Rational Numbers — Exercise 1(C)

Find the product :

$\dfrac{4}{9} \times \dfrac{7}{12}$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$\dfrac{4}{9} \times \dfrac{7}{12} = \dfrac{4 \times 7}{9 \times 12} \\[1em] = \dfrac{28}{108} \\[1em] = \dfrac{7}{27}$

Hence, $\mathbf{\dfrac{4}{9} \times \dfrac{7}{12} = \dfrac{7}{27}}$.

Find the product :

$-9 \times \dfrac{7}{18}$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$-9 \times \dfrac{7}{18} = \dfrac{-9}{1} \times \dfrac{7}{18} \\[1em] = \dfrac{(-9) \times 7}{1 \times 18} \\[1em] = \dfrac{-63}{18} = \dfrac{-7}{2} \\[1em] = -3\dfrac{1}{2}$

Hence, $\mathbf{-9 \times \dfrac{7}{18} = -3\dfrac{1}{2}}$.

Find the product :

$\dfrac{-3}{16} \times \dfrac{8}{-15}$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$\dfrac{-3}{16} \times \dfrac{8}{-15} = \dfrac{(-3) \times 8}{16 \times (-15)} \\[1em] = \dfrac{-24}{-240} \\[1em] = \dfrac{1}{10}$

Hence, $\mathbf{\dfrac{-3}{16} \times \dfrac{8}{-15} = \dfrac{1}{10}}$.

Find the product :

$\dfrac{6}{7} \times \dfrac{-21}{12}$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$\dfrac{6}{7} \times \dfrac{-21}{12} = \dfrac{6 \times (-21)}{7 \times 12} \\[1em] = \dfrac{-126}{84} = \dfrac{-3}{2} \\[1em] = -1\dfrac{1}{2}$

Hence, $\mathbf{\dfrac{6}{7} \times \dfrac{-21}{12} = -1\dfrac{1}{2}}$.

Find the product :

$\dfrac{5}{-18} \times \dfrac{-9}{20}$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$\dfrac{5}{-18} \times \dfrac{-9}{20} = \dfrac{5 \times (-9)}{(-18) \times 20} \\[1em] = \dfrac{-45}{-360} \\[1em] = \dfrac{1}{8}$

Hence, $\mathbf{\dfrac{5}{-18} \times \dfrac{-9}{20} = \dfrac{1}{8}}$.

Find the product :

$\dfrac{-13}{15} \times \dfrac{-25}{26}$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$\dfrac{-13}{15} \times \dfrac{-25}{26} = \dfrac{(-13) \times (-25)}{15 \times 26} \\[1em] = \dfrac{325}{390} \\[1em] = \dfrac{5}{6}$

Hence, $\mathbf{\dfrac{-13}{15} \times \dfrac{-25}{26} = \dfrac{5}{6}}$.

Find the product :

$\dfrac{7}{24} \times (-48)$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$\dfrac{7}{24} \times (-48) = \dfrac{7}{24} \times \dfrac{-48}{1} \\[1em] = \dfrac{7 \times (-48)}{24 \times 1} \\[1em] = \dfrac{-336}{24} \\[1em] = -14$

Hence, $\mathbf{\dfrac{7}{24} \times (-48) = -14}$.

Find the product :

$\dfrac{-13}{5} \times (-10)$

Answer

The product of two rational numbers $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ is given by $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$.

Multiplying,

$\dfrac{-13}{5} \times (-10) = \dfrac{-13}{5} \times \dfrac{-10}{1} \\[1em] = \dfrac{(-13) \times (-10)}{5 \times 1} \\[1em] = \dfrac{130}{5} \\[1em] = 26$

Hence, $\mathbf{\dfrac{-13}{5} \times (-10) = 26}$.

