RD Sharma 2014 Solutions for Class 7 Math Chapter 5 Operations On Rational Numbers are provided here with simple step-by-step explanations. These solutions for Operations On Rational Numbers are extremely popular among class 7 students for Math Operations On Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2014 Book of class 7 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s RD Sharma 2014 Solutions. All RD Sharma 2014 Solutions for class 7 Math are prepared by experts and are 100% accurate.

Multiply:
(i)
(ii)
(iii)
(iv)

#### Question 2:

Multiply:
(i)
(ii)
(iii)
(iv)

(i) $\frac{-5}{17}×\frac{51}{-60}=\frac{-5}{17}×\frac{17×3}{-5×3×4}=\frac{1}{4}$

(ii) $\frac{-6}{11}×\frac{-55}{36}=\frac{-6}{11}×\frac{-5×11}{6×6}=\frac{5}{6}$

(iv) $\frac{6}{7}×\frac{-49}{36}=\frac{6}{7}×\frac{-7×7}{6×6}=\frac{-7}{6}$

#### Question 3:

Simplify peach of the following and express the result as a rational number in standard from:
(i) $\frac{-16}{21}×\frac{14}{5}$
(ii) $\frac{7}{6}×\frac{-3}{28}$
(iii) $\frac{-19}{36}×16$
(iv) $\frac{-13}{9}×\frac{27}{-26}$

#### Question 4:

Simplify:
(i) $\left(-5×\frac{2}{15}\right)-\left(-6×\frac{2}{9}\right)$
(ii) $\left(\frac{-9}{4}×\frac{5}{3}\right)+\left(\frac{13}{2}×\frac{5}{6}\right)$

#### Question 5:

Simplify:
(i) $\left(\frac{13}{9}×\frac{-15}{2}\right)+\left(\frac{7}{3}×\frac{8}{5}\right)+\left(\frac{3}{5}×\frac{1}{2}\right)$
(ii) $\left(\frac{3}{11}×\frac{5}{6}\right)-\left(\frac{9}{12}×\frac{4}{3}\right)+\left(\frac{5}{13}×\frac{6}{15}\right)$

Divide:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

#### Question 2:

Find the value and express as a rational number in standard form:
(i) $\frac{2}{5}÷\frac{26}{15}$
(ii) $\frac{10}{3}÷\frac{-35}{12}$
(iii) $-6÷\left(\frac{-8}{17}\right)$
(iv) $\frac{40}{98}÷\left(-20\right)$

#### Question 3:

The product of two rational numbers is 15. If one of the numbers is −10, find the other.

Let the first rational number = x.
Second number = −10
Their product = 15

Then, we have
$x×-10=15\phantom{\rule{0ex}{0ex}}⇒x=15×\frac{1}{-10}=5×3×\frac{-1}{2×5}=\frac{-3}{2}$

#### Question 4:

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

Let the first rational number = x
Second number             = $\frac{-4}{15}$
Their product              = $\frac{-8}{9}$

Then, we have
$x×\frac{-4}{15}=\frac{-8}{9}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-8}{9}×\frac{-15}{4}=\frac{-2×4}{3×3}×\frac{-3×5}{4}=\frac{10}{3}$

#### Question 5:

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

Let x be the number by which we should multiply
Then, according to the question, we have
$\frac{-1}{6}×x=\frac{-23}{9}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-23}{9}×\left(-6\right)=\frac{46}{3}$

#### Question 6:

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

Let be the number by which we multiply
Then, we have

$x×\frac{-15}{28}=\frac{-5}{7}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-5}{7}×\frac{28}{-15}=\frac{-5}{7}×\frac{7×4}{-5×3}=\frac{4}{3}$

#### Question 7:

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

Let x be the number required. Then, we have
$x×\frac{-8}{13}=24\phantom{\rule{0ex}{0ex}}⇒x=24×\frac{-13}{8}=-13×3=-39$

#### Question 8:

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

Let be the number by which we should multiply
Then, we have
$\frac{-3}{4}×x=\frac{2}{3}⇒x=\frac{2}{3}×\frac{4}{-3}=\frac{-8}{9}$

#### Question 9:

Find (x + y) ÷ (x + y), if
(i)
(ii)
(iii)

(i) x =
Then, (x+y)  = $\frac{2}{3}+\frac{3}{2}=\frac{2×2}{3×2}+\frac{3×3}{2×3}=\frac{4}{6}+\frac{9}{6}=\frac{13}{6}$

Then, .

