NCERT Solutions Class 7 Maths Chapter 9

NCERT Solutions for Class 7 Mathematics Chapter 9 Rational Numbers

The NCERT Solutions Class 7 Mathematics Chapter 9 of Extramarks provides detailed answers to the exercises of Rational Numbers Class 7. These solutions will give students an edge by assisting them in preparing for their exams. The best approach to each question in NCERT Class 7th Mathematics Chapter 9 is discussed in detail by the subject matter experts. Students can go through the NCERT Solutions to strengthen their mathematical concepts and learn more about the subject.

NCERT Solutions for Class 7 Mathematics Chapter 9 Rational Numbers -

(include NCERT Solutions for Class 7 Mathematics Chapter 9 Rational Numbers)

Access NCERT Solutions for Class 7 Mathematics Chapter 9 – Rational Numbers

NCERT Solutions for Class 7 Maths Chapter 9 –

Chapter 9 of Class 7 Mathematics NCERT covers the very important topic of Rational Numbers. The following topics are covered in this chapter:

• Introduction of the chapter
• Need for Rational Numbers
• What are Rational Numbers?
• Positive and Negative Rational Numbers
• Rational Numbers on a Number Line
• Rational Numbers in Standard Form
• Comparison of Rational Numbers
• Rational Numbers Between Two Rational Numbers
• Operations on Rational Numbers (including Addition, Subtraction, Multiplication and Division of Rational Numbers)

Chapter 9 – Rational Numbers

9.1 Introduction

You must be familiar with types of numbers like natural numbers, whole numbers and integers. Natural numbers start with 1 and go up to infinity. If you add 0 to the natural numbers, the numbers obtained are whole numbers. If you add negative numbers to the above set, it will result in a set of integers. Similarly, you must have covered fractions in your previous classes; Fractions are numbers in the form of a numerator/denominator.

This chapter is an extension of the number system which introduces the concept of rational numbers.

9.2 Need for Rational Numbers

Rational numbers are a special set of numbers that can be written in a fractional form. In other words, they can be written in the way of a ratio of p and q where both of them are integers and q is not equal to zero.

These rational numbers are often used daily while solving problems as there are many measures of quantities which cannot be adequately described by integers or natural numbers alone. Thus, there is a need for rational numbers to represent such numbers.

9.3 What are Rational Numbers?

As mentioned earlier, a rational number is a number that can be expressed as a ratio of p/q, where both p and q are integers and q is not equal to zero (the numerator and the denominator of a rational number being expressed as p and q). It should be noted that 0 can also be a rational number as a rational number can be represented as 0/5 or 0/100 etc.

9.4 Positive and Negative Rational Numbers

If both the numerator and the denominator of a rational number have the same sign, it is called a positive rational number. They are always greater than 0.

A negative rational number has either the numerator or the denominator with an opposite sign. They are all less than zero and can be in the form of (-p/q) or (p/-q).

9.5 Rational Numbers on a Number Line

You must be familiar with the concept of representing numbers on a number line. The points that lie to the right side of zero are the positive integers and have a plus (+) sign and the points that lie to the left side of zero are the negative integers that have a minus (–) sign.

Speaking in terms of rational numbers, when you write a natural number on the number line, they can be thought of as rational numbers with a denominator 1.

9.6 Rational Numbers in Standard Form

A rational number is in its standard form only if it has a positive integer in its denominator and there are no other common factors between its numerator and its denominator (instead of 1).

For example, 1/4 is a rational number and the only common factor between the divisor and the dividend is 1. In other words, this implies that 1/4 is in its standard form.

When it comes to a rational number in its standard form, it can also have a negative number, however, it can only be in its numerator. If there is a rational number that is not in its standard form, it can be brought to its standard form by reducing the fraction.

9.7 Comparison of Rational Numbers

In this section of NCERT Solutions Class 7 Mathematics Chapter 9, you will learn how to compare two given rational numbers. The following steps are to be followed for such comparison:

• Write the two rational numbers to be compared in such a way that their denominators are positive.
• Find the LCM of the positive denominators and express each one of the rational numbers with the LCM as the common denominator.
• Compare the two rational numbers obtained by comparing their numerators. The rational number with a greater numerator is the higher rational number.
• If you need to compare two negative rational numbers, do so by ignoring their negative sign and in the end reverse the comparison.
• If you need to compare a negative rational number with a positive one, it is obvious that the negative one will be lower than the positive one.

9.8 Rational Numbers Between Two Rational Numbers

To insert rational numbers between any two rational numbers, you need to go back to the concepts of integers and their properties on various operations. If there are two non-consecutive integers (x, y), there can be (x - y - 1) integers between them. However, there can be no integers in between two consecutive integers. As per another general rule, there can be infinite rational numbers between any two non-consecutive rational numbers. This property is called dense property.

If there are two rational numbers, say ‘a’ and ‘b’, and they do not have the same denominator, you need to convert the denominator of the fractions using the LCM method to the same denominator. If there is a number that can be written between them, the rational numbers obtained by this method will be the rational number between them.

If there can be no numbers between the numerators, the numerator and the denominator should be multiplied by 10. This will let you know the rational numbers between them. You can continue multiplying it with multiples of 10 like 100, 1000 etc to get more rational numbers between them.

9.9 Operations on Rational Numbers

This section explains how one can add, subtract, multiply and divide any two given rational numbers or more.

If there are two rational numbers and their denominators are the same, the addition operation between them is computed by adding their numerators and keeping the denominator the same.

The following steps need to be followed to add any two rational numbers with different denominators:

• If the denominator of the rational numbers are different and not positive, they need to be adjusted to have positive denominators.
• Find the LCM of the denominator of such rational numbers.
• Convert the rational numbers so that they have the same denominator. This is done by multiplying the numerator and the denominator with a common multiple.
• The last step is to add the two numbers (of the converted rational numbers) and simplify if possible to obtain the final answer.