Find the multiplicative inverse (or reciprocal) of:

$\dfrac{-17}{12}$

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

Interchanging the numerator and the denominator of $\dfrac{-17}{12}$, we get $\dfrac{12}{-17} = \dfrac{-12}{17}$

$[\text{Check : } \dfrac{-17}{12} \times \dfrac{-12}{17} = 1]$

Hence, the multiplicative inverse of $\mathbf{\dfrac{-17}{12} \text{ is } \dfrac{-12}{17}}$.

Find the multiplicative inverse (or reciprocal) of:

-16

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

Writing -16 in the form $\dfrac{p}{q}$ and interchanging the numerator and the denominator,

$\Rightarrow -16 = \dfrac{-16}{1} \\[1em] \Rightarrow \text{Multiplicative inverse} = \dfrac{1}{-16} = \dfrac{-1}{16}$

Hence, the multiplicative inverse of -16 is $\mathbf{\dfrac{-1}{16}}$.

Find the multiplicative inverse (or reciprocal) of:

$\dfrac{0}{2}$

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

Here $\dfrac{0}{2} = 0$, and there is no rational number which when multiplied by 0 gives 1.

Hence, the multiplicative inverse of $\mathbf{\dfrac{0}{2}}$ does not exist.

Find the multiplicative inverse (or reciprocal) of:

$\dfrac{-3}{-5}$

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

First express the rational number with a positive denominator.

$\dfrac{-3}{-5} = \dfrac{-3 \times (-1)}{-5 \times (-1)} = \dfrac{3}{5}$

Interchanging the numerator and the denominator of $\dfrac{3}{5}$, we get $\dfrac{5}{3}$.

$[\text{Check : } \dfrac{3}{5} \times \dfrac{5}{3} = 1]$

Hence, the multiplicative inverse of $\mathbf{\dfrac{-3}{-5} \text{ is } \dfrac{5}{3}}$.

Find the multiplicative inverse (or reciprocal) of:

$\dfrac{2}{-5}$

Answer

To find the multiplicative inverse (or reciprocal) of a non-zero rational number, interchange its numerator and denominator.

First express the rational number with a positive denominator.

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

Interchanging the numerator and the denominator of $\dfrac{-2}{5}$, we get $\dfrac{5}{-2} = \dfrac{-5}{2}$

$[\text{Check : } \dfrac{-2}{5} \times \dfrac{-5}{2} = 1]$

Hence, the multiplicative inverse of $\mathbf{\dfrac{2}{-5} \text{ is } \dfrac{-5}{2}}$.

Find the quotient:

$\dfrac{17}{8} \div \dfrac{51}{4}$

Answer

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$\dfrac{17}{8} \div \dfrac{51}{4} = \dfrac{17}{8} \times \dfrac{4}{51} \\[1em] = \dfrac{17 \times 4}{8 \times 51} \\[1em] = \dfrac{68}{408} \\[1em] = \dfrac{1}{6}$

Hence, $\mathbf{\dfrac{17}{8} \div \dfrac{51}{4} = \dfrac{1}{6}}$.

Find the quotient:

$\dfrac{-16}{35} \div \dfrac{15}{14}$

Answer

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$\dfrac{-16}{35} \div \dfrac{15}{14} = \dfrac{-16}{35} \times \dfrac{14}{15} \\[1em] = \dfrac{(-16) \times 14}{35 \times 15} \\[1em] = \dfrac{-224}{525} \\[1em] = \dfrac{-32}{75}$

Hence, $\mathbf{\dfrac{-16}{35} \div \dfrac{15}{14} = \dfrac{-32}{75}}$.

Find the quotient:

$\dfrac{-12}{7} \div (-16)$

Answer

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$\dfrac{-12}{7} \div (-16) = \dfrac{-12}{7} \div \dfrac{-16}{1} \\[1em] = \dfrac{-12}{7} \times \dfrac{1}{-16} \\[1em] = \dfrac{(-12) \times 1}{7 \times (-16)} \\[1em] = \dfrac{-12}{-112} = \dfrac{3}{28}$

Hence, $\mathbf{\dfrac{-12}{7} \div (-16) = \dfrac{3}{28}}$.

