ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices is a useful resource for students to understand and prepare for the ICSE board examinations. These solutions help students obtain a clear conceptual knowledge of all the chapters of ML Aggarwal textbooks and also help them solve problems in an effective manner. Further, the experts at BYJU’S have provided the ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices PDF, in order to help students with their exam preparation.

The 8th chapter contains problems on simplifying the powers and exponents of numbers and algebraic expressions. The solutions can be utilised by students for any doubt clearance or quick reference during the process of self-study. The ML Aggarwal solutions are primarily created to help students with their exam preparations and boost their confidence in solving any difficult problems. All these solutions are according to the latest ICSE guidelines thus, ensuring the high possibilities of securing excellent marks in the examinations.

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices Download PDF

 

ml aggarwal solutions for class 9 maths chapter 8 01
ml aggarwal solutions for class 9 maths chapter 8 02
ml aggarwal solutions for class 9 maths chapter 8 03
ml aggarwal solutions for class 9 maths chapter 8 04
ml aggarwal solutions for class 9 maths chapter 8 05
ml aggarwal solutions for class 9 maths chapter 8 06
ml aggarwal solutions for class 9 maths chapter 8 07
ml aggarwal solutions for class 9 maths chapter 8 08
ml aggarwal solutions for class 9 maths chapter 8 09
ml aggarwal solutions for class 9 maths chapter 8 10
ml aggarwal solutions for class 9 maths chapter 8 11
ml aggarwal solutions for class 9 maths chapter 8 12
ml aggarwal solutions for class 9 maths chapter 8 13
ml aggarwal solutions for class 9 maths chapter 8 14
ml aggarwal solutions for class 9 maths chapter 8 15
ml aggarwal solutions for class 9 maths chapter 8 16
ml aggarwal solutions for class 9 maths chapter 8 17
ml aggarwal solutions for class 9 maths chapter 8 18
ml aggarwal solutions for class 9 maths chapter 8 19
ml aggarwal solutions for class 9 maths chapter 8 20
ml aggarwal solutions for class 9 maths chapter 8 21
ml aggarwal solutions for class 9 maths chapter 8 22
ml aggarwal solutions for class 9 maths chapter 8 23
ml aggarwal solutions for class 9 maths chapter 8 24
ml aggarwal solutions for class 9 maths chapter 8 25
ml aggarwal solutions for class 9 maths chapter 8 26
ml aggarwal solutions for class 9 maths chapter 8 27
ml aggarwal solutions for class 9 maths chapter 8 28
ml aggarwal solutions for class 9 maths chapter 8 29
ml aggarwal solutions for class 9 maths chapter 8 30
ml aggarwal solutions for class 9 maths chapter 8 31
ml aggarwal solutions for class 9 maths chapter 8 32
ml aggarwal solutions for class 9 maths chapter 8 33

 

Access ML Aggarwal Solutions for Class 9 Maths Chapter 8 Indices

Exercise 8

Simplify the following (1 to 20):

1. (i) (81/16)-3/4

Solution:

(81/16)-3/4

= [(34/24)]-3/4

= [(3/2)4]-3/4

= (3/2)-3/4 x 4

= (3/2)-3

= (2/3)3

= 23/33

= (2 x 2 x 2)/(3 x 3 x 3)

= 8/27

(ii) ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 1

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 2

= (5/4)3 x -2/3

= (5/4)-2

= (4/5)2

= 16/25

2. (i) (2a-3b2)3

Solution:

(2a-3b2)3

= 23 a -3×3 b 2×3

= 8a-1b6

(ii) (a-1 + b-1)/(ab)-1

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 3

3. (i) (x-1 y-1)/(x-1 + y-1)

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 4

(ii) (4 x 107) (6 x 10-5)/(8 x 1010)

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 5

4. (i) 3a/b-1 + 2b/a-1

Solution:

3a/b-1 + 2b/a-1

= 3a/(1/b) + 2b/(1/a)

= (3a x b)/1 + (2b x a)/1

= 3ab + 2ab = 5ab

(ii) 50 x 4-1 + 81/3

Solution:

50 x 4-1 + 81/3

= 1 x (1/4) + (2)3 x 1/3

= ¼ + 2

= (1 + 8)/4

= 9/4 = 2¼

5. (i) (8/125)-1/3

Solution:

(8/125)-1/3

= [(2 x 2 x 2)/(5 x 5 x 5)]-1/3

= (23/53)-1/3

= (2/5)3 x -1/3

= (2/5)-1

= 5/2 = 2½

(ii) (0.027)-1/3

Solution:

(0.027)-1/3

= (27/1000)-1/3

= [(3 x 3 x 3)/(10 x 10 x 10)]-1/3

= (33/103)-1/3

= (3/10)3 x -1/3

= (3/10)-1

= 10/3

6. (i) (-1/27)-2/3

Solution:

(-1/27)-2/3

= (-1/33)-2/3

= (-1/3)3 x -2/3

= (-1/3)-2

= (-3)2

= 9

(ii) (64)-2/3 ÷ 9-3/2

Solution:

(64)-2/3 ÷ 9-3/2

We can write it as

= (43)-2/3 ÷ (32)-3/2

By further calculation

= 4 3 ×- 2/3 ÷ 3 2 × -3/2

So we get

= 4-2 ÷ 3-3

= 4-2/ 3-3

It can be written as

= 1/42 / 1/33

= 33/42

We get

= 27/16

= 1 11/16

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 6

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 7

It can be written as

= (3)2n × (3)n

= 3 2n + n

= 33n

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 8

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 9

= 100/600

= 1/6

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 10

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 11

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 12

= (1/2)1

= ½

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 13

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 14

9. (i) (3x2)-3 × (x9)2/3

(ii) (8x4)1/3 ÷ x1/3.

Solution:

(i) (3x2)-3 × (x9)2/3

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 15

(ii) (8x4)1/3 ÷ x1/3

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 16

= 2 × x3/3

So we get

= 2 × x1

= 2 × x

= 2x

10. (i) (32)0 + 3-4 × 36 + (1/3)-2

(ii) 95/2 – 3.(5)0 – (1/81)-1/2

Solution:

(i) (32)0 + 3-4 × 36 + (1/3)-2

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 17

So we get

= 1 + 9 + 9

= 19

(ii) 95/2 – 3.(5)0 – (1/81)-1/2

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 18

Here

= 243 – 3 – (9 × 1)/1

= 240 – 9

= 231

11. (i) 163/4 + 2 (1/2)-1 (3)0

(ii) (81)3/4 – (1/32)-2/5 + (8)1/3 (1/2)-1 (2)0.

Solution:

(i) 163/4 + 2 (1/2)-1 (3)0

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 19

So we get

= (2)3 + 4

= 2 × 2 × 2 + 4

= 8 + 4

= 12

(ii) (81)3/4 – (1/32)-2/5 + (8)1/3 (1/2)-1 (2)0

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 20

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 21

= 27 – 4 + 4

= 27

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 22

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 23

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 24

= 9/4

= 2 ¼

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 25

= 19

13. (i) [(64)-2/3 2-2 + 80]-1/2

(ii) 3n × 9n + 1 ÷ (3n – 1 × 9n – 1).

Solution:

(i) [(64)-2/3 2-2 + 80]-1/2

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 26

= [4 × 1 × 1]-1/2

= (4)-1/2

Here

= (2 × 2)-1/2

= (2)2 × -1/2

= (2)-1

= 1/(2)1

= ½

(ii) 3n × 9n + 1 ÷ (3n – 1 × 9n – 1)

We can write it as

= 3n × (3 × 3)n + 1 ÷ (3n – 1 × (3 × 3)n – 1)

By further calculation

= 3n × (3)2 × (n + 1) ÷ (3n – 1 × (3)2(n-1)])

= 3n × (3)2n + 2 ÷ (3n – 1 × (3)2n – 2)

So we get

= (3)n + 2n + 2 ÷ (3)n – 1 + 2n – 2

= (3)3n + 2 ÷ (3)3n – 3

Here

= (3)3n + 2 – 3n + 3

= (3)5

We get

= 3 × 3 × 3 × 3 × 3

= 243

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 27

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 28

= 2 – 4

= – 2

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 29

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 30

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 31

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 32

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 33

= 4

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 34

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 35

= 56x – 2 – 6x

= 5-2

= 1/(5)2

= 1/25

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 36

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 37

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 38

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 39

= 7 – 7 × 7

= 7 – 49

= – 42

(ii) (27)4/3 + (32)0.8 + (0.8)-1

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 40

= 98.25

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 41

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 42

= (3)1

= 3

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 43

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 44

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 45

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 46

We can write it as

= (xm – n)l. (xn – 1)m. (x1-m)n

By further calculation

= (x)(m – n)l. (x)(n – 1)m. (x)(l – m)n

= xml – nl. xnm – lm. xln – mn

So we get

= xml – nl + nm – lm + ln – mn

= x0

= 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 47

We can write it as

= (xa + b – c)a – b. (xb + c – a)b – c. (xc + a – b)c – a

By further calculation

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

So we getML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 48

= x0

= 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 49

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 50

Solution:

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 51

= x0

= 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 52

= x0

= 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 53

20. (i) (a-1 + b-1) ÷ (a-2 – b-2)

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 54

Solution:

(i) (a-1 + b-1) ÷ (a-2 – b-2)

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 55

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 56

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 57

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 58

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 59

= 1

21. Prove the following:

(i) (a + b)-1 (a-1 + b-1) = 1/ab

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 60

Solution:

(i) (a + b)-1 (a-1 + b-1) = 1/ab

Here

LHS = (a + b)-1 (a-1 + b-1)

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 61

= RHS

Hence, proved.

