Rational Numbers - Exercise 4(E)

Multiply :

(i) $\dfrac{2}{3}$ by $\dfrac{4}{5}$

(ii) $\dfrac{7}{6}$ by $\dfrac{9}{2}$

(iii) $\dfrac{5}{6}$ by 30

(iv) $\dfrac{-3}{4}$ by $\dfrac{8}{7}$

(v) $\dfrac{-16}{9}$ by $\dfrac{12}{-5}$

(vi) $\dfrac{35}{-8}$ by $\dfrac{12}{-5}$

(vii) $\dfrac{-3}{10}$ by $\dfrac{-40}{9}$

(viii) $\dfrac{-32}{5}$ by $\dfrac{15}{-16}$

(ix) $\dfrac{-8}{15}$ by $\dfrac{-25}{32}$

Answer

(i) $\dfrac{2}{3}$ by $\dfrac{4}{5}$

We have:

$\phantom{=} \dfrac{2}{3} \times \dfrac{4}{5} \\[1em] = \dfrac{2 \times 4}{3 \times 5} \\[1em] = \dfrac{8}{15}$

Hence the answer is $\dfrac{8}{15}$

(ii) $\dfrac{7}{6}$ by $\dfrac{9}{2}$

We have:

$\phantom{=} \dfrac{7}{6} \times \dfrac{9}{2} \\[1em] = \dfrac{7 \times 9}{6 \times 2} \\[1em] = \dfrac{63}{12} \\[1em] = \dfrac{21}{4} \quad \text{[Dividing both by 3]}$

Hence the answer is $\dfrac{21}{4}$

(iii) $\dfrac{5}{6}$ by 30

We have:

$\phantom{=} \dfrac{5}{6} \times 30 \\[1em] = \dfrac{5}{6} \times \dfrac{30}{1} \\[1em] = \dfrac{5 \times 30}{6 \times 1} \\[1em] = \dfrac{150}{6} \\[1em] = 25$

Hence the answer is 25

(iv) $\dfrac{-3}{4}$ by $\dfrac{8}{7}$

We have:

$\phantom{=} \dfrac{-3}{4} \times \dfrac{8}{7} \\[1em] = \dfrac{-3 \times 8}{4 \times 7} \\[1em] = \dfrac{-24}{28} \\[1em] = \dfrac{-6}{7} \quad \text{[Dividing both by 4]}$

Hence the answer is $\dfrac{-6}{7}$

(v) $\dfrac{-16}{9}$ by $\dfrac{12}{-5}$

We have:

$\phantom{=} \dfrac{-16}{9} \times \dfrac{12 \times (-1)}{-5 \times (-1)} \\[1em] \text{[Making denominator positive]} \\[1em] = \dfrac{-16}{9} \times \dfrac{-12}{5} \\[1em] = \dfrac{(-16) \times (-12)}{9 \times 5} \\[1em] = \dfrac{192}{45} \\[1em] = \dfrac{64}{15} \quad \text{[Dividing both by 3]}$

Hence the answer is $\dfrac{64}{15}$

(vi) $\dfrac{35}{-8}$ by $\dfrac{12}{-5}$

We have:

$\phantom{=} \dfrac{35 \times (-1)}{-8 \times (-1)} \times \dfrac{12 \times (-1)}{-5 \times (-1)} \\[1em] \text{[Making denominator positive]} \\[1em] = \dfrac{-35}{8} \times \dfrac{-12}{5} \\[1em] = \dfrac{35 \times 12}{(8) \times (5)} \\[1em] = \dfrac{420}{40} \\[1em] = \dfrac{42}{4} \\[1em] = \dfrac{21}{2} \quad \text{[Dividing both by 2]}$

Hence the answer is $\dfrac{21}{2}$

(vii) $\dfrac{-3}{10}$ by $\dfrac{-40}{9}$

We have:

$\phantom{=} \dfrac{-3}{10} \times \dfrac{-40}{9} \\[1em] = \dfrac{(-3) \times (-40)}{10 \times 9} \\[1em] = \dfrac{120}{90} \\[1em] = \dfrac{12}{9} \\[1em] = \dfrac{4}{3} \quad \text{[Dividing both by 3]}$

Hence the answer is $\dfrac{4}{3}$

(viii) $\dfrac{-32}{5}$ by $\dfrac{15}{-16}$

We have:

$\phantom{=} \dfrac{-32}{5} \times \dfrac{15 \times (-1)}{-16 \times (-1)} \\[1em] = \dfrac{(-32) \times -15}{5 \times (16)} \\[1em] = \dfrac{480}{80} \\[1em] = 6 \hspace{2cm}\text{[Dividing both by 80]}$

Hence the answer is 6

(ix) $\dfrac{-8}{15}$ by $\dfrac{-25}{32}$

We have:

$\phantom{=} \dfrac{-8}{15} \times \dfrac{-25}{32} \\[1em] = \dfrac{(-8) \times (-25)}{15 \times 32} \\[1em] = \dfrac{200}{480} \\[1em] = \dfrac{20}{48} \\[1em] = \dfrac{5}{12} \quad \text{[Dividing both by 4]}$

Hence the answer is $\dfrac{5}{12}$

Simplify :

(i) $\dfrac{7}{15} \times \dfrac{5}{6}$

(ii) $\dfrac{-5}{24} \times \dfrac{6}{25}$

(iii) $\dfrac{7}{-18} \times \dfrac{-9}{14}$

(iv) $\dfrac{-9}{5} \times \dfrac{-10}{3}$

(v) $-28 \times \dfrac{-8}{7}$

(vi) $\dfrac{8}{-21} \times \dfrac{-14}{3}$

Answer

(i) $\dfrac{7}{15} \times \dfrac{5}{6}$

We have:

$\phantom{=} \dfrac{7}{15} \times \dfrac{5}{6} \\[1em] = \dfrac{7}{3} \times \dfrac{1}{6} \\[1em] = \dfrac{7 \times 1}{3 \times 6} \\[1em] = \dfrac{7}{18}$

Hence the answer is $\dfrac{7}{18}$

(ii) $\dfrac{-5}{24} \times \dfrac{6}{25}$

We have:

$\phantom{=} \dfrac{-5}{24} \times \dfrac{6}{25} \\[1em] = \dfrac{-1}{24} \times \dfrac{6}{5} \\[1em] = \dfrac{-1}{4} \times \dfrac{1}{5} \\[1em] = \dfrac{-1 \times 1}{4 \times 5} \\[1em] = \dfrac{-1}{20}$

Hence the answer is $\dfrac{-1}{20}$

(iii) $\dfrac{7}{-18} \times \dfrac{-9}{14}$

First, express $\dfrac{7}{-18}$ with a positive denominator: $\dfrac{7 \times (-1)}{18 \times (-1)} = \dfrac{-7}{18}$.

We have:

$\phantom{=} \dfrac{-7}{18} \times \dfrac{-9}{14} \\[1em] = \dfrac{-1}{18} \times \dfrac{-9}{2} \\[1em] = \dfrac{(-1) \times (-9)}{18 \times 2} \\[1em] = \dfrac{(1) \times (9)}{18 \times 2} \\[1em] = \dfrac{9}{36} \\[1em] = \dfrac{1}{4}$

Hence the answer is $\dfrac{1}{4}$

(iv) $\dfrac{-9}{5} \times \dfrac{-10}{3}$

We have:

$\phantom{=} \dfrac{-9}{5} \times \dfrac{-10}{3} \\[1em] = \dfrac{-3}{5} \times \dfrac{-10}{1} \\[1em] = \dfrac{-3}{1} \times \dfrac{-2}{1} \\[1em] = \dfrac{(-3) \times (-2)}{1 \times 1} \\[1em] = \dfrac{3 \times 2}{1 \times 1} \\[1em] = \dfrac{6}{1} = 6$

Hence the answer is 6

(v) $-28 \times \dfrac{-8}{7}$

Express -28 as $\dfrac{-28}{1}$

We have:

$\phantom{=} \dfrac{-28}{1} \times \dfrac{-8}{7} \\[1em] = \dfrac{-4}{1} \times \dfrac{-8}{1} \\[1em] = \dfrac{(-4) \times (-8)}{1 \times 1} \\[1em] = \dfrac{(4) \times (8)}{1 \times 1} \\[1em] = \dfrac{32}{1} = 32$

Hence the answer is 32

(vi) $\dfrac{8}{-21} \times \dfrac{-14}{3}$

First, express $\dfrac{8}{-21}$ with a positive denominator: $\dfrac{8 \times (-1)}{-21 \times (-1)} = \dfrac{-8}{21}$.

