Rational Numbers - Exercise 4(F)

Find the multiplicative inverse (or reciprocal) of each of the following rational numbers :

(i) $\dfrac{6}{25}$

(ii) $-2\dfrac{3}{11}$

(iii) $\dfrac{-8}{9}$

(iv) $\dfrac{-23}{16}$

(v) 12

(vi) $\dfrac{1}{10}$

(vii) -6

(viii) -1

(ix) $\dfrac{-1}{5}$

(x) $\dfrac{-7}{-9}$

Answer

(i) $\dfrac{6}{25}$

Since, $\dfrac{6}{25}$ x $\dfrac{25}{6}$ = 1

∴ The multiplicative inverse of $\dfrac{6}{25}$ is $\dfrac{25}{6}$.

(ii) $-2\dfrac{3}{11}$

First, convert the mixed fraction to an improper fraction: $\dfrac{-(2 \times 11 + 3)}{11} = \dfrac{-25}{11}$.

Since, $\dfrac{-25}{11}$ x $\dfrac{-11}{25}$ = 1

∴ The multiplicative inverse of $\dfrac{-25}{11}$ is $\dfrac{-11}{25}$.

(iii) $\dfrac{-8}{9}$

Since, $\dfrac{-8}{9}$ x $\dfrac{-9}{8}$ = 1

∴ The multiplicative inverse of $\dfrac{-8}{9}$ is $\dfrac{-9}{8}$.

(iv) $\dfrac{-23}{16}$

Since, $\dfrac{-23}{16}$ x $\dfrac{-16}{23}$ = 1

∴ The multiplicative inverse of $\dfrac{-23}{16}$ is $\dfrac{-16}{23}$.

(v) 12

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

Since, $\dfrac{12}{1}$ x $\dfrac{1}{12}$ = 1

∴ The multiplicative inverse of $\dfrac{12}{1}$ is $\dfrac{1}{12}$.

(vi) $\dfrac{1}{10}$

Since, $\dfrac{1}{10}$ x $\dfrac{10}{1}$ = 1

∴ The multiplicative inverse of $\dfrac{1}{10}$ is $\dfrac{10}{1}$, which is 10.

(vii) -6

Express -6 as $\dfrac{-6}{1}$.

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

∴ The multiplicative inverse of $\dfrac{-6}{1}$ is $\dfrac{-1}{6}$.

(viii) -1

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

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

∴ The multiplicative inverse of $\dfrac{-1}{1}$ is $\dfrac{-1}{1}$, which is still -1.

(ix) $\dfrac{-1}{5}$

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

∴ The multiplicative inverse of $\dfrac{-1}{5}$ is $\dfrac{-5}{1}$, which is -5.

(x) $\dfrac{-7}{-9}$

First, simplify the fraction: $\dfrac{-7}{-9} = \dfrac{7}{9}$.

Since, $\dfrac{7}{9}$ x $\dfrac{9}{7}$ = 1

∴ The multiplicative inverse of $\dfrac{7}{9}$ is $\dfrac{9}{7}$.

Evaluate :

(i) $\dfrac{7}{12} ÷ \dfrac{-4}{3}$

(ii) $\dfrac{-12}{25} ÷ \dfrac{-5}{6}$

(iii) $\dfrac{-27}{32} ÷ \dfrac{-9}{16}$

(iv) $-2\dfrac{4}{7} ÷ \dfrac{6}{35}$

(v) $26 ÷ \dfrac{-1}{13}$

(vi) $\dfrac{1}{25} ÷ -5$

Answer

(i) $\dfrac{7}{12} ÷ \dfrac{-4}{3}$

We have:

$\dfrac{7}{12} \div \dfrac{-4}{3} \\[1em] = \dfrac{7}{12} \times \dfrac{3}{-4} \quad \left[\text{Reciprocal of } \dfrac{-4}{3} \text{ is } \dfrac{3}{-4}\right] \\[1em] =\dfrac{7}{12} \times \dfrac{3 \times (-1)}{-4 \times (-1)} \\[1em] = \dfrac{7}{12} \times \dfrac{-3}{4} \\[1em] = \dfrac{7 \times (-3)}{12 \times 4} \\[1em] = \dfrac{7 \times (-1)}{4 \times 4} \\[1em] = \dfrac{-7}{16}$

