Exercise 1.1

Solution :

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

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

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

$\left| {\,\,{\rm{by}}\,{\rm{commutativity}}\,{\rm{ of}}\,{\rm{ multiplication}}} \right.$

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

$\left| {\,{\rm{by}}\,{\rm{ commutativity }}\,{\rm{of }}\,{\rm{addition}}} \right.$

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

$\,\,\,\,\,\,\,\left| {{\rm{ by\,\, distributivity}}} \right.$

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

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

$= \frac{{ - 1}}{2} + \frac{5}{2} = \frac{{( - 1) + 5}}{2} = \frac{4}{2} = 2.$

Solution :

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

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

$= \frac{2}{5} \times \left( { - \frac{3}{7}} \right) - \frac{1}{6} \times \frac{3}{2} + \frac{2}{5} \times \frac{1}{{14}}$

$\left| {{\rm{ by\,\, commutativity}}\,{\rm{ of}}\,{\rm{ multiplication}}} \right.$

$= \frac{2}{5} \times \left( { - \frac{3}{7}} \right) + \frac{2}{5} \times \frac{1}{{14}} - \frac{1}{6} \times \frac{3}{2}$

$\,\,\,\,\left| {\,{\rm{by\,\, commutativity}}\,{\rm{ of}}\,{\rm{ addition}}} \right.$

$= \frac{2}{5} \times \left\{ {\left( { - \frac{3}{7}} \right) + \frac{1}{{14}}} \right\} - \frac{1}{6} \times \frac{3}{2}$

$\,\,\,\,\,\,\,\,\,\,\left| {\,{\rm{by}}\,{\rm{ distributivity}}} \right.$

$= \frac{2}{5} \times \left\{ {\frac{{( - 6) + 1}}{{14}}} \right\} - \frac{1}{6} \times \frac{3}{2}$

$= \frac{2}{5} \times \left\{ {\frac{{ - 5}}{{14}}} \right\} - \frac{1}{6} \times \frac{3}{2} = \frac{{ - 1}}{7} - \frac{1}{4}$

$= \frac{{ - 4 - 7}}{{28}} = \frac{{ - 11}}{{28}}$

Solution :

$\frac{2}{8}$

${\rm{Additive}}\,{\rm{inverse}}\,{\rm{of}}\,\,\,\frac{2}{8}\,\,\,{\rm{is}}\,\,\, - \frac{2}{8}.$

Solution :

$- \frac{5}{9}$

${\rm{Additive}}\,{\rm{inverse}}\,{\rm{of}}\,\,\, - \frac{5}{9}\,\,\,{\rm{is}}\,\,\,\frac{5}{9}.$

Solution :

$\frac{{ - 6}}{{ - 5}}$

$\frac{{ - 6}}{{ - 5}} = \frac{6}{5}$

${\rm{Additive}}\,{\rm{inverse}}\,{\rm{of}}\frac{{ - 6}}{{ - 5}}\,{\rm{is}}\,\frac{{ - 6}}{5}$

Solution :

$\frac{2}{{ - 9}}$

${\rm{Additive}}\,{\rm{inverse}}\,{\rm{of}}\,\frac{2}{{ - 9}}\,{\rm{is}}\,\frac{2}{9}$

Solution :

$\,\frac{{19}}{{ - 6}}\,$

${\rm{Additive}}\,{\rm{inverse}}\,{\rm{of}}\,\frac{{19}}{{ - 6}}\,{\rm{is}}\,\frac{{19}}{6}$

Solution :

$x = \frac{{11}}{{15}}$

${\rm{LHS}} = - ( - x)$

$= - \left( { - \frac{{11}}{{15}}} \right) = \frac{{11}}{{15}}$

$= x = {\rm{RHS}}$

Solution :

$x = - \frac{{13}}{{17}}$

${\rm{LHS}} = - ( - x)$

$= - \left\{ { - \left( {\frac{{ - 13}}{{17}}} \right)} \right\}$

$= - \frac{{13}}{{17}} = x = {\rm{RHS}}$

Solution :

$\, - 13$

${\rm{The}}\,{\rm{multiplicative}}\,{\rm{inverse}}\,{\rm{of}}\, - 13\,{\rm{is}}\,\frac{{ - 1}}{{13}}.$

Solution :

$\frac{{ - 13}}{{19}}$

${\rm{The}}\,{\rm{multiplicative}}\,{\rm{inverse}}\,{\rm{of}}\,\frac{{ - 13}}{9}\,{\rm{is}}\,\frac{{ - 19}}{{13}}.$

Solution :

$\frac{1}{5}$

${\rm{The}}\,{\rm{multiplicative}}\,{\rm{inverse}}\,{\rm{of}}\,\frac{1}{5}\,{\rm{is}}\,5.$

Solution :

$\frac{{ - 5}}{8} \times \frac{{ - 3}}{7}$

$\frac{{ - 5}}{8} \times \frac{{ - 3}}{7} = \frac{{( - 5) \times ( - 3)}}{{8 \times 7}} = \frac{{15}}{{56}}$

${\rm{Therefore,}}\,{\rm{the}}\,{\rm{multiplicative}}\,{\rm{inverse}}\,{\rm{of}}$

$\frac{{ - 5}}{8} \times \frac{{ - 3}}{7}\,{\rm{is}}\,\frac{{56}}{{15}}$

Solution :

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

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

${\rm{Therefore,}}\,{\rm{the}}\,{\rm{multiplicative}}\,{\rm{inverse}}\,{\rm{of}}$

$- 1 \times \frac{{ - 2}}{5}\,{\rm{is}}\,\frac{5}{2}.$

Solution :

$- 1$

${\rm{The}}\,{\rm{multiplicative}}\,{\rm{inverse}}\,{\rm{of}}\, - 1\,{\rm{is}}\, - 1.$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,|\,\,\,\,( - 1) \times ( - 1) = 1$

Solution :

1 is the multiplicative identity

Solution :

Commutativity of multiplication

Solution :

Multiplicative inverse.

