples of 7 and verify the

following :

‘The cube of a multiple of 7 is a multiple

of 73’.

Solution—

5 natural numbers which are multiple of 7,

are 7, 14, 21, 28, 35

(7)3 = (7)3 which is multiple of 73

(14)3 = (2 × 7)3 = 23 × 73, which is multiple

of 73


of 73


of 73

(35)3 = (5 × 7)3 = 53 × 73 which is multiple of 73

Hence proved.

8. Which of the following are perfect cubes?

(i) 64 (ii) 216

(iii) 243 (iv) 1000

(v) 1728 (vi) 3087

(vii) 4608 (viii) 106480

(ix) 166375 (x) 456533

Solution—

(i) 64 = 2 ×2 × 2 × 2 × 2 × 2

2 642 322 162 82 42 2

1

Grouping the factors in triplets of equal

factors, we see that no factor is left

64 is a perfect cube

(ii) 216 = 2 × 2 × 2 × 3 × 3 × 3

2 2162 1082 543 273 93 3

1



216 is a perfect cube.

(iii) 243 = 3 × 3 × 3 × 3 × 3

3 2433 813 273 93 3

1

Grouping the factors in triplets, we see that

two factors 3 × 3 are left

243 is not a perfect cube.

(iv) 1000 = 2 × 2 × 2 × 5 × 5 × 5

2 10002 5002 2505 1255 255 5

1


105




(v) 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

2 17282 8642 4322 2162 1082 543 273 93 3

1

Grouping the factors in triplets of the equal



(vi) 3087 = 3 × 3 × 7 × 7 × 7

Grouping the factors in triplets of the equal

factors, we see that two factor 3 × 3 are left


(vii) 4608 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×

3 × 3


factors, we see that two factors 3, 3 are left


(viii)106480 = 2 × 2 × 2 × 2 × 5 × 11 × 11 × 11

2 1064802 532402 266202 133105 6655

11 133111 12111 11

1


factors, we see that factors 2, 5 are left


(ix) 166375 = 5 × 5 × 5 × 11 × 11 × 11

5 1663755 332755 6655

11 133111 12111 11

1




(x) 456533 = 7 × 7 × 7 × 11 × 11 × 11

7 4565337 652197 9317

11 133111 12111 11

1




9. Which of the following are cubes of even

natural numbers ?

216, 512, 729, 1000, 3375, 13824

Solution—

We know that the cube of an even natural

number is also an even natural number

2 46082 23042 11522 5762 2882 1442 722 362 183 93 3

1

3 30873 10297 3437 497 7

1


106

216, 512, 1000, 13824 are even natural

numbers.

These can be the cubes of even natural

number.

10. Which of the following are cubes of odd

natural numbers ?

125, 343, 1728, 4096, 32768, 6859

Solution—

We know that the cube of an odd natural

number is also an odd natural number.

_ 125, 343, 6859 are the odd natural numbers

These can be the cubes of odd natural

numbers.

11. What is the smallest number by which

the following numbers must be multiplied,

so that the products are perfect cubes ?

(i) 675 (ii) 1323

(iii) 2560 (iv) 7803

(v) 107811 (vi) 35721

Solution—

(i) 675 = 3 × 3 × 3 × 5 × 5

3 6753 2253 755 255 5

1

Grouping the factors in triplet of equal

factors, 5 × 5 are left without triplet

So, by multiplying by 5, the triplet will be

completed.

Least number to be multiplied = 5

(ii) 1323 = 3 × 3 × 3 × 7 × 7

3 13233 4413 1477 497 7

1


factors. We find that 7 × 7 has been left

So, multiplying by 7, we get a triplet

The least number to be multiplied = 7

(iii) 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

× 5

2 25602 12802 6402 3202 1602 802 402 202 105 5

1


factors, 5 is left.

To complete a triplet 5 × 5 is to multiplied

Least number to be multiplied = 5 × 5 = 25

(iv) 7803 = 3 × 3 × 3 × 17 × 17

3 78033 26013 867

17 28917 17

1


factors, we find the 17 × 17 are left

So, to complete the triplet, we have to multiply

by 17


(v) 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

3 1078113 359373 119793 3993

11 133111 12111 11

1


factors, factor 3 is left


107

So, to complete the triplet 3 × 3 is to be

multiplied

Least number to be multiplied = 3 × 3 = 9

(vi) 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7


factors, we find that 7 × 7 is left

So, in order to complete the triplets, we have

to multiplied by 7


12. By which smallest number must the

following numbers be divided so that the

quotient is a perfect cube ?

(i) 675 (ii) 8640 (iii) 1600

(iv) 8788 (v) 7803 (vi) 107811

(vii) 35721 (viii) 243000

Solution—

(i) 675 = 3 × 3 × 3 × 5 × 5


factors, 5 × 5 is left

5 × 5 is to be divided so that the quotient will

be a perfect cube.

The least number to be divided = 5 × 5 = 25

(ii) 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

× 5


factors, 5 is left

In order to get a perfect cube, 5 is to divided

Least number to be divided = 5

(iii) 1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5

2 16002 8002 4002 2002 1002 505 255 5

1


factors, we find that 5 × 5 is left

In order to get a perfect cube 5 × 5 = 25 is to

be divided.

Least number to be divide = 25

(iv) 8788 = 2 × 2 × 13 × 13 × 13

2 87882 4394

13 219713 16913 13

1


factors, we find that 2 × 2 has been left

In order to get a perfect cube, 2 × 2 is to be

divided


(v) 7803 = 3 × 3 × 3 × 17 × 17

3 78033 26013 867

17 28917 17

1


factors, we see that 17 × 17 has been left.

So, in order to get a perfect cube, 17 × 17 is

be divided

3 357213 119073 39693 13233 4413 1477 497 7

1

3 6753 2253 755 255 5

1

2 86402 43202 21602 10802 5402 2703 1353 453 155 5

1


108

Least number to be divided = 17 × 17 = 289

(vi) 107811 = 3 × 3 × 3 × 3 × 11 × 11 × 11

3 1078113 359373 119793 3993

11 133111 12111 11

1


factors, 3 is left

In order to get a perfect cube, 3 is to be

divided


(vii) 35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7

3 357213 119073 39693 13233 4413 1477 497 7

1


factors, we see that 7 × 7 is left

So, in order to get a perfect cube, 7 × 7 = 49

is to be divided


(viii)243000 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 ×

5 × 5 × 5


factors, 3 × 3 is left

By dividing 3 × 3, we get a perfect cube

Least number to be divided = 3 × 3 = 9

13. Prove that if a number is trebled then its

cube is 27 times the cube of the given

number.