9.9.2 Subtraction

If there are two fractions of rational numbers, say p/q and c/d and you subtract c/d from p/q, it will amount equally to adding the additive inverse of c/d to p/q. Thus, when c/d needs to be subtracted from p/q, the subtraction is written as (p/q) + (-c/d). Once the rational numbers are written in this form, the same method of addition is applied to obtain the result.

9.9.3 Multiplication

This section explains how to multiply any two given rational numbers. For example, if you need to multiply a/b and c/d, all you need to do is multiply the numerators and the denominators of these two rational numbers to get the result.

9.9.4 Division

The division operation of rational numbers is the inverse of multiplication. So the same concept of multiplication is applied to dividing rational numbers. If there are two rational numbers, say w/a and p/z and p/z does not equal 0, w/a ÷ p/z = w/a × z/p.

Key Features of the NCERT solution for Class 7 Mathematics Chapter 9

The NCERT Solutions of Rational Numbers Class 7 need to be practised and solved thoroughly to be understood completely by the students. Let’s look at the reasons why students should refer to NCERT Solutions for Class 7 Mathematics Chapter:

• It will help you build a  strong foundation as it provides enough exercises from the chapter for regular practice.
• The solutions are reliable as all the exercises are solved by the subject matter experts.
•  Guidance and practice tips provided in our solutions will step up your level of preparation. You can even take the test to assess your preparation ahead of the exam. This will enhance your confidence level and it will definitely boost your score.

NCERT Solutions for Class 7 Mathematics

You can refer to  NCERT Solutions for Class 7 Mathematics from Extramarks and practise the questions given in the NCERT textbook from the link given below.

NCERT Solutions for Class 7

In case you found the NCERT Solutions Class 7 Mathematics Chapter 9 useful, you can click on the link below to find solutions for all chapters of other subjects.

Q.1

$\begin{array}{l}\mathrm{List}\text{ }\mathrm{five}\text{ }\mathrm{rational}\text{ }\mathrm{numbers}\text{ }\mathrm{between}:\\ \left(\mathrm{i}\right)\text{ }-1\text{ }\mathrm{and}\text{ }0\text{ }\left(\mathrm{ii}\right)\text{ }-2\mathrm{and}-1\\ \left(\mathrm{iii}\right)\text{ }\frac{-4}{5}\text{ }\mathrm{and}\text{ }\frac{-2}{3}\text{ }\left(\mathrm{iv}\right)\text{ }–\frac{1}{2}\text{ }\mathrm{and}\text{ }\frac{2}{3}\end{array}$

Ans

$\begin{array}{l}\text{(i) Five rational numbers between}-\text{1 and 0 are:}\\ \frac{-2}{3},\frac{-1}{2},\frac{-2}{5},\frac{-1}{3},\frac{-2}{7}\\ \text{(ii)}-\text{2 and}-\text{1}\\ -2=\frac{-12}{6}\text{\hspace{0.17em}}\text{and}\text{\hspace{0.17em}}\text{-1=}\frac{-6}{6}\\ \text{Five rational numbers between}-\text{2 and}-\text{1 are:}\\ \frac{-11}{6},\frac{-10}{6},\frac{-9}{6},\frac{-8}{6},\frac{-7}{6}\end{array}$ $\begin{array}{l}\left(\text{iii)}\frac{-4}{5}\text{and}\frac{-2}{3}\\ \frac{-4}{5}=\frac{-4×9}{5×9}=\frac{-36}{45}\\ \frac{-2}{3}=\frac{-2×15}{3×15}=\frac{-30}{45}\\ \text{Five rational numbers between}\frac{-4}{5}\text{and}\frac{-2}{3}\text{are:}\\ \frac{-35}{45},\frac{-34}{45},\frac{-33}{45},\frac{-32}{45},\frac{-31}{45}\\ \text{(iv) –}\frac{\text{1}}{2}\text{and}\frac{2}{3}\\ –\frac{1}{2}=–\frac{1×18}{2×18}=–\frac{18}{36}\\ \frac{2}{3}=\frac{2×12}{3×12}=\frac{24}{36}\\ \text{Five rational numbers between –}\frac{\text{1}}{2}\text{and}\frac{2}{3}\text{are:}\\ \frac{19}{36},\frac{20}{36},\frac{21}{36},\frac{22}{36},\frac{23}{36}\end{array}$

