NCERT Solutions Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1

Exercise 1.1 of Class 8 Maths Chapter 1 introduces basic properties of rational numbers like commutative, associative, identity, and inverse properties. You will be solving problems to determine if the properties hold true for particular rational numbers. These properties provide the basis for many of the mathematical concepts you will learn and know how to apply.

This exercise is aligned to the latest NCERT syllabus and fits into the CBSE Class 8 Maths curriculum. Practicing these solutions will improve your logical reasoning and enable you to solve questions that are based on properties of numbers. The NCERT Solutions are answered in a step by step manner for better conceptual understanding.

With continued practice and application, you will develop a better grip on rational numbers, become more accurate, and be better off for your school exams. These properties will help you handle more advanced concepts in future chapters and in competitive exams.

1.0Download NCERT Solutions Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1: Free PDF

Download the free PDF of NCERT Solutions for Class 8 Maths Chapter 1 Exercise 1.1 from below:

2.0Key Concepts in Exercise 1.1 of Class 8 Maths Chapter 1

The key concepts covered in this exercise are as follows:

• Commutative property of multiplication
• Associative property of multiplication
• Additive inverse (for subtraction)
• Multiplicative inverse (the reciprocal)
• Recognizing identity elements (0 for addition, 1 for multiplication)

3.0NCERT Class 8 Maths Chapter 1: Other Exercises

4.0NCERT Class 8 Maths Chapter 1 Exercise 1.1: Detailed Solutions

• Using appropriate properties find (i) $-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$ (ii) $\frac{2}{5}\times \left(-\frac{3}{7}\right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$ Sol. (i) $-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}=-\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}$ $$\begin{gathered} =\left(-\frac{3}{5}\right)\times \frac{2}{3}+\left(-\frac{3}{5}\right)\times \frac{1}{6}+\frac{5}{2} \\ =\left(-\frac{3}{5}\right)\times \left(\frac{2}{3}+\frac{1}{6}\right)+\frac{5}{2} \\ =\left(-\frac{3}{5}\right)\times \frac{4+1}{6}+\frac{5}{2}=-\frac{3}{5}\times \frac{5}{6}+\frac{5}{2} \\ =-\frac{3\times 5}{5\times 6}+\frac{5}{2}=\frac{-1}{2}+\frac{5}{2}=\frac{-1+5}{2}=\frac{4}{2}=2. \end{gathered}$$ (ii) $\frac{2}{5}\times \left(-\frac{3}{7}\right)-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14}\times \frac{2}{5}$ $=\frac{2}{5}\times \left(-\frac{3}{7}\right)+\frac{1}{14}\times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}$ $=\frac{2}{5}\times \left(-\frac{3}{7}\right)+\frac{2}{5}\times \frac{1}{14}-\frac{1}{6}\times \frac{3}{2}$ $=\frac{2}{5}\times \left[\left(-\frac{3}{7}\right)+\frac{1}{14}\right]-\frac{1}{6}\times \frac{3}{2}$ $=\frac{2}{5}\times \left(\frac{-6+1}{14}\right)-\frac{3}{12}$ $=\frac{2}{5}\times \frac{-5}{14}-\frac{1}{4}=\frac{-1}{7}-\frac{1}{4}=\frac{-4-7}{28}=\frac{-11}{28}$.
• Write the additive inverse of each of the following (i) $\frac{2}{8}$ (ii) $\frac{-5}{9}$ (iii) $\frac{-6}{-5}$ (iv) $\frac{2}{-9}$ (v) $\frac{19}{-6}$ Sol. (i) The additive inverse of $\frac{2}{8}$ is $\left(\frac{-2}{8}\right)=\frac{-2}{8}$. (ii) The additive inverse of $\frac{-5}{9}$ is $-\left(\frac{-5}{9}\right)=\frac{5}{9}$. (iii) The additive inverse of $\frac{-6}{-5}$ is $\frac{-6}{5}$. (iv) The additive inverse of $\frac{2}{-9}$ is $\frac{2}{9}$. (v) The additive inverse of $\frac{19}{-6}$ is $\frac{19}{6}$.
• Verify that $-(-\mathrm{x})=\mathrm{x}$ for (i) $x=\frac{11}{15}$ (ii) $\mathrm{x}=-\frac{13}{17}$ Sol. (i) $\mathrm{x}=\frac{11}{15}$ $\therefore -(-\mathrm{x})=-\left(-\frac{11}{15}\right)=\frac{11}{15}=\mathrm{x}$ (ii) $\mathrm{x}=\frac{-13}{17}$ $\therefore -(-\mathrm{x})=-\left[-\left(\frac{-13}{17}\right)\right]=-\left[\frac{13}{17}\right]=-\frac{13}{17}$
