Peer review of OP Malhotra Class 9 Maths Solutions Chapter 6 Indices/Exponents Chapter Test can encourage collaborative learning.

## S Chand Class 9 ICSE Maths Solutions Chapter 6 Indices/Exponents Chapter Test

Question 1.
Determine whether each equation is true or false. Change the right side of the equation to make a true equation.
(i) (2a)-3 = $$\frac{2}{a^3}$$
(ii) $$\left(\left(a^{-1}\right)^{-1}\right)^{-1}=\frac{1}{a}$$
(iii) (2 + 3)-1 = 2-1 + 3-1
(iv) If x ≠ $$\frac { 1 }{ 3 }$$, then (3x – 1)0 = (1 – 3x)0
Why is the condition x ≠ $$\frac { 1 }{ 3 }$$ given in part (iv) above?
Solution:

Which is not possible

Simplify:

Question 2.
$$\frac{\left(5 x^3 y^{-3} z\right)^{-2}}{y^4 z^{-2}}$$
Solution:

Question 3.
$$\left(\frac{8 a^3 b^{-4}}{64 a^{-9} b^2}\right)^{\frac{2}{3}}$$
Solution:

Question 4.
– $$\sqrt[4]{16 a^4 b^8}$$
Solution:
– $$\sqrt[4]{16 a^4 b^8}$$
= – $$\left(2^{4 \times \frac{1}{4}} a^{4 \times \frac{1}{4}} \cdot b^{8 \times \frac{1}{4}}\right)$$
= – (21a1b2) = – 2ab²

Question 5.
$$\left(\frac{y^{\frac{2}{3}} \cdot y^{-\frac{5}{6}}}{y^{\frac{1}{9}}}\right)^9$$
Solution:

Question 6.
$$\sqrt[3]{\sqrt{x^6}}$$
Solution:
$$\sqrt[3]{\sqrt{x^6}}=\left[\left(x^6\right)^{\frac{1}{2}}\right]^{\frac{1}{3}}=x^{6 \times \frac{1}{2} \times \frac{1}{3}}$$
= x1 = x

Question 7.
Which of the following is (are) equivalent to 16–$$\frac { 1 }{ 2 }$$?
(a) – 8
(b) $$\frac { 1 }{ 4 }$$
(c) – 4
(d) 4-1
Solution:
(b) $$\frac { 1 }{ 4 }$$
$$16^{-\frac{1}{2}}=\left(4^2\right)^{\frac{-1}{2}}=4^{2 \times\left(\frac{-1}{2}\right)}$$
= 4-1

Question 8.
Which of the following is undefined?
(a) – 25$$\frac { 1 }{ 2 }$$
(b) 25$$\frac { 1 }{ 2 }$$
(c) – 25–$$\frac { 1 }{ 2 }$$
(d) (-25)$$\frac { 1 }{ 2 }$$
Solution:
(d) (-25)$$\frac { 1 }{ 2 }$$
(- 25)$$\frac { 1 }{ 2 }$$ is not defined as square root of negative term is not possible.

Question 9.
True or False?
(a) $$\frac{a^{4 n}}{a^n}=a^4$$
(b) $$\frac{1}{a^{m-n}}=d^{n-m}$$
(c) a-n.an = 1
(d) $$\frac{a^n}{b^m}=\left(\frac{a}{b}\right)^{n-m}$$
Solution:
(a) $$\frac{a^{4 n}}{a^n}=a^{4 n-n}=a^{3 n} \neq a^4$$ False

(b) $$\frac{1}{a^{m-n}}=a^{-(m-n)}=a^{-m+n}$$
= an-m = an-m True

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

(d) $$\frac{a^n}{b^m} \neq\left(\frac{a}{b}\right)^{n-m}$$ False

Question 10.
(i) Solve : (- 4.8)k = 1
(ii) $$\sqrt[3]{\sqrt{0.000064}}$$ is equal to
(a) 0.0002
(b) 0.002
(c) 0.02
(d) 0.2
Solution:
(d) 0.2
(i) (- 4.8)k = 1
⇒ (- 4.8)k = (- 4.8)0 (∵ a0 = 1)
Comparing, we get
K = 0

(ii) $$\sqrt[3]{\sqrt{0.000064}}$$
= (0.04 x 0.04 x 0.04)$$\frac { 1 }{ 2 }$$x$$\frac { 1 }{ 2 }$$
= (0.2 x 0.2 x 0.2 x 0.2 x 0.2 x 0.2)$$\frac { 1 }{ 6 }$$
= [(0.2)6]$$\frac { 1 }{ 6 }$$
= (0.2)1
= 0.2