# RD Sharma Solutions Class 8 Rational Numbers Exercise 1.7

## RD Sharma Solutions Class 8 Chapter 1 Exercise 1.7

#### Exercise 1.7

Q-1. Divide:

(i) $1\; by \; \frac{1}{2}$

(ii) $5 \; by\; \frac{-5}{7}$

(iii) $\frac{ -3 }{ 4 } \; by \; \frac{1}{2}$

(iv) $\frac{ -7 }{ 8 } \; by \; \frac{-21}{ 16 }$

(v) $\frac{ 7 }{ -4 } \; by \; \frac{ 63 }{ 64 }$

(vi) $0 \; by \; \frac{ -7 }{ 5 }$

(vii) $\frac{ -3 }{ 4 } \; by \; 6$

(viii) $\frac{ 2 }{ 3 } \; by \; \frac{ -7 }{ 12 }$

(ix) $-4 \; by \; \frac{ -3 }{ 5 }$

(x) $\frac{ -3 }{ 13 } \; by \; \frac{ -4 }{ 65 }$

Solution:

(i) $1 \div \frac{ 1 }{ 2 } = 1 \times \frac{ 2 }{ 1 } = 2$

(ii) $5 \div \frac{ -5 }{ 7 } = 5 \times \frac{ 7 }{ -5 } = -7$

(iii) $\frac{ -3 }{ 4 } \div \frac{ 9 }{ -16} = \frac{ -3 }{ 4 } \times \frac{ -16 }{ 9 } = \frac{ 4 }{ 3 }$

(iv) $\frac{ -7 }{ 8 } \div \frac{ -21 }{ 16} = \frac{ -7 }{ 8 } \times \frac{ -16 }{ 21 } = \frac{ 2 }{ 3 }$

(v) $\frac{ -7 }{ 4 } \div \frac{ 63 }{ 64 } = \frac{ 7 }{ -4 } \times \frac{ 64 }{ 63 } = \frac{ -16 }{ 9 }$

(vi) $0 \div \frac{ -7 }{ 5 } = 0 \times \frac{ -5 }{ 7 } = 0$

(vii) $\frac{ -3 }{ 4 } \div -6 = \frac{ -3 }{ 4 } \times \frac{ 1 }{ -6 } = \frac{ 1 }{ 8 }$

(viii) $\frac{ 2 }{ 3 } \div \frac{ -7 }{ 12 } = \frac{ 2 }{ 3 } \times \frac{ 12 }{ -7 } = \frac{ -8 }{ 7 }$

(ix) $-4 \div \frac{ -3 }{ 5 } = -4 \times \frac{ 5 }{ -3 } = \frac{ 20 }{ 3 }$

(x) $\frac{ -3 }{ 13 } \div \frac{ -4 }{ 65 } = \frac{ -3 }{ 13 } \times \frac{ 65 }{ -4 } = \frac{ 15 }{ 4 }$

Q-2. Find the value and express as a rational number in standard form:

(i) $\frac{ 2 }{ 5 } \div \frac{ 26 }{ 15 }$

(ii) $\frac{ 10 }{ 3 } \div \frac{ -35 }{ 12 }$

(iii) $-6 \div \frac{ -8 }{ 17 }$

(iv) $\frac{ -40 }{ 99 } \div \left( -20 \right )$

(v) $\frac{ -22 }{ 27 } \div \frac{ -110 }{ 18 }$

(vi) $\frac{ -36 }{ 125 } \div \frac{ -3 }{ 75 }$

Solution:

(i) $\frac{ 2 }{ 5 } \div \frac{ 26 }{ 15 } = \frac{ 2 }{ 5 } \times \frac{ 15 }{ 26 } = \frac{ 3 }{ 13 }$

(ii) $\frac{ 10 }{ 3 } \div \frac{ -35 }{ 12 } = \frac{ 10 }{ 3 } \times \frac{ 12 }{ -35 } = \frac{ -8 }{ 7 }$

(iii) $-6 \div \frac{ -8 }{ 17 } = -6 \times \frac{ 17 }{ -8 } = \frac{ 51 }{ 4 }$

(iv) $\frac{ -40 }{ 99 } \div \left( -20 \right ) = \frac{ -40 }{ 99 } \times \frac{ 1 }{ -20 } = \frac{ 2 }{ 99 }$

(v) $\frac{ -22 }{ 27 } \div \frac{ -110 }{ 18 } = \frac{ -22 }{ 27 } \times \frac{ 18 }{ -110 } = \frac{ 2 }{ 15 }$

(vi) $\frac{ -36 }{ 125 } \div \frac{ -3 }{ 75 } = \frac{ -36 }{ 125 } \times \frac{ 75 }{ -3 } = \frac{ 36 }{ 5 }$

Q-3. The product of two rational numbers is 15. If one  of the numbers  is -10. Find the other number.

Solution:

Let, the other number be x.

So, $x \times \left ( -10 \right ) = 15$

$\Rightarrow x = \frac{ 15 }{ -10 } = \frac{ 3 }{ -2 }$

So, the other number is $\frac{ -3 }{ 2 }$.