• Find the multiplicative inverse of the following: (i) -13 (ii) $\frac{-13}{19}$ (iii) $\frac{1}{5}$ (iv) $\frac{-5}{8}\times \frac{-3}{7}$ (v) $-1\times \frac{-2}{7}$ (vi) -1 Sol. (i) The multiplicative inverse of -13 is $(-13{)}^{-1}=\frac{1}{-13}$ (ii) The multiplicative inverse of $\frac{-13}{19}\text{ is }{\left(\frac{-13}{19}\right)}^{-1}=\frac{19}{-13}=\frac{-19}{13}.$ (iii) The multiplicative inverse of $\frac{1}{5}$ is 5 . (iv) We have, $\frac{-5}{8}\times \frac{-3}{7}=\frac{-5\times -3}{8\times 7}=\frac{15}{56}$ The multiplicative inverse of $\frac{15}{56}$ is ${\left(\frac{15}{56}\right)}^{-1}=\frac{56}{15}$ (v) $-1\times \frac{-2}{7}=\frac{2}{7}$ The multiplicative inverse of $\frac{2}{7}=\frac{7}{2}$ (vi) The multiplicative inverse of -1 is -1 .
• Name the property under multiplication used in each of the following (i) $\frac{-4}{5}\times 1=1\times \frac{-4}{5}=-\frac{4}{5}$ (ii) $-\frac{13}{17}\times \frac{-2}{7}=\frac{-2}{7}\times \frac{-13}{17}$ (iii) $\frac{-19}{29}\times \frac{29}{-19}=1$ Sol. (i) Existence of multiplicative identity. (ii) Commutative property of multiplication. (iii) Existence of multiplicative inverse.
• Multiply $\frac{6}{13}$ by the reciprocal of $\frac{-7}{16}$. Sol. $\frac{6}{13}\times ($ the reciprocal of $\frac{-7}{16})$ $=\frac{6}{13}\times {\left(\frac{-7}{16}\right)}^{-1}$ $=\frac{6}{13}\times \frac{16}{-7}=-\frac{96}{91}$.
• Tell what property allows you to compute $\frac{1}{3}\times \left(6\times \frac{4}{3}\right)$ as $\left(\frac{1}{3}\times 6\right)\times \frac{4}{3}$. Sol. Associative property of multiplication over rational numbers allows us to compute : $\frac{1}{3}\times \left(6\times \frac{4}{3}\right)$ as $\left(\frac{1}{3}\times 6\right)\times \frac{4}{3}$.
• Is $\frac{8}{9}$ the multiplicative inverse of $-1\frac{1}{8}$ ? Why or why not? Sol. No, $\frac{8}{9}$ is not the multiplicative inverse of $-1\frac{1}{8}$. Because $\frac{8}{9}\times \left(-1\frac{1}{8}\right)=\frac{8}{9}\times \frac{-9}{8}=-1\ne 1$.
• Is 0.3 the multiplicative inverse of $3\frac{1}{3}$ ? Why or why not? Sol. Yes, 0.3 is multiplicative inverse of $3\frac{1}{3}$. Because $0.3\times 3\frac{1}{3}=\frac{3}{10}\times \frac{10}{3}=1$.
• Write (i) The rational number that does not have a reciprocal. (ii) The rational numbers that are equal to their reciprocals. (iii) The rational number that is equal to its negative. Sol. (i) We know that there is no rational number which when multiplied with 0 , gives 1. Therefore, the rational number 0 has no reciprocal. (ii) We know that the reciprocal of 1 is 1 and the reciprocal of -1 is -1 . Therefore 1 and -1 are the only rational numbers which are equal to their reciprocals. (iii) The rational number 0 is equal to its negative.
• Fill in the blanks (i) Zero has reciprocal. (ii) The numbers and are their own reciprocals. (iii) The reciprocal of - 5 is _. (iv) Reciprocal of $\frac{1}{x}$, where $x\ne 0$ is - (v) The product of two rational numbers is always a . (vi) The reciprocal of a positive rational number is -. Sol. (i) No (ii) $1,-1$ (iii) $\frac{-1}{5}$ (iv) x (v) Rational number (vi) Positive

5.0Key Features and Benefits of Class 8 Maths Chapter 1 Exercise 1.1

• Covers important properties of rational numbers found in the NCERT syllabus. 
• Focused questions and solutions makes it easy to practise under exam timed conditions. 
• Provides clear, step-by-step solutions that help to build understanding. 
• Helps to develop greater speed of solving questions on rational numbers for CBSE school and Board tests. 
• Helps develop a stronger sense of numbers that is useful for Olympiads and other competitive exams.