Find the quotient:

$-9 \div \left(\dfrac{-5}{18}\right)$

Answer

To divide one rational number by another, multiply the dividend by the reciprocal of the divisor.

Dividing,

$-9 \div \left(\dfrac{-5}{18}\right) = \dfrac{-9}{1} \times \dfrac{18}{-5} \\[1em] = \dfrac{(-9) \times 18}{1 \times (-5)} \\[1em] = \dfrac{-162}{-5} = \dfrac{162}{5} \\[1em] = 32\dfrac{2}{5}$

Hence, $\mathbf{-9 \div \left(\dfrac{-5}{18}\right) = 32\dfrac{2}{5}}$.

Name the property of multiplication illustrated by the following statement:

$\dfrac{-8}{9} \times \dfrac{-13}{7} = \dfrac{-13}{7} \times \dfrac{-8}{9}$

Answer

Here the order of the two rational numbers has been changed but the product remains the same.

If $\dfrac{a}{b} \text{ and } \dfrac{c}{d}$ are any two rational numbers, then $\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{c}{d} \times \dfrac{a}{b}$

Hence, the given statement illustrates the commutative law of multiplication.

Name the property of multiplication illustrated by the following statement:

$\left(\dfrac{-3}{4} \times \dfrac{5}{7}\right) \times \dfrac{-9}{11} = \dfrac{-3}{4} \times \left(\dfrac{5}{7} \times \dfrac{-9}{11}\right)$

Answer

Here the grouping of the three rational numbers has been changed but the product remains the same.

If $\dfrac{a}{b}, \dfrac{c}{d} \text{ and } \dfrac{e}{f}$ are any three rational numbers, then

$\left(\dfrac{a}{b} \times \dfrac{c}{d}\right) \times \dfrac{e}{f} = \dfrac{a}{b} \times \left(\dfrac{c}{d} \times \dfrac{e}{f}\right)$

Hence, the given statement illustrates the associative law of multiplication.

Name the property of multiplication illustrated by the following statement:

$\dfrac{-2}{3} \times \left(\dfrac{-5}{6} + \dfrac{7}{8}\right) = \left(\dfrac{-2}{3} \times \dfrac{-5}{6}\right) + \left(\dfrac{-2}{3} \times \dfrac{7}{8}\right)$

Answer

Here a rational number has been multiplied by the sum of two rational numbers, which is the same as the sum of the two separate products.

If $\dfrac{a}{b}, \dfrac{c}{d} \text{ and } \dfrac{e}{f}$ are any three rational numbers, then

$\dfrac{a}{b} \times \left(\dfrac{c}{d} + \dfrac{e}{f}\right) = \left(\dfrac{a}{b} \times \dfrac{c}{d}\right) + \left(\dfrac{a}{b} \times \dfrac{e}{f}\right)$

Hence, the given statement illustrates the distributive law of multiplication over addition.

Name the property of multiplication illustrated by the following statement:

$\dfrac{-18}{7} \times 1 = 1 \times \dfrac{-18}{7} = \dfrac{-18}{7}$

Answer

Here the rational number remains unchanged on being multiplied by 1.

For every rational number $\dfrac{a}{b}$, we have $\dfrac{a}{b} \times 1 = 1 \times \dfrac{a}{b} = \dfrac{a}{b}$.

Hence, the given statement illustrates the existence of multiplicative identity, 1 being the multiplicative identity for rational numbers.

Name the property of multiplication illustrated by the following statement:

$\dfrac{-13}{17} \times \dfrac{17}{-13} = \dfrac{17}{-13} \times \dfrac{-13}{17} = 1$

Answer

Here the product of the rational number and its reciprocal is 1.

For every non-zero rational number $\dfrac{a}{b}$, we have $\dfrac{a}{b} \times \dfrac{b}{a} = \dfrac{b}{a} \times \dfrac{a}{b} = 1$

Hence, the given statement illustrates the existence of multiplicative inverse (or reciprocal).