Hence, the answer is $\dfrac{-7}{16}$

(ii) $\dfrac{-12}{25} ÷ \dfrac{-5}{6}$

We have:

$\dfrac{-12}{25} \div \dfrac{-5}{6} \\[1em] = \dfrac{-12}{25} \times \dfrac{6}{-5} \quad \left[\text{Reciprocal of } \dfrac{-5}{6} \text{ is } \dfrac{6}{-5}\right] \\[1em] = \dfrac{-12}{25} \times \dfrac{6 \times (-1)}{-5 \times (-1)} \\[1em] = \dfrac{-12}{25} \times \dfrac{-6}{5} \\[1em] = \dfrac{(-12) \times (-6)}{25 \times 5} \\[1em] = \dfrac{72}{125}$

Hence, the answer is $\dfrac{72}{125}$

(iii) $\dfrac{-27}{32} ÷ \dfrac{-9}{16}$

We have:

$\dfrac{-27}{32} \div \dfrac{-9}{16} \\[1em] = \dfrac{-27}{32} \times \dfrac{16}{-9} \quad \left[\text{Reciprocal of } \dfrac{-9}{16} \text{ is } \dfrac{16}{-9}\right] \\[1em] = \dfrac{-27}{32} \times \dfrac{16 \times (-1)}{-9 \times (-1)} \\[1em] = \dfrac{-27}{32} \times \dfrac{-16}{9} \\[1em] = \dfrac{-3}{32} \times \dfrac{-16}{1} \\[1em] = \dfrac{-3}{2} \times \dfrac{-1}{1} \\[1em] = \dfrac{(-3) \times (-1)}{2 \times 1} \\[1em] = \dfrac{3}{2}$

Hence, the answer is $\dfrac{3}{2}$

(iv) $-2\dfrac{4}{7} ÷ \dfrac{6}{35}$

We have:

$\phantom{=} -2\dfrac{4}{7} ÷ \dfrac{6}{35} \\[1em] = -\dfrac{18}{7} ÷ \dfrac{6}{35} \\[1em] = \dfrac{-18}{7} \times \dfrac{35}{6} \quad \left[\text{Reciprocal of } \dfrac{6}{35} \text{ is } \dfrac{35}{6}\right] \\[1em] = \dfrac{-3}{7} \times \dfrac{35}{1} \\[1em] = \dfrac{-3}{1} \times \dfrac{5}{1} \\[1em] = \dfrac{-3 \times 5}{1 \times 1} \\[1em] = \dfrac{-15}{1} \\[1em] = -15$

Hence, the answer is -15

(v) $26 ÷ \dfrac{-1}{13}$

Express 26 as $\dfrac{26}{1}$

We have:

$\dfrac{26}{1} \div \dfrac{-1}{13} \\[1em] = \dfrac{26}{1} \times \dfrac{13}{-1} \quad \left[\text{Reciprocal of } \dfrac{-1}{13} \text{ is } \dfrac{13}{-1}\right] \\[1em] \text{Make denominator positive} \\[1em] = \dfrac{26}{1} \times \dfrac{13 \times (-1)}{-1 \times (-1)} \\[1em] = \dfrac{26}{1} \times \dfrac{-13}{1} \\[1em] = \dfrac{26 \times (-13)}{1 \times 1} \\[1em] = \dfrac{-338}{1} \\[1em] = -338$

Hence, the answer is -338

(vi) $\dfrac{1}{25} ÷ -5$

Express -5 as $\dfrac{-5}{1}$

$\dfrac{1}{25} \div \dfrac{-5}{1} \\[1em] = \dfrac{1}{25} \times \dfrac{1}{-5} \quad \left[\text{Reciprocal of } \dfrac{-5}{1} \text{ is } \dfrac{1}{-5}\right] \\[1em] = \dfrac{1}{25} \times \dfrac{1 \times (-1)}{-5 \times (-1)} \\[1em] = \dfrac{1}{25} \times \dfrac{-1}{5} \\[1em] = \dfrac{1 \times (-1)}{25 \times (5)} \\[1em] = \dfrac{-1}{125} \\[1em]$

Hence, the answer is $\dfrac{-1}{125}$

The product of two rational numbers is $\dfrac{2}{5}$. If one of them is $\dfrac{-8}{25}$, find the other.

Answer

Let p and q be two rational numbers.

One rational number = p = $\dfrac{-8}{25}$

Other rational number = ?

