NCERT Solutions for Class 9 Maths Chapter 1 Number Systems (Ex 1.6) Exercise 1.6

Exercise (1.6)

1. Compute the value of each of the following expressions:

(i) $ {6}{{ {4}}^{\dfrac{ {1}}{ {2}}}}$

Ans: The given number is \[{{64}^{\dfrac{1}{2}}}\].

By the laws of indices,

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$, where$a>0$.

Therefore,

$64^{\frac{1}{2}}=\sqrt[2]{64}$

$=\sqrt[2]{8\times 8}$

$=8$

Hence, the value of ${{64}^{\dfrac{1}{2}}}$ is $8$.

(ii) $ {3}{{ {2}}^{\dfrac{ {1}}{ {5}}}}$

Ans: The given number is ${{32}^{\dfrac{1}{5}}}$.

By the laws of indices,

${{a}^{\dfrac{m}{n}}}=\sqrt[m]{{{a}^{m}}}$, where $a>0$

$32^{\frac{1}{5}}=\sqrt[5]{32}$

$=\sqrt[5]{2\times 2\times 2\times 2\times 2}$

$=\sqrt[5]{2^{5}}$

$=2$

Alternative Method:

By the law of indices ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, then it gives

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

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

$=2^{\frac{5}{5}}$

$=2$

Hence, the value of the expression ${{32}^{\dfrac{1}{5}}}$ is $2$.

(iii) $ {12}{{ {5}}^{\dfrac{ {1}}{ {5}}}}$

Ans: The given number is ${{125}^{\dfrac{1}{3}}}$.

By the laws of indices

  ${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where$a>0$.

Therefore,

$125^{\frac{1}{3}}=\sqrt[3]{125}$

$\sqrt[3]{5\times 5\times 5}$

$=5$

Hence, the value of the expression ${{125}^{\dfrac{1}{3}}}$ is $5$.

2. Compute the value of each of the following expressions:

(i) ${{ {9}}^{\dfrac{ {3}}{ {2}}}}$

Ans: The given number is ${{9}^{\dfrac{3}{2}}}$.

By the laws of indices,

 ${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where$a>0$.

Therefore,

$9^{\frac{3}{2}}=\sqrt[2]{(9)^{3}}$

$=\sqrt[2]{9\times 9\times 9}$

$=\sqrt[2]{3\times 3\times 3\times 3\times 3\times 3}$

$=3\times 3\times 3$

$=27$

Alternative Method:

By the laws of indices, ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$, then it gives

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

$=(3^{2})^{\frac{3}{2}}$

$=3^{2\times \frac{3}{2}}$

$=3^{3}$

That is,

${{9}^{\dfrac{3}{2}}}=27$.

Hence, the value of the expression ${{9}^{\dfrac{3}{2}}}$ is $27$.

(ii) $ {3}{{ {2}}^{\dfrac{ {2}}{ {5}}}}$

Ans: We know that ${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$ where $a>0$.

We conclude that ${{32}^{\dfrac{2}{5}}}$ can also be written as

$\sqrt[5]{(32)^{2}}=\sqrt[5]{(2\times 2\times 2\times 2\times 2)\times (2\times 2\times 2\times 2\times 2)}$

$=2\times 2$

$=4$

Therefore, the value of ${{32}^{\dfrac{2}{5}}}$ is $4$.

(iii) $ {1}{{ {6}}^{\dfrac{ {3}}{ {4}}}}$

Ans: The given number is ${{16}^{\dfrac{3}{4}}}$.

By the laws of indices, 

${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$, where $a>0$.

Therefore,

$16^{\frac{3}{4}}=\sqrt[4]{(16)^{3}}$

$=\sqrt[4]{(2\times 2\times 2\times 2)\times (2\times 2\times 2\times 2)\times (2\times 2\times 2\times 2)}$

$=2\times 2\times 2$

$=8$

Hence, the value of the expression ${{16}^{\dfrac{3}{4}}}$ is $8$.

Alternative Method:

By the laws of indices,

${{({{a}^{m}})}^{n}}={{a}^{mn}}$, where $a>0$.

Therefore,

$16^{\frac{3}{4}}=(4\times 4)^{\frac{3}{4}}$

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

$=(4)^{2\times \frac{3}{4}}$

$=(2^{2})^{2\times \frac{3}{4}}$

$=2^{2\times 2\times \frac{3}{4}}$

$=2^{3}$

$=8$

Hence, the value of the expression is ${{16}^{\dfrac{3}{4}}}=8$.

(iv) $ {12}{{ {5}}^{ {-}\dfrac{ {1}}{ {3}}}}$

Ans: The given number is ${{125}^{-\dfrac{1}{3}}}$.

By the laws of indices, it is known that 

${{a}^{-n}}=\dfrac{1}{{{a}^{^{n}}}}$, where $a>0$.