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 62

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 63

= xyz

= RHS

Hence, proved.

22. If a = cz, b = ax and c = by, prove that xyz = 1.

Solution:

It is given that

a = cz, b = ax and c = by

We can write it as

a = (by)z where c = by

So we get

a = byz

Here

a = (ax)yz

a1 = axyz

By comparing both

xyz = 1

Therefore, it is proved.

23. If a = xyp – 1, b = xyq – 1 and c = xyr – 1, prove that

aq – r. br – p. cp – q = 1.

Solution:

It is given that

a = xyp – 1

Here

aq – r = (xyb – 1)q – r = xq – r. y(q – r) (p – 1)

b = xyq – 1

Here

br – p = (xyq – 1)r – p = xr – p. y(q – 1) (r – p)

c = xyr – 1

Here

cp – q = (xyr – 1)p – q = xp – q. y(r – 1) (p – q)

Consider

LHS = aq – r. br – p. cp – q

Substituting the values

= xq – r. y(q – r) (p – 1). xr – p. y(q – 1) (r – p). xp – q. y(r – 1) (p – q)

By further calculation

= xq – r + r – p – q. y(p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q)

So we get

= x0. ypq – pr – q + r + qr – pr – r + p + rp – qr – p + q

= x0. y0

= 1 × 1

= 1

= RHS

24. If 2x = 3y = 6-z, prove that 1/x + 1/y + 1/z = 0.

Solution:

Consider

2x = 3y = 6-z = k

Here

2x = k

We can write it as

2 = (k)1/x

3y = k

We can write it as

3 = (k)1/y

6-z = k

We can write it as

6 = (k)-1/z

So we get

2 × 3 = 6

(k)1/x × (k)1/y = (k)-1/z

By further calculation

(k)1/x + 1/y = (k)-1/z

We get

1/x + 1/y = – 1/z

1/x + 1/y + 1/z = 0

Therefore, it is proved.

25. If 2x = 3y = 12z, prove that x = 2yz/y – z.

Solution:

It is given that

2x = 3y = 12z

Consider

2x = 3y = 12z = k

Here

2x = k where 2 = (k)1/x

3y = k where 3 = (k)1/y

12z = k where 12 = (k)-1/z

We know that

12 = 2 × 2 × 3

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 64

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 65

Therefore, it is proved.

26. Simplify and express with positive exponents:

(3x2)0, (xy)-2, (-27a9)2/3.

Solution:

We know that

(3x2)0 = 1

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 66

27. If a = 3 and b = – 2, find the values of:

(i) aa + bb

(ii) ab + ba.

Solution:

It is given that

a = 3 and b = – 2

(i) aa + bb = (3)3 + (-2)-2

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 67

(ii) ab + ba = (3)-2 + (-2)3

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 68

28. If x = 103 × 0.0099, y = 10-2 × 110, find the value of ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 69

Solution:

It is given that

x = 103 × 0.0099, y = 10-2 × 110

We know that

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 70

= √9

= √(3 × 3)

= 3

29. Evaluate x1/2. y-1. z2/3 when x = 9, y = 2 and z = 8.

Solution:

It is given that

x = 9, y = 2 and z = 8

We know that

x1/2. y-1. z2/3 = (9)1/2. (2)-1. (8)2/3

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 71

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 72

= 6

30. If x4y2z3 = 49392, find the values of x, y and z, where x, y and z are different positive primes.

Solution:

It is given that

x4y2z3 = 49392

We can write it as

x4y2z3 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7

x4y2z3 = (2)4 (3)2 (7)3 ……. (1)

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 73

Now compare the powers of 4, 2 and 3 on both sides of equation (1)

x = 2, y = 3 and z = 7

31. If ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 74, find x and y, where a, b are different positive primes.

Solution:

It is given that

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 75

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 76

By comparing the base on both sides

2 = x

x = 2

– 4/3 = 2y

2y = – 4/3

By further calculation

y = – 4/3 × ½ = – 2/3

32. If (p + q)-1 (p-1 + q-1) = paqb, prove that a + b + 2 = 0, where p and q are different positive primes.

Solution:

It is given that

(p + q)-1 (p-1 + q-1) = paqb

We can write it as

ML Aggarwal Solutions for Class 9 Maths Chapter 8 Image 77

By cross multiplication

p-1q-1 = paqb

By comparing the powers

a = – 1 and b = – 1

Here

LHS = a + b + 2

Substituting the values

= – 1 – 1 + 2

= 0

= RHS