Q-4. The product of two rational numbers is $\frac{-8 }{ 9 }$. If one of the number is $\frac{ -4 }{ 15 }$, Find the other number.

Solution: Let, the other number be x.

So, $x \times \frac{ -4 }{ 15 } = \frac{ -8 }{ 9 }$

$\Rightarrow x = \frac{ -8 }{ 9 } \div \frac{ -4 }{ 15 }$

$\Rightarrow x = \frac{ -8 }{ 9 } \times \frac{ 15 }{ -4 }$

$\Rightarrow x = \frac{ 10 }{ 3 }$

Thus, the other number is $\frac{ 10 }{ 3 }$

Q-5. By what number should we multiply $\frac{ -1 }{ 6 }$ so that the product may be $\frac{ -23 }{ 9 }$?

Solution:

Let, the number be x.

$x \times \frac{ -1 }{ 6 } = \frac{ -23 }{ 9 }$

$\Rightarrow x = \frac{ -23 }{ 9 } \div \frac{ -1 }{ 6 }$

$\Rightarrow x = \frac{ -23 }{ 9 } \times \frac{ 6 }{ -1 }$

$\Rightarrow x = \frac{ 46 }{ 3 }$

Thus, the other number is $\frac{ 46 }{ 3 }$

Q-6. By what number should we multiply $\frac{ -15 }{ 28 }$ so that the product may be $\frac{ -5 }{ 7 }$?

Solution:

Let, the number be x.

$x \times \frac{ -15 }{ 28 } = \frac{ -5 }{ 7 }$

$\Rightarrow x = \frac{ -5 }{ 7 } \div \frac{ -15 }{ 28 }$

$\Rightarrow x = \frac{ -5 }{ 7 } \times \frac{ 28 }{ -15 }$

$\Rightarrow x = \frac{ 4 }{ 3 }$

Thus, the other number is $\frac{ 4 }{ 3 }$

Q-7. By what number should we multiply $\frac{ -8 }{ 13 }$ so that the product may be 24?

Solution:

Let, the number be x.

$x \times \frac{ -8 }{ 13 } = 24$

$\Rightarrow x = 24 \div \frac{ -8 }{ 13 }$

$\Rightarrow x = 24 \times \frac{ 13 }{ -8 }$

$\Rightarrow x = -39$

Thus, the other number is -39.

Q-8. By what number should $\frac{ -3 }{ 4 }$ be multiplied in order to produce  $\frac{ 2 }{ 3 }$?

Solution:

Let, the other number that should be multiplied with $\frac{ -3 }{ 4 }$ to produce $\frac{ 2 }{ 3 }$ be x.

$x \times \frac{ -3 }{ 4 } = \frac{ 2 }{ 3 }$

$\Rightarrow x = \frac{ 2 }{ 3 } \div \frac{ -3 }{ 4 }$

$\Rightarrow x = \frac{ 2 }{ 3 } \times \frac{ 4 }{ -3 }$

$\Rightarrow x = \frac{ -8 }{ 9 }$

Thus, the other number is $\frac{ -8 }{ 9 }$

Q-9. Find $\left ( x + y \right ) \div \left ( x – y \right )$, if

(i) $x = \frac{ 2 }{ 3 } , y = \frac{ 3 }{ 2 }$

(ii) $x = \frac{ 2 }{ 5 } , y = \frac{ 1 }{ 2 }$

(iii) $x = \frac{ 5 }{ 4 } , y = \frac{ -1 }{ 3 }$

(iv) $x = \frac{ 2 }{ 7 } , y = \frac{ 4 }{ 3 }$

(v) $x = \frac{ 1 }{ 4 } , y = \frac{ 3 }{ 2 }$

Solution:

(i) $\left ( x + y \right ) \div \left ( x – y \right )$

= $\left ( \frac{ 2 }{ 3 } + \frac{ 3 }{ 2 } \right ) \div \left ( \frac{ 2 }{ 3 } – \frac{ 3 }{ 2 } \right )$

= $\frac{13}{6} \times \frac{6}{-5}$ = $\frac{-13}{5}$

Thus, $\left ( x + y \right ) \div \left ( x – y \right ) = \frac{ -13 }{ 5 }$

(ii) $\left ( x + y \right ) \div \left ( x – y \right )$

= $\left ( \frac{ 2 }{ 5 } + \frac{ 1 }{ 2 } \right ) \div \left ( \frac{ 2 }{ 5 } – \frac{ 1 }{ 2 } \right )$

= $\frac{9}{10} \times \frac{10}{-1}$ = -9

Thus, $\left ( x + y \right ) \div \left ( x – y \right ) = -9$

(iii) $\left ( x + y \right ) \div \left ( x – y \right )$

= $\left ( \frac{ 5 }{ 4 } + \frac{ -1 }{ 3 } \right ) \div \left ( \frac{ 5 }{ 4 } – \frac{ -1 }{ 3 } \right )$