Therefore, 

$125^{-\frac{1}{3}}=\frac{1}{125^{\frac{1}{3}}}$

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

$\sqrt[3]{(\frac{1}{125})}$

$\sqrt[3]{(\frac{1}{5}\times \frac{1}{5}\times \frac{1}{5})}$

$=\frac{1}{5}$

Hence, the value of the expression ${{125}^{-\dfrac{1}{3}}}$ is  $\dfrac{1}{5}$.

3. Simplify and evaluate each of the expressions:

(i)${{ {2}}^{\dfrac{ {2}}{ {3}}}} {.}{{ {2}}^{\dfrac{ {1}}{ {5}}}}$

Ans: The given expression is ${{2}^{\dfrac{2}{3}}}{{.2}^{\dfrac{1}{5}}}$.

By the laws of indices, it is known that

${{a}^{m}}\cdot {{a}^{n}}={{a}^{m+n}}$, where $a>0$.

Therefore,

$2^{\frac{2}{3}}.2^{\frac{1}{5}}=(2)^{\frac{2}{3} +\frac{1}{5}}$

$=(2)^{\frac{10+3}{15}}$

$=2^{\frac{13}{15}}$

Hence, the value of the expression ${{2}^{\dfrac{2}{3}}}{{.2}^{\dfrac{1}{5}}}$ is ${{2}^{\dfrac{13}{15}}}$.

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

Ans: The given expression is ${{\left( {{3}^{\frac{1}{3}}} \right)}^{7}}$.

It is known by the laws of indices that,

 ${{({{a}^{m}})}^{n}}={{a}^{mn}}$, where $a>0$.

Therefore,

${{\left( {{3}^{\frac{1}{3}}} \right)}^{7}}={{3}^{\frac{7}{3}}}.$

Hence, the value of the expression ${{\left( {{3}^{\frac{1}{3}}} \right)}^{7}}$is  ${{3}^{\frac{7}{3}}}$.

(iii) $\dfrac{ {1}{{ {1}}^{\dfrac{ {1}}{ {2}}}}}{ {1}{{ {1}}^{\dfrac{ {1}}{ {4}}}}}$

Ans: The given number is $\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}$.

It is known by the Laws of Indices that

 $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, where $a>0$.

Therefore,

$\frac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}=11^{\frac{1}{2}-\frac{1}{4}}$

$=11^{\frac{2-1}{4}}$

$=11^{\frac{1}{4}}$

Hence, the value of the expression $\dfrac{{{11}^{\dfrac{1}{2}}}}{{{11}^{\dfrac{1}{4}}}}$ is  ${{11}^{\dfrac{1}{4}}}$.

(iv) ${{ {7}}^{\dfrac{ {1}}{ {2}}}} {.}{{ {8}}^{\dfrac{ {1}}{ {2}}}}$

Ans: The given expression is ${{7}^{\dfrac{1}{2}}}\cdot {{8}^{\dfrac{1}{2}}}$.

It is known by the Laws of Indices that

${{a}^{m}}\cdot {{b}^{m}}={{(a\cdot b)}^{m}}$, where $a>0$.

Therefore,

$7^{\frac{1}{2}}.8^{\frac{1}{2}}=(7\times 8)^{\frac{1}{2}}$

$=56^{\frac{1}{2}}$

Hence, the value of the expression ${{7}^{\dfrac{1}{2}}}\cdot {{8}^{\dfrac{1}{2}}}$ is ${{(56)}^{\dfrac{1}{2}}}$.

NCERT Solutions for Class 9 Maths Chapter 1 Number System

The Number System Solutions are prepared by expert teachers in the field of Maths. It is prepared in such a way that it helps you gain a fundamental understanding of Maths. When you understand the basics, it becomes easy to score good marks in the exam. The Class 9 NCERT Maths are going to play a crucial part in shaping your understanding of further chapters in Maths.

You get a brief glance into the different Number Systems explained as follows: 

• Natural Numbers- any number that is used for counting purpose, starting from one, is considered a natural number.

• Whole Numbers- the union set of all Natural numbers with including zero is the set of whole numbers.

• Integers- the set of all the whole numbers including their negative terms is called as the set of integers.

• Rational Numbers- any number that can be written as a ratio of two natural numbers is called a rational number.

• Irrational Numbers- any number that cannot be written as a ratio of two natural numbers is called an irrational number.
  
  

Maths NCERT Solutions Class 9 Chapter 1 Exercise 1.6

NCERT Maths Class 9 Chapter 1 solutions contain around three problems with various sub-problems. These problems are aimed at testing your understanding of Number System. The Class 9 Maths Chapter 1 solutions by Vedantu explains in detail the answers to all the problems given in the exercise. These elaborate explanations of the chapter ensure better understanding of the chapter and help you to gain more marks.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number Systems

• Chapter 2 - Polynomials

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introduction to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelograms and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Heron's Formula

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Access Other Exercise Solutions of Class 9 Maths Chapter 1 - Number Systems

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