Solution—

Let x be the number, then trebled number of

x = 3x

Cubing, we get :

(3x)3 = (3)3 x3 = 27x3

27x3 is 27 times the cube of x i.e., of x3

14. What happenes to the cube of a number

if the number is multiplied by

(i) 3 ? (ii) 4 ? (iii) 5 ?

Solution—

Let x be the given number

(x)3 = x3

(i) If x is multiplied by 3, then the cube of

(3x)3 = (3)3 × x3 = 27x3

The cube of the resulting number is 27 times

of cube of the given number

(ii) If x is multiplied by 4, then the cube of (4x)3

= (4)3 × x3 = 64x3


of the cube of the given number

(iii) If x is multiplied by 5, then the cube of

(5x)3 = (5)3 × x3 = 125x3


of the cube of the given number

15. Find the volume of a cube, one face of

which has an area of 64 m2.

Solution—

Area of one face of a cube = 64 m2

Side (edge) of cube = 64 m

= 64 = 8 m

Volume of the cube = (side)3 = (8 m)3

= 512 m3

16. Find the volume of a cube whose surface

area is 384 m2.

Solution—

2 2430002 1215002 607503 303753 101253 33753 11253 3755 1255 255 5

1


109

Surface area of a cube = 384 m2

Let side = a

Then 6a2 = 384 a2 = 6

384 = 64 = (8)2

a = 8 m

Now volume = a3 = (8)3 m3 = 512 m3

17. Evaluate the following :

(i) 3

2

122

125

(ii)

3

2

122

86

Solution—

(i) 3

2

122

125

=

3

2

1

14425

= 3

2

1

169

=

3

2

12

13

= 32

12

13 = 133 = 2197

(ii) 3

2

122

86

=

3

2

1

6436

= 3

2

1

100

=

3

2

12

10

= 32

12

10 = 103

= 1000

18. Write the units digit of the cube of each

of the following numbers :

31, 109, 388, 833, 4276, 5922, 77774, 44447,

125125125.

Solution—

We know that if unit digit of a number n is

= 1, then units digit of its cube = 1






= 7, then the units digit of its cube = 3




Now units digit of the cube of 31 = 1

Units digit of the cube of 109 = 9

Units digits of the cube of 388 = 2







19. Find the cubes of the following numbers by column method :

(i) 35 (ii) 56 (iii) 72

Solution—

(i) (35)3

I column II column III column IV column

a3 3 × a2 × b 3 × a × b2 b3

(3)3 3 × (3)2 × 5 3 × 3 × (5)2 (5)3

= 27 = 135 = 225 = 125

+ 15 + 23 + 12

42 158 237

42 8 7 5


110

(35)3 = 42875

(ii) (56)3


a3 3 × a2 × b 3 × a × b2 b3

(5)3 3 × 52 × 6 3 × 5 × 62 (6)3

= 125 = 450 = 540 = 216

+ 50 + 56 + 21

175 506 561

175 6 1 6

(56)3 = 175616

(iii) (72)3


a3 3 × a2 × b 3 × a × b2 b3

(7)3 3 × 72 × 2 3 × 7 × 22 (2)3

= 343 = 294 = 84 = 8

+ 30 + 8

373 302

373 2 4 8

(72)3 = 373248

20. Which of the following numbers are not

perfect cubes ?

(i) 64 (ii) 216

(iii) 243 (iv) 1728

Solution—

(i) 64 = 2 × 2 × 2 × 2 × 2 × 2

Grouping the factors in triplets, of equal



(ii) 216 = 2 × 2 × 2 × 3 × 3 × 3

2 2162 1082 543 273 93 3

1




(iii) 243 = 3 × 3 × 3 × 3 × 3


factors, we see that 3 × 3 are left


(iv) 1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

2 642 322 162 82 42 2

1

3 2433 813 273 93 3

1

2 17282 8642 4322 2162 1082 543 273 93 3

1


111


factors, we see that no factor is left.


21. For each of the non-perfect cubes, in Q.

20, find the smallest number by which it

must be

(a) multiplied so that the product is a perfect

cube.

(b) divided so that the quotient is a perfect

cube.

Solution—

In qustion 20, 243 is not a perfect cube and

243 = 3 × 3 × 3 × 3 × 3


factors, we see that 3 × 3 is left.

(a) In order to make it a perfect cube, 3 is to be

multiplied which makes a triplet.

(b) In order to make it a perfect cube, 3 × 3 or 9

is to be divided.

22. By taking three different values of n

verify the truth of the following

statements :

(i) If n is even, then n3 is also even.

(ii) If n is odd, then n3 is also odd.

(iii) If n leaves remainder 1 when divided by

3, then n3 also leaves 1 as remainder

when divided by 3.

(iv) If a natural number n is of the form 3p +

2 then n3 also a number of the same type.

Solution—

(i) n is even number.

Let n = 2, 4, 6 then

(a) n3 = (2)3 = 2 × 2 × 2 = 8, which is an even

number.

(b) (n)3 = (4)3 = 4 × 4 × 4 = 64, which is an even

number.

(c) (n)3 = (6)3 = 6 × 6 × 6 = 216, which is an

even number.

(ii) n is odd number.

Let x = 3, 5, 7

(a) (n)3 = (3)3 = 3 × 3 × 3 = 27, which is an odd

number.

(b) (n)3 = (5)3 = 5 × 5 × 5 = 125, which is an

odd number.

(c) (n)3 = (7)3 = 7 × 7 × 7 = 343, which is an

odd number.

(iii) If n leaves remainder 1 when divided by 3,

then n3 is also leaves 1 as remainder,

Let n = 4, 7, 10

If n = 4, then

n3 = (4)3 = 4 × 4 × 4 = 64

= 64 3 = 21, remainder = 1

If n = 7, then

n3 = (7)3 = 7 × 7 × 7 = 343

343 3 = 114, remainder = 1

If n = 10, then

(n)3 = (10)3 = 10 × 10 × 10 = 1000

1000 3 = 333, remainder = 1

(iv) If the natural number is of the form 3p + 2,

then n3 is also of the same type

Let p = 1, 2, 3, then

(a) If p = 1, then

n = 3p + 2 = 3 × 1 + 2 = 3 + 2 = 5

n3 = (5)3 = 5 × 5 × 5 = 125

125 = 3 × 41 + 2 = 3p + 2

(b) If p = 2, then

n = 3p + 2 = 3 × 2 + 2 = 6 + 2 = 8

n3 = (8)3 = 8 × 8 × 8 = 512

512 = 3 × 170 + 2 = 3p + 2

(c) If p = 3, then

n = 3p + 2 = 3 × 3 + 2 = 9 + 2 = 11

n3 = (11)3 = 11 × 11 × 11 = 1331

and 1331 = 3 × 443 + 2 = 3p + 2

Hence proved.

