# Exponents ICSE Class-7th Concise Selina Maths Solutions

Exponents ICSE Class-7th Concise Selina Maths Solutions Chapter-5. We provide step by step Solutions of Exercise / lesson-5 Exponents for ICSE Class-7 Concise Selina Mathematics. Our Solutions contain all type Questions with Exe-5 A and Exe-5 B, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7.

## Exponents ICSE Class-7th Concise Selina Maths Solutions Chapter-5

–: Select Topics :–

Exe-5 A,

Exe-5 B,

### Exercise – 5 A Exponents ICSE Class-7th Concise Selina Maths Solutions

#### Question 1 :-

Find the value of:
(i) 6²
(ii) 73
(iii) 44
(iv) 55
(v) 83
(vi) 75

Evaluate:
(i) 23 x 42
(ii) 23 x 52
(iii) 33 x 52
(iv) 22 x 33
(v) 32 x 52
(vi) 53 x 24
(vii) 3x 42
(ix) (5 x 4)2

Evaluate:

Evaluate :

#### Question 5 :-

Which is greater :
(i) 23 or 32
(ii) 25 or 52
(iii) 43 or 34
(iv) 54 or 45

#### Question 6 :-

Express each of the following in exponential form :
(i) 512
(ii) 1250
(iii) 1458
(iv) 3600
(v) 1350
(vi) 1176

#### Question 7 :-

If a = 2 and b = 3, find the value of:
(i) (a + b)2
(ii) (b – a)3
(iii) (a x b)a
(iv) (a x b)b

#### Question 8 :-

Express:
(i) 1024 as a power of 2.
(ii) 343 as a power of 7.
(iii) 729 as a power of 3.

#### Question 9 :-

If 27 x 32 = 3x x 2y; find the values of x and y.

#### Question 10 :-

If 64 x 625 = 2a x 5b; find :
(i) the values of a and b.
(ii) 2b x 5a

### Exercise – 5 B Exponents ICSE Class-7th Concise Selina Mathematics

#### Question 1 :-

Fill in the blanks:
(i) In 52 = 25, base = ……… and index = ……….
(ii) If index = 3x and base = 2y, the number = ………

(i) In 52 = 25, base = 5 and index = 2.

(ii) If index = 3x and base = 2y, the number = 2y3x.

#### Question 2 :-

Evaluate:
(i) 28 ÷ 23
(ii) 2 28
(iii) (26)0
(iv) (3o)6
(v) 83 x 8-5 x 84
(vi) 5 x 53 + 55
(vii) 54 ÷ 53 x 55
(viii) 44 ÷ 43 x 40
(ix) (35 x 47 x 58)0

#### Question 3 :-

Simplify, giving Solutions with positive index:

(xvii)

(2a3)4 (4a2)2

= (2a3)4 (22a2)2

= 24 a3×4 . 22×2 a2×2

= 24a12 . 24a4

= 24+4 a12+4

= 28a16

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × a16

= 256 a16

(xix)

#### Question 4 :-

Simplify and express the Solution in the positive exponent form :

(i)

(ii)

(iii)

(iv)

-128/2187

 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1
 3 2187 3 729 3 243 3 81 3 27 3 9 3 3 1

(v)

(vi)

#### Question 5 :-

Evaluate

(iii)

53 × 32 + (17)0 × 73

= 5 × 5 × 5 × 3 × 3 + (17)0 × 7 × 7 × 7   (∵ a0 = 1)

= 125 × 9 + 1 × 343

= 1125 + 343 = 1468

(iv)

(v)

= 1 + 4 + 8 = 13

(vi)

#### Question 6 :-

If m2 = -2 and n = 2; find the values of:
(i) m + r2 – 2mn
(ii) mn + nm
(iii) 6m-3 + 4n2
(iv) 2n3 – 3m

(iii)

6m-3 + 4n2

m = – 2, n =2

= 6(-2)-3 + 4(2)2

(iv)

2n3 – 3m

m = -2, n = 2

= 2(2)3 – 3(-2)

= 2 × (2 × 2 × 2) – 3 × (- 2)

= 16 – 3 × (-2)

= 16 + 6 = 22

— End of Exponents Solutions :–

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