NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5 - Number Systems

Q1 (i) Find : $64^{\frac{1}{2}}$

Solution:
Given number is $64^{\frac{1}{2}}$

Now, on simplifying it, we will get

$\Rightarrow 64^{\frac{1}{2}} = (8^2)^\frac{1}{2} = 8$

Therefore, the answer is 8.

Q1 (ii) Find : $32^{\frac{1}{5}}$

Solution:
Given number is $32^{\frac{1}{5}}$

Now, on simplifying it, we will get

$\Rightarrow 32^{\frac{1}{5}} = (2^5)^\frac{1}{5} = 2$

Therefore, the answer is 2.

Q1 (iii) Find : $125^{\frac{1}{3}}$

Solution:
Given number is $125^{\frac{1}{3}}$

Now, on simplifying it, we will get

$\Rightarrow 125^{\frac{1}{3}} = (5^3)^\frac{1}{3} = 5$

Therefore, the answer is 5.

Q2 (i) Find : $9^{\frac{3}{2}}$

Solution:

Given number is $9^{\frac{3}{2}}$

Now, on simplifying it, we will get

$\Rightarrow 9^{\frac{3}{2}} = (3^2)^\frac{3}{2} = 3^3 = 27$

Therefore, the answer is 27.

Q2 (ii) Find : $32^{\frac{2}{5}}$

Solution:
Given number is $32^{\frac{2}{5}}$

Now, on simplifying it, we will get

$\Rightarrow 32^{\frac{2}{5}} = (2^5)^\frac{2}{5} = 2^2 = 4$

Therefore, the answer is 4.

Q2 (iii) Find : $16^{\frac{3}{4}}$

Solution:
Given number is $16^{\frac{3}{4}}$

Now, on simplifying it, we will get

$\Rightarrow 16^{\frac{3}{4}} = (2^4)^\frac{3}{4} = 2^3 = 8$

Therefore, the answer is 8.

Q2 (iv) Find : $125^{\frac{-1}{3}}$

Solution:
Given number is $125^{\frac{-1}{3}}$

Now, on simplifying it, we will get

$\Rightarrow 125^{\frac{-1}{3}} = (5^3)^\frac{-1}{3} = 5^{-1} = \frac{1}{5}$

Therefore, the answer is $\frac{1}{5}$ 15

Q3 (i) Simplify : $2^{\frac{2}{3}}.2^{\frac{1}{5}}$

Solution:
Given number is $2^{\frac{2}{3}}.2^{\frac{1}{5}}$

Now, on simplifying it, we will get

$\Rightarrow 2^{\frac{2}{3}}.2^{\frac{1}{5}} = 2^{\frac{2}{3}+\frac{1}{5}} = 2^{\frac{10+3}{15}} = 2^\frac{13}{15}$ $\left ( \because a^m.a^n = a^{m+n} \right )$

Therefore, the answer is $2^{\frac{13}{15}}$

Q3 (ii) Simplify : $\left (\frac{1}{3^{3}} \right )^{7}$

Solution:
Given number is $\left (\frac{1}{3^{3}} \right )^{7}$

Now, on simplifying it, we will get

$\Rightarrow \left ( \frac{1}{3^3} \right )^7= \frac{1^7}{3^{3\times7}} = \frac{1}{3^{21}} = 3^{-21}$ $\left ( \because (a^m)^n = a^{m.n} \ and \ \frac{1}{a^m}= a^{-m}\right )$

Therefore, the answer is $3^{-21}$

Q3 (iii) Simplify : $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

Solution:
Given number is $\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}$

Now, on simplifying it, we will get

$\Rightarrow \frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}} = 11^{\frac{1}{2}-\frac{1}{4}}= 11^{\frac{2-1}{4}} = 11^\frac{1}{4}$ $\left ( \because \frac{a^m}{a^n}= a^{m-n}\right )$

Therefore, the answer is $11^{\frac{1}{4}}$

Q3 (iv) Simplify : $7^{\frac{1}{2}}.8^{\frac{1}{2}}$

Solution:
Given number is $7^{\frac{1}{2}}.8^{\frac{1}{2}}$

Now, on simplifying it we will get
$\Rightarrow 7^{\frac{1}{2}}.8^{\frac{1}{2}}= (7\times8)^{\frac{1}{2}} = 56^{\frac{1}{2}}$ $\left ( \because a^m.b^m=(a.b)^m\right )$

Therefore, the answer is $56^{\frac{1}{2}}$