23. Write true (T) or false (F) for the

following statements :

(i) 392 is a perfect cube.

(ii) 8640 is not a perfect cube.

(iii) No cube can end with exactly two zeros.

(iv) There is no perfect cube which ends in 4.

(v) For an integer a, a3 is always greater than a2.

(vi) If a and b are integers such that a2 > b2,

then a3 > b3.

(vii) If a divides b, then a3 divides b3.

(viii) If a2 ends in 9, then a3 ends in 7.

(ix) If a2 ends in 5, then a3 ends in 25.


112

(x) I f a2 ends in an even number of zeros.

then a3 ends in an odd number of zeros.

Solution—

(i) False _ 392 = 2 × 2 × 2 × 7 × 7

_ Its all factors are not in triplets of equal factors.

(ii) True

_ 8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3

× 5

2 86402 43202 21602 10802 5402 2703 1353 453 155 5

1

_ 5 is left.

(iii) True : A number ending three zeros can be a

perfect cube.

(iv) False : _ (4)3 = 4 × 4 × 4 = 64, which ends

with 4.

(v) False : If n is a proper fraction, it is not

possible.

(vi) False : It is not true if a and b are proper

fraction.

(vii) True.

(viii)False, as a2 ends in 9, then a3 does not

necessarily ends in 7. It ends in 3 also.

(ix) False, it is not necessarily that a3 ends in 25

it can end also in 75.

(x) False : If a2 ends with even zeros, then a3

will ends with odd zeros but of multiple of

3.

2 3922 1962 987 497 7

1

1. Find the cubes of :

(i) –11 (ii) –12 (iii) –21

Solution—

(i) (–11)3 = – (11)3 = – (11 × 11 × 11) = –1331

(ii) (–12)3 = – (12)3 = – (12 × 12 × 12) = –1728

(iii) (–21)3 = – (21)3 = – (21 × 21 × 21) = –9261

2. Which of the following numbers are cubes

of negative integers.

(i) –64 (ii) –1056 (iii) –2197

(iv) –2744 (v) –42875

Solution—

(i) –64 = – (2 × 2 × 2 × 2 × 2 × 2)

EXERCISE 4.2

2 642 322 162 82 42 2

1

_ All factors of 64 can be grouped in triplets

of the equal factors completely.

–64 is a perfect cube of negative integer.

(ii) –1056 = – (2 × 2 × 2 × 2 × 2 × 3 × 11)

2 10562 5282 2642 1322 663 33

11 111

_ All the factors of 1056 can be grouped in

triplets of equal factors grouped completely

–1058 is not a perfect cube of negative

integer.


113

(iii) –2197 = – (2197) = – (13 × 13 × 13)

13 219713 16913 13

1

_ All the factors of –2197 can be grouped in

triplets of equal factors completely


(iv) –2744 = – (2744) = – (2 × 2 × 2 × 7 × 7 × 7)

2 27442 13722 6867 3437 497 7

1



–2744 is a perfect cube of negative integer

(v) –42875 = – (42875)

= – (5 × 5 × 5 × 7 × 7 × 7)

5 428755 85755 17157 3437 497 7

1




3. Show that the following integers are cubes

of negative integers. Also, find the

integer whose cube is the given integer :

(i) –5832 (ii) –2744000

Solution—

(i) –5832 = – (5832)

= – (2 × 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3)

2 58322 29162 14583 7293 2433 813 273 93 3

1



–5832 is a perfect cube

Now taking one factor from each triplet we

find that

–5832 is a cube of – (2 × 3 × 3) = –18

Cube root of –5832 = –18

(ii) –2744000 = – (2744000)

= 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7

× 7

2 27440002 13720002 6860002 3430002 1715002 857505 428755 85755 17157 3437 497 7

1


factors, we see that no factor is left.

Therefore it is a perfect cube.

Now taking one factor from each triplet, we

find that.

–2744000 is a cube of – (2 × 2 × 5 × 7) i.e.

–140

Cube root of –2744000 = –140


114

4. Find the cube of :

(i)9

7(ii)

11

8(iii)

7

12

(iv)8

13(v) 2

5

2(vi) 3

4

1

(vii) 0.3 (viii) 1.5 (ix) 0.08

(x) 2.1

Solution—

(i)

3

9

7

=

999

777

= 729

343

(ii)

3

11

8

=

11

8 ×

11

8 ×

11

8

= –111111

888

= –1331

512

(iii)

3

7

12

=

7

12 ×

7

12 ×

7

12

= 777

121212

= 343

1728

(iv)

3

8

13

=

8

13 ×

8

13 ×

8

13

888

131313

= –512

2197

(v)

3

5

22

=

3

5

12

=

5

12 ×

5

12 ×

5

12

= 555

121212

= 125

1728

(vi)

3

4

13

=

3

4

13

=

4

13 ×

4

13 ×

4

13

= 444

131313

= 64

2197

(vii) (0.3)3 =

3

10

3

=

10

3 ×

10

3 ×

10

3

= 1000

27 = 0.027

(viii)(1.5)3 =

3

10

15

=

10

15 ×

10

15 ×

10

15

= 101010

151515

= 1000

3375 = 3.375

(ix) (0.08)3 =

3

100

8

=

100

8 ×

100

8 ×

100

8

= 1000000

512 = 0.000512

(x) (2.1)3 =

3

10

21

=

10

21 ×

10

21 ×

10

21

= 101010

212121

= 1000

9261 = 9.261

5. Which of the following numbers are cubes

of rational numbers :

(i)64

27(ii)

128

125

(iii) 0.001331 (iv) 0.04

Solution—

(i)64

27 =

444

333

=

3

4

3

64


(ii)128

125 =

2222222

555

=

222

533

3

We see that 128 is not a perfect cube

128



115

(iii) 0.001331 = 1000000

1331

= 101010101010

111111

=

3

1010

11

=

3

100

11

It is a perfect cube.