Name the property of multiplication illustrated by the following statement:

$\dfrac{-9}{7} \times 0 = 0$

Answer

Here the product of the rational number and 0 is 0.

For every rational number $\dfrac{a}{b}$, we have $\dfrac{a}{b} \times 0 = 0 \times \dfrac{a}{b} = 0$

Hence, the given statement illustrates the multiplicative property of zero.

Verify whether the given statement is true or false:

$\dfrac{-8}{9} \div \dfrac{-4}{3} = \dfrac{-4}{3} \div \dfrac{-8}{9}$

Answer

Solve the L.H.S. and the R.H.S. separately by multiplying the dividend with the reciprocal of the divisor, and then compare them.

Solving L.H.S.,

$\Rightarrow \dfrac{-8}{9} \div \dfrac{-4}{3} \\[1em] \Rightarrow \dfrac{-8}{9} \times \dfrac{3}{-4} \\[1em] \Rightarrow \dfrac{(-8) \times 3}{9 \times (-4)} = \dfrac{-24}{-36} = \dfrac{2}{3}$

Solving R.H.S.,

$\Rightarrow \dfrac{-4}{3} \div \dfrac{-8}{9} \\[1em] \Rightarrow \dfrac{-4}{3} \times \dfrac{9}{-8} \\[1em] \Rightarrow \dfrac{(-4) \times 9}{3 \times (-8)} = \dfrac{-36}{-24} = \dfrac{3}{2}$

As L.H.S. $≠$ R.H.S., division of rational numbers is not commutative.

Hence, the given statement is false.

Verify whether the given statement is true or false:

$\dfrac{-7}{24} \div \dfrac{3}{-16} = \dfrac{3}{-16} \div \dfrac{-7}{24}$

Answer

Solve the L.H.S. and the R.H.S. separately by multiplying the dividend with the reciprocal of the divisor, and then compare them.

Solving L.H.S.,

$\Rightarrow \dfrac{-7}{24} \div \dfrac{3}{-16} \\[1em] \Rightarrow \dfrac{-7}{24} \times \dfrac{-16}{3} \\[1em] \Rightarrow \dfrac{(-7) \times (-16)}{24 \times 3} = \dfrac{112}{72} = \dfrac{14}{9}$

Solving R.H.S.,

$\Rightarrow \dfrac{3}{-16} \div \dfrac{-7}{24} \\[1em] \Rightarrow \dfrac{3}{-16} \times \dfrac{24}{-7} \\[1em] \Rightarrow \dfrac{3 \times 24}{(-16) \times (-7)} = \dfrac{72}{112} = \dfrac{9}{14}$

As L.H.S. $≠$ R.H.S., division of rational numbers is not commutative.

Hence, the given statement is false.

Verify whether the given statement is true or false:

$\left(\dfrac{-3}{5} \div \dfrac{-12}{35}\right) \div \dfrac{1}{4} = \dfrac{-3}{5} \div \left(\dfrac{-12}{35} \div \dfrac{1}{4}\right)$

Answer

Solve the L.H.S. and the R.H.S. separately by first simplifying the bracket, and then compare them.

Solving L.H.S.,

$\Rightarrow \dfrac{-3}{5} \div \dfrac{-12}{35} = \dfrac{-3}{5} \times \dfrac{35}{-12} \\[1em] \Rightarrow \dfrac{(-3) \times 35}{5 \times (-12)} = \dfrac{-105}{-60} = \dfrac{7}{4} \\[1em] \therefore \left(\dfrac{-3}{5} \div \dfrac{-12}{35}\right) \div \dfrac{1}{4} = \dfrac{7}{4} \div \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{7}{4} \times \dfrac{4}{1} = 7$

Solving R.H.S.,

$\Rightarrow \dfrac{-12}{35} \div \dfrac{1}{4} = \dfrac{-12}{35} \times \dfrac{4}{1} = \dfrac{-48}{35} \\[1em] \therefore \dfrac{-3}{5} \div \left(\dfrac{-12}{35} \div \dfrac{1}{4}\right) = \dfrac{-3}{5} \div \dfrac{-48}{35} \\[1em] \Rightarrow \dfrac{-3}{5} \times \dfrac{35}{-48} = \dfrac{-105}{-240} = \dfrac{7}{16}$

As L.H.S. $≠$ R.H.S., division of rational numbers is not associative.

