# RBSE Maths Class 8 Chapter 1: Important Questions and Solutions

RBSE Maths Chapter 1 – Rational Numbers Class 8 Important questions and solutions are provided here. The important questions and solutions of Chapter 1 available at BYJU’S contain detailed explanations. All these important questions are given, based on the new pattern prescribed by the RBSE. Students can also get RBSE Class 8 solutions for all the concepts here.

Chapter 1 of the RBSE Class 8 Maths will help the students to solve problems related to rational numbers, operations on rational numbers on the number line, properties of rational numbers, operations of zero with rational numbers, determining rational numbers between two rational numbers.

### RBSE Maths Chapter 1: Exercise 1.1 Textbook Important Questions and Solutions

Question 1: Fill in the blanks with appropriate answers.

(i) Product of two rational numbers is always……………(rational / integers).

(ii) Additive inverse of any negative rational number is ……… (negative/positive).

(iii) The inverse of zero is ……… (zero / indeterminate).

(iv) Additive identity of a rational number is ……. (zero / one)

(v) Multiplicative identity for rational numbers is ……… .(zero / one).

(vi) Reciprocal of a rational number is ………….of that (inverse/same).

(vii) The negative rational number on the number line always lies on ……… of zero (right / left).

(viii) The positive rational number on the number line always lies on ……… of zero ( right / left).

(ix) When a rational number is added with its additive inverse then the result is always ……… (zero / same)

(x) When a rational number is divided by the same rational number then the result is always ……… (zero / one)

Solution:

(i) Rational (ii) positive (iii) undefined (iv) zero (v) one (vi) reciprocal (vii) left (viii) right (ix) zero (x) one.

Question 2: Add the following rational numbers. [Two problems are solved using the number line]

$$\begin{array}{l}[i] \frac{5}{2}+[\frac{-3}{4}]\\ [ii] \left(-\frac{2}{3}\right)+\left(-\frac{4}{5}\right)+\left(\frac{5}{6}\right)\\ [iii] 0 + \frac{-2}{3}\\ [iv] -2\frac{1}{3}+4\frac{3}{5}\\ [v] \left(-\frac{6}{5}\right)+\left(-\frac{13}{7}\right)\\ [vi] \left(-\frac{8}{19}\right)+\left(-\frac{4}{57}\right)\\\end{array}$$

Solution:

$$\begin{array}{l}[i] \left(\frac{5}{2}\right)+\left(-\frac{3}{4}\right)\\ =\frac{5}{2}-\frac{3}{4}\\ \mathrm{Least\:Common\:Multiplier\:of\:}2,\:4:\quad 4\\ \frac{5}{2}=\frac{5\cdot \:2}{2\cdot \:2}=\frac{10}{4} =\frac{10}{4}-\frac{3}{4}\\ =\frac{10-3}{4}\\ =\frac{7}{4}\end{array}$$

$$\begin{array}{l}[ii] \left(-\frac{2}{3}\right)+\left(-\frac{4}{5}\right)+\left(\frac{5}{6}\right)\\ \left(-\frac{2}{3}\right)+\left(-\frac{4}{5}\right)+\left(\frac{5}{6}\right)\\ =-\frac{2}{3}-\frac{4}{5}+\frac{5}{6}\\ \mathrm{Least\:Common\:Multiplier\:of\:}3,\:5,\:6:\quad 30\\ \frac{2}{3}=\frac{2\cdot \:10}{3\cdot \:10}=\frac{20}{30}\\ \frac{4}{5}=\frac{4\cdot \:6}{5\cdot \:6}=\frac{24}{30}\\ \frac{5}{6}=\frac{5\cdot \:5}{6\cdot \:5}=\frac{25}{30}\\ =-\frac{20}{30}-\frac{24}{30}+\frac{25}{30}\\ =\frac{-19}{30}\\ [iii] 0 + \frac{-2}{3}\\ 0+\frac{-2}{3}\\ =0-\frac{2}{3}\\ =-\frac{2}{3}\\\end{array}$$