(iv) 0.04 = 100

4 =

1010

22

= 10

2 ×

10

2 or 0.2 ×

0.2

It is clear that it is not a perfect cube as 10

2

or 0.2 is not in triplet.

1. Find the cube roots of the following

numbers by successive subtraction of

numbers : 1, 7, 19, 37, 61, 91, 127, 169,

217, 271, 331, 397, ........

(i) 64 (ii) 512

(iii) 1728

Solution—

(i) 64

64 – 1 = 63

63 – 7 = 56

56 – 19 = 37

37 – 37 = 0

64 = (4)3

Cube root of 64 = 4

(ii) 512

512 – 1 = 511

511 – 7 = 504

504 – 19 = 485

485 – 37 = 448

448 – 61 = 387

387 – 91 = 296

296 – 127 = 169

169 – 169 = 0

512 = (8)3

Cube root of 512 = 8

(iii) 1728

1728 – 1 = 1727

1727 – 7 = 1720

EXERCISE 4.3

1720 – 19 = 1701

1701 – 37 = 1664

1664 – 61 = 1603

1603 – 91 = 1512

1512 – 127 = 1385

1385 – 169 = 1216

1216 – 217 = 999

999 – 271 = 728

728 – 331 = 397

397 – 397 = 0

1728 = (12)3

Cube root of 1728 = 12

2. Using the method of successive

subtraction, examine whether or not the

following numbers are perfect cubes :

(i) 130 (ii) 345

(iii) 792 (iv) 1331

Solution—

(i) 130

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

We see that 5 is left


(ii) 345

345 – 1 = 344


116

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

We see that 2 is left


(iii) 792

792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

We see 280 is left as 280 < 217


(iv) 1331

1331 – 1 = 1330

1330 – 7 = 1323

1323 – 19 = 1304

1304 – 37 = 1267

1267 – 61 = 1206

1206 – 91 = 1115

1115 – 127 = 988

988 – 169 = 819

819 – 217 = 602

602 – 271 = 331

331 – 331 = 0


3. Find the smallest number that must be

subtracted from those of the numbers in

question 2, which are not perfect cubes,

to make them perfect cubes. What are

the corresponding cube roots ?

Solution—

We have examined in Question 2, the

numbers 130, 345 and 792 are not perfect

cubes. Therefore

(i) 130

130 – 1 = 129

129 – 7 = 122

122 – 19 = 103

103 – 37 = 66

66 – 61 = 5

Here 5 is left _ 5 < 91

5 is to be subtracted to get a perfect cube.

Cube root of 130 – 5 = 125 is 5

(ii) 345

345 – 1 = 344

344 – 7 = 337

337 – 19 = 318

318 – 37 = 281

281 – 61 = 220

220 – 91 = 129

129 – 127 = 2

Here 2 is left _ 2 < 169

Cube root of 345 – 2 = 343 is 7

2 is to be subtracted to get a perfect cube.

(iii) 792

792 – 1 = 791

791 – 7 = 784

784 – 19 = 765

765 – 37 = 728

728 – 61 = 667

667 – 91 = 576

576 – 127 = 449

449 – 169 = 280

280 – 217 = 63

63 < 217

63 is to be subtracted

Cube root of 792 – 63 = 729 is 9

4. Find the cube root of each of the following

natural numbers :

(i) 343 (ii) 2744

(iii) 4913 (iv) 1728

(v) 35937 (vi) 17576

(vii) 134217728 (viii) 48228544

(ix) 74088000 (x) 157464

(xi) 1157625 (xii) 33698267

Solution—

(i) 3 343 = 3 777 = 3 37

= 7

7 3437 497 7

1


117

(ii) 3 2744 2 27442 13722 6867 3437 497 7

1

= 3 777222 = 3 33 72

= 2 × 7 = 14

(iii) 3 4913 = 3 171717 = 3 317

17 491317 28917 17

1

= 17

(iv) 3 1728

2 17282 8642 4322 2162 1082 543 273 93 3

1

= 3 333222222

= 3 333 322

= 2 × 2 × 3 = 12

(v) 3 35937

3 359373 119793 3993

11 133111 12111 11

1

= 3 111111333

= 3 33 113 = 3 × 11 = 33

(vi) 3 17576

2 175762 87882 4394

13 219713 16913 13

1

= 3 131313222

= 3 33 132 = 2 × 13 = 26

(vii) 3 134217728

2 1342177282 671088642 335544322 167772162 83886082 41943042 20971522 10485762 5242882 2621442 1310722 655362 327682 163842 81922 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2

1


118

= 3

222

222222222222

222222222222

= 3 333333333 222222222 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512

(viii) 3 48228544

= 3

131313

777222222

= 3 3333 13722

= 2 × 2 × 7 × 13 = 364

(ix) 3 74088000

= 3

777555

333222222

= 3 33333 75322

= 2 × 2 × 3 × 5 × 7 = 420

(x) 3 157464

2 1574642 787322 393663 196833 65613 21873 7293 2433 813 273 93 3

1

= 3 333333333222

= 3 3333 3332

= 2 × 3 × 3 × 3 = 54

(xi) 3 1157625

3 11576253 3858753 1286255 428755 85755 17157 3437 497 7

1

= 3 777555333

= 3 333 753

= 3 × 5 × 7 = 105

2 740880002 370440002 185220002 92610002 46305002 23152505 11576255 2315255 463053 92613 30873 10297 3437 497 7

1

2 482285442 241142722 120571362 60285682 30142842 15071427 7535717 1076537 15379

13 219713 16913 13

1


119

(xii) 3 33698267

17 3369826717 198225117 11660319 685919 36119 19

1

= 3 191919171717

= 3 33 1917

= 17 × 19 = 323

5. Find the smallest number which when

multiplied with 3600 will make the

product a perfect cube. Further, find the

cube root of the product.

Solution—

3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

2 36002 18002 9002 4503 1253 755 255 5

1


factors, we see that 2, 3 × 3 and 5 × 5 are

left

In order to complete the triplets, we have to

multiply it by 2, 3 and 5.

The smallest number to be multiplied = 2 × 2

× 3 × 5 = 60

Now product = 3600 × 60 = 216000

and cube root of 216000

= 555333222222

= 3333 5322

= 2 × 2 × 3 × 5 = 60

6. Multiply 210125 by the smallest number

so that the product is a perfect cube. Also,

find out the cube root of the product.