Hence, the given statement is false.

Fill in the blanks:

(i) ......... is the rational number that does not have a multiplicative inverse.

(ii) ......... and ......... are the rational numbers which are equal to their own reciprocals.

(iii) The product of a rational number and its reciprocal is .........

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

(v) The reciprocal of a negative rational number is .........

(vi) ......... is not the reciprocal of any number.

(vii) The multiplicative inverse of a rational number $\dfrac{1}{a}, a ≠ 0$, is .........

Answer

(i) There is no rational number which when multiplied by 0 gives 1.

Hence, 0 is the rational number that does not have a multiplicative inverse.

(ii) As 1 × 1 = 1 and (-1) × (-1) = 1, each of these numbers is its own reciprocal.

Hence, 1 and -1 are the rational numbers which are equal to their own reciprocals.

(iii) For every non-zero rational number $\dfrac{a}{b}$, we have $\dfrac{a}{b} \times \dfrac{b}{a} = 1$.

Hence, the product of a rational number and its reciprocal is 1.

(iv) The reciprocal is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change. For example, the reciprocal of $\dfrac{3}{5} \text{ is } \dfrac{5}{3}$.

Hence, the reciprocal of a positive rational number is a positive rational number.

(v) The reciprocal is obtained by interchanging the numerator and the denominator, so the sign of the rational number does not change. For example, the reciprocal of $\dfrac{-3}{5} \text{ is } \dfrac{5}{-3} = \dfrac{-5}{3}$.

Hence, the reciprocal of a negative rational number is a negative rational number.

(vi) As the product of any rational number and 0 is 0 and never 1, no rational number has 0 as its reciprocal.

Hence, 0 is not the reciprocal of any number.

(vii) Interchanging the numerator and the denominator of $\dfrac{1}{a}$, we get $\dfrac{a}{1} = a$.

Check : $\dfrac{1}{a} \times a = 1$

Hence, the multiplicative inverse of a rational number $\bm{\dfrac{1}{a}}, a ≠ 0$ is $\bm{a}$.

The product of two rational numbers is -7. If one of the numbers is $\dfrac{-8}{11}$, find the other.

Answer

Let the other rational number be $x$.

As the product of the two rational numbers is -7,

$\Rightarrow \dfrac{-8}{11} \times x = -7 \\[1em] \Rightarrow x = -7 \div \dfrac{-8}{11} \\[1em] \Rightarrow x = \dfrac{-7}{1} \times \dfrac{11}{-8} \\[1em] \Rightarrow x = \dfrac{(-7) \times 11}{1 \times (-8)} \\[1em] \Rightarrow x = \dfrac{-77}{-8} = \dfrac{77}{8} \\[1em] \Rightarrow x = 9\dfrac{5}{8}$

Hence, the other rational number is $\mathbf{\dfrac{77}{8}}$.

The product of two rational numbers is $\dfrac{-16}{9}$. If one of the numbers is $\dfrac{-4}{3}$, find the other.

Answer

Let the other rational number be $x$.

As the product of the two rational numbers is $\dfrac{-16}{9}$,

$\Rightarrow \dfrac{-4}{3} \times x = \dfrac{-16}{9} \\[1em] \Rightarrow x = \dfrac{-16}{9} \div \dfrac{-4}{3} \\[1em] \Rightarrow x = \dfrac{-16}{9} \times \dfrac{3}{-4} \\[1em] \Rightarrow x = \dfrac{(-16) \times 3}{9 \times (-4)} \\[1em] \Rightarrow x = \dfrac{-48}{-36} = \dfrac{4}{3} \\[1em] \Rightarrow x = 1\dfrac{1}{3}$

Hence, the other rational number is $\mathbf{\dfrac{4}{3}}$.