Product of two rational numbers = p x q = $\dfrac{2}{5}$

q = $\dfrac{2}{5}$ ÷ p

Substituting the values in above, we get:

$\text{q} = \dfrac{2}{5} \div \dfrac{-8}{25} \\[1em] = \dfrac{2}{5} \times \dfrac{25}{-8} \quad \left[\text{Reciprocal of } \dfrac{-8}{25} \text{ is } \dfrac{25}{-8}\right] \\[1em] = \dfrac{\overset{1}{\cancel{2}}}{\underset{1}{\cancel{5}}} \times \dfrac{\overset{-5}{\cancel{-25}}}{\underset{4}{\cancel{8}}} \quad \left[\because \dfrac{25}{-8} = \dfrac{25 \times (-1)}{-8 \times (-1)} = \dfrac{-25}{8}\right] \\[1em] = \dfrac{1}{1} \times \dfrac{-5}{4} \\[1em] = \dfrac{1 \times -5}{1 \times 4} \\[1em] = \dfrac{-5}{4}$

The other rational number q is $\dfrac{-5}{4}$.

The product of two rational numbers is $\dfrac{-2}{3}$. If one of them is $\dfrac{16}{39}$, find the other.

Answer

Let p and q be two rational numbers.

One rational number = p = $\dfrac{16}{39}$

Other rational number = q = ?

Product of two rational numbers = p x q = $\dfrac{-2}{3}$

q = $\dfrac{-2}{3}$ ÷ p

Substituting the values in above, we get:

$\text{q} = \dfrac{-2}{3} \div \dfrac{16}{39} \\[1em] = \dfrac{-2}{3} \times \dfrac{39}{16} \quad \left[\text{Reciprocal of } \dfrac{16}{39} \text{ is } \dfrac{39}{16}\right] \\[1em] = \dfrac{\overset{-1}{\cancel{-2}}}{\underset{1}{\cancel{3}}} \times \dfrac{\overset{13}{\cancel{39}}}{\underset{8}{\cancel{16}}} \\[1em] = \dfrac{-1}{1} \times \dfrac{13}{8} \\[1em] = \dfrac{-1 \times 13}{1 \times 8} \\[1em] = \dfrac{-13}{8}$

The other rational number q is $\dfrac{-13}{8}$.

By what rational number should $\dfrac{-9}{35}$ be multiplied to get $\dfrac{3}{5}$ ?

Answer

Let the required number be x. Then,

$\dfrac{-9}{35} \times x = \dfrac{3}{5} \\[1em] \Rightarrow x = \dfrac{3}{5} \div \dfrac{-9}{35} \\[1em] = \dfrac{3}{5} \times \dfrac{35}{-9} \quad \left[\text{Reciprocal of } \dfrac{-9}{35} \text{ is } \dfrac{35}{-9}\right] \\[1em] = \dfrac{3}{5} \times \dfrac{-35}{9} \quad \left[\because \dfrac{35}{-9} = \dfrac{35 \times (-1)}{-9 \times (-1)} = \dfrac{-35}{9}\right] \\[1em] = \dfrac{\overset{1}{\cancel{3}}}{\underset{1}{\cancel{5}}} \times \dfrac{\overset{-7}{\cancel{-35}}}{\underset{3}{\cancel{9}}} \\[1em] = \dfrac{1}{1} \times \dfrac{-7}{3} \quad \text{[Dividing 3 and 9 by 3, 35 and 5 by 5]} \\[1em] = \dfrac{1 \times (-7)}{1 \times 3} \\[1em] = \dfrac{-7}{3}$

Hence, the required number is $\dfrac{-7}{3}$.

By what rational number should $\dfrac{25}{8}$ be multiplied to get $\dfrac{-20}{7}$ ?

Answer

Let the required number be x. Then,

$\dfrac{25}{8} \times x = \dfrac{-20}{7} \\[1em] \Rightarrow x = \dfrac{-20}{7} \div \dfrac{25}{8} \\[1em] = \dfrac{-20}{7} \times \dfrac{8}{25} \quad \left[\text{Reciprocal of } \dfrac{25}{8} \text{ is } \dfrac{8}{25}\right] \\[1em] = \dfrac{-4}{7} \times \dfrac{8}{5} \quad \text{[Dividing 20 and 25 by 5]} \\[1em] = \dfrac{-4 \times 8}{7 \times 5} \\[1em] = \dfrac{-32}{35}$

The required number is $\dfrac{-32}{35}$.

