# NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (EX 9.1) Exercise 9.1

The solutions to the questions provided in Exercise 9.1 are enclosed in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 based on Rational Numbers. These will help students learn how to find the rational numbers between the given pair of numbers and also how to find equivalent rational numbers as well as their representations on the number line. Students will also get to know how to find out if one rational number is greater than, lesser than, or equal to the other, thus arranging them in ascending or descending order. A few questions also involve reducing rational numbers to their simplest form. The problems in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 based on Rational Numbers are 10 in total and are relatively simple to solve if the methods discussed in the chapter are followed.

If the strategies discussed in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 are used, then all the questions are reasonably easy to solve.

One of the main lessons learned by students from these NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 is that 0 is neither a positive rational number nor a negative rational number. A rational number is also said to be in its standard form if the denominator is an integer that is positive and the numerator and denominator share only the number one as their only factor, according to  NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1.Furthermore, students should be reminded that there are an infinite number of rational numbers that can exist between any two rational numbers. Through the link provided on the Extramarks website and mobile application, students can obtain solutions to Chapter 9 Exercise 9.1 Rational Numbers in the form of NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1.

## NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (EX 9.1) Exercise 9.1

The solutions to the questions encapsulated in the exercise are available in NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1. Understanding the number system in depth and performing simple arithmetic operations correctly are prerequisites for working with rational numbers. After studying NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1, students should be able to handle ratios and fractions without difficulty in order to proceed with the calculations and simplify them.

Students must carefully read the entire text as well as the examples in the book, paying close attention to the highlights because these encompass the key ideas. These highlights in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 also allow for a quick review during the examination period, as and when necessary. The NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1, as its name suggests, are based on the chapter titled Rational Numbers. In the chapter, Rational Numbers are explained in great detail.

### Access NCERT Solution for Class 7 Mathematics Chapter 9 – Rational Numbers

The fraction of a rational number has a non-zero denominator. It is employed in daily life since more than one measure of a quantity cannot be adequately expressed by an integer or natural number. As a part of the solutions to one of the most important chapters of Mathematics, themes such as the number system, rational number properties, operations, and applications are covered in these NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1. Some tips and simple techniques are also demonstrated in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 to help make this chapter a little bit easier. Topics covered in NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 include comparisons of rational numbers, positive and negative rational numbers, rational numbers on the number line, rational numbers in standard form, and rational numbers between two rational numbers.

### Topics Covered in NCERT Class 7 Maths Exercise 9.1 Rational Numbers

There are conveniently accessible PDF versions of the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 centred on Rational Numbers on the Extramarks mobile application and website. In this activity from NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1, students will learn about important concepts. The themes covered in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 include comparisons between rational numbers, positive and negative rational numbers, rational numbers on the number line, rational numbers in standard form, and rational numbers between two rational numbers. These NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 have been created by top educators of Extramarks to assist students in achieving high grades in Mathematics.

In the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1, Rational Numbers are explained in extensive detail. The following main ideas and subjects are covered in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 which constitute the first part of Rational Numbers:

• The necessity of rational numbers
• How do rational numbers work?
• A rational number is one that can be written as p/q, where p and q are both integers and q≠0. Rational numbers include all integers and fractions.

### Important Points to Remember to Solve Questions of Exercise 9.1 of NCERT Maths Class 7

While curating the NCERT Solutions for Class 7 Maths, Chapter 9, Exercise 9.1, mathematics experts went to great lengths to understand the complex needs of Class 7 students. There are some very important topics and concepts that have been discussed in extensive detail in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 for students to better comprehend the chapter. Below is a list of all such topics discussed in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1:

• When p and q are integers and q≠0, a number is said to be rational if it can be represented in the form p/q.
• Natural numbers, whole numbers, integers, fractions, decimals (terminating, non-terminating, and repeating), and negative fractional values are all examples of rational numbers.
• When one of two rational numbers can be produced from the other by multiplying or dividing the other by the same non-zero integer, the two are said to be equivalent.
• Positive rational numbers are those in which the numerator and denominator are both positive or negative integers. A rational number is referred to as a negative rational number if either the numerator or the denominator is a negative integer.
• A rational number that is neither positive nor negative, is 0. On a number line, rational numbers can be displayed.
• If the denominator of a rational number is a positive integer and there is only one other common factor between the numerator and denominator, the number is said to be in the standard form.
• Any two rational numbers can be found between an unlimited or limitless number of other rational numbers.
• Comparing rational numbers can be done using a number line or similar to comparing fractions.
• When adding rational numbers with the same denominator, add their numerators while maintaining the same denominator. Take the L.C.M. of the denominators of the rational numbers and convert them to equivalent forms with L.C.M. as the denominator, and then add the resulting rational numbers. The number that, when added to a, results in “zero” is known as the additive inverse, or the opposite, of the number “a”. ‘-a’ stands for ‘a’ as additive inverse. When subtracting two rational numbers, one is added to the other together with the additive inverse of the other rational number.
• Calculating the product of two rational integers involves dividing the product of the denominators by the product of the numerators.
• By switching the numerator and denominator of a given number, the reciprocal of that number can be created. A number’s product with its reciprocal is always one.
• When dividing two rational numbers, divide the reciprocal of the rational number that is serving as the divisor by the other rational number.

The aforementioned topics are discussed in great detail in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1.

### NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Exercise 9.1

Extramarks makes sure to provide students with educational content that can make learning engaging and interactive. Since Class 7 is a foundational stage in a student’s career, it is important for them to retain the concepts and subjects they study for future studies. This is where Extramarks’ NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 step in. Extramarks understands the needs of students today, and hence the subject-matter experts have made sure to curate the best possible NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1. The answers in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 are very apt and to the point, making it easy for students to understand exactly what to write in their answer sheets. The NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 by Extramarks are thus a very helpful tool to practise the Class 7th Maths Chapter 9 Exercise 9.1.

The NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 for Maths Class 7 Chapter 9 Exercise 9.1 are available for access on the Extramarks website and mobile application.

Extramarks’ subject-matter experts have made sure to provide students with video lessons and interactive live sessions for Class 7 Chapter 9 Maths Exercise 9.1. Mentors of Extramarks understand how crucial the topic is for Class 7 students and have thus made sure to provide the Class 7 Maths Chapter 9 Exercise 9.1 Solution for the same for students’ reference.

### NCERT Solution Class 7 Maths Chapter 9 Other Exercise

There are a total of 2 exercises in the 9th Chapter of the Mathematics CBSE syllabus. There are a total of 10 questions in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 for Class 7 Maths Ch 9 Ex 9.1 and a total of 4 questions in the second exercise of the chapter the solutions to which are accessible on the Extramarks website and mobile application. More subjects are covered in the first half of the chapter, including rational numbers in standard form, rational number comparisons, and rational numbers between two rational numbers.

• It is necessary to study the standard form of rational numbers, which specifies that a rational number is in the standard form if its denominator is a positive integer and the numerator and denominator have only one additional common factor other than 1.
• There are countless rational numbers that can be found between any two rational numbers.

Furthermore, topics like the Introduction, Equivalent Rational Numbers, Rational Numbers in Standard form, Rational numbers between Rational Numbers, Properties of Rational Numbers, Properties of Rational Numbers, Addition, Subtraction and Multiplication and Division of Rational Numbers, Negatives and Reciprocals, Additive Inverse of a Rational Number, Representing on a Number Line and Comparison of Rational Numbers are a few of the many topics discussed in the NCERT Solutions Class 7 Maths Chapter 9 Exercise 9.1 by Extramarks.

The complete range of K-12 subjects is covered by Extramarks’ curriculum, which spans grades KG through 12. The Extramarks educational website is well known for offering solutions for a wide range of subjects, including Science, Math, Social Studies, English (core and elective), Hindi, Sanskrit, Economics, History, Geography, Political Science, Accounting, Business Studies, Physics, Chemistry, And Biology, as well as Computer Science (C++ and Python), Informatics Practises, Multimedia Web and Technology, and more. All kinds of reference materials, including NCERT solutions, are covered in the practise and study materials of the learning application, which adheres to the format of NCERT books. The educational platform offers feedback, scorecards, a summary of the student’s performance, a challenge section where users can participate in quiz competitions, and more. Students can click on the link provided to download credible NCERT Solutions For Class 7 Maths Chapter 9 Exercise 9.1.

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

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

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

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

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.9

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

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.10

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

Ans

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