We have:

$\phantom{=} \dfrac{-8}{21} \times \dfrac{-14}{3} \\[1em] = \dfrac{-8}{3} \times \dfrac{-2}{3} \\[1em] = \dfrac{(-8) \times (-2)}{3 \times 3} \\[1em] = \dfrac{8 \times 2}{3 \times 3} \\[1em] = \dfrac{16}{9}$

Hence the answer is $\dfrac{16}{9}$

Simplify :

(i) $\dfrac{5}{12} \times (-36)$

(ii) $\dfrac{-17}{18} \times 12$

(iii) $\dfrac{-5}{6} \times \dfrac{6}{5}$

(iv) $-14 \times \dfrac{9}{28}$

(v) $-4\dfrac{4}{5} \times \Big(-7\dfrac{1}{2}\Big)$

(vi) $\dfrac{-8}{15} \times \dfrac{-25}{32}$

Answer

(i) $\dfrac{5}{12} \times (-36)$

Express -36 as $\dfrac{-36}{1}$

We have:

$\phantom{=} \dfrac{5}{12} \times \dfrac{-36}{1} \\[1em] = \dfrac{5}{1} \times \dfrac{-3}{1} \\[1em] = \dfrac{5 \times (-3)}{1 \times 1} \\[1em] = \dfrac{-15}{1} = -15$

Hence the answer is -15

(ii) $\dfrac{-17}{18} \times 12$

Express 12 as $\dfrac{12}{1}$

We have:

$\phantom{=} \dfrac{-17}{18} \times \dfrac{12}{1} \\[1em] = \dfrac{-17}{3} \times \dfrac{2}{1} \\[1em] = \dfrac{-17 \times 2}{3 \times 1} \\[1em] = \dfrac{-34}{3}$

Hence the answer is $\dfrac{-34}{3}$

(iii) $\dfrac{-5}{6} \times \dfrac{6}{5}$

We have:

$\phantom{=} \dfrac{-5}{6} \times \dfrac{6}{5} \\[1em] = \dfrac{-6}{1} \times \dfrac{1}{6} \\[1em] = \dfrac{-1}{1} \times \dfrac{1}{1} \\[1em] = \dfrac{-1 \times 1}{1 \times 1} \\[1em] = \dfrac{-1}{1} = -1$

Hence the answer is -1

(iv) $-14 \times \dfrac{9}{28}$

Express -14 as $\dfrac{-14}{1}$

We have:

$\phantom{=} \dfrac{-14}{1} \times \dfrac{9}{28} \\[1em] = \dfrac{-1}{1} \times \dfrac{9}{2} \\[1em] = \dfrac{-1 \times 9}{1 \times 2} \\[1em] = \dfrac{-9}{2}$

Hence the answer is $\dfrac{-9}{2}$

(v) $-4\dfrac{4}{5} \times \Big(-7\dfrac{1}{2}\Big)$

We have:

$\phantom{=} \dfrac{-24}{5} \times \dfrac{-15}{2} \\[1em] \text{[Converting mixed to improper fraction]} \\[1em] = \dfrac{-12}{5} \times \dfrac{-15}{1} \\[1em] = \dfrac{-12}{1} \times \dfrac{-3}{1} \\[1em] = \dfrac{-12 \times (-3)}{1 \times 1} \\[1em] = \dfrac{12 \times 3}{1 \times 1} \\[1em] = \dfrac{36}{1} = 36$

Hence the answer is 36

(vi) $\dfrac{-8}{15} \times \dfrac{-25}{32}$

We have:

$\phantom{=} \dfrac{-8}{15} \times \dfrac{-25}{32} \\[1em] = \dfrac{-1}{15} \times \dfrac{-25}{4} \\[1em] = \dfrac{-1}{3} \times \dfrac{-5}{4} \\[1em] = \dfrac{-1 \times (-5)}{3 \times 4} \\[1em] = \dfrac{1 \times 5}{3 \times 4} \\[1em] = \dfrac{5}{12}$

Hence the answer is $\dfrac{5}{12}$

Simplify :

(i) $\Big(\dfrac{2}{5} \times \dfrac{5}{8}\Big) + \Big(\dfrac{-3}{7} \times \dfrac{14}{-15}\Big)$

(ii) $\Big(\dfrac{-14}{3} \times \dfrac{-12}{7}\Big) + \Big(\dfrac{-6}{25} \times \dfrac{15}{8}\Big)$

(iii) $\Big(\dfrac{6}{25} \times \dfrac{-15}{8}\Big) - \Big(\dfrac{13}{100} \times \dfrac{-25}{26}\Big)$

(iv) $\Big(\dfrac{-14}{5} \times \dfrac{-10}{7}\Big) - \Big(\dfrac{-8}{9} \times \dfrac{3}{16}\Big)$

Answer

(i) $\Big(\dfrac{2}{5} \times \dfrac{5}{8}\Big) + \Big(\dfrac{-3}{7} \times \dfrac{14}{-15}\Big)$

First, express $\dfrac{14}{-15}$ with a positive denominator: $\dfrac{14 \times (-1)}{-15 \times (-1)} = \dfrac{-14}{15}$.

