August 9, 2024

# NCERT-class-8-maths-chapter-1-exercise-1.1 Solution

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) $\mathbf{-\frac{2}{3}\times&space;\frac{3}{5}+\frac{5}{2}-\frac{3}{5}\times&space;\frac{1}{6}}$

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

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

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

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

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

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

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

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

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

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

= $-\frac{1}{7}-\frac{1}{4}&space;=&space;\frac{-4-7}{28}$

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

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

(i) $\frac{2&space;}{8}$

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

(ii)  $-\frac{5}{9}$

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

(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 $-\frac{2}{9}+\frac{2}{9}&space;=0$ , so  additive inverse of $\frac{2}{-9}$   is  $\frac{2}{9}$ .

(v)  $\mathbf{\frac{19}{-6}}$

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

3. Verify that $-&space;(-&space;x)&space;=&space;x$  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 $-\frac{11}{15}+\frac{11}{15}&space;=0$ .

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

(ii) We have $x=-\frac{13}{17}$ ,

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

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

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 $\frac{a}{b}\times&space;\frac{c}{d}&space;=1$ .

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

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

Sol. Since  $\frac{(-13)}{19}\times&space;\frac{19}{(-13)}&space;=1$ , 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 $\frac{1}{5}\times&space;5=1$.

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

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

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

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

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

Since $\frac{2}{5}\times&space;\frac{5}{2}=1$,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) $\mathbf{\frac{-4}{5}\times&space;1=&space;1\times&space;\frac{-4}{5}&space;=\frac{-4}{5}}$

Sol.  1 is the multiplicative identity.

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

Sol. Commutativity of rational numbers.

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

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 $\frac{-7}{16}\times&space;\frac{16}{-7}=1$ .

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

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

Sol. Associativity of rational numbers.

8.  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 their product is not 1.

9. Is 0.3 the multiplicative inverse of $3\frac{1}{3}$ ? Why or why not?

Sol.  Yes, 0.3  is the multiplicative inverse of $3\frac{1}{3}$  because $0.3\times&space;3\frac{1}{3}&space;=&space;\frac{3}{10}&space;\times&space;\frac{10}{3}=1$ .

10. Write.

(i) The rational number that does not have a reciprocal.    Ans. 0

(ii) The rational numbers that are equal to their reciprocals.   Ans. 1 and -1

(iii) The rational number that is equal to its negative.  Ans.  0

11. Fill in the blanks.

(i) Zero has ____no____ reciprocal.

(ii) The numbers ___1_____ and __-1______ are their own reciprocals

(iii) The reciprocal of – 5 is ___$\frac{-1}{5}$_____.

(iv) Reciprocal of  $\frac{1}{x}$where $x\neq&space;0$  is   ….$x$……….

(v) The product of two rational numbers is always a ____rational number___.

(vi) The reciprocal of a positive rational number is __positive______.