(ii) x =
Then, (x+y) = $\frac{2}{5}+\frac{1}{2}=\frac{2×2}{5×2}+\frac{1×5}{2×5}=\frac{4}{10}+\frac{5}{10}=\frac{9}{10}$

Then,

(iii) x =
Then, (x+y) = $\frac{5}{4}+\frac{-1}{3}=\frac{5×3}{4×3}+\frac{-1×4}{3×4}=\frac{15}{12}+\frac{-4}{12}=\frac{11}{12}$

Then,

#### Question 10:

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

The cost of $7\frac{2}{3}=\frac{23}{3}\mathrm{met}\text{res}$  of rope = Rs. $12\frac{3}{4}=\frac{51}{4}$.
Then, the cost of 1 metre of rope = Rs.

#### Question 11:

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

The cost of = .
The cost of 1 metre of cloth = Rs. $\frac{301}{4}÷\frac{7}{3}=\frac{301}{4}×\frac{3}{7}=\frac{43×7}{4}×\frac{3}{7}=\frac{129}{4}=32\frac{1}{4}.$

#### Question 12:

By what number should $\frac{-33}{16}$ be divided to got $\frac{-11}{4}$?

Let be the number required.

Then, we have

$\frac{-33}{16}÷x=\frac{-11}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{-33}{16}×\frac{1}{x}=\frac{-11}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{-33}{16}×\frac{4}{-11}=x\phantom{\rule{0ex}{0ex}}x=\frac{-3×11}{4×4}×\frac{4}{-11}=\frac{3}{4}$

#### Question 13:

Divide the sum of $\frac{-13}{5}$ and $\frac{12}{7}$ by the product of

Then, according to the question, we have

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

#### Question 14:

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

According to the question, we need to divide the first figure by the second:

$\frac{97}{12}÷\frac{33}{12}=\frac{97}{12}×\frac{12}{33}=\frac{97}{33}$

#### Question 15:

If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?

Total cloth given = 54 metres
Total number of pairs of trousers made = 24
Length of cloth required for each pair of trousers =

#### Question 1:

Find six rational numbers between $\frac{-4}{8}$ and $\frac{3}{8}$.

#### Question 2:

Find 10 rational numbers between $\frac{7}{13}$ and $\frac{-4}{13}$.

Since

#### Question 3:

State true or false:
(i) Between any two distinect integers there is always an linteger.
(ii) Between any two distinct rational numbers there is always a rational number.
(iii) Between any two distinct rational numbers there is always a rational number.

(i) False, because there is no integer between 2 and 3.
(ii) True
(iii) True

(i)
(ii)
(iii)
(iv)

(i)
(ii)
(iii)
(iv)

(i)

(ii)

(iii)

(iv)

#### Question 3:

Simplify:
(i) $\frac{8}{9}+\frac{-11}{6}$
(ii) $\frac{-5}{16}+\frac{7}{24}$
(iii) $\frac{1}{-12}+\frac{2}{-15}$
(iv) $\frac{-8}{19}+\frac{-4}{57}$

#### Question 4:

Add and express the sum as a mixed fraction:
(i) $\frac{-12}{5}+\frac{43}{10}$
(ii) $\frac{24}{7}+\frac{-11}{4}$
(iii) $\frac{-31}{6}+\frac{-27}{8}$

#### Question 1:

Subtract the first rational number from the second in each of the following:
(i)
(ii)
(iii)
(iv)

#### Question 2:

Evaluate each of the following:
(i) $\frac{2}{3}-\frac{3}{5}$
(ii) $-\frac{4}{7}-\frac{2}{-3}$
(iii) $\frac{4}{7}-\frac{-5}{-7}$
(iv) $-2-\frac{5}{9}$

#### Question 3:

The sum of the two numbers is $\frac{5}{9}$. If one of the numbers is $\frac{1}{3}$, find the other.

First number = $\frac{1}{3}$
Let the second number = x.
According to the question, we have
$\frac{1}{3}+x=\frac{5}{9}\phantom{\rule{0ex}{0ex}}⇒x=\frac{5}{9}-\frac{1}{3}=\frac{5-1×3}{9}=\frac{5-3}{9}=\frac{2}{9}$

#### Question 4:

The sum of two numbers is $\frac{-1}{3}$. If one of the numbers is $\frac{-12}{3}$, find the other.