= $\frac{11}{12} \times \frac{12}{11}$ = $\frac{ 11 }{ 19 }$

Thus, $\left ( x + y \right ) \div \left ( x – y \right ) = \frac{11 }{ 19 }$

(iv) $\left ( x + y \right ) \div \left ( x – y \right )$

= $\left ( \frac{ 2 }{ 7 } + \frac{ 4 }{ 3 } \right ) \div \left ( \frac{ 2 }{ 7 } – \frac{ 4 }{ 3 } \right )$

= $\frac{34}{21} \times \frac{21}{-22}$ = $\frac{-17}{11}$

Thus, $\left ( x + y \right ) \div \left ( x – y \right ) = \frac{ -17 }{ 11 }$

(v) $\left ( x + y \right ) \div \left ( x – y \right )$

= $\left ( \frac{ 1 }{ 4 } + \frac{ 3 }{ 2 } \right ) \div \left ( \frac{ 1 }{ 4 } – \frac{ 3 }{ 2 } \right )$

= $\frac{7}{4} \times \frac{4}{-5}$ = $\frac{ -7 }{ 5 }$

Thus, $\left ( x + y \right ) \div \left ( x – y \right ) = \frac{ -7 }{ 5 }$

Q-10: The cost of $7\frac{2}{3}$ metres of rope is Rs $12\frac{3}{4}$. Find its cost per metres.

Solution: The cost of $7\frac{2}{3}$ metres of rope is Rs. $7\frac{2}{3}$.

Therefore,

Cost per metre = $7\frac{2}{3} \div 7\frac{2}{3}$

$= \frac{ 51 }{ 4 } \div \frac{ 23 }{ 3 }$

$= \frac{ 51 }{ 4 } \times \frac{ 3 }{ 23 }$

$= \frac{ 153 }{ 92 }$ = Rs. $1 \frac{ 61 }{ 92 }$

Hence, the cost of rope per metres = Rs. $1 \frac{ 61 }{ 92 }$

Q-11. The cost of $2\frac{1}{3}$ metres of cloth is Rs. $75\frac{1}{4}$. Find the cost of cloth per metres.

Solution: The cost of $2\frac{1}{3}$ metres of cloth is Rs. $75\frac{1}{4}$.

Therefore,

Cost per metre = $75\frac{1}{4} \div 2\frac{1}{3}$

$= \frac{ 301 }{ 4 } \div \frac{ 7 }{ 3 }$

$= \frac{ 301 }{ 4 } \times \frac{ 3 }{ 7 }$

$= \frac{ 129 }{ 4 }$ = Rs. $32 \frac{ 1 }{ 4 }$

Thus, Rs. $32 \frac{ 1 }{ 4 }$ or Rs. 32.25 is the cost of cloth per metre.

Q-12. By what number should $\frac{ -33 }{ 16 }$ be divided to get $\frac{ -11 }{ 4 }$?

Solution:

Let, the other number  be x.

$\frac{ -33 }{ 16 } \div x = \frac{ -11 }{ 4 }$

$\Rightarrow \frac{ -33 }{ 16 } \times \frac{ 1 }{ x } = \frac{ -11 }{ 4 }$

$\Rightarrow \frac{1} {x} = \frac{ -11 }{ 4 } \times \frac{ 16 }{ -33 }$

$\Rightarrow \frac{ 1 }{ x } = \frac{ 4 }{ 3 }$

$\Rightarrow x = \frac{ 3 }{ 4 }$

Thus, the other number is $\frac{ 3 }{ 4 }$

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

Solution:

$\left ( \frac{ -13 }{ 5 } + \frac{ 12 }{ 7 } \right ) \div \left ( \frac{ -31 }{ 7 } \times \frac{ -1 }{ 2 } \right )$

= $\frac{ -13 \times 7 + 12 \times 5 }{ 35 } \div \frac{ 31 }{ 14 }$

= $\frac{ -91 + 60 }{ 35 } \div \frac{31 }{ 14 }$

= $\frac{ -31 }{ 35 } \times \frac{14 }{ 31 }$ = $\frac{ -2 }{ 5 }$

Q-14. Divide the sum of $\frac{ 65 }{ 12 }$ and $\frac{ 12 }{ 7 }$ by their differences.

Solution:

$\left ( \frac{ 65 }{ 12} + \frac{ 12 }{ 7 } \right ) \div \left ( \frac{ 65 }{ 12 } – \frac{ 12 }{ 7 } \right )$

= $\frac{65 \times 7 + 12 \times 12 }{ 84 } \div \frac{ 65 \times 7 – 12 \times 12 }{ 84 }$

= $\frac{455 + 144 }{ 84 } \div \frac{ 455 – 144 }{ 84 }$

= $\frac{ 599 }{ 84 } \div \frac{ 311 }{ 84 }$

= $\frac{ 599 }{ 84 } \times \frac{ 84 }{ 311 }$ = $\frac{ 599 }{ 311 }$

Q-15. If 24 trousers of equal size can be prepared in 54 meters of cloth, what length of cloth is required for each trouser?

Solution:

Cloth needed to prepare 24 trousers = 54 m

So,

Length of the cloth required for each trousers = $54 \div 24$ = $\frac{ 54 }{ 24 }$ = $\frac{ 9 }{ 4 }$ m = $2 \frac{ 1 }{ 4 }$ metres.