Q.2

$\begin{array}{l}\text{Write four more rational numbers in each of the following}\\ \text{patterns:}\\ \left(\text{i}\right)\text{ }\frac{-\text{3}}{\text{5}}\text{,}\frac{-\text{6}}{\text{10}}\text{,}\frac{-\text{9}}{\text{15}}\text{,}\frac{-\text{12}}{\text{20}}\text{,… }\left(\text{ii}\right)\text{ }\frac{-\text{1}}{\text{4}}\text{,}\frac{-\text{2}}{\text{8}}\text{,}\frac{-\text{3}}{\text{12}}\text{,…}\\ \left(\text{iii}\right)\text{ }\frac{-\text{1}}{\text{6}}\text{,}\frac{\text{2}}{-\text{12}}\text{,}\frac{\text{3}}{-\text{18}}\text{,}\frac{\text{4}}{-\text{24}}\text{,… }\left(\text{iv}\right)\text{ }\frac{-\text{2}}{\text{3}}\text{,}\frac{\text{2}}{-\text{3}}\text{,}\frac{\text{4}}{-\text{6}}\text{,}\frac{\text{6}}{-\text{9}}\text{,…}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20},\dots \\ \frac{-3}{5},\frac{-3×2}{5×2},\frac{-3×3}{5×3},\frac{-3×4}{5×4},\dots \\ \text{Clearly, the numerator is a multiple of 3 and denominator is}\\ \text{a multiple of 5}\text{.}\\ \text{Therefore, next four rational numbers in this pattern are}\\ \frac{-3×5}{5×5},\frac{-3×6}{5×6},\frac{-3×7}{5×7},\frac{-3×8}{5×8},\dots \\ \frac{-15}{25},\frac{-18}{30},\frac{-21}{35},\frac{-24}{40},\dots \\ \\ \text{(ii)}\frac{-1}{4},\frac{-2}{8},\frac{-3}{12},\dots \\ \frac{-1}{4},\frac{-1×2}{4×2},\frac{-1×3}{4×3},\dots \\ \text{The next four rational number in this pattern are}\\ \frac{-1×4}{4×4},\frac{-1×5}{4×5},\frac{-1×6}{4×6},\frac{-1×7}{4×7},\dots \end{array}$ $\begin{array}{l}\frac{-4}{16},\frac{-5}{20},\frac{-6}{24},\frac{-7}{28},\dots \\ \\ \text{(iii)}\frac{-1}{6},\frac{2}{-12},\frac{3}{-18},\frac{4}{-24},\dots \\ \frac{-1}{6},\frac{1×2}{-6×2},\frac{1×3}{-6×3},\frac{1×4}{-6×4},\dots \\ \text{The next four rational number in this pattern are}\\ \frac{1×5}{-6×5},\frac{1×6}{-6×6},\frac{1×7}{-6×7},\frac{1×8}{-6×8},\dots \\ \frac{5}{-30},\frac{6}{-36},\frac{7}{-42},\frac{8}{-48},\dots \\ \\ \text{(iv)}\frac{-2}{3},\frac{2}{-3},\frac{4}{-6},\frac{6}{-9}\dots \\ =\frac{-2}{3},\frac{2}{-3},\frac{2×2}{-3×2},\frac{2×3}{-3×3}\dots \\ \text{The next four rational number in this pattern are}\\ \frac{2×4}{-3×4},\frac{2×5}{-3×5},\frac{2×6}{-3×6},\frac{2×7}{-3×7},\dots \\ \frac{8}{-12},\frac{10}{-15},\frac{12}{-18},\frac{14}{-21},\dots \end{array}$

Q.3

$\begin{array}{l}\mathrm{Give}\text{ }\mathrm{four}\text{ }\mathrm{rational}\text{ }\mathrm{numbers}\text{ }\mathrm{equivalent}\text{ }\mathrm{to}:\\ \left(\mathrm{i}\right)\text{ \hspace{0.17em}}\frac{-2}{7}\text{ }\left(\mathrm{ii}\right)\text{ }\frac{5}{-3}\text{ }\left(\mathrm{iii}\right)\text{ \hspace{0.17em}}\frac{4}{9}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\frac{-2}{7}\\ \text{Four rational numbers are}\\ \frac{-2×2}{7×2},\frac{-2×3}{7×3},\frac{-2×4}{7×4},\frac{-2×5}{7×5}\\ =\frac{-4}{14},\frac{-6}{21},\frac{-8}{28},\frac{-10}{35}\\ \text{(ii)}\frac{5}{-3}\\ \text{Four rational numbers are}\\ \frac{5×2}{-3×2},\frac{5×3}{-3×3},\frac{5×4}{-3×4},\frac{5×5}{-3×5}\\ =\frac{10}{-6},\frac{15}{-9},\frac{20}{-12},\frac{25}{-15}\\ \text{(iii)}\text{\hspace{0.17em}}\frac{4}{9}\\ \text{Four rational numbers are}\\ \frac{4×2}{9×2},\frac{4×3}{9×3},\frac{4×4}{9×4},\frac{4×5}{9×5}\\ =\frac{8}{18},\frac{12}{27},\frac{16}{36},\frac{20}{45}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Draw}\text{ }\mathrm{the}\text{ }\mathrm{number}\text{ }\mathrm{line}\text{ }\mathrm{and}\text{ }\mathrm{represent}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{rational}\\ \mathrm{numbers}\text{ }\mathrm{on}\text{ }\mathrm{it}:\\ \left(\mathrm{i}\right)\frac{3}{4}\left(\mathrm{ii}\right)\frac{-5}{8}\left(\mathrm{iii}\right)\frac{-7}{4}\left(\mathrm{iv}\right)\frac{7}{8}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{3}{4}\\ \text{This fraction represent 3 parts out of 4 equal parts}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 4 equal parts}\text{.}\\ \text{It can be represented on the number line as:}\end{array}$

$\begin{array}{l}\text{(ii)}\frac{-5}{8}\\ \text{This fraction represent 5 parts out of 8 equal parts}\text{.}\\ \text{Negative sign represents that it is on the negative side of the}\\ \text{number line}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 8 equal parts}\text{.}\\ \text{It can be represented on the number line as:}\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{(iii)}\frac{-7}{4}=-1\frac{3}{4}\\ \text{This mixed fraction represent 1 full part and 3 parts out of}\\ \text{4 equal parts}\text{. Negative sign represents that it is on the}\\ \text{negative side of the number line}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 4 equal parts}.\\ \text{It can be represented on the number line as:}\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{(iv)}\frac{7}{8}\\ \text{This fraction represent 7 parts out of 8 equal parts}\text{.}\\ \text{Therefore each space between two integers on number}\\ \text{line must be divided into 8 equal parts}\text{.}\\ \text{It can be represented on the number line as:}\end{array}$