$$\begin{array}{l}[iv] -2\frac{1}{3}+4\frac{3}{5}\\ =-\left(4\frac{3}{5}+2\frac{1}{3}\right)\\ \mathrm{Subtract\:whole\:numbers}\:4-2:\quad 2\\ \mathrm{Combine\:fractions}\:\frac{3}{5}-\frac{1}{3}\\ \mathrm{Least\:Common\:Multiplier\:of\:}5,\:3:\quad 15\\ \frac{3}{5}=\frac{3\cdot \:3}{5\cdot \:3}=\frac{9}{15}\\ \frac{1}{3}=\frac{1\cdot \:5}{3\cdot \:5}=\frac{5}{15}\\ =\frac{9}{15}-\frac{5}{15}\\ =\frac{9-5}{15}\\ =\frac{4}{15}\\ =2+\frac{4}{15}\\ =2+\frac{4}{15}\\ [v] \left(-\frac{6}{5}\right)+\left(-\frac{13}{7}\right)\\ =-\frac{6}{5}-\frac{13}{7}\\ \mathrm{Least\:Common\:Multiplier\:of\:}5,\:7:\quad 35\\ \frac{6}{5}=\frac{6\cdot \:7}{5\cdot \:7}=\frac{42}{35}\\ \frac{13}{7}=\frac{13\cdot \:5}{7\cdot \:5}=\frac{65}{35}\\ =-\frac{42}{35}-\frac{65}{35}\\ =\frac{-42-65}{35}\\ =\frac{-107}{35}\\ =-\frac{107}{35}\\ [vi] \left(-\frac{8}{19}\right)+\left(-\frac{4}{57}\right)\\ =-\frac{8}{19}-\frac{4}{57}\\ \mathrm{Least\:Common\:Multiplier\:of\:}19,\:57:\quad 57\\ \frac{8}{19}=\frac{8\cdot \:3}{19\cdot \:3}=\frac{24}{57}\\ =-\frac{24}{57}-\frac{4}{57}\\ =\frac{-24-4}{57}\\ =\frac{-28}{57}\\ =-\frac{28}{57}\end{array}$$

Question 3: Solve the following. [Two problems are solved using the number line]

$$\begin{array}{l}[i]\frac{2}{3}+\frac{5}{4}\\ [ii]\:-2\:\frac{1}{9}+7\\ [iii]\:-\frac{7}{16}+\left[-\frac{3}{48}\right]\\ [iv]\:4\:\frac{3}{5}-\left[-2\:\frac{1}{3}\right]\end{array}$$

Solution:

$$\begin{array}{l}[i] \frac{2}{3}+\frac{5}{4}\\ \mathrm{Least\:Common\:Multiplier\:of\:}3,\:4:\quad 12\\ \frac{2}{3}=\frac{2\cdot \:4}{3\cdot \:4}=\frac{8}{12}\\ \frac{5}{4}=\frac{5\cdot \:3}{4\cdot \:3}=\frac{15}{12}\\ =\frac{8}{12}+\frac{15}{12}\\ =\frac{8+15}{12}\\ =\frac{23}{12}\end{array}$$

$$\begin{array}{l}[ii] -2\frac{1}{9}+7\\ =-\left(7+2\frac{1}{9}\right)\\ \mathrm{Break\:off}\:1\:\mathrm{from}\:7\\ 7=6+1\\ =6+\frac{9}{9}\\ =6\frac{9}{9}\\ =6\frac{9}{9}-2\frac{1}{9}\\ \mathrm{Subtract\:whole\:numbers}\:6-2:\quad 4\\ \mathrm{Combine\:fractions}\:\frac{9}{9}-\frac{1}{9}:\quad \frac{8}{9}\\ =4+\frac{8}{9}\\ =4\frac{8}{9}\\ [iii] -\frac{7}{16}+\left(-\frac{3}{48}\right)\\ =-\frac{7}{16}-\frac{3}{48}\\ \mathrm{Cancel\:}\frac{3}{48}:\quad \frac{1}{16}\\ =-\frac{7}{16}-\frac{1}{16}\\ =\frac{-7-1}{16}\\ =\frac{-8}{16}\\ =-\frac{8}{16}\\ =-\frac{1}{2}\\ [iv] 4\frac{3}{5}-\left[-2\frac{1}{3}\right]\\ \mathrm{Convert\:mixed\:numbers\:to\:improper\:fraction:}\\ 4\frac{3}{5}=\frac{4\cdot 5+3}{5}=\frac{23}{5} \\ \frac{23}{5}-\left[-2\cdot \frac{1}{3}\right]\\ 2\frac{1}{3}=\frac{2\cdot 3+1}{3}=\frac{7}{3}\\ \frac{23}{5}-\left[-\frac{7}{3}\right]\\ =\frac{23}{5}+\frac{7}{3}\\ \mathrm{Least\:Common\:Multiplier\:of\:}5,\:3:\quad 15\\ \frac{23}{5}=\frac{23\cdot \:3}{5\cdot \:3}=\frac{69}{15}\\ \frac{7}{3}=\frac{7\cdot \:5}{3\cdot \:5}=\frac{35}{15}\\ =\frac{69}{15}+\frac{35}{15}\\ =\frac{69+35}{15}\\ =\frac{104}{15}\\ \mathrm{Convert\:improper\:fractions\:to\:mixed\:numbers}\\ \frac{104}{15}=6\frac{14}{15}\\ =6\frac{14}{15}\end{array}$$

