Ncert Class 8 Maths Chapter 1 exercise 1.1

NCERT solutions for class 8 maths| Chapter 1 Rational numbers

A detailed and step-wise solutions to all the questions at the end of the chapter from the NCERT Maths book are given below:

Ncert-class-8-maths-chapter-1-exercise-1.1

1. Using appropriate properties find

(i) $\mathbf{-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}}$

Sol. $-\frac{2}{3}\times \frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times \frac{1}{6}$

= $-\frac{2}{3}\times \frac{3}{5}-\frac{3}{5}\times \frac{1}{6}+\frac{5}{2}$ (by commutativity)

= $\frac{3}{5}\times \left ( (-\frac{2}{3})+(-\frac{1}{6})\right )+\frac{5}{2}$ (by distributivity)

= $\frac{3}{5} \times \left ( \frac{(-4)+(-1)}{6} \right )+\frac{5}{2}$

= $\frac{3}{5}\times (-\frac{5}{6})+\frac{5}{2} = -\frac{1}{2}+\frac{5}{2}$

= $\frac{4}{2}=2$

(ii) $\mathbf{\frac{2}{5}\times (-\frac{3}{7})-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}}$

Sol. $\frac{2}{5}\times (-\frac{3}{7})-\frac{1}{6}\times \frac{3}{2}+\frac{1}{14} \times \frac{2}{5}=\frac{2}{5}\times (-\frac{3}{7})+\frac{1}{14} \times \frac{2}{5}-\frac{1}{6}\times \frac{3}{2}$ (by commutativity)

= $\frac{2}{5}\times \left ( (-\frac{3}{7}) +\frac{1}{14}\right )-\frac{1}{4}$

= $\frac{2}{5 }\times \left ( \frac{-6+1}{14} \right )-\frac{1}{4}$ = $\frac{2}{5 }\times \left ( \frac{-5}{14} \right )-\frac{1}{4}$

= $-\frac{1}{7}-\frac{1}{4} = \frac{-4-7}{28}$

= $\frac{-11}{28}$

2. Write the additive inverse of each of the following:

(i) $\frac{2 }{8}$

Sol. $-\frac{2}{8}$ is the additive inverse of $\frac{2 }{8}$ because $-\frac{2}{8}+\frac{2}{8} =\frac{-2+2}{8}=\frac{0}{8} =0$ .

(ii) $-\frac{5}{9}$

Sol. $\frac{5}{9}$ is the additive inverse of $-\frac{5 }{9}$ because $-\frac{5}{9}+\frac{5}{9} =\frac{-5+5}{9}=\frac{0}{9} =0$ .

(iii) $\frac{-6}{-5}$

Sol. $\frac{-6}{-5}=\frac{6}{5}$ . So additive inverse of $\frac{6}{5}$ is $-\frac{6}{5}$ .

(iv) $\frac{2}{-9}$

Sol. Since $-\frac{2}{9}+\frac{2}{9} =0$ , so additive inverse of $\frac{2}{-9}$ is $\frac{2}{9}$ .

(v) $\mathbf{\frac{19}{-6}}$

Sol. Since $-\frac{19}{6}+\frac{19}{6} =0$ , so additive inverse of $\frac{19}{-6}$ is $\frac{19}{6}$ .

3. Verify that $- (- x) = x$ for (i) $x=\frac{11}{15}$ (ii) $x=-\frac{13}{17}$

Sol. (i) We have $x=\frac{11}{15}$ ,

The additive inverse of $x=\frac{11}{15}$ is $-x=-\frac{11}{15}$ since $-\frac{11}{15}+\frac{11}{15} =0$ .

The same equality $-\frac{11}{15}+\frac{11}{15} =0$ , shows that the additive inverse of $-\frac{11}{15}$ is $\frac{11}{15}$ or $-(-\frac{11}{15}) =\frac{11}{15}$ , i.e. $- (- x) = x$ .

(ii) We have $x=-\frac{13}{17}$ ,

The additive inverse of $x=-\frac{13}{17}$ is $-x=\frac{13}{17}$ since $-\frac{13}{17}+\frac{13}{17} =0$ .

The same equality $-\frac{13}{17}+\frac{13}{17} =0$ , shows that the additive inverse of $\frac{13}{17}$ is $-\frac{13}{17}$ , i.e. $- (- x) = x$ .

4. Find the multiplicative inverse of the following.

(i) – 13

Sol. We say that a rational number $\frac{c}{d}$ is called the reciprocal or multiplicative inverse of another non-zero rational number $\frac{a}{b}$
if $\frac{a}{b}\times \frac{c}{d} =1$ .

Since $(-13)\times \frac{1}{(-13)} =1$ , so multiplicative inverse of -13 is $-\frac{1}{13}$ .

(ii) $\frac{-13}{19}$

Sol. Since $\frac{(-13)}{19}\times \frac{19}{(-13)} =1$ , so multiplicative inverse of – $\frac{(-13)}{19}$ is $-\frac{19}{13}$ .

(iii) $\mathbf{\frac{1}{5}}$

Sol. Multiplicative inverse of $\frac{1}{5}$ is 5 , because $\frac{1}{5}\times 5=1$ .

(iv) $\mathbf{\frac{-5}{8}}\times \frac{-3}{7}$

Sol. $\frac{-5}{8}\times \frac{-3}{7} =\frac{15}{56}$

Since $\frac{15}{56}\times \frac{56}{15}=1$ ,so multiplicative inverse of $\frac{15}{56}$ is $\frac{56}{15}$ .

(v) $-1\times \frac{-2}{5}$

Sol $-1\times \frac{-2}{5} =\frac{2}{5}$

Since $\frac{2}{5}\times \frac{5}{2}=1$ ,so multiplicative inverse of $\frac{2}{5}$ is $\frac{5}{2}$ .

(vi) – 1

Sol. Multiplicative inverse of -1 is -1 , because (-1) x (-1) =1 .

5. Name the property under multiplication used in each of the following

(i) $\mathbf{\frac{-4}{5}\times 1= 1\times \frac{-4}{5} =\frac{-4}{5}}$

Sol. 1 is the multiplicative identity.

(ii) $\mathbf{-\frac{13}{17}\times \frac{-2}{7} = \frac{-2}{7}\times -\frac{13}{17} }$

Sol. Commutativity of rational numbers.

(iii) $\mathbf{\frac{-19}{29}\times \frac{29}{-19}=1}$

Sol. Multiplicative inverse .

6. Multiply $\mathbf{\frac{6}{13}}$ by the reciprocal of $\mathbf{\frac{-7}{16}}$ .

Sol. The reciprocal of $\mathbf{\frac{-7}{16}}$ is $\frac{16}{-7}$ because $\frac{-7}{16}\times \frac{16}{-7}=1$ .

Now the product of $\mathbf{\frac{6}{13}}$ and $\frac{16}{-7}$ = $\frac{6}{13}\times \frac{16}{-7}= -\frac{96}{91}$ .

7. Tell what property allows you to compute $\mathbf{\frac{1}{3}\times (6\times \frac{4}{3})}$ as $\mathbf{(\frac{1}{3}\times 6)\times \frac{4}{3}}$ .