By what rational number must $\dfrac{1}{26}$ be divided to get $\dfrac{-8}{39}$?

Answer

Let the required rational number be $x$.

According to the given condition,

$\Rightarrow \dfrac{1}{26} \div x = \dfrac{-8}{39} \\[1em] \Rightarrow x = \dfrac{1}{26} \div \dfrac{-8}{39} \\[1em] \Rightarrow x = \dfrac{1}{26} \times \dfrac{39}{-8} \\[1em] \Rightarrow x = \dfrac{1 \times 39}{26 \times (-8)} \\[1em] \Rightarrow x = \dfrac{39}{-208} = \dfrac{-3}{16}$

Hence, the required rational number is $\mathbf{\dfrac{-3}{16}}$.

Divide the sum of $\dfrac{13}{5} \text{ and } \dfrac{-12}{7}$ by the product of $\dfrac{-31}{7} \text{ and } \dfrac{1}{-2}$.

Answer

First find the sum of $\dfrac{13}{5} \text{ and } \dfrac{-12}{7}$ and the product of $\dfrac{-31}{7} \text{ and } \dfrac{1}{-2}$.

Note that LCM of 5 and 7 is 35.

Finding the sum,

$\dfrac{13}{5} + \dfrac{-12}{7} = \dfrac{13 \times 7 + (-12) \times 5}{35} \\[1em] = \dfrac{91 - 60}{35} \\[1em] = \dfrac{31}{35}$

Finding the product,

$\dfrac{-31}{7} \times \dfrac{1}{-2} = \dfrac{(-31) \times 1}{7 \times (-2)} \\[1em] = \dfrac{-31}{-14} = \dfrac{31}{14}$

Dividing the sum by the product,

$\dfrac{31}{35} \div \dfrac{31}{14} = \dfrac{31}{35} \times \dfrac{14}{31} \\[1em] = \dfrac{31 \times 14}{35 \times 31} \\[1em] = \dfrac{14}{35} = \dfrac{2}{5}$

Hence, the required quotient is $\mathbf{\dfrac{2}{5}}$.

Divide the sum of $\dfrac{65}{12} \text{ and } \dfrac{8}{3}$ by their difference.

[Hint: Here we have to find $\left(\dfrac{65}{12} + \dfrac{8}{3}\right) \div \left(\dfrac{65}{12} - \dfrac{8}{3}\right)$]

Answer

First find the sum and the difference of the given rational numbers.

LCM of 12 and 3 = 12

Finding the sum,

$\dfrac{65}{12} + \dfrac{8}{3} = \dfrac{65 + 8 \times 4}{12} \\[1em] = \dfrac{65 + 32}{12} \\[1em] = \dfrac{97}{12}$

Finding the difference,

$\dfrac{65}{12} - \dfrac{8}{3} = \dfrac{65 - 8 \times 4}{12} \\[1em] = \dfrac{65 - 32}{12} \\[1em] = \dfrac{33}{12} = \dfrac{11}{4}$

Dividing the sum by the difference,

$\dfrac{97}{12} \div \dfrac{11}{4} = \dfrac{97}{12} \times \dfrac{4}{11} \\[1em] = \dfrac{97 \times 4}{12 \times 11} \\[1em] = \dfrac{388}{132} = \dfrac{97}{33} \\[1em] = 2\dfrac{31}{33}$

Hence, the required quotient is $\mathbf{2\dfrac{31}{33}}$.

Find the cost of $3\dfrac{2}{5}$ metres of cloth at ₹ $63\dfrac{3}{4}$ per metre.

Answer

We know that total cost = cost of 1 metre × length of cloth.