Q.5

$\begin{array}{l}\mathrm{The}\text{ }\mathrm{points}\text{ }\mathrm{P},\mathrm{Q},\mathrm{R},\mathrm{S},\mathrm{T},\mathrm{U},\mathrm{A}\text{ }\mathrm{and}\text{ }\mathrm{B}\text{ }\mathrm{on}\text{ }\mathrm{the}\text{ }\mathrm{number}\text{ }\mathrm{line}\text{ }\mathrm{are}\\ \mathrm{such}\text{ }\mathrm{that},\text{ }\mathrm{TR}=\mathrm{RS}=\mathrm{S}\text{ }\mathrm{U}\text{ }\mathrm{and}\text{ }\mathrm{AP}=\mathrm{PQ}=\mathrm{QB}.\text{ }\mathrm{Name}\text{ }\mathrm{the}\text{ }\mathrm{rational}\\ \mathrm{numbers}\text{ }\mathrm{represented}\text{ }\mathrm{by}\text{ }\mathrm{P},\mathrm{Q},\mathrm{R}\text{ }\mathrm{and}\text{ }\mathrm{S}.\end{array}$

Ans

$\begin{array}{l}\text{Distance between U and T is 1 unit.}\\ \text{It is divided into 3 equal parts.}\\ \text{TR}=\text{RS}=\text{SU=}\frac{1}{3}\\ \mathrm{R}=-1-\frac{1}{3}=\frac{-3}{3}-\frac{1}{3}=\frac{-4}{3}\\ \mathrm{S}=-1-\frac{2}{3}=\frac{-3}{3}-\frac{2}{3}=\frac{-5}{3}\\ \text{Similarly,}\\ \text{AB}=\text{1unit}\\ \text{It is divided into 3 equal parts.}\\ \text{P}=\text{2+}\frac{1}{3}=\frac{6}{3}+\frac{1}{3}=\frac{7}{3}\\ \text{Q}=\text{2+}\frac{2}{3}=\frac{6}{3}+\frac{2}{3}=\frac{8}{3}\end{array}$

Q.6

$\begin{array}{l}\mathrm{Which}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{pairs}\text{ }\mathrm{represent}\text{ }\mathrm{the}\text{ }\mathrm{same}\\ \mathrm{rationa}\text{ }\mathrm{lnumber}?\\ \left(\mathrm{i}\right)\text{ }\frac{-7}{21}\mathrm{and}\frac{3}{9}\text{ }\left(\mathrm{ii}\right)\text{ }\frac{-16}{20}\mathrm{and}\frac{20}{-25}\text{ }\left(\mathrm{iii}\right)\text{ }\frac{-2}{-3}\mathrm{and}\frac{2}{3}\\ \left(\mathrm{iv}\right)\text{ }\frac{-3}{5}\mathrm{and}\frac{-12}{20}\text{ }\left(\mathrm{v}\right)\text{ }\frac{8}{-5}\mathrm{and}\frac{-24}{15}\text{ }\left(\mathrm{vi}\right)\text{ }\frac{1}{3}\mathrm{and}\frac{-1}{9}\\ \left(\mathrm{vii}\right)\text{ }\frac{-5}{-9}\mathrm{and}\frac{5}{-9}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\frac{-7}{21}\text{and}\frac{3}{9}\\ \frac{-7}{21}=\frac{-1×7}{3×7}=\frac{-1}{3}\\ \frac{3}{9}=\frac{1×3}{3×3}=\frac{1}{3}\\ \text{Since}\frac{1}{3}\ne \frac{-1}{3}\\ \text{So, given pair does not represent same rational number}\text{.}\\ \left(\text{ii}\right)\frac{-16}{20}\text{and}\frac{20}{-25}\\ \frac{-16}{20}=\frac{-4×4}{4×5}=\frac{-4}{5}\\ \frac{20}{-25}=\frac{4×5}{-5×5}=\frac{-4}{5}\\ \text{So, given pair represent same rational number}\text{.}\end{array}$ $\begin{array}{l}\left(\text{iii}\right)\frac{-2}{-3}\text{and}\frac{2}{3}\\ \frac{-2}{-3}=\frac{2}{3}\\ \text{So, given pair represent same rational number}.\\ \left(\text{iv}\right)\frac{-3}{5}\text{and}\frac{-12}{20}\\ \frac{-12}{20}=\frac{-3×4}{5×4}=\frac{-3}{5}\\ \text{So, given pair represent same rational number}\\ \left(\text{v}\right)\frac{8}{-5}\text{and}\frac{-24}{15}\\ \frac{-24}{15}=\frac{-8×3}{5×3}=\frac{-8}{5}\\ \text{So, given pair represent same rational number}\text{.}\\ \left(\text{vi}\right)\frac{\text{1}}{3}\text{and}\frac{-1}{9}\\ \text{Since}\frac{1}{3}\ne \frac{-1}{9}\\ \text{So, given pair does not represent same rational number}\text{.}\\ \left(\text{vii}\right)\frac{-5}{-9}\text{and}\frac{5}{-9}\\ \frac{-5}{-9}=\frac{-1×5}{-1×9}=\frac{5}{9}\\ \frac{5}{-9}=\frac{5}{-9}\\ \text{Since}\frac{5}{9}\ne \frac{5}{-9}\\ \text{So, given pair does not represent same rational number}\text{.}\end{array}$