Solution—

210125 = 5 × 5 × 5 × 41 × 41

5 2101255 420255 8405

41 168141 41

1


factors, we see that 41 × 41 is left

In order to complete the triplet, we have to

multiply it by 41

Smallest number to be multiplied = 41

Product = 210125 × 41 = 8615125

Cube root of 8615125

= 3 414141555

= 3 33 415 = 5 × 41 = 205

7. What is the smallest number by which

8192 must be divided so that quotient is

a perfect cube ? Also, find the cube root

of the quotient so obtained.

Solution—

8192 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 ×

2 × 2 × 2 × 2

2 81922 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2

1


120


factors, we see that 2 is left

Dividing by 2, we get the quotient a perfect

cube

Perfect cube = 8192 2 = 4096

and cube root = 3 4096

= 3 3333 2222 = 2 × 2 × 2 × 2 = 16

8. Three numbers are in the ratio 1 : 2 : 3.

The sum of their cubes is 98784. Find the

numbers.

Solution—

Ratio in numbers = 1 : 2 : 3

Let first number = x

Then second number = 2x

and third number = 3x

Sum of cubes of there numbers = (x)3 + (2x)3

+ (3x)3

x3 + 8x3 + 27x3 = 98784 36x3 = 98784

x3 = 36

98784 = 2744

x = 3 2744

= 3 777222

= 3 33 72 = 2 × 7 = 14

Number will be 14, 2 × 14, 3 × 14

i.e. 14, 28, 42

9. The volume of a cube is 9261000 m3. Find

the side of the cube.

Solution—

Volume of cube = 9261000 m3

Side = 3 Volume

= 3 9261000

= 3 10009261

3 92613 30873 10297 3437 497 7

1

10 100010 10010 10

1

= 3 101010777333

= 3 333 1073 = 3 × 7 × 10 m = 210 m

2 27442 13722 6867 3437 497 7

1

1. Find the cube roots of each of the

following integers :

(i) –125 (ii) –5832

(iii) –2744000 (iv) –753571

(v) –32768

Solution—

(i) 3 125 = – 3 125

= – 3 555 = – 3 35 = –5

(ii) 3 5832 = – 3 5832

EXERCISE 4.4

2 58322 29162 14583 7293 2433 813 273 93 3

1

= – 3 333333222


121

= – 3 333 332

= – (2 × 3 × 3) = –18

(iii) 3 2744000 = – 3 2744000

= – 3 10002744

= – 3 101010777222

= – 3 333 1072

= – (2 × 7 × 10) = –140

(iv) 3 753571 = – 3 753571

= – 3 131313777

= – 3 33 137

= – (7 × 13) = –91

(v) 3 32768 = – 3 32768

= – 3

222222

222222222

= – 3 33333 22222 = – (2 × 2 × 2 × 2 × 2) = –32

2. Show that :

(i) 3 27 × 3 64 = 3 6427

(ii) 3 72964 = 3 64 × 3 729

(iii) 3 216125 = 3 125 × 3 216

(iv) 3 1000125 = 3 125 × 3 1000

Solution—

(i) 3 27 × 3 64 = 3 6427

L.H.S. = 3 27 × 3 64

= 3 333 × 3 444

= 3 33 × 3 34 = 3 × 4 = 12

R.H.S. = 3 6427 = 3 444333

= 3 33 43 = 3 × 4 = 12

L.H.S. = R.H.S.

(ii) 3 72964 = 3 64 × 3 729

L.H.S. = 3 72964

= 3 999444

= 3 33 94 = 4 × 9 = 36

R.H.S. = 3 64 × 3 729

= 3 444 × 3 999

= 3 34 × 3 39 = 4 × 9 = 36

L.H.S. = R.H.S.

(iii) 3 216125 = 3 125 × 3 216

L.H.S. = 3 216125

= – 3 216125 = – 3 666555

= – 3 33 65 = –5 × 6 = –30

R.H.S. = 3 125 × 3 216

= – 3 125 × 3 216

2 27442 13722 6867 3437 497 7

1

7 7535717 1076537 15379

13 219713 16913 13

1

2 327682 163842 81922 40962 20482 10242 5122 2562 1282 642 322 162 82 42 2

1


122

= – 3 555 × 3 666

= – 3 35 × 3 36 = –5 × 6 = –30

L.H.S. = R.H.S.

(iv) 3 1000125 = 3 125 × 3 1000

L.H.S. = 3 1000125 = 3 1000125

= 3 101010555 = 3 33 105

= 5 × 10 = 50

R.H.S. = 3 125 × 3 1000

= 3 125 × 3 1000

= 3 555 × 3 101010

=

3 35 ×

3 310

= (–5) × (–10) = 50

L.H.S. = R.H.S.


numbers :

(i) 8 × 125 (ii) –1728 × 216

(iii) –27 × 2744 (iv) –729 × –15625

Solution—

(i) 3 1258 = 3 555222

= 3 33 52 = 2 × 5 = 10

(ii) 3 2161728 = – 3 2161728

= – 3 1728 × 3 216

= 3 333222222

× 3 333222

= – 3 333 322 × 3 33 32

= – (2 × 2 × 3) × (2 × 3)

= –12 × 6 = –72

(iii) 3 274427 = – 3 274427

= – 3 27 × 3 2744

2 27442 13722 6867 3437 497 7

1

= – 3 333 × 3 777222

= –

3 333 3 723

= – (3 × 2 × 7) = –42

(iv) 3 15625729

= 3 15625729

= 3 729 × 3 15625

3 7293 2433 813 273 93 3

1

5 156255 31255 6255 1255 255 5

1

= 3 333333 × 3 555555

= 3 33 33 × 3 33 55

= (3 × 3) × (5 × 5) = 9 × 25

= 225

4. Evaluate :

(i) 3 3364 (ii) 3 1717178

(iii) 3 5492700

(iv) 125 3 6a – 3 6125a

Solution—

(i) 3 33 64 = 4 × 6 = 24

(ii) 3 1717178 = 3 171717222

= 3 33 172 = 2 × 17 = 34


123

(iii) 3 5492700

= 3 577275522

= 3 333 752 = 2 × 5 × 7 = 70

(iv) 125 3 6a – 3 6125a

= 125 3 222 aaa

= – 3 222555 aaa

= 125 3 32a – 3 3235 a

= 125a2 – 5a2 = 120a2


rational numbers.