The cost of 17 pencils is ₹ $59\dfrac{1}{2}$. Find the cost of each pencil.

Answer

Given:

Total cost = ₹ $59\dfrac{1}{2}$ = ₹ $\dfrac{119}{2}$

Total pencils = 17

Cost of 1 pencil = ?

Cost of 1 pencil = Total cost ÷ Total pencils

Substituting the values in above, we get:

Cost of 1 pencil = ₹ $\dfrac{119}{2} ÷ 17$

$= ₹ \dfrac{119}{2} \div \dfrac{17}{1} \\[1em] = ₹ \dfrac{119}{2} \times \dfrac{1}{17} \quad \left[\text{Reciprocal of } \dfrac{17}{1} \text{ is } \dfrac{1}{17}\right] \\[1em] = ₹ \dfrac{7}{2} \times \dfrac{1}{1} \quad \text{[Dividing 119 by 17]} \\[1em] = ₹ \dfrac{7 \times 1}{2 \times 1} \\[1em] = ₹ \dfrac{7}{2} = ₹ 3\dfrac{1}{2}$  

The cost of each pencil is ₹ $3\dfrac{1}{2}$.

The cost of 20 metres of ribbon is ₹335. Find the cost of each metre of it.

Answer

Given:

Total cost = ₹ 335

Total length = 20 metres

Cost of 1 metre of ribbon = ?

Cost of 1 metre of ribbon = Total cost ÷ Total length

Substituting the values in above, we get:

Cost of 1 metre of ribbon = ₹ 335 ÷ 20 metres

$= ₹ \dfrac{335}{20} \\[1em] = ₹ \dfrac{67}{4} \quad \text{[Dividing 335 and 20 by 5]} \\[1em] = ₹ 16\dfrac{3}{4} \quad \text{[Converting to mixed fraction]}$

The cost of each metre is ₹ $16\dfrac{3}{4}$.

How many pieces, each of length $2\dfrac{3}{4}$m, can be cut from a rope of length 66 m ?

Answer

Given:

Total length of rope = 66 m

Length of each piece = $2\dfrac{3}{4}$ m = $\dfrac{11}{4}$ m

Number of pieces = ?

Number of pieces = (Total length ÷ length of each piece)

Substituting the values in above, we get:

Number of pieces = 66 m ÷ $\dfrac{11}{4}$ m

$= \dfrac{66}{1} \times \dfrac{4}{11} \quad \left[\text{Reciprocal of }\dfrac{11}{4} \text{ is } \dfrac{4}{11}\right] \\[1em] = \dfrac{6}{1} \times \dfrac{4}{1} \quad \text{[Dividing 66 by 11]} \\[1em] = \dfrac{6 \times 4}{1 \times 1} \\[1em] = \dfrac{24}{1} = 24$

Hence, 24 pieces can be cut from the rope.

Fill in the blanks :

(i) (...............) ÷ $\Big(\dfrac{-5}{6}\Big)$ = -30

(ii) (...............) ÷ (-8) = $\Big(\dfrac{-3}{4}\Big)$

(iii) $\Big(\dfrac{-15}{14}\Big)$ ÷ (...............) = $\dfrac{5}{2}$

(iv) (-16) ÷ (...............) = 6

Answer

(i) 25 ÷ $\Big(\dfrac{-5}{6}\Big)$ = -30

(ii) 6 ÷ (-8) = $\Big(\dfrac{-3}{4}\Big)$

(iii) $\Big(\dfrac{-15}{14}\Big)$ ÷ $\Big(\dfrac{\bold{-3}}{\bold{7}}\Big)$ = $\dfrac{5}{2}$

(iv) (-16) ÷ $\Big(\dfrac{\bold{-8}}{\bold{3}}\Big)$ = 6

Explanation

(i) To find the dividend, multiply the quotient by the divisor: $-30 \times \dfrac{-5}{6} = 25$.

(ii) The dividend is found by multiplying the quotient and divisor: $\dfrac{-3}{4} \times -8 = 6$.

(iii) To find the divisor, divide the dividend by the quotient: $\dfrac{-15}{14} \div \dfrac{5}{2} = \dfrac{-3}{7}$.

(iv) The missing divisor is calculated by dividing the dividend by the result: $-16 \div 6 = \dfrac{-8}{3}$.