Q.7

$\begin{array}{l}\mathrm{Rewrite}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{rational}\text{ }\mathrm{numbers}\text{ }\mathrm{in}\text{ }\mathrm{the}\\ \mathrm{simplest}\text{ }\mathrm{form}:\\ \left(\mathrm{i}\right)\text{ }\frac{-8}{6}\text{ }\left(\mathrm{ii}\right)\text{ }\frac{25}{45}\text{ }\left(\mathrm{iii}\right)\text{ }\frac{-44}{72}\text{ }\left(\mathrm{iv}\right)\text{ }\frac{-8}{10}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\frac{-8}{6}\\ \frac{-8}{6}=\frac{-4×\overline{)2}}{3×\overline{)2}}=\frac{-4}{3}\\ \left(\text{ii}\right)\frac{\text{25}}{45}\\ \frac{25}{45}=\frac{\overline{)5}×5}{\overline{)5}×9}=\frac{5}{9}\\ \left(\text{iii}\right)\frac{-\text{44}}{72}\\ \frac{-44}{72}=\frac{-11×\overline{)4}}{18×\overline{)4}}=\frac{-11}{18}\\ \left(\text{iv}\right)\frac{-8}{10}\\ -\frac{8}{10}=\frac{-4×\overline{)2}}{5×\overline{)2}}=\frac{-4}{5}\end{array}$

Q.8

$\begin{array}{l}\mathrm{Which}\text{ }\mathrm{is}\text{ }\mathrm{greater}\text{ }\mathrm{in}\text{ }\mathrm{each}\text{ }\mathrm{of}\text{ }\mathrm{the}\text{ }\mathrm{following}:\\ \left(\mathrm{i}\right)\text{ }\frac{2}{3},\frac{5}{2}\text{ }\left(\mathrm{ii}\right)\text{ }\frac{-5}{6},\frac{-4}{3}\text{ }\left(\mathrm{iii}\right)\text{ }\frac{-3}{4},\frac{2}{-3}\\ \left(\mathrm{iv}\right)\text{ }\frac{-1}{4},\frac{1}{4}\text{ }\left(\mathrm{v}\right)\text{ }-3\frac{2}{7},-3\frac{4}{5}\end{array}$

Ans

$\begin{array}{l}\left(\text{i}\right)\frac{2}{3},\frac{5}{2}\\ \frac{2}{3}=\frac{2×2}{3×2}=\frac{4}{6}\\ \frac{5}{2}=\frac{5×3}{2×3}=\frac{15}{6}\\ \text{Since, 1}5>4\\ \text{So,}\frac{5}{2}\text{is greater than}\frac{2}{3}.\\ \left(\text{ii}\right)\frac{-5}{6},\frac{-4}{3}\\ \frac{-5}{6}=\frac{-5×3}{6×3}=\frac{-15}{18}\\ \frac{-4}{3}=\frac{-4×6}{3×6}=\frac{-24}{18}\\ \text{Since,}-15>-24\\ \text{So,}\frac{-5}{6}\text{is greater than}\frac{-4}{3}.\end{array}$ $\begin{array}{l}\left(\text{iii}\right)\frac{-3}{4},\frac{2}{-3}\\ \frac{-3}{4}=\frac{-3×3}{4×3}=\frac{-9}{12}\\ \frac{2}{-3}=\frac{2×4}{-3×4}=\frac{-8}{12}\\ \text{Since,}-8>-9\\ \text{So,}\frac{2}{-3}\text{is greater than}\frac{-3}{4}.\\ \left(\text{iv}\right)\frac{-1}{4},\frac{1}{4}\\ \text{Since,}1>-1\\ \text{So,}\frac{1}{4}\text{is greater than}\frac{-1}{4}.\\ \left(\text{v}\right)-3\frac{2}{7},-3\frac{4}{5}\\ -3\frac{2}{7}=\frac{-23}{7}=\frac{-23×5}{7×5}=\frac{-115}{35}\\ -3\frac{4}{5}=\frac{-19}{5}=\frac{-19×7}{5×7}=\frac{-133}{35}\\ \text{Since,}-\text{115>}-\text{133}\\ \text{So,}-\text{3}\frac{2}{7}\text{is greater than}-\text{3}\frac{4}{5}.\end{array}$

Q.9

$\begin{array}{l}\mathrm{Write}\text{ }\mathrm{the}\text{ }\mathrm{following}\text{ }\mathrm{rational}\text{ }\mathrm{numbers}\text{ }\mathrm{in}\text{ }\mathrm{a}\text{ }\mathrm{scending}\text{ }\mathrm{order}:\\ \left(\mathrm{i}\right)\text{ }\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\text{ }\left(\mathrm{ii}\right)\text{ }\frac{-1}{3},\frac{-2}{9},\frac{-4}{3}\text{ }\left(\mathrm{iii}\right)\text{ }\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{-3}{5},\frac{-2}{5},\frac{-1}{5}\\ \text{Since,}-1>-2>-3\\ \text{So,}\frac{-3}{5}<\frac{-2}{5}<\frac{-1}{5}.\\ \\ \text{(ii)}\frac{-1}{3},\frac{-2}{9},\frac{-4}{3}\\ \frac{-1}{3}=\frac{-1×3}{3×3}=\frac{-3}{9}\\ \frac{-4}{3}=\frac{-4×3}{3×3}=\frac{-12}{9}\\ \text{Since,}-2>-3>-12\\ \text{So,}\frac{-4}{3}<\frac{-1}{3}<\frac{-2}{9}.\end{array}$ $\begin{array}{l}\text{(iii)}\frac{-3}{7},\frac{-3}{2},\frac{-3}{4}\\ \frac{-3}{7}=\frac{-3×4}{7×4}=\frac{-12}{28}\\ \frac{-3}{2}=\frac{-3×14}{2×14}=\frac{-42}{28}\\ \frac{-3}{4}=\frac{-3×7}{4×7}=\frac{-21}{28}\\ \text{Since,}-12>-21>-42\\ \text{So,}\frac{-3}{2}<\frac{-3}{4}<\frac{-3}{7}.\end{array}$

