Ch- Test of Indices/Exponents Class 9 OP Malhotra ICSE Maths Solutions Ch-6. We Provide Step by Step Solutions / Answer of Indices/Exponents OP Malhotra Maths. Visit official Website CISCE  for detail information about ICSE Board Class-9 Mathematics.

Indices/Exponents Class 9 OP Malhotra Ch-Test ICSE Maths Solutions Ch-6

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-6	Indices/Exponents
Writer	OP Malhotra
Ch-Test	Extra Practice Questions
Edition	2026-2027

Ch-Test on Indices/Exponents

Indices/Exponents Class 9 OP Malhotra ICSE Maths Solutions Ch-6

Que-1: Determine whether each equation is true or false. Change the right side of the equation to make a true equation.
(i) (2a)‾³ = 2/a³
(ii) ((a‾¹)‾¹)‾¹ = 1/a
(iii) (2 + 3)‾¹ = 2‾¹ + 3‾¹
(iv) If x ≠ 1/3, then (3x – 1)^0 = (1 – 3x)^0
Why is the condition x ≠ 1/3 given in part (iv) above?

Sol: (i) (2a)‾³ = 2/a³
(2a)‾³ = 1/(2a)³
= 1/8a³ ≠ 2/a³  False

(ii) ((a‾¹)‾¹)‾¹ = 1/a
((a‾¹)‾¹)‾¹ = a(‾¹)(‾¹)(‾¹)
= a‾¹ = 1/a = 1/a
It is true.

(iii) (2 + 3)‾¹ = 2‾¹ + 3‾¹
(2 + 3)‾¹ = 5‾¹ = 1/5
But 2‾¹ + 3‾¹ = 1/2 + 1/3
= (3+2)/6 = 5/6
1/5 ≠ 5/6
It is false.

(iv) x ≠ 1/3
(3x – 1)^0 = (1 – 3x)^0
(3x – 1)^0 and (1 – 3x)^0 = 1   (x^0 = 1)
1 = 1
It is true.

If x = 1/3, then
(3x – 1)^0
= [3(1/3) – 1]^0
= (1-1)^0 = 0^0
Which is not possible

Simplify:

Que-2: (5x³y‾³z)‾² / (y^4)z‾²

Sol: (5x³y‾³z)‾² / (y^4)z‾²
= [(5‾²)(x^-6)(y^6)(z‾²)] / (y^4)z‾²
= (1/25) {1/(x^6)} × y^(6-4) z^(-2+2)
= (1/25) {y²/(x^6)} × (z^0)
= y² / 25x^6

Que-3: [{(8a³(b^−4)}/{(64a^−9)b²}]^(2/3)

Sol: 

Que-4: – 4√{16(a^4)(b^8)}

Sol: – 4√{16(a^4)(b^8)}
= – [2^{4×(1/4)} a^{4×(1/4)} ⋅ b{^8×(1/4)}]
= – (2¹a¹b²) = – 2ab²

Que-5: [{y^(2/3) ⋅ {y^(−5/6)} / {y^(1/9)}]^9

Sol: [{y^(2/3) ⋅ {y^(−5/6)} / {y^(1/9)}]^9
= y^[(2/3)-(5/6)-(1/9)]^9
= y^{(12-15-2)/18}^9
= y^(-15/18)×9
= y^(-5/2) = 1/y^(5/2)

Que-6: 3√(√(x^6))

Sol: 3√(√(x^6))
= [(x^6)^1/2]^13
= x^[6×(1/2)×(1/3)]
= x¹ = x

Que-7: Which of the following is (are) equivalent to 16^(–1/2) ?

(a) – 8
(b) 1/4
(c) – 4
(d) 4¯¹

Solution: (b) 1/4
16^(−1/2) = (4²)^(−1/2)
= 4^[2×(−1/2)]
= 4¯¹

Que-8: Which of the following is undefined?

(a) – 25^(1/2)
(b) 25^(1/2)
(c) – 25^(–1/2)
(d) (-25)^(1/2)

Sol: (d) (-25)^(1/2)
(- 25)^(1/2) is not defined as square root of negative term is not possible.

Que-9: True or False?

(a) (a^4n)/(a^n) = a^4
(b) 1/a^(m−n) = d^(n−m)
(c) a^-n. aⁿ = 1
(d) a^n / b^m = (a/b)^(n−m)

Sol: (a) (a^4n)/(a^n) = a^(4n-n)
= a^3n ≠ a4 False

(b) 1/a^(m−n) = a^{−(m−n)}
= a^(−m+n)
= a^(n-m) = a^(n-m) True

(c) a^-n. aⁿ = 1
a^-n. aⁿ = a ^-n+n = a⁰ = 1 = 1 True

(d) (a^n)/(b^m) ≠ (a/b)^(n−m) False

Que-10: (i)  Solve : (- 4.8)^k = 1
(ii) 3√(√0.000064) is equal to

(a) 0.0002
(b) 0.002
(c) 0.02
(d) 0.2

Sol: (d) 0.2
(i) (- 4.8)^k = 1
⇒ (- 4.8)^k = (- 4.8)⁰ (∵ a⁰ = 1)
Comparing, we get
K = 0

(ii) 3√(√0.000064)
= (0.04 x 0.04 x 0.04)^{(1/2)x(1/2)}
= (0.2 x 0.2 x 0.2 x 0.2 x 0.2 x 0.2)^(1/6)
= [(0.2)^6]^(1/6)
= (0.2)¹
= 0.2

— : End of Indices/Exponents 9 OP Malhotra Ch-Test ICSE Maths Step by step Solutions :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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