(i)729

125(ii)

12167

10648

(iii)24389

19683(iv)

3456

686

(v)42875

39304

Solution—

(i) 3

729

125 = 3

3

729

125

3

3

729

125 = 3

3

333333

555

= 3 33

3 3

33

5

=

33

5

= 9

5

(ii) 3

12167

10648 = 3

3

12167

10648

2 106282 53242 2662

11 133111 12111 11

1

23 1216723 52923 23

1

= 3

3

232323

111111222

= 3 3

3 33

23

112 =

23

112 =

23

22

(iii) 3

24389

19683 = 3

3

24389

19683

= 3

3

24389

19683

3 196833 65613 21873 7293 2433 813 273 93 3

1

29 2438929 84129 29

1

= 3

3

292929

333333333

= 3 3

3 333

29

333 =

29

333 =

29

27

(iv) 3

3456

686

= 3

17282

3432

= 3

1728

343

= 3

3

1728

343

= 3

3

1728

343

7 3437 497 7

1

2 17282 8642 4322 2162 1082 543 273 93 3

1


124

= 3

3

333222222

777

= 3 333

3 3

322

7

= 322

7

= 12

7

(v) 3

42875

39304

= 3

42875

39304

= 3

3

42875

39304

2 393042 196522 9826

17 491317 28917 17

1

5 428755 85755 17157 3437 497 7

1

= 3

3

777555

171717222

= 3 33

3 33

75

172

=

75

172

= 35

34


rational numbers :

(i) 0.001728 (ii) 0.003375

(iii) 0.001 (iv) 1.331

Solution—

(i) 3 001728.0 = 3

1000000

1728 = 3

3

1000000

1728

= 3

3

101010101010

333222222

= 3 33

3 333

1010

322

=

1010

322

= 100

12 = 0.12

(ii) 3 003375.0 = 3

1000000

3375 = 3

3

1000000

3375

3 33753 11253 3755 1255 255 5

1

= 3

3

101010101010

555333

= 3 33

3 33

1010

53

=

1010

53

= 100

15 = 0.15

(iii) 3 001.0 = 3

1000

1 = 3

3

1000

1

= 3

3

101010

111

= 3 310

1 =

10

1 = 0.1

(iv) 3 331.1 = 3

1000

1331 = 3

3

1000

1331

11 133111 12111 11

1

= 3

3

101010

111111

=

3 3

3 3

10

11 =

10

11 = 1.1

7. Evaluate each of the following :

(i) 3 27 + 3 0080. + 3 0640.

(ii) 3 1000 + 3 0080. – 3 1250.

(iii) 3

216

729 ×

9

6

(iv) 3

0080

0270

.

.

040

090

.

. – 1


125

(v) 3 131313101010 ...

Solution—

(i) 3 27 + 3 008.0 + 3 064.0

= 3 333 + 3

1000

8 + 3

1000

64

= 3 33 + 3

101010

222

+ 3

101010

444

= 3 + 10

2 +

10

4 = 3 + 0.2 + 0.4

= 3.6

(ii) 3 1000 + 3 008.0 – 3 125.0

= 3 101010 + 3

1000

8 – 3

1000

125

= 3 310 + 3

101010

222

– 3

101010

555

= 3 310 + 33

3

10

2 – 3

3

3

10

5

= 10 + 10

2 –

10

5 = 10 + 0.2 – 0.5

= 10.2 – 0.5 = 9.7

(iii) 3

216

729 ×

9

6 = 3

3

216

729 ×

9

6

= 3

3

666

999

×

9

6

= 3 3

3 3

6

9 ×

9

6 =

6

9 ×

9

6 = 1

(iv) 3

008.0

027.0

04.0

09.0 – 1

= 3

3

008.0

027.0

04.0

09.0 – 1

=

3

3

1000

8

1000

27

100

4

100

9

– 1

=

3

3

3

3

1000

8

1000

27

100

4

100

9

– 1

=

3

3

3

3

101010

222

101010

333

1010

22

1010

33

– 1

=

3 3

3 3

3 3

3 3

10

2

10

3

2

2

2

2

10

2

10

3

– 1

=

10

210

3

10

210

3

– 1 = 10

3 ×

2

10

10

3 ×

2

10 – 1

= 2

3

2

3 – 1

= 2

3 ×

3

2 – 1 = 1 – 1 = 0

(v) 3 1313131.01.01.0

3 33131.0 = 0.1 × 13 = 1.3

8. Show that :

(i)3

3

1000

729 = 3

1000

729

(ii)3

3

343

512 = 3

343

512


126

Solution—

(i) 3

3

1000

729 = 3

1000

729

L.H.S. = 3

3

1000

729 = 3

3

101010

999

= 3 3

3 3

10

9 =

10

9 = 0.9

R.H.S. = 3

1000

729 = 3

101010

999

= 3

10

9

10

9

10

9

= 3

3

10

9

=

10

9 = 0.9

L.H.S. = R.H.S.

(ii) 3

3

343

512 = 3

343

512

L.H.S. = 3

3

343

512 = 3

3

343

512

= 3

3

777

222222222

= 3 3

3 333

7

222 =

7

222 =

7

8

R.H.S. = 3

343

512

= 3

343

512 = 3

777

888

= 3

7

8

7

8

7

8

= 3

3

7

8

=

7

8

L.H.S. = R.H.S.


(i) 3 27125 = 3 × ____

(ii) 3 ___8 = 8

(iii) 3 1728 = 4 × ____

(iv) 3 480 = 3 3 × 2 × 3 ___

(v) 3 ___ = 3 7 × 3 8

(vi) 3 ___ = 3 4 × 3 5 × 3 6

(vii) 3

125

27 =

5

___

(viii) 3

1331

729 =

___

9

(ix) 3

___

512 =

13

8

Solution—

(i) 3 27125 = 3 333555

= 3 33 35 = 5 × 3 = 3 × 5

(ii) 3 ___8 = 8 = 3 38 = 3 888

3 888 = 8

(iii) 3 1728

2 17282 8642 4322 2162 1082 543 273 93 3

1


127

= 3 333222222

= 3 333 322

= 2 × 2 × 3 = 4 × 3

3 1728 = 4 × 3

(iv) 3 480

2 4802 2402 1202 602 303 155 5

1

= 3 5322222

= 3 3 53222

= 3 32 × 3 5322

= 2 × 3 3 × 3 20

3 480 = 3 3 × 2 × 3 20

(v) 3 ___ = 3 7 × 3 8

3 7 × 3 8 = 3 87 = 3 56

3 56 = 3 7 × 3 8

(vi) 3 ___ = 3 4 × 3 5 × 3 6

= 3 4 × 3 5 × 3 6 = 3 654 = 3 120

3 120 = 3 4 × 3 5 × 3 6

(vii) 3

125

27 =

5

___

= 3

125

27 = 3

555

333

= 3

5

3

5

3

5

3

= 3

3

5

3

=

5

3

3

125

27 =

5

3

(viii) 3

1331

729 =

___

9

= 3

1331

729 = 3

111111

999

= 33

3

11

9 =

11

9

3

1331

729 =

11

9

(ix) 3

___

512 =

13

8

13

8 =

3

13

8

= 3

131313

888

= 3

2197

512

3

2197

512 =

13

8

10. The volume of a cubical box is 474.552

cubic metres. Find the length of each side

of the box.