We have:

$\phantom{=} \Big(\dfrac{2}{5} \times \dfrac{5}{8}\Big) + \Big(\dfrac{-3}{7} \times \dfrac{-14}{15}\Big) \\[1em] = \Big(\dfrac{1}{1} \times \dfrac{1}{4}\Big) + \Big(\dfrac{-1}{1} \times \dfrac{-2}{5}\Big) \\[1em] = \Big(\dfrac{1}{1} \times \dfrac{1}{4}\Big) + \Big(\dfrac{-1}{1} \times \dfrac{-2}{5}\Big) \\[1em] = \dfrac{1}{4} + \dfrac{2}{5}$

L.C.M. of 4 and 5 is 20.

Now, expressing each fraction with denominator 20:

$= \dfrac{1 \times 5}{4 \times 5} + \dfrac{2 \times 4}{5 \times 4} \\[1em] = \dfrac{5}{20} + \dfrac{8}{20} \\[1em] = \dfrac{5 + 8}{20} \\[1em] = \dfrac{13}{20}$

Hence the answer is $\dfrac{13}{20}$

(ii) $\Big(\dfrac{-14}{3} \times \dfrac{-12}{7}\Big) + \Big(\dfrac{-6}{25} \times \dfrac{15}{8}\Big)$

We have:

$\phantom{=} \Big(\dfrac{-14}{3} \times \dfrac{-12}{7}\Big) + \Big(\dfrac{-6}{25} \times \dfrac{15}{8}\Big) \\[1em] = \Big(\dfrac{-2}{1} \times \dfrac{-4}{1}\Big) + \Big(\dfrac{-3}{5} \times \dfrac{3}{4}\Big) \\[1em] = \dfrac{8}{1} + \dfrac{-9}{20}$

L.C.M. of 1 and 20 is 20.

Now, expressing each fraction with denominator 20:

$= \dfrac{8 \times 20}{1 \times 20} + \dfrac{-9}{20} \\[1em] = \dfrac{160}{20} + \dfrac{-9}{20} \\[1em] = \dfrac{160 - 9}{20} \\[1em] = \dfrac{151}{20}$

Hence the answer is $\dfrac{151}{20}$

(iii) $\Big(\dfrac{6}{25} \times \dfrac{-15}{8}\Big) - \Big(\dfrac{13}{100} \times \dfrac{-25}{26}\Big)$

We have:

$\phantom{=} \Big(\dfrac{6}{25} \times \dfrac{-15}{8}\Big) - \Big(\dfrac{13}{100} \times \dfrac{-25}{26}\Big) \\[1em] = \Big(\dfrac{3}{5} \times \dfrac{-3}{4}\Big) - \Big(\dfrac{1}{4} \times \dfrac{-1}{2}\Big) \\[1em] = \Big(\dfrac{3 \times (-3)}{5 \times 4}\Big) - \Big(\dfrac{1 \times (-1)}{4 \times 2}\Big) \\[1em] = \Big(\dfrac{-9}{20}\Big) - \Big(\dfrac{-1}{8}\Big)$

L.C.M. of 20 and 8 is 40.

Now, expressing each fraction with denominator 40:

$= \dfrac{-9 \times 2}{20 \times 2} - \dfrac{-1 \times 5}{8 \times 5} \\[1em] = \dfrac{-18}{40} - \Big(\dfrac{-5}{40}\Big) \\[1em] = \dfrac{-18}{40} + \dfrac{5}{40} \\[1em] = \dfrac{-18 + 5}{40} \\[1em] = \dfrac{-13}{40}$

Hence the answer is $\dfrac{-13}{40}$.

(iv) $\Big(\dfrac{-14}{5} \times \dfrac{-10}{7}\Big) - \Big(\dfrac{-8}{9} \times \dfrac{3}{16}\Big)$

We have:

$\phantom{=} \Big(\dfrac{-14}{5} \times \dfrac{-10}{7}\Big) - \Big(\dfrac{-8}{9} \times \dfrac{3}{16}\Big) \\[1em] = \Big(\dfrac{-2}{1} \times \dfrac{-2}{1}\Big) - \Big(\dfrac{-1}{3} \times \dfrac{1}{2}\Big) \\[1em] = \Big(\dfrac{-2 \times (-2)}{1 \times 1}\Big) - \Big(\dfrac{-1 \times 1}{3 \times 2}\Big) \\[1em] = \dfrac{4}{1} - \Big(\dfrac{-1}{6}\Big) \\[1em] = \dfrac{4}{1} + \dfrac{1}{6}$

L.C.M. of 1 and 6 is 6.

