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

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

NCERT Solutions for Class 9 Maths Chapter 1 exercise 1.6 is one of the best study material for students aiming to score high in their exams. Experts from many reputed and renowned universities have contributed to the creation of NCERT solutions for Class 9 Maths Chapter 1. These solutions have a detailed explanation of all the topics given in a step by step manner. Get the NCERT solutions for Class 9 Maths Chapter 1 available as free PDF downloads to enhance your exam preparation. Download free Maths NCERT Solution Class 9 to amp up your preparations and to score well in your examinations. Subjects like Science, Maths, Engish will become easy to study if you have access to NCERT Solution Class 9 Science, Maths solutions, and solutions of other subjects that are available on Vedantu only.

### Major Concepts Covered Under NCERT Class 9 Math Chapter 1 Number Systems

Chapter 1 Number Systems of Class 9 NCERT Math is a foundational concept of mathematics. The chapter on Number Systems is also quite important from the examination point of view. Therefore, we have curated a list of all the important concepts that you’ll learn in Class 9 Math Chapter 1 Number Systems here.

 Sl. No. Important Topics 1 Rational Numbers 2 Irrational Numbers 3 Real Numbers and Their Decimal Expansions 4 Representing Real Numbers on the Number Line 5 Laws of Exponents for Real Numbers
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## Access NCERT Solutions for Class 9 Maths Chapter 1 – Number System

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.

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

 Chapter 8 Number Systems All Exercises in PDF Format Exercise 1.1 4 Question and Solutions Exercise 1.2 3 Questions and Solutions Exercise 1.3 9 Questions and Solutions Exercise 1.4 2 Question and Solutions Exercise 1.5 5 Questions and Solutions

Vedantu realizes students’ struggle in understanding academic curriculums. It thus addresses and provides required solutions for these academic hurdles that a candidate faces. Subsequently, NCERT Solutions for Class 9th Maths Chapter 1 will help you understand the curriculum better. We provide comprehensive guidance to students so that they do not have to hover from one site to another to obtain the best study guide.

Students can take help from our PDF downloads and learn answers to questions they find difficult. One can utilize the provided study material for reference purposes and have better preparation for this subject. Our offering in terms of subject-wise study material for all classes is designed to help students excel in their academic ventures. The provided PDF for the NCERT Exercise solution is absolutely free and can be easily downloaded online. Vedantu intends to make learning easy by providing students with various study materials online.

### Why Should You Choose Vedantu NCERT Solutions for Math Class 9?

Countless students have benefitted from the NCERT Solutions provided by Vedantu. But if you have any doubts regarding Vedantu’s NCERT Solutions, read these points to understand them better.

• Vedantu NCERT Solutions for Math Class 9 are curated by experienced subject matter experts.

• We provide Math NCERT solutions in a very simple way, demonstrating every problem in a stepwise manner so as to make the solutions self-explanatory.

• Make Vedantu your all-in-one resource for NCERT materials as we provide solutions to every question covered in the chapter, along with chapter review and important topics.

1. Where can I find NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6?

You can find the best NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.6 on Vedantu. Vedantu is a leading ed-tech company that provides well-curated NCERT Solutions. These are available in the free PDF to help students in exam preparation. Class 9 Maths NCERT Solutions for Chapter 1 Number Systems Exercise 1.6 and other exercises are designed by Math experts. These solutions include stepwise explanations to all the problems given in the exercise. These solutions are prepared as per the latest syllabus and NCERT guidelines. Visit Vedantu.com if you are searching for online exercise-wise NCERT Solutions for Class 9 Maths Chapter 1 Number Systems.

If you are in Class 9 CBSE Board, you must download NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.6. The solutions are helpful for exam preparation. NCERT Solutions for Class 9 Mathematics Chapter 1 are designed to assist students in understanding the chapter, doubt clearance, revision, etc. These solutions are designed by subject matter experts. It follows the latest NCERT norms and exam pattern. Students will also be provided with important tips and tricks in the free PDF material to score well in the exam. By referring to the online NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.6 or downloading the free PDF of the same, students will be able to easily learn the chapter.

3. How does Vedantu’s platform help in learning Class 9 Maths Chapter 1 Number Systems Exercise 1.6?

Vedantu is an ed-tech company. It provides various materials related to the chapter including online NCERT Solutions for Class 9 Maths Chapter 1Numner Systems. It is available for free access for students. Students can access well-designed NCERT Solutions for Class 9 Maths Chapter 1 Ex 1.6 and exercises. These are prepared by subject matter experts. These solutions are beneficial in exam preparation since these are designed as per the latest syllabus and exam pattern. Exercise-wise NCERT Solutions for Class 9 Maths Chapter 1 provide complete coverage of all important concepts and NCERT questions. Apart from NCERT Solutions, it also provides revision notes for the chapter and solved question papers for free.

4. What type of questions are included in Class 9 Maths Chapter 1 Number Systems Exercise 1.6?

Students will learn about the laws of exponents for Real Numbers in Exercise 1.6 of Class 9 math Chapter 1 Number Systems. It will help them to simplify the exponents for real numbers. They will be taught to present the large exponential values into their simplified form. The questions included in Chapter 1 Ex 1.6 of Class 10 Maths CBSE are solved by subject experts at Vedantu. These are solved in the free PDF as per the methods explained in the example questions. Students can avail the complete solutions for chapter 1 of Class 9 Maths Exercise 1.6 and other exercises. It will guide students to solve all problems related to Number Systems in the best manner.

5. What are some important questions of Exercise 1.6 of Class 9 Maths?

Important questions of Exercise 1.6 expected for exams are:

1. Find the value of (256)^0.16 × (256)^0.09
2. Find the value of x if x^(1/2) = (0.36)^0.5
3. If 5^(2x-1) – (25)^(x-1) = 2500, then find the value of x.

NCERT Solutions for Exercise 1.6 of Chapter 1 Number Systems Class 9 Maths in hindi and english medium are available for free download. We have updated our solutions as per new CBSE academic session 2021-22 based on the recent CBSE Curriculum.

6. Is Exercise 1.6 of Class 9 Maths interesting?

Yes, Exercise 1.6 is interesting and NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.6 is one of the best study materials for students aiming to score high in their exams. These solutions have a detailed explanation of all the topics given in a step by step manner.  In the solutions of NCERT Maths exercise 1.6 on Vedantu, you will find explanations for all the concepts. The NCERT solutions are available in PDF format at Vedantu.

7. What is the chapter Number System about?

The solutions  of chapter Number System are created by math professors who are experts in the area. It is written in such a way that it will assist you in gaining a basic grasp of mathematics. It becomes much easier to achieve good exam results if you have a firm grasp on the fundamentals. It gives you a glance on different types of numbers. This chapter tells us about the different types of numbers and helps us to differentiate between them.

8. What is the difference between rational and irrational numbers?

The term "rational number" refers to numbers that are either finite or repeating in nature. A rational number is an integer that can be expressed as a ratio of two natural numbers. These are numbers that are non-terminating and non-repeating by definition. A real number that cannot be expressed as a simple fraction is called an irrational number. An irrational number is one that cannot be expressed as a ratio of two natural numbers. It is impossible to express in terms of a ratio.

9. What are natural and whole numbers according to Class 9 Maths?

Natural Numbers - A natural number is every number that comes after one in the counting system. It is any number that is used for counting purposes. Numbers that begin with one and have no fractional components are known as natural numbers. It is usually denoted by the letter N.

Whole Numbers - The set of whole numbers is the union of all natural numbers, including zero. This is the system of natural numbers, which includes the number zero, also known as the entire number system. It is usually denoted by the letter W.