Solution—

Volume of cubical box = 474.552 cu m

Each side = 3 Volume

= 3 552.474

= 3

1000

474552 = 3

3

1000

474552

2 4745522 2372762 1186383 593193 197733 6591

13 219713 16913 13

1


128

= 3

3

101010

131313333222

= 3 3

3 333

10

1332 =

10

1332

= 10

78 = 7.8 m

11. Three numbers are to one another 2 : 3 :

4. The sum of their cubes is 0.334125. Find

the numbers.

Solution—

Ratio in three numbers = 2 : 3 : 4

Let first number = 2x

Second number = 3x

and third number = 4x

(2x)3 + (3x)3 + (4x)3 = 0.334125

8x2 + 27x3 + 64x3 = 1000000

334125

99x3 = 1000000

334125 x3 =

991000000

334125

x3 = 1000000

3375

3 33753 11253 3755 1255 255 5

1

= 100100100

555333

= 3

33

100

53

x = 33

33

100

53

= 100

53 =

100

15 = 0.15

First number = 2x = 2 × 0.15 = 0.30

Second number = 3x = 3 × 0.15 = 0.45

Third number = 4x = 4 × 0.15 = 0.60

Numbers are 0.3, 0.45, 0.6

12. Find side of a cube whose volume is

216

24389 m3.

Solution—

Volume of cube = 216

24389 m3

Side = 3 Volume

= 3

216

24389 = 3

3

216

24389

29 2438929 84129 29

1

6 2166 366 6

1

= 3

3

666

292929

=

3 3

3 3

6

29

= 6

29 m

13. Evaluate :

(i) 3 36 × 3 384 (ii) 3 96 × 3 144

(iii) 3 100 × 3 270 (iv) 3 121 × 3 297

Solution—

(i) 3 36 × 3 384 = 3 38436

2 362 183 93 3

1

2 3842 1922 962 482 242 122 63 3

1

= 3 322222223322


129

= 3 222222222333

= 3 3333 2223 = 3 × 2 × 2 × 2 = 24

(ii) 3 96 × 3 144 = 3 14496

2 962 482 242 122 63 3

1

2 1442 722 362 183 93 3

1

= 3 332222322222

= 3 333222222222

= 3 3333 3222

= 2 × 2 × 2 × 3 = 24

(iii) 3 100 × 3 270 = 3 270100

= 3 27000 = 3 271000

= 3 333101010 = 3 33 310

= 10 × 3 = 30

(iv) 3 121 × 3 297

= 3 297121

11 2973 273 93 3

1

= 3 333111111

= 3 33311 = 11 × 3 = 33

14. Find the cube root of the numbers :

2460375, 20346417, 210644875, 57066625

using the fact that

(i) 2460375 = 3375 × 729

(ii) 20346417 = 9261 × 2197

(iii) 210644875 = 42875 × 4913

(iv) 57066625 = 166375 × 343

Solution—

(i) 3 2460375 = 3 7293375

= 3 3375 × 3 729

3 33753 11253 3755 1255 255 5

1

3 7293 2433 813 273 93 3

1

= 3 555333 × 3 333333

= 3 33 53 × 3 33 33

= 3 × 5 × 3 × 3 = 135

(ii) 3 20346417 = 3 21979261

= 3 9261 × 3 2197

3 92613 30873 10297 3437 497 7

1

13 219713 16913 13

1

= 3 777333 × 3 131313

= 3 33 73 × 3 313

= (3 × 7) × 13 = 21 × 13 = 273

(iii) 3 210644875 = 3 491342875

= 3 42875 × 3 4913

5 428755 85755 17157 3437 497 7

1

17 491317 28917 17

1


130

= 3 777555 × 3 171717

= 3 33 75 × 3 317

= (5 × 7) × 17 = 35 × 17 = 595

(iv) 3 57066625 = 3 343166375

= 3 166375 × 3 343

5 1663755 332755 6655

11 133111 12111 11

1

7 3437 497 7

1

= 3 111111555 × 3 777

= 3 33 115 × 3 37

= (5 × 11) × 7 = 55 × 7 = 385

15. Find the units digit of the cube root of

the following numbers ?

(i) 226981 (ii) 13824

(iii) 571787 (iv) 175616

Solution—

(i) 226981

In it unit digit is 1

The units digit of its cube root will be = 1

(_ 1 × 1 × 1 = 1)

Tens digit of the cube root will be = 6

(ii) 13824

_ The units digit of 13824 = 4

(_ 4 × 4 × 4 = 64)

Units digit of the cube root of it = 4

(iii) 571787


The units digit of its cube root = 3

(_ 3 × 3 × 3 = 27)

(iv) 175616


The units digit of its cube root = 6

(_ 6 × 6 × 6 = 216)

16. Find the tens digit of the cube root of each

of the numbers in Question No. 15.

Solution—

(i) In 226981

_ Units digit is 1

Units digit of its cube root = 1

We have 226

(Leaving three digits number 981)

63 = 216 and 73 = 343

63 226 73

The ten’s digit of cube root will be 6

(ii) In 13824

Leaving three digits number 824, we have 13

_ (2)3 = 8, (3)3 = 27

23 13 33

Tens digit of cube root will be 2

(iii) In 571787

Leaving three digits number 787, we have 571

83 = 512, 93 = 729

83 571 93


(iv) In 175616

Leaving three digit number 616, we have 175

_ 53 = 125, 63 = 216

53 175 63


Making use of the cube root table, find

the cube roots of the following (correct

to three decimal places) (1 - 22)

1. 7 2. 70 3. 700

EXERCISE 4.5

4. 7000 5. 1100 6. 780

7. 7800 8. 1346 9. 250

10. 5112 11. 9800 12. 732

13. 7342 14. 133100 15. 37800


131

16. 0.27 17. 8.6 18. 0.86

19. 8.65 20. 7532 21. 833

22. 34.2

Solution—

1. 3 7 = 1.913 (From the table)