Question 4: Find the product of the following rational numbers.

$$\begin{array}{l}[i]\frac{13}{15}\times 5\\ [ii]\:\frac{4}{-5}\times -\frac{5}{4}\\ [iii]\:-\frac{2}{5}\times \left[-\frac{3}{7}\right]\\ [iv]\:\frac{15}{18}\times \frac{5}{6}\times \frac{21}{5}\\ [v]\:\frac{9}{4}\times -\frac{7}{5}\times -\frac{6}{21}\\ [vi]\:2\:\frac{1}{9}\times -3\:\frac{1}{2}\end{array}$$

Solution:

$$\begin{array}{l}[i] \frac{13}{15}\times \:5\\ =\frac{5}{1}\times \frac{13}{15}\\ =\frac{13}{3}\\ [ii] \:\frac{4}{-5}\times -\frac{5}{4}\\ \frac{4}{-5}\left(-\frac{5}{4}\right)\\ =-\frac{4}{-5}\times \frac{5}{4}\\ =-\frac{5}{-5}\\ =-\left(-1\right)\\ =1\\ [iii] \:-\frac{2}{5}\times \left[-\frac{3}{7}\right]\\ -\frac{2}{5}\left(-\frac{3}{7}\right)\\ =\frac{2}{5}\times \frac{3}{7}\\ =\frac{2\times \:3}{5\times \:7}\\ =\frac{6}{5\times \:7}\\ =\frac{6}{35}\\ [iv] \:\frac{15}{18}\times \frac{5}{6}\times \frac{21}{5}\\ =\frac{15\times \:5\times \:21}{18\times \:6\times \:5}\\ =\frac{15\times \:21}{18\times \:6}\\ =\frac{315}{18\times \:6}\\ =\frac{315}{108}\\ =\frac{35}{12}\\ [v] \:\frac{9}{4}\times -\frac{7}{5}\times -\frac{6}{21}\\ =\frac{9}{4}\times \frac{7}{5}\times \frac{6}{21}\\ \frac{6}{21}=\frac{2}{7}\\ =\frac{9}{4}\times \frac{7}{5}\times \frac{2}{7}\\ =\frac{9\times \:7\times \:2}{4\times \:5\times \:7}\\ =\frac{9\times \:2}{4\times \:5}\\ =\frac{18}{4\times \:5}\\ =\frac{18}{20}\\ =\frac{9}{10}\\ [vi] \:2\:\frac{1}{9}\times -3\:\frac{1}{2}\\ Convert\:mixed\:numbers\:to\:improper\:fractions\\ 2\frac{1}{9}=\frac{2\times 9+1}{9}=\frac{19}{9}\\ \frac{19}{9}\times \:-3\frac{1}{2}\\ Convert\:mixed\:numbers\:to\:improper\:fractions\\ 3\frac{1}{2}=\frac{3\times 2+1}{2}=\frac{7}{2}\\ \frac{19}{9}\times \:-\frac{7}{2}\\ =-\frac{19}{9}\times \frac{7}{2}\\ =-\frac{133}{9\times \:2}\\ =-\frac{133}{18}\\ Convert\:improper\:fractions\:to\:mixed\:numbers\\ \frac{133}{18}=7\frac{7}{18}\\ =-7\frac{7}{18}\end{array}$$

Question 5: Find the quotient of the following.

$$\begin{array}{l}[i]\left[-6\div \frac{3}{5}\right]\\ [ii]\:-\frac{27}{5}\div \left[-\frac{54}{10}\right]\\ [iii]\:\frac{21}{36}\div \left[-\frac{7}{18}\right]\\ [iv]\:-\frac{7}{12}\div \left[-\frac{3}{13}\right]\\ [v]\:-2\:\frac{1}{9}\div 6\:\frac{1}{9}\\ [vi]\:\frac{2}{15}\div \left[-\frac{8}{45}\right]\end{array}$$

Solution:

$$\begin{array}{l}[i] \left[-6\div \frac{3}{5}\right]\\ =-\frac{6}{\frac{3}{5}}\\ =\frac{6\cdot \:5}{3}\\ =\frac{30}{3}\\ =-\frac{30}{3}\\ =-10\\ [ii] \:-\frac{27}{5}\div \left[-\frac{54}{10}\right]\\ =-\frac{\frac{27}{5}}{-\frac{54}{10}}\\ =-\frac{\frac{27}{5}}{\frac{54}{10}}\\ =-\frac{27\cdot \:10}{5\cdot \:54}\\ =-\frac{27\cdot \:10}{270}\\ =-\frac{270}{270}\\ =-\left(-\frac{270}{270}\right)\\ =-\left(-1\right)\\ =1\\ [iii] \:\frac{21}{36}\div \left[-\frac{7}{18}\right]\\ =\frac{\frac{21}{36}}{-\frac{7}{18}}\\ =-\frac{\frac{21}{36}}{\frac{7}{18}}\\ =-\frac{21\cdot \:18}{36\cdot \:7}\\ =-\frac{21\cdot \:18}{252}\\ =-\frac{378}{252}\\ =-\frac{3}{2}\\ [iv] \:-\frac{7}{12}\div \left[-\frac{3}{13}\right]\\ =-\frac{\frac{7}{12}}{-\frac{3}{13}}\\ =-\frac{\frac{7}{12}}{\frac{3}{13}}\\ =-\frac{7\cdot \:13}{12\cdot \:3}\\ =-\frac{7\cdot \:13}{36}\\ =-\frac{91}{36}\\ =-\left(-\frac{91}{36}\right)\\ =\frac{91}{36}\\ [v] \:-2\:\frac{1}{9}\div 6\:\frac{1}{9}\\ Convert\:mixed\:numbers\:to\:improper\:fractions\\ 2\frac{1}{9}=\frac{2\times 9+1}{9}=\frac{19}{9}\\ -\frac{19}{9}\div \:6\times \frac{1}{9}\\ =-\frac{19}{9}\div \:6\frac{1}{9}\\ Convert\:mixed\:numbers\:to\:improper\:fractions\\ 6\frac{1}{9}=\frac{6\times 9+1}{9}=\frac{55}{9}\\ -\frac{19}{9}\div \frac{55}{9}\\ =-\frac{19}{9}\div \frac{55}{9}\\ =-\frac{19}{9}\times \frac{9}{55}\\ =-\frac{19}{55}\\ [vi] \:\frac{2}{15}\div \left[-\frac{8}{45}\right]\\ =\frac{\frac{2}{15}}{-\frac{8}{45}}\\ =-\frac{\frac{2}{15}}{\frac{8}{45}}\\ =-\frac{2\cdot \:45}{15\cdot \:8}\\ =-\frac{2\cdot \:45}{120}\\ =-\frac{90}{120}\\ =-\frac{3}{4}\end{array}$$

Question 6: Simplify the following.

$$\begin{array}{l}[i]\frac{3}{5}+\frac{7}{10}+\left[-\frac{8}{12}\right]+\frac{4}{3}\\ [ii]\:2\:\frac{1}{2}+\left[-3\:\frac{1}{2}\right]+\left[-2\:\frac{1}{3}\right]+\left[2\:\frac{1}{9}\right]\\ [iii]\:\left[-\frac{7}{5}\right]\times \frac{2}{3}\times \frac{15}{16}\times \left[-\frac{8}{9}\right]\\ [iv]\:\frac{1}{2}\div \left[\left(-\frac{1}{3}\right)\div \left(\frac{2}{7}\right)\right]\end{array}$$

Solution:

$$\begin{array}{l}[i] \frac{3}{5}+\frac{7}{10}+\left[-\frac{8}{12}\right]+\frac{4}{3}\\ =\frac{3}{5}+\frac{7}{10}-\frac{8}{12}+\frac{4}{3}\\ =\frac{3}{5}+\frac{7}{10}-\frac{2}{3}+\frac{4}{3}\\ \mathrm{Combine\:the\:fractions\:}-\frac{2}{3}+\frac{4}{3}:\quad \frac{2}{3}\\ =\frac{3}{5}+\frac{7}{10}+\frac{2}{3}\\ \mathrm{Least\:Common\:Multiplier\:of\:}5,\:10,\:3:\quad 30\\ \frac{3}{5}=\frac{3\cdot \:6}{5\cdot \:6}=\frac{18}{30}\\ \frac{7}{10}=\frac{7\cdot \:3}{10\cdot \:3}=\frac{21}{30}\\ \frac{2}{3}=\frac{2\cdot \:10}{3\cdot \:10}=\frac{20}{30}\\ =\frac{18}{30}+\frac{21}{30}+\frac{20}{30}\\ =\frac{18+21+20}{30}\\ =\frac{59}{30}\\ [ii] \:2\:\frac{1}{2}+\left[-3\:\frac{1}{2}\right]+\left[-2\:\frac{1}{3}\right]+\left[2\:\frac{1}{9}\right]\\ \mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad 2\frac{1}{2}=\frac{5}{2}\\ \mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad 3\frac{1}{2}=\frac{7}{2}\\ \mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad 2\frac{1}{3}=\frac{7}{3}\\ \mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad 2\frac{1}{9}=\frac{19}{9}\\ \frac{5}{2}+\left[-\frac{7}{2}\right]+\left[-\frac{7}{3}\right]+\left[\frac{19}{9}\right]\\ \mathrm{Least\:Common\:Multiplier\:of\:}2,\:2,\:3,\:9:\quad 18\\ \frac{5}{2}=\frac{5\cdot \:9}{2\cdot \:9}=\frac{45}{18}\\ \frac{7}{2}=\frac{7\cdot \:9}{2\cdot \:9}=\frac{63}{18}\\ \frac{7}{3}=\frac{7\cdot \:6}{3\cdot \:6}=\frac{42}{18}\\ \frac{19}{9}=\frac{19\cdot \:2}{9\cdot \:2}=\frac{38}{18}\\ =\frac{45}{18}-\frac{63}{18}-\frac{42}{18}+\frac{38}{18}\\ =\frac{45-63-42+38}{18}\\ =\frac{-22}{18}\\ =-\frac{22}{18}\\ =-\frac{11}{9}\\ \mathrm{Convert\:improper\:fractions\:to\:mixed\:numbers}:\quad \frac{11}{9}=1\frac{2}{9}\\ =-1\frac{2}{9}\\ [iii] \:\left[-\frac{7}{5}\right]\times \frac{2}{3}\times \frac{15}{16}\times \left[-\frac{8}{9}\right]\\ =\frac{7}{5}\times \frac{2}{3}\times \frac{15}{16}\times \frac{8}{9}\\ =\frac{7\times \:2\times \:15\times \:8}{5\times \:3\times \:16\times \:9}\\ =\frac{1680}{5\times \:3\times \:16\times \:9}\\ =\frac{1680}{2160}\\ =\frac{7}{9}\\ [iv] \:\frac{1}{2}\div \left[\left(-\frac{1}{3}\right)\div \left(\frac{2}{7}\right)\right]\\ =\frac{\frac{1}{2}}{\frac{-\frac{1}{3}}{\frac{2}{7}}}\\ =-\frac{\frac{1}{3}}{\frac{2}{7}}\\ =-\frac{1\cdot \:7}{3\cdot \:2}\\ =-\frac{7}{3\cdot \:2}\\ =-\frac{7}{6}\\ \Rightarrow \frac{\frac{1}{2}}{-\frac{7}{6}}\\ =-\frac{\frac{1}{2}}{\frac{7}{6}}\\ =-\frac{1\cdot \:6}{2\cdot \:7}\\ =-\frac{6}{2\cdot \:7}\\ =-\frac{6}{14}\\ =-\frac{3}{7}\end{array}$$

Question 7: Using the properties of rational numbers, find the value of each of them.

$$\begin{array}{l}[i] \frac{3}{5}\times \left[-\frac{3}{7}\right]-\frac{2}{7}\times \frac{3}{2}+\frac{3}{15}\times \frac{5}{9}\\ [ii] \:\frac{5}{2}-\frac{3}{5}\times \frac{7}{2}+\frac{3}{5}\times \left[-\frac{2}{3}\right]\end{array}$$

Solution:

$$\begin{array}{l}[i] \frac{3}{5}\times \left[-\frac{3}{7}\right]-\frac{2}{7}\times \frac{3}{2}+\frac{3}{15}\times \frac{5}{9}\\ =-\frac{3}{5}\times \frac{3}{7}-\frac{2}{7}\times \frac{3}{2}+\frac{3}{15}\times \frac{5}{9}\\ \frac{3}{5}\times \frac{3}{7}=\frac{9}{35}\\ \frac{2}{7}\times \frac{3}{2}=\frac{3}{7}\\ \frac{3}{15}\times \frac{5}{9}=\frac{1}{9}\\ =-\frac{9}{35}-\frac{3}{7}+\frac{1}{9}\\ \mathrm{Least\:Common\:Multiplier\:of\:}35,\:7,\:9:\quad 315\\ \frac{9}{35}=\frac{9\times \:9}{35\times \:9}=\frac{81}{315}\\ \frac{3}{7}=\frac{3\times \:45}{7\times \:45}=\frac{135}{315}\\ \frac{1}{9}=\frac{1\times \:35}{9\times \:35}=\frac{35}{315}\\ =-\frac{81}{315}-\frac{135}{315}+\frac{35}{315}\\ =\frac{-81-135+35}{315}\\ =\frac{-181}{315}\\ =-\frac{181}{315}\\ [ii] \:\frac{5}{2}-\frac{3}{5}\times \frac{7}{2}+\frac{3}{5}\times \left[-\frac{2}{3}\right]\\ =\frac{5}{2}-\frac{3}{5}\times \frac{7}{2}-\frac{3}{5}\times \frac{2}{3}\\ \frac{3}{5}\times \frac{7}{2}=\frac{21}{10}\\ \frac{3}{5}\times \frac{2}{3}=\frac{2}{5}\\ =\frac{5}{2}-\frac{21}{10}-\frac{2}{5}\\ \mathrm{Least\:Common\:Multiplier\:of\:}2,\:10,\:5:\quad 10\\ \frac{5}{2}=\frac{5\times \:5}{2\times \:5}=\frac{25}{10}\\ \frac{2}{5}=\frac{2\times \:2}{5\times \:2}=\frac{4}{10}\\ =\frac{25}{10}-\frac{21}{10}-\frac{4}{10}\\ =\frac{25-21-4}{10}\\ =\frac{0}{10}\\ =0\end{array}$$