Solution :

Reciprocal of $\frac{{ - 7}}{{16}}\,{\rm{is}}\,\frac{{ - 16}}{7}$

Now,

$\frac{6}{{13}} \times \frac{{ - 16}}{7} = \frac{{6 \times ( - 16)}}{{13 \times 7}} = \frac{{ - 96}}{{91}}$

Solution :

Associativity.

Solution :

$- 1\frac{1}{8} = - \frac{9}{8}$

Now, $\frac{8}{9} \times \frac{{ - 9}}{8} = - 1 \ne 1$

So, No; $\frac{8}{9}$ is not the multiplicative inverse of $- 1\frac{1}{8}\left( { = - \frac{9}{8}} \right)$ because the product of $\frac{8}{9}$ and $- 1\frac{1}{8}\left( { = - \frac{9}{8}} \right)$ is not 1.

Solution :

Yes; 0.3 is the multiplicative inverse of $\frac{{10}}{3}$ because

$\frac{3}{{10}} \times \frac{{10}}{3} = \frac{{3 \times 10}}{{10 \times 3}} = \frac{{30}}{{30}} = 1$

Solution :

(i) The rational number ‘0’ does not have a reciprocal.

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

(iii) The rational number 0 is equal to its

Solution :

(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 $x \ne 0$ is x.

(v) The product of two rational numbers is always a rational number

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

Class 8th Ncert Solutions

Chapter List

Rational Numbers

Exercise 1.1

Class 8th Ncert Solutions

Chapter List

Rational Numbers

Exercise 1.1

Q 1.(i) : Using appropriate properties find: −25×35+52−35×16 - \frac{2}{5} \times \frac{3}{5} + \frac{5}{2} - \frac{3}{5} \times \frac{1}{6}

Q 1.(ii) : Using appropriate properties find: 25×(−37)−16×32+114×25\frac{2}{5} \times \left( { - \frac{3}{7}} \right) - \frac{1}{6} \times \frac{3}{2} + \frac{1}{{14}} \times \frac{2}{5}

Q 2.(i) : Write the additive inverse of the following: 28\frac{2}{8}

Q 2.(ii) : Write the additive inverse of the following: −59\frac{{ - 5}}{9}

Q 2.(iii) : Write the additive inverse of the following: −6−5\frac{{ - 6}}{{ - 5}}

Q 2.(iv) : Write the additive inverse of the following: 2−9\frac{2}{{ - 9}}

Q 2.(v) : Write the additive inverse of the following: 19−6.\frac{{19}}{{ - 6}}.

Q 3.(i) : Verify that −(−x)=x - ( - x) = x for: x=1115x = \frac{{11}}{{15}}

Q 3.(ii) : Verify that −(−x)=x - ( - x) = x for: x=−1317x = - \frac{{13}}{{17}}

Q 4.(i) : Find the multiplicative inverse of the following: −13 - 13\,\,\,\,\,

Q 4.(ii) : Find the multiplicative inverse of the following: −1319\frac{{ - 13}}{{19}}

Q 4.(iii) : Find the multiplicative inverse of the following: 15\frac{1}{5}\,\,\,

Q 4.(iv) : Find the multiplicative inverse of the following: −58×−37\frac{{ - 5}}{8} \times \frac{{ - 3}}{7}

Q 4.(v) : Find the multiplicative inverse of the following: −1×−25\, - 1 \times \frac{{ - 2}}{5}\,\,

Q 4.(vi) : Find the multiplicative inverse of the following: −1.\, - 1.

Q 5.(i) : Name the property under multiplication used in the following: −45×(1)=1×−45=−45\frac{{ - 4}}{5} \times (1) = 1 \times \frac{{ - 4}}{5} = - \frac{4}{5}

Q 5.(ii) : Name the property under multiplication used in the following: −1317×−27=−27×−1317\, - \frac{{13}}{{17}} \times \frac{{ - 2}}{7} = \frac{{ - 2}}{7} \times \frac{{ - 13}}{{17}}

Q 5.(iii) : Name the property under multiplication used in the following: −1929×29−19=1\frac{{ - 19}}{{29}} \times \frac{{29}}{{ - 19}} = 1

Q 6. : Multiply 613\frac{6}{{13}} by the reciprocal of −716\frac{{ - 7}}{{16}}

Q 7. : Tell what property allows you to compute: 13×(6×43)as(13×6)×43\frac{1}{3} \times \left( {6 \times \frac{4}{3}} \right)\,as\,\left( {\frac{1}{3} \times 6} \right) \times \frac{4}{3}

Q 8. : Is 89\frac{8}{9} the multiplicative inverse of −118 - 1\frac{1}{8}? Why or why not?

Q 9. : Is 0.3 the multiplicative inverse of 3133\frac{1}{3}? Why or why not?

Q 10. : 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.