Now, expressing each fraction with denominator 6:

$= \dfrac{4 \times 6}{1 \times 6} + \dfrac{1}{6} \\[1em] \dfrac{24}{6} + \dfrac{1}{6} \\[1em] = \dfrac{24 + 1}{6} \\[1em] = \dfrac{25}{6}$

Hence the answer is $\dfrac{25}{6}$.

Find the cost of $3\dfrac{1}{3}$ kg of rice at ₹ $40\dfrac{1}{2}$ per kg.

Answer

Given:

Quantity of rice = $3\dfrac{1}{3}$ kg = $\dfrac{10}{3}$ kg

Cost per kg = ₹ $40\dfrac{1}{2}$ = ₹ $\dfrac{81}{2}$

Cost of $3\dfrac{1}{3}$ kg of rice = ?

Cost of $3\dfrac{1}{3}$ kg of rice = (Quantity of rice) x (Cost per kg)

Substituting the values in above, we get:

Cost of $3\dfrac{1}{3}$ kg of rice = $\dfrac{10}{3}$ kg x ₹ $\dfrac{81}{2}$

$= ₹ \dfrac{10}{3} \times \dfrac{81}{2} \\[1em] = ₹ \dfrac{5}{1} \times \dfrac{27}{1} \\[1em] = ₹ \dfrac{5 \times 27}{1 \times 1} \\[1em] = ₹ \dfrac{135}{1} \\[1em] = ₹ 135$

Hence, the cost of $3\dfrac{1}{3}$ kg of rice is ₹ 135.

Find the distance covered by a car in $2\dfrac{2}{5}$ hours at a speed of $46\dfrac{2}{3}$ km per hour.

Answer

Given:

Time taken = $2\dfrac{2}{5}$ hours = $\dfrac{12}{5}$ hours

Speed of the car = $46\dfrac{2}{3}$ km per hour = $\dfrac{140}{3}$ km per hour

Total distance = ?

We know the formula,

Distance = Speed x Time

Substituting the values in above, we get:

Distance = $\dfrac{140}{3}$ km per hour x $\dfrac{12}{5}$ hours

$= \dfrac{140}{3} \times \dfrac{12}{5} \text{ km} \\[1em] = \dfrac{28}{1} \times \dfrac{4}{1} \text{ km} \\[1em] = \dfrac{28 \times 4}{1 \times 1} \text{ km} \\[1em] = \dfrac{112}{1} \text{ km} \\[1em] = 112 \text{ km}$

Hence, the distance covered by the car is 112 km.

Write the multiplicative inverse of :

(i) $\dfrac{5}{6}$

(ii) $\dfrac{-3}{7}$

(iii) -8

(iv) $\dfrac{-11}{3}$

(v) $\dfrac{-1}{8}$

Answer

(i) $\dfrac{5}{6}$

Since, $\dfrac{5}{6}$ x $\dfrac{6}{5}$ = 1

∴ The multiplicative inverse of $\dfrac{5}{6}$ is $\dfrac{6}{5}$.

(ii) $\dfrac{-3}{7}$

Since, $\dfrac{-3}{7}$ x $\dfrac{-7}{3}$ = 1

∴ The multiplicative inverse of $\dfrac{-3}{7}$ is $\dfrac{-7}{3}$.

(iii) -8

Express -8 as $\dfrac{-8}{1}$

Since, $\dfrac{-8}{1}$ x $\dfrac{-1}{8}$ = 1

∴ The multiplicative inverse of $\dfrac{-8}{1}$ is $\dfrac{-1}{8}$.

(iv) $\dfrac{-11}{3}$

Since, $\dfrac{-11}{3}$ x $\dfrac{-3}{11}$ = 1

∴ The multiplicative inverse of $\dfrac{-11}{3}$ is $\dfrac{-3}{11}$.

(v) $\dfrac{-1}{8}$

Since, $\dfrac{-1}{8}$ x $\dfrac{-8}{1}$ = 1

∴ The multiplicative inverse of $\dfrac{-1}{8}$ is -8.