Question 8: Write the additive inverse of the following.

$$\begin{array}{l}[i] \frac{7}{19}\\ [ii] \frac{-9}{5}\\ [iii] \frac{-3}{-7}\\ [iv] \frac{5}{-9}\\ [v] \frac{-13}{-17}\\ [vi] \frac{21}{-31}\end{array}$$

Solution:

$$\begin{array}{l}[i] \frac{7}{19}\\\end{array}$$
.

$$\begin{array}{l}\frac{7}{19}\\\end{array}$$
is
$$\begin{array}{l}\frac{-7}{19}\\\end{array}$$
.

$$\begin{array}{l}[ii] \frac{-9}{5}\\\end{array}$$
.

$$\begin{array}{l}\frac{-9}{5}\\\end{array}$$
is
$$\begin{array}{l}\frac{9}{5}\\\end{array}$$
.

$$\begin{array}{l}[iii] \frac{-3}{-7}\\\end{array}$$
.

$$\begin{array}{l}\frac{-3}{-7}\\\end{array}$$
is
$$\begin{array}{l}\frac{-3}{7}\\\end{array}$$
.

$$\begin{array}{l}[iv] \frac{5}{-9}\\\end{array}$$
.

$$\begin{array}{l}\frac{5}{-9}\\\end{array}$$
is
$$\begin{array}{l}\frac{5}{9}\\\end{array}$$
.

$$\begin{array}{l}[v] \frac{-13}{-17}\\\end{array}$$
.

$$\begin{array}{l}\frac{-13}{-17}\\\end{array}$$
is
$$\begin{array}{l}\frac{-13}{17}\\\end{array}$$
.

$$\begin{array}{l}[vi] \frac{21}{-31}\\\end{array}$$
.

$$\begin{array}{l}\frac{21}{-31}\\\end{array}$$
is
$$\begin{array}{l}\frac{21}{31}\\\end{array}$$
.

Question 9: Find the multiplicative inverse of the rational numbers given below.

$$\begin{array}{l}[i] -17\\ [ii] \frac{-11}{17}\\ [iii] -1\times \frac{-3}{5}\\ [iv] \frac{13}{-19}\end{array}$$

Solution:

$$\begin{array}{l}[i] -17\\\end{array}$$
.

The multiplicative inverse of

$$\begin{array}{l}-17\\\end{array}$$
is
$$\begin{array}{l}\frac{-1}{17}\end{array}$$
.

$$\begin{array}{l}[ii] \frac{-11}{17}\\\end{array}$$
.

The multiplicative inverse of

$$\begin{array}{l}\frac{-11}{17}\\\end{array}$$
is
$$\begin{array}{l}\frac{-17}{11}\\\end{array}$$
.

$$\begin{array}{l}[iii] -1\times \frac{-3}{5}\\\end{array}$$
.

The multiplicative inverse of

$$\begin{array}{l}-1\times \frac{-3}{5}\\\end{array}$$
is
$$\begin{array}{l}\frac{5}{3}\\\end{array}$$
.

[iv]
$$\begin{array}{l}\frac{13}{-19}\\\end{array}$$
.

The multiplicative inverse of

$$\begin{array}{l}\frac{13}{-19}\\\end{array}$$
is
$$\begin{array}{l}\frac{-19}{-13}\\\end{array}$$
.

Question 10: Write the product of

$$\begin{array}{l}\frac{5}{7}\\\end{array}$$
and the inverse of
$$\begin{array}{l}\frac{-7}{15}\\\end{array}$$

Solution:

$$\begin{array}{l}\frac{5}{7}\end{array}$$
reciprocal of
$$\begin{array}{l}\frac{-7}{15}\\ =\frac{5}{7} \times \frac{15}{-7}\\ =\frac{75}{-49}\\ =\frac{-75}{49}\end{array}$$

Question 11: Using mean method,

[i] Write any five rational numbers between -3 and 0.

[ii] Write any four rational numbers larger than 0 and smaller than

$$\begin{array}{l}\frac{5}{6}\end{array}$$
.