Q.10

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{sum}:\\ \left(\mathrm{i}\right)\text{ }\frac{5}{4}+\left(\frac{-11}{4}\right)\text{ }\left(\mathrm{ii}\right)\text{ }\frac{5}{3}+\frac{3}{5}\text{ }\left(\mathrm{iii}\right)\text{ }\frac{-9}{10}+\frac{22}{15}\\ \left(\mathrm{iv}\right)\text{ }\frac{-3}{-11}+\frac{5}{9}\text{ }\left(\mathrm{v}\right)\text{ }\frac{-8}{19}+\frac{\left(-2\right)}{57}\text{ }\left(\mathrm{vi}\right)\text{ }\frac{-2}{3}+0\\ \left(\mathrm{vii}\right)\text{ }-2\frac{1}{3}+4\frac{3}{5}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{5}{4}+\left(\frac{-11}{4}\right)\\ =\frac{5}{4}-\frac{11}{4}=\frac{-6}{4}=\frac{-3}{2}\\ \\ \text{(ii)}\frac{5}{3}+\frac{3}{5}\\ \text{L.C.M of 3 and 5 is 15.}\\ \text{So,}\frac{5}{3}+\frac{3}{5}=\frac{5×5}{3×5}+\frac{3×3}{5×3}\\ =\frac{25}{15}+\frac{9}{15}\\ =\frac{25+9}{15}=\frac{34}{15}\end{array}$ $\begin{array}{l}\text{(iii)}\frac{-9}{10}+\frac{22}{15}\\ \text{L}\text{.C}\text{.M of 10 and 15 is 30.}\\ \frac{-9}{10}+\frac{22}{15}=\frac{-9×3}{10×3}+\frac{22×2}{15×2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-27}{30}+\frac{44}{30}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-27+44}{30}=\frac{17}{30}\\ \\ \text{(iv)}\frac{-3}{-11}+\frac{5}{9}\\ \text{L}\text{.C}\text{.M of 9 and 11 is 99}\text{.}\\ \text{So,}\text{\hspace{0.17em}}\frac{-3}{-11}+\frac{5}{9}=\frac{3}{11}+\frac{5}{9}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{3×9}{11×9}+\frac{5×11}{9×11}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{27}{99}+\frac{55}{99}=\frac{82}{99}\\ \\ \text{(v)}\frac{-8}{19}+\frac{\left(-2\right)}{57}\\ \text{L}\text{.C}\text{.M of 19 and 57 is 57.}\\ \text{So,}\frac{-8}{19}+\frac{\left(-2\right)}{57}=\frac{-8×3}{19×3}-\frac{2}{57}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-24}{57}-\frac{2}{57}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-24-2}{57}=-\frac{26}{57}\\ \end{array}$ $\begin{array}{l}\text{}\end{array}$ $\begin{array}{l}\text{(vi)}\frac{-2}{3}+0\\ =\frac{-2}{3}+0×\frac{3}{3}\\ =-\frac{2}{3}+0=\frac{-2}{3}\\ \\ \text{(vii) -2}\frac{1}{3}+4\frac{3}{5}\\ =-\frac{7}{3}+\frac{23}{5}\\ \text{L}\text{.C}\text{.M of 3 and 5 is 15}\text{.}\\ \text{So,}-\frac{7}{3}+\frac{23}{5}=\frac{-7×5}{3×5}+\frac{23×3}{5×3}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-35}{15}+\frac{69}{15}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{-35+69}{15}=\frac{34}{15}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Find}\\ \left(\mathrm{i}\right)\frac{7}{24}-\frac{17}{36}\left(\mathrm{ii}\right)\frac{5}{63}-\left(\frac{-6}{21}\right)\left(\mathrm{iii}\right)\frac{-6}{13}-\left(\frac{-7}{15}\right)\\ \left(\mathrm{iv}\right)\frac{-3}{8}-\frac{7}{11}\left(\mathrm{v}\right)-2\frac{1}{9}-6\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{7}{24}-\frac{17}{36}\\ \text{Since, L.C.M of 24 and 36 is 72.}\\ \text{So,}\frac{7}{24}-\frac{17}{36}=\frac{7×3}{24×3}-\frac{17×2}{36×2}\\ =\frac{21}{72}-\frac{34}{72}=\frac{21-34}{72}=\frac{-13}{72}\\ \\ \text{(ii)}\frac{5}{63}-\left(\frac{-6}{21}\right)\\ =\frac{5}{63}+\frac{6}{21}\\ \text{Since, L.C.M of 21 and 63 is 63.}\\ \text{So,\hspace{0.17em}}\frac{5}{63}+\frac{6}{21}=\frac{5}{63}+\frac{6×3}{21×3}\\ =\frac{5}{63}+\frac{18}{63}=\frac{5+18}{63}=\frac{23}{63}\\ \\ \text{(iii)}\frac{-6}{13}-\left(\frac{-7}{15}\right)\\ =\frac{-6}{13}+\frac{7}{15}\\ \text{Since, L.C.M of 13 and 15 is 195.}\\ \text{So,}\frac{-6×15}{13×15}+\frac{7×13}{13×15}=\frac{-90}{195}+\frac{91}{195}\\ =\frac{-90+91}{195}=\frac{1}{195}\end{array}$ $\begin{array}{l}\text{(iv)}\frac{-3}{8}-\frac{7}{11}\\ \text{Since, L.C.M of 8 and 11 is 88.}\\ \text{So,}\frac{-3×11}{8×11}-\frac{7×8}{11×8}=\frac{-33}{88}-\frac{56}{88}\\ =\frac{-33-56}{88}=\frac{-89}{88}\\ \\ \text{(v)}-2\frac{1}{9}-6\\ =-\frac{19}{9}-6\\ \text{Since, L.C.M of 9 and 1 is 9.}\\ \text{So,}\frac{-19}{9}-6=\frac{-19}{9}-\frac{6×9}{1×9}\\ =\frac{-19}{9}-\frac{54}{9}\\ =\frac{-73}{9}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{product}:\\ \left(\mathrm{i}\right)\frac{9}{2}×\left(\frac{-7}{4}\right)\left(\mathrm{ii}\right)\frac{3}{10}×\left(-9\right)\left(\mathrm{iii}\right)\frac{-6}{5}×\frac{9}{11}\\ \left(\mathrm{iv}\right)\frac{3}{7}×\left(\frac{-2}{5}\right)\left(\mathrm{v}\right)\frac{3}{11}×\frac{2}{5}\left(\mathrm{vi}\right)\frac{3}{-5}×\frac{-5}{3}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{9}{2}×\left(\frac{-7}{4}\right)\\ =\frac{9}{2}×\frac{-7}{4}=\frac{-63}{8}\\ \text{(ii)}\frac{3}{10}×\left(-9\right)\\ =-\frac{3}{10}×9=\frac{-27}{10}\\ \text{(iii)}\frac{-6}{5}×\frac{9}{11}\\ =\frac{-6×9}{5×11}=\frac{-54}{55}\\ \text{(iv)}\frac{3}{7}×\left(\frac{-2}{5}\right)\\ =\frac{3×\left(-2\right)}{7×5}=\frac{-6}{35}\\ \text{(v)}\frac{3}{11}×\frac{2}{5}\\ =\frac{3×2}{11×5}=\frac{6}{55}\\ \text{(vi)}\frac{3}{-5}×\frac{-5}{3}\\ =\frac{3×\left(-5\right)}{-5×3}=\frac{-15}{-15}=1\end{array}$