Length of the cloth $= 3\dfrac{2}{5} \text{ m} = \dfrac{17}{5} \text{ m}$

Cost of 1 metre of cloth $= ₹ 63\dfrac{3}{4} = ₹ \dfrac{255}{4}$

Finding the total cost,

$\dfrac{255}{4} \times \dfrac{17}{5} = \dfrac{255 \times 17}{4 \times 5} \\[1em] = \dfrac{4335}{20} \\[1em] = \dfrac{867}{4} = 216\dfrac{3}{4}$

Hence, the cost of the cloth is $\mathbf{₹ 216\dfrac{3}{4} \text{ or } ₹ 216.75}$.

Find the area of a rectangular part which is $36\dfrac{3}{5}$ m long and $16\dfrac{2}{5}$ m broad.

Answer

We know that area of a rectangle = length × breadth.

Length of the rectangular part $= 36\dfrac{3}{5} \text{ m} = \dfrac{183}{5} \text{ m}$

Breadth of the rectangular part $= 16\dfrac{2}{5} \text{ m} = \dfrac{82}{5} \text{ m}$

Finding the area,

$\dfrac{183}{5} \times \dfrac{82}{5} = \dfrac{183 \times 82}{5 \times 5} \\[1em] = \dfrac{15006}{25} \\[1em] = 600\dfrac{6}{25} \text{ sq. m}$

Hence, the area of the rectangular part is $\mathbf{600\dfrac{6}{25} \text{ sq. m}}$.

Find the area of a square plot of land whose each side measures $8\dfrac{1}{2}$ metres.

[Hint: Area of the plot = side² = $\left(8\dfrac{1}{2}\right)^2$ sq. m = $\left(8\dfrac{1}{2} \times 8\dfrac{1}{2}\right)$ sq. m]

Answer

We know that area of a square = side × side.

Side of the square plot $= 8\dfrac{1}{2} \text{ m} = \dfrac{17}{2} \text{ m}$

Finding the area,

$\dfrac{17}{2} \times \dfrac{17}{2} = \dfrac{17 \times 17}{2 \times 2} \\[1em] = \dfrac{289}{4} \\[1em] = 72\dfrac{1}{4} \text{ sq. m}$

Hence, the area of the square plot is $\mathbf{72\dfrac{1}{4} \text{ sq. m}}$.

A cord of length $71\dfrac{1}{2}$ m has been cut into 26 pieces of equal length. What is the length of each piece?

[Hint: Length of each piece = $71\dfrac{1}{2}$ m $\div$ 26]

Answer

To find the length of each piece, divide the total length of the cord by the number of pieces.

Total length of the cord $= 71\dfrac{1}{2} \text{ m} = \dfrac{143}{2} \text{ m}$

Number of pieces = 26

Finding the length of each piece,

$\dfrac{143}{2} \div 26 = \dfrac{143}{2} \times \dfrac{1}{26} \\[1em] = \dfrac{143 \times 1}{2 \times 26} \\[1em] = \dfrac{143}{52} = \dfrac{11}{4} \\[1em] = 2\dfrac{3}{4} \text{ m}$

Hence, the length of each piece is $\mathbf{2\dfrac{3}{4} \text{ m}}$.

The area of a room is $65\dfrac{1}{4}$ sq. metres. If its breadth is $5\dfrac{1}{16}$ metres, what is its length?

Answer

We know that area of a rectangle = length × breadth, so length = area ÷ breadth.

Area of the room $= 65\dfrac{1}{4} \text{ sq. m} = \dfrac{261}{4} \text{ sq. m}$

Breadth of the room $= 5\dfrac{1}{16} \text{ m} = \dfrac{81}{16} \text{ m}$

Finding the length,

$\dfrac{261}{4} \div \dfrac{81}{16} = \dfrac{261}{4} \times \dfrac{16}{81} \\[1em] = \dfrac{261 \times 16}{4 \times 81} \\[1em] = \dfrac{4176}{324} = \dfrac{116}{9} \\[1em] = 12\dfrac{8}{9} \text{ m}$

Hence, the length of the room is $\mathbf{12\dfrac{8}{9} \text{ m}}$.