[iii] Find any three rational numbers between

$$\begin{array}{l}\frac{-3}{4}\end{array}$$
and
$$\begin{array}{l}\frac{5}{6}\end{array}$$

Solution:

[i] Write any five rational numbers between -3 and 0.

$$\begin{array}{l}=\frac{-3+0}{2}\\ -3<\frac{-3}{2}<0\end{array}$$

Rational number between -3 and

$$\begin{array}{l}\frac{-3}{2}\end{array}$$
.

$$\begin{array}{l}=\frac{-3+[\frac{-3}{2}]}{2}\\ =\frac{\frac{-6 +[-3]}{2}}{2}\\ =\frac{\frac{-6-3}{2}}{2}\\ =\frac{-9}{4}\\ -3<\frac{-9}{4}<\frac{-3}{2}<0\end{array}$$

Rational number between -3 and

$$\begin{array}{l}\frac{-9}{4}\end{array}$$
.

$$\begin{array}{l}=\frac{-3+[\frac{-9}{4}]}{2}\\ =\frac{\frac{-3 \times 4+[-9]}{4}}{2}\\ =\frac{-12-9}{4 \times 2}\\ =\frac{-21}{8} -3<\frac{-21}{8}<\frac{-9}{4}<\frac{-3}{2}<0\end{array}$$

Rational number between

$$\begin{array}{l}\frac{-3}{2}\end{array}$$
and 0.

$$\begin{array}{l}=\frac{[\frac{-3}{2}]+0}{2}\\ =\frac{-3}{4} -3<\frac{-21}{8}<\frac{-9}{4}<\frac{-3}{2}<\frac{-3}{4}<0\end{array}$$

Rational number between

$$\begin{array}{l}\frac{-3}{4}\end{array}$$
and 0.

$$\begin{array}{l}=\frac{[\frac{-3}{4}]+0}{2}\\ =\frac{-3}{8} -3<\frac{-21}{8}<\frac{-9}{4}<\frac{-3}{2}<\frac{-3}{4}<\frac{-3}{8}<0\end{array}$$

Hence, the five rational numbers between -3 and 0 are

$$\begin{array}{l}\frac{-21}{8},\frac{-9}{4},\frac{-3}{2},\frac{-3}{4},\frac{-3}{8}<0\\\end{array}$$
.

[ii] Write any four rational numbers larger than 0 and smaller than
$$\begin{array}{l}\frac{5}{6}\end{array}$$
.

Rational number between 0 and

$$\begin{array}{l}\frac{5}{6}\end{array}$$
.

$$\begin{array}{l}=0+\frac{\frac{5}{6}}{2}\\ =\frac{5}{12}\\ 0<\frac{5}{12}<\frac{5}{6}\end{array}$$

Rational number between

$$\begin{array}{l}\frac{5}{6}\end{array}$$
and
$$\begin{array}{l}\frac{5}{12}\end{array}$$
.

$$\begin{array}{l}=\frac{\frac{5}{6}+\frac{5}{12}}{2}\\ =\frac{\frac{10+5}{12}}{2}\\ =\frac{15}{2 \times 2}\\ =\frac{5}{8}\\ 0<\frac{5}{12}<\frac{5}{8}<\frac{5}{6}\end{array}$$

Rational number between

$$\begin{array}{l}\frac{5}{8}\end{array}$$
and
$$\begin{array}{l}\frac{5}{6}\end{array}$$
.

$$\begin{array}{l}=\frac{\frac{5}{8}+\frac{5}{6}}{2}\\ =\frac{\frac{15+20}{24}}{2}\\ =\frac{35}{24 \times 2}\\ =\frac{35}{48}\\ 0<\frac{5}{12}<\frac{5}{8}<\frac{35}{48}<\frac{5}{6}\end{array}$$

Rational number between

$$\begin{array}{l}\frac{35}{48}\end{array}$$
and
$$\begin{array}{l}\frac{5}{6}\end{array}$$
.

$$\begin{array}{l}=\frac{\frac{35}{48}+\frac{5}{6}}{2}\\ =\frac{\frac{35+40}{48}}{2}\\ =\frac{75}{48 \times 2}\\ =\frac{75}{96}\\ 0<\frac{5}{12}<\frac{5}{8}<\frac{35}{48}<\frac{75}{96}<\frac{5}{6}\end{array}$$
.

Hence, the four rational numbers larger than 0 and smaller than

$$\begin{array}{l}\frac{5}{6}\end{array}$$
are
$$\begin{array}{l}\frac{5}{12},\frac{5}{8},\frac{35}{48},\frac{75}{96}\end{array}$$
.