2. 3 70 = 4.121 (From the table)

3. 3 700 = 3 1007 = 8.879 (From 3 10x )

4. 3 7000 = 3 10070 = 19.13

(From 3 100x )

5. 3 1100 = 3 10011 = 10.32

(From 3 100x )

6. 3 780 = 3 1078 = 9.205

(From 3 10x )

7. 3 7800 = 3 10078 = 19.83

(From 3 100x )

8. 3 1346 = 3 6732 = 3 203.67

= 3 3.67 × 3 20

Now from the tables 3 67 = 4.062 and 3 68

= 4.082

Difference after 1, 4.082 – 4.062 = 0.020

and difference after .3 = 0.020 × .3 = 0.060

= 0.06

3 3.67 = 4.062 + 0.006

= 4.068

and 3 20 = 2.714 (from the table)

3 1346 = 3 3.67 × 3 20

= 4.067 × 2.714

= 11.040

9. 3 250 = 3 2125

= 3 2555 = 3 3 25

5 3 2 = 5 × 1.260 (from the table)

= 6.300 = 6.3

10. 3 5112 = 3 639222 = 2 3 639

= 2 3 109.63

= 2 × 3 9.63 × 3 10

Now, from the table

3 63 = 3.979 and 3 64 = 4.000

Difference after 64 – 63 = 1 = 4.000 – 3.979

= 0.021

Difference after 9 = 0.021 × 0.9 = 0.0189

3 9.63 = 3.979 + 0.0189 = 3.9979

and 3 5112 = 2 × 10 × 9.63

= 2 × 2.154 × 3.9979

= 17.2229 = 17.223

11. 3 9800 = 3 10098 = 21.40

(from x100 )

12. 3 732 = 3 102.73

= 3 2.73 × 3 10

Now 3 73 = 4.179 and 3 74 = 4.198

Difference after 74 – 73 = 1

= 4.198 – 4.179 = 0.019

and difference off 0.2 = 0.2 × 0.019 = 0.0038

3 2.73 = 4.179 + 0.0038 = 4.1828

and 3 10 = 2.154

3 732 = 3 2.73 × 3 10

= 4.1828 × 2.154

= 9.0097 = 9.01

13. 3 7342 = 3 10042.73

= 3 42.73 × 3 100

Now 3 73 = 4.179 and 3 74 = 4.198


= 0.019


132

Difference after 0.42 = 0.42 × 0.019

= 0.00798

3 42.73 = 4.179 + 0.00798 = 4.18698

and 3 100 = 4.642

and 3 7342 = 3 42.73 × 3 100

= 4.18698 × 4.642 = 19.436

14. 3 133100 = 3 1001331

= 3 100111111 = 3 3 10011

= 11 3 100 = 11 × 4.642

= 51.062

15. 3 37800 = 3 10008.37 = 3 3 8.3710

= 3 310 × 3 8.37

= 10 × 3 8.37

3 37 = 3.332 and 3 38 = 3.362


= 0.030

and difference after 0.8 = 0.030 × 0.8

= 0.0240 = 0.024

3 8.37 = 3.332 + 0.024 = 3.356

and 3 37800 = 3 8.37 × 10

= 3.356 × 10 = 33.56

16. 3 27.0 = 3

100

27 = 3

1000

270 = 3 10

1000

27

= 3 1010

3

10

3

10

3

= 10

3 3 10

= 10

3 × 2.154 = 3 × 0.2154

= 0.6462 = 0.646

17. 3 6.8

3 8 = 2.000 and 3 9 = 2.080

Difference after 1 (9 – 8) = 2.080 – 2.000

= 0.080

Difference after 0.6 = 0.6 × 0.08 = 0.048

3 6.8 = 2.000 + 0.048 = 2.048

18. 3 86.0 = 3

100

86 = 3

3

100

86

= 642.4

414.4(from the table)

= 4642

4414 = 0.95088

= 0.9509 = 0.951

19. 3 65.8

8 < 8.65 < 9

3 8 < 3 65.8 < 3 9

3 8 = 2.000 , 9 = 2.080

Difference of (9 – 8) = 1 is = 0.080

difference of 0.65 = 0.080 × 0.65

= 0.08 × 0.65 = 0.052

3 65.8 = 2.000 + 0.052 = 2.052

20. 3 7532 = 3 10032.75

= 3 32.75 × 3 100

75 < 75.32 < 76

3 75 < 3 32.75 < 3 76

3 75 = 4.217 and 3 76 = 4.236

Difference on (76 – 75) = 1 = 4.236 – 4.217

= 0.019

Difference on 0.32 = 0.019 × 0.32

= 0.00608 = 0.0061

3 32.75 = 4.217 + 0.0061 = 4.2231 = 4.223

Now 3 7532 = 3 32.75 × 3 100 = 4.223 ×

4.642 = 19.603

21. 3 833 = 3 103.83

= 3 3.83 × 3 10


133


3 3.83 lies between 3 83 and 3 84

Now 3 83 = 4.362 and 3 84 = 4.380

Difference on 84 – 83 = 1 is 4.380 – 4.362 =

0.018

Difference on 0.3 = 0.018 × 0.3 = 0.0054

3 3.83 = 4.362 + 0.0054 = 4.3674

and 3 10 = 2.154

3 833 = 3 3.83 × 3 10

= 4.3674 × 2.154 = 9.407

22. 3 2.34


or 3 2.34 lies between 3 34 and 3 35

Now 3 34 = 3.240 and 3 35 = 3.271

Difference on 35 – 34 = 1 is = 3.271 – 3.240

= 0.031

Difference on 0.2 = 0.031 × 0.2 = 0.0062

3 2.34 = 3.240 + 0.0062 = 3.2462

= 3.246

23. What is the length of the side of a cube

whose volume is 275 cm3. Make use of

the table for the cube root.

Solution—

Volume of cube = 275 cm3

Length of side = 3 Volume = 3 275

= 3 105.27

But 27.5 lies between 27 and 28

or 3 5.27 lies between 3 27 and 3 28

Now 3 27 = 3.000 and 3 28 = 3.037

Difference on 28 – 27 = 1 is 0.037

Difference on 0.5 = 0.037 × 0.5 = 0.0185

3 5.27 = 3.000 + 0.0185 = 3.0185

and 3 10 = 2.154

3 275 = 3 5.27 × 3 10

= 3.0185 × 2.154

= 6.5018 = 6.502 cm


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