First number = $\frac{-12}{3}$
Let the second number = x.
Then, according to the question, we have
$\frac{-12}{3}+x=\frac{-1}{3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-1}{3}-\frac{-12}{3}=\frac{-1+12}{3}=\frac{11}{3}$

#### Question 5:

The sum of two numbers is $\frac{-4}{3}$. If one of the numbers is −5, find the other.

First number = $-5$
Let the second number = x.
Then, according to the question, we have
$-5+x=\frac{-4}{3}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-4}{3}-\frac{-5}{1}=\frac{-4+5×3}{3}=\frac{-4+15}{3}=\frac{11}{3}$

#### Question 6:

The sum of two rational numbers is −8. If one of the numbers is $\frac{-15}{7}$, find the other.

First number = $\frac{-15}{7}$
Let the second number = x.
Then, according to the question, we have
$\frac{-15}{7}+x=-8\phantom{\rule{0ex}{0ex}}⇒x=\frac{-8}{1}-\frac{-15}{7}=\frac{-8×7+15}{7}=\frac{-56+15}{7}=\frac{-41}{7}$

#### Question 7:

What should be added to $\frac{-7}{8}$ so as to get $\frac{5}{9}?$

Then, according to the question, we have

#### Question 8:

What number should be added to $\frac{-5}{11}$ so as to get $\frac{26}{33}?$

Then, according to the question, we have

#### Question 9:

What number should be added to $\frac{-5}{7}$ to get $\frac{-2}{3}?$

Then, according to the question, we have

#### Question 10:

What number should be suvtracted from $\frac{-5}{3}$ to get $\frac{5}{6}?$

Let x be the number that should be subtracted from
Then, according to the question, we have

#### Question 11:

What number should be subtracted from $\frac{3}{7}$ to get $\frac{5}{4}?$

Let x be the number that should be subtracted from
Then, according to the question, we have
$\frac{3}{7}-x=\frac{5}{4}\phantom{\rule{0ex}{0ex}}⇒x=\frac{3}{7}-\frac{5}{4}=\frac{3×4-5×7}{28}=\frac{12-35}{28}=\frac{-23}{28}$

#### Question 12:

What should be added to $\left(\frac{2}{3}+\frac{3}{5}\right)$ to get $\frac{-2}{15}?$

$\frac{2}{3}+\frac{3}{5}=\frac{2×5}{3×5}+\frac{3×3}{5×3}=\frac{10}{15}+\frac{9}{15}=\frac{19}{15}$
Let x be the number that should be added to $\frac{19}{15}$ to get $\frac{-2}{15}$
Then, we have

#### Question 13:

What should be added to $\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}\right)$ to get 3?

Then, we have
$\frac{31}{30}+x=3\phantom{\rule{0ex}{0ex}}⇒x=3-\frac{31}{30}=\frac{3×30}{1×30}-\frac{31}{30}=\frac{90}{30}-\frac{31}{30}=\frac{59}{30}$

#### Question 14:

What should be subtracted from $\left(\frac{3}{4}-\frac{2}{3}\right)$ to get $\frac{-1}{6}?$

Let  x be the number that should be subtracted from

Then, we have
$\frac{1}{12}-x=\frac{-1}{6}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{12}-\frac{-1}{6}=\frac{1}{12}-\frac{-1×2}{6×2}=\frac{1}{12}-\frac{-2}{12}=\frac{3}{12}=\frac{1}{4}$

#### Question 15:

Simplify:
(i) $\frac{-3}{2}+\frac{5}{4}-\frac{7}{4}$
(ii) $\frac{5}{3}-\frac{7}{6}+\frac{-2}{3}$
(iii) $\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}$
(iv) $\frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7}$

#### Question 16:

Fill in the blanks:
(i) $\frac{-4}{13}-\frac{-3}{26}=....$
(ii) $\frac{-9}{14}+....=-1$
(iii) $\frac{-7}{9}+....=3$
(iv) $....+\frac{15}{23}=4$

(i) $\frac{-4}{13}-\frac{-3}{26}=\frac{-4×2}{13×2}-\frac{-3}{26}=\frac{-8+3}{26}=\frac{-5}{26}$
$x+\frac{15}{23}=4\phantom{\rule{0ex}{0ex}}⇒x=4-\frac{15}{23}=\frac{4×23}{1×23}-\frac{15}{23}=\frac{92-15}{23}=\frac{77}{23}$