[iii] Find any three rational numbers between
$$\begin{array}{l}\frac{-3}{4}\end{array}$$
and
$$\begin{array}{l}\frac{5}{6}\end{array}$$

Rational number between

$$\begin{array}{l}\frac{-3}{4}\end{array}$$
and
$$\begin{array}{l}\frac{5}{6}\end{array}$$
.

$$\begin{array}{l}=\frac{\frac{-3}{4}+\frac{5}{6}}{2}\\ =\frac{\frac{-9+10}{12}}{2}\\ =\frac{1}{12 \times 2}\\ =\frac{1}{24} \frac{-3}{4}<\frac{1}{24}<\frac{5}{6}\end{array}$$

Rational number between

$$\begin{array}{l}\frac{1}{24}\end{array}$$
and
$$\begin{array}{l}\frac{5}{6}\end{array}$$
.

$$\begin{array}{l}=\frac{\frac{-1}{24}+\frac{5}{6}}{2}\\ =\frac{\frac{1+20}{24}}{2}\\ =\frac{21}{24 \times 2}\\ =\frac{7}{16} \frac{-3}{4}<\frac{1}{24}<\frac{7}{16}<\frac{5}{6}\end{array}$$

Rational number between

$$\begin{array}{l}\frac{-3}{4}\end{array}$$
and
$$\begin{array}{l}\frac{1}{24}\end{array}$$
.

$$\begin{array}{l}=\frac{\frac{-3}{4}+\frac{1}{24}}{2}\\ =\frac{\frac{-18+1}{24}}{2}\\ =\frac{-17}{24 \times 2}\\ =\frac{-17}{48} \frac{-3}{4}<\frac{-17}{48}<\frac{1}{24}<\frac{7}{16}<\frac{5}{6}\end{array}$$
.

Hence, the three rational numbers between

$$\begin{array}{l}\frac{-3}{4}\end{array}$$
and
$$\begin{array}{l}\frac{5}{6}\end{array}$$
are
$$\begin{array}{l}\frac{-17}{48},\frac{1}{24},\frac{7}{16}\end{array}$$
.

### RBSE Maths Chapter 1: Additional Important Questions and Solutions

Question 1: Find the sum of the following rational numbers.

$$\begin{array}{l}[i] -\frac{11}{7}+\frac{4}{7}\\ [ii] \frac{3}{5}+\left[-\frac{2}{5}\right]\\ [iii] -\frac{3}{4}+\left[-\frac{5}{4}\right]\end{array}$$

Solution:

$$\begin{array}{l}[i] -\frac{11}{7}+\frac{4}{7}\\ =\frac{-11+4}{7}\\ =\frac{-7}{7}\\ =-1\\ [ii] \frac{3}{5}+\left(-\frac{2}{5}\right)\\ =\frac{3}{5}-\frac{2}{5}\\ =\frac{3-2}{5}\\ =\frac{1}{5}\\ [iii] -\frac{3}{4}+\left(-\frac{5}{4}\right)\\ =-\frac{3}{4}-\frac{5}{4}\\ =\frac{-8}{4}\\ =-\frac{8}{4}\\ =-2\end{array}$$

Question 2: Find the value of the following.

$$\begin{array}{l}[i] 4\times \left[-\frac{1}{3}\right]\\ [ii] \:\left[-\frac{3}{5}\times 7\right]\\\end{array}$$

Solution:

$$\begin{array}{l}[i] 4\left(-\frac{1}{3}\right)\\ =-4\times \frac{1}{3}\\ =-\frac{1\times \:4}{3}\\ =-\frac{4}{3}\\ [ii] -\frac{3}{5}\times \:7\\ =\frac{3}{5}\times \frac{7}{1}\\ =-\frac{3\times \:7}{5\times \:1}\\ =-\frac{21}{5\times \:1}\\ =-\frac{21}{5}\end{array}$$

Question 3: Solve the following problem using the properties of rational numbers.

Solution:

$$\begin{array}{l}\frac{5}{8}\times \left[-\frac{3}{7}\right]+\frac{5}{8}\times \left[-\frac{7}{6}\right]\\ =-\frac{5}{8}\times \frac{3}{7}-\frac{5}{8}\times \frac{7}{6}\\ \frac{5}{8}\times \frac{3}{7}=\frac{15}{56}\\ \frac{5}{8}\times \frac{7}{6}=\frac{35}{48}\\ =-\frac{15}{56}-\frac{35}{48}\\ \mathrm{Least\:Common\:Multiplier\:of\:}56,\:48:\quad 336\\ \frac{15}{56}=\frac{15\times \:6}{56\times \:6}=\frac{90}{336}\\ \frac{35}{48}=\frac{35\times \:7}{48\times \:7}=\frac{245}{336}\\ =-\frac{90}{336}-\frac{245}{336}\\ =\frac{-335}{336}\\ =-\frac{335}{336}\end{array}$$