# NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (EX 9.2) Exercise 9.2

Every time distinct quantities are involved, arithmetic operations are always performed on them in order to find a solution. Rational numbers work similarly. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 include exercise questions. A major focus is placed on operations like addition, subtraction, multiplication, and division when dealing with rational numbers as a part of NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. The practice questions and solutions in the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 require students to carry out the arithmetic operations indicated above on rational numbers. There are solutions to four problems in NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. Students can easily understand rational numbers if they pay attention to the concepts and logic outlined in the NCERT Solutions for Class 9 Maths, Chapter 9, Exercise 9.2.

The exercise will teach the students that when adding rational numbers with the same denominators, the numerators must be added while maintaining the denominators’ equality. Additionally, when a positive integer is multiplied by a rational number, the denominator value is unaffected; just the numerator has changed.  Students can find the PDF version of the NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 on the Extramarks website and mobile application.The ideas of fractions, integers, and how mathematical operations are applied to them in the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 must be reviewed by students. As a result, they will be better prepared to approach the NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2.After reviewing the material from previous chapters, students should practise the examples in the NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2.This will help them keep their computation speed consistent, which will be useful in exams as well. NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (EX 9.2) Exercise 9.2 [ Add a PDF of NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers (EX 9.2) Exercise 9.2]

Students are taught the key ideas that will help them answer problems with ease in the following sessions as part of the NCERT Solutions for Class 9 Maths, Chapter 9, Exercise 9.2.To reach the solution, students must concentrate on the listed procedures. This will make it very simple for them to address challenges of a similar nature. Students can discover answers to each question in the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 available on Extramarks, using the most recent course materials. Students have access to all of Extramarks’ questions for practice.

Below are a few points to remember before practising the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2:

• Although a rational number need not be an integer, every integer is a rational number.
• If two rational numbers have the same standard form, they are equivalent.
• Divide or multiply a rational number numerator and denominator by a non-zero integer to get an equivalent rational number.
• If there are two rational integers with the same denominator, one with the larger numerator is larger than the other.
• A rational positive number is one that is higher than zero.
• Every irrational negative number is smaller than 0 in value.
• The number line can display rational numbers.

For Class 7, experts recommend NCERT textbooks.  Extramarks’ experts have compiled NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 for the benefit of students.It takes a lot of practice, reasoning, and conceptual clarity to master Mathematics. As a result, Extramarks is providing students with NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 that address all the fundamental ideas mentioned in Class 7 Maths Chapter 9 Exercise 9.2.

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

More subjects are covered in the first half of the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, 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: 1. There are countless rational numbers that can be found between any two rational numbers. The Extramarks website and mobile application provide NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 in an easy-to-read PDF format. These NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 focus on operations on rational numbers, specifically addition, subtraction, multiplication, and division of rational numbers.The NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.2 are ideal reference materials for students who want to excel in arithmetic.

### NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers Exercise 9.2

Students can effortlessly overcome challenging problems with the assistance of the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 offered by Extramarks. It is advised that students practise the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, which are offered in PDF versions. Additionally, the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 make basic mathematical ideas simple for students to comprehend. For students taking the CBSE Class 7 Mathematics exam, the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 are a reliable study tool. To enhance students’ academic performance and knowledge, proficient instructors of Extramarks have compiled the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 to the questions from the Mathematics NCERT textbook. To help them better understand the concepts, students are advised to practise  NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. Their confidence will grow as a result of rigorous practice, which will help them get better grades.

Students may find the chapter to be enjoyable with Extramarks’ NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2. Consistent practise will help students perform better in Class 8, and beyond. Students can benefit a lot by laying the groundwork for a concrete foundation in Mathematics. Experts at Extramarks have created step-by-step NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 with thorough justifications. To assist students in performing well on the Mathematics exam, the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 are organised per practice. Below is an overview of all the important topics covered in the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2:

### Definition of Rational Numbers:

Any integer can be used for p and q, and q must be greater than zero to be considered rational. Rational numbers, therefore, comprise whole numbers, natural numbers, integers, fractions of integers, and decimals (terminating decimals and recurring decimals). In this lesson, students can learn more about rational numbers, including examples, how to recognise them, and how to use them.

### What are Rational Numbers?

There is a connection between the words “ratio” and the word “rational”. Rational numbers are therefore closely tied to the idea of fractions, which are in turn intimately connected to ratios. In other terms, a number is a rational number if it can be written as a fraction in which the numerator and denominator are both integers.

### Types of Rational Numbers:

The various categories of rational numbers are listed below:

• -2, 0, 3, and other integers are examples of rational numbers.
• Rational numbers include fractions like 3/7, -6/5, etc. that have integer numerators and denominators.
• Rational numbers include those with terminating decimals, such as 0.35, 0.7116, 0.9768, etc.
• Rational numbers include non-terminating decimals with some repeating patterns (after the decimal point), such as 0.333, 0.141414, etc. Commonly referred to as repeating non-terminating decimals.

### Identification of Rational Numbers:

The following qualities make it simple to recognise rational numbers:

• Rational numbers include all integers, whole numbers, natural numbers, and fractions containing integers.
• It can be said that a number is rational if its decimal form is ending or recurring, as in the cases of 5.6 and 2.141414.
• Irrational numbers are those in which it appears that the decimals never finish or don’t repeat. Checking to see if a number can be represented in the form p/q, where p and q are integers and q is not equal to 0, is another method for determining whether or not a number is rational.

When individuals start describing the objects in their environment in numerical terms, Arithmetic is said to have begun. Natural or counting numbers are what people often use to count the objects in their immediate environment. Whole numbers result from the inclusion of a zero, while integers result from the inclusion of negative numbers. These fundamentals about how the number system developed from the natural to the whole to integers are where the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 centred on Rational Numbers begin. In these NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2, the number system is extended to include rational numbers and how their arithmetic operations are handled.

Students will discover that certain amounts occasionally need to be represented by a value that is neither an integer nor a fraction, leading to the creation of a category of rational numbers. Additionally, the term “rational number” derives from the word “ratio”. Students should thus recall the fundamental concept of rational numbers from Class 7 Maths 9.2, which states that rational numbers are those that can be represented in the form of a/b, where ‘a’ and ‘b’ are integers, and ‘b’ is not equal to zero. The link on the Extramarks website and mobile application will allow students to obtain the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 for further information on rational numbers.

Readers may think of arithmetic as an outcome of the basic operation of counting. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 on Rational Numbers are much more helpful for learning about numbers since they let students express the quantity or measure by counting. On the Extramarks website and mobile application, students may find the NCERT material in PDF format. Contrary to popular belief, counting is just the process of generating new numbers from older ones. This process produces an unending series of integers, which may be divided into categories such as prime numbers, rational and irrational numbers, whole numbers, and so on. Students will have a deeper understanding of rational numbers by working through the practise problems in the NCERT solutions Class 7 Mathematics Chapter 9 based on Rational numbers.

Chapter 9 to which the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.2 pertain, discusses the number system in general, with a particular emphasis on rational numbers, their types, such as equivalent rational numbers, positive and negative rational numbers, rational numbers in standard form, and how to represent rational numbers on the number line. Additionally, this chapter describes the comparison of rational numbers as well as how to find rational numbers between any two provided rational numbers. There are a total of 14 problems in Class 7 Mathematics Chapter 9. With the aid of the fundamental techniques covered in the chapter, they are all often simple and quick to resolve. To get at the answers, students only need to concentrate on the calculation portion of their work.

Q.1

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

Ans

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

Q.2

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

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.3

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

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.4

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

Ans

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