Q.13

$\begin{array}{l}\mathrm{Find}\text{ }\mathrm{the}\text{ }\mathrm{value}\text{ }\mathrm{of}:\\ \left(\mathrm{i}\right)\text{ }\left(-4\right)÷\frac{2}{3}\text{ }\left(\mathrm{ii}\right)\text{ }\frac{-3}{5}÷2\text{ }\left(\mathrm{iii}\right)\text{ }\frac{-4}{5}÷\left(-3\right)\\ \left(\mathrm{iv}\right)\text{ }\frac{-1}{8}÷\frac{3}{4}\text{ }\left(\mathrm{v}\right)\text{ }\frac{-2}{13}÷\frac{1}{7}\text{ }\left(\mathrm{vi}\right)\text{ }\frac{-7}{12}÷\left(\frac{-2}{13}\right)\\ \left(\mathrm{vii}\right)\text{ }\frac{3}{13}÷\left(\frac{-4}{65}\right)\end{array}$

Ans

$\begin{array}{l}\text{(i)}\left(-4\right)÷\frac{2}{3}\\ =-4×\frac{3}{2}=\frac{-12}{2}=-6\\ \text{(ii)}\frac{-3}{5}÷2\\ =\frac{-3}{5}×\frac{1}{2}=\frac{-3}{10}\\ \text{(iii)}\frac{-4}{5}÷\left(-3\right)\\ =\frac{-4}{5}×\frac{1}{-3}=\frac{-4}{-15}=\frac{4}{15}\end{array}$ $\begin{array}{l}\text{(iv)}\frac{-1}{8}÷\frac{3}{4}\\ =\frac{-1}{8}×\frac{4}{3}=\frac{-4}{24}=\frac{-1}{6}\\ \text{(v)}\frac{-2}{13}÷\frac{1}{7}\\ =\frac{-2}{13}×\frac{7}{1}=\frac{-14}{13}\\ \text{(vi)}\frac{-7}{12}÷\left(\frac{-2}{13}\right)\\ =\frac{-7}{12}×\frac{13}{-2}=\frac{-91}{-24}=\frac{91}{24}\\ \text{(vii)}\frac{3}{13}÷\left(\frac{-4}{65}\right)\\ =\frac{3}{13}×\frac{65}{-4}=\frac{3×65}{13×\left(-4\right)}\\ =\frac{3×\overline{)13}×5}{\overline{)13}×\left(-4\right)}=\frac{15}{-4}=\frac{-15}{4}\end{array}$

Q.14

$\begin{array}{l}\mathrm{Fill}\text{ }\mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{boxes}\text{ }\mathrm{with}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{symbol}\text{ }\mathrm{out}\text{ }\mathrm{of}\text{ }>,<,\text{ }\mathrm{and}\text{ }=\\ \left(\mathrm{i}\right)\text{ }\frac{-5}{7}\overline{)\text{ }}\frac{2}{3}\left(\mathrm{ii}\right)\text{ }\frac{-4}{5}\overline{)\text{ }}\frac{-5}{7}\left(\mathrm{iii}\right)\text{ }\frac{-7}{8}\overline{)\text{ }}\frac{14}{-16}\\ \left(\mathrm{iv}\right)\text{ }\frac{-8}{5}\overline{)\text{ }}\frac{-7}{4}\left(\mathrm{v}\right)\text{ }\frac{1}{-3}\overline{)\text{ }}\frac{-1}{4}\left(\mathrm{vi}\right)\text{ }\frac{5}{-11}\overline{)\text{ }}\frac{-5}{11}\\ \left(\mathrm{vii}\right)\text{ }0\overline{)\text{ }}\frac{-7}{6}\end{array}$

Ans

$\begin{array}{l}\text{(i)}\frac{-5}{7}\overline{)\text{ }}\frac{2}{3}\\ \frac{-5}{7}=\frac{-5×3}{7×3}=\frac{-15}{21}\\ \frac{2}{3}=\frac{2×7}{3×7}=\frac{14}{21}\\ \text{Since, 14>}-\text{15}\\ \text{So, }\frac{-5}{7}\overline{)<}\frac{2}{3}\end{array}$ $\begin{array}{l}\text{(ii)}\frac{-4}{5}\overline{)\text{ }}\frac{-5}{7}\\ \frac{-4}{5}=\frac{-4×7}{5×7}=\frac{-28}{35}\\ \frac{-5}{7}=\frac{-5×5}{7×5}=\frac{-25}{35}\\ \text{Since,}-25>-28\\ \text{So, }\frac{-4}{5}\overline{)<}\frac{-5}{7}\\ \text{(iii)}\frac{-7}{8}\overline{)\text{ }}\frac{14}{-16}\\ \frac{14}{-16}=\frac{7×2}{-8×2}=\frac{-7}{8}\\ \text{So, }\frac{-7}{8}\overline{)=}\frac{14}{-16}\\ \text{(iv)}\frac{-8}{5}\overline{)\text{ }}\frac{-7}{4}\\ \frac{-8}{5}=\frac{-8×4}{5×4}=\frac{-32}{20}\\ \frac{-7}{4}=\frac{-7×5}{4×5}=\frac{-35}{20}\\ \text{Since,}-32>-\text{35}\\ \text{So, }\frac{-8}{5}\overline{)>}\frac{-7}{4}\end{array}$

Q.15

$\begin{array}{l}\mathrm{Fill}\text{ }\mathrm{in}\text{ }\mathrm{the}\text{ }\mathrm{boxes}\text{ }\mathrm{with}\text{ }\mathrm{the}\text{ }\mathrm{correct}\text{ }\mathrm{symbol}\text{ }\mathrm{out}\text{ }\mathrm{of}\text{ }>,\text{ }<,\text{ }\mathrm{and}\text{\hspace{0.17em}}=.\\ \left(\mathrm{i}\right)\frac{-5}{7}\overline{)\text{ }}\frac{2}{3}\text{ }\left(\mathrm{ii}\right)\frac{-4}{5}\overline{)\text{ }}\frac{-5}{7}\text{ }\left(\mathrm{iii}\right)\frac{-7}{8}\overline{)\text{ }}\frac{14}{-16}\\ \left(\mathrm{iv}\right)\frac{-8}{5}\overline{)\text{ }}\frac{-7}{4}\text{ }\left(\mathrm{v}\right)\frac{1}{-3}\overline{)\text{ }}\frac{-1}{4}\text{ \hspace{0.17em} }\left(\mathrm{vi}\right)\frac{5}{-11}\overline{)\text{ }}\frac{-5}{11}\\ \left(\mathrm{vii}\right)0\overline{)\text{ }}\frac{-7}{6}\end{array}$

Ans

$\begin{array}{l}\text{(i) }\frac{-5}{7}\overline{)\text{ }}\frac{2}{3}\\ \frac{-5}{7}=\frac{-5×3}{7×3}=\frac{-15}{21}\\ \frac{2}{3}=\frac{2×7}{3×7}=\frac{14}{21}\\ \text{Since, 14>}-\text{15}\\ \text{So, }\frac{-5}{7}\overline{)<}\frac{2}{3}\end{array}$ $\begin{array}{l}\text{(ii) }\frac{-4}{5}\overline{)\text{ }}\frac{-5}{7}\\ \frac{-4}{5}=\frac{-4×7}{5×7}=\frac{-28}{35}\\ \frac{-5}{7}=\frac{-5×5}{7×5}=\frac{-25}{35}\\ \text{Since,}-25>-28\\ \text{So, }\frac{-4}{5}\overline{)<}\frac{-5}{7}\\ \text{(iii) }\frac{-7}{8}\overline{)\text{ }}\frac{14}{-16}\\ \frac{14}{-16}=\frac{7×2}{-8×2}=\frac{-7}{8}\\ \text{So, }\frac{-7}{8}\overline{)=}\frac{14}{-16}\\ \text{(iv) }\frac{-8}{5}\overline{)\text{ }}\frac{-7}{4}\\ \frac{-8}{5}=\frac{-8×4}{5×4}=\frac{-32}{20}\\ \frac{-7}{4}=\frac{-7×5}{4×5}=\frac{-35}{20}\\ \text{Since,}-32>-\text{35}\\ \text{So, }\frac{-8}{5}\overline{)>}\frac{-7}{4}\end{array}$ $\begin{array}{l}\text{(v) }\frac{1}{-3}\overline{)\text{ }}\frac{-1}{4}\\ \frac{1}{-3}=\frac{1×4}{-3×4}=\frac{4}{-12}\\ \frac{-1}{4}=\frac{-1×3}{4×3}=\frac{-3}{12}\\ \text{Since,}-3>-\text{4}\\ \text{So,}\frac{1}{-3}\overline{)\text{ }<\text{ }}\frac{-1}{4}\\ \text{(vi) }\frac{5}{-11}\overline{)\text{ }}\frac{-5}{11}\\ \text{since}\frac{5}{-11}=\frac{-5}{11}\\ \text{So,}\frac{5}{-11}\overline{)=}\frac{-5}{11}\\ \text{(vii) }0\overline{)\text{ }}\frac{-7}{6}\\ \text{since}0>\frac{-7}{6}\\ \text{So,}0\overline{)>}\frac{-7}{6}\end{array}$

1. What are the different concepts covered in Chapter 9 of Class 7 Mathematics NCERT?

The following topics are covered  in this chapter:

1. Need For Rational Numbers
2. What Are Rational Numbers
3. Positive And Negative Rational Numbers
4. Rationals Numbers On a Number Line
5. Rational Numbers in Standard Form
6. Comparison of Rational Numbers
7. Rational Numbers Between Two Rational Numbers
8. Operations On Rational Numbers