Rd Sharma 2019 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 2019 Book of Class 7 Math Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2019 Solutions. All Rd Sharma 2019 Solutions for class 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

#### Question 1:

Mark the correct alternative in each of the following:

What should be added to $\frac{-7}{9}$ to get 2?

(a) $\frac{11}{9}$                                 (b) $\frac{-11}{9}$                                 (c) $\frac{-25}{9}$                                 (d) $\frac{25}{9}$

Sum of the given number and the required number = 2

Given number = $\frac{-7}{9}$

∴ Required number = Sum of the numbers − Given number

$=2-\left(\frac{-7}{9}\right)\phantom{\rule{0ex}{0ex}}=\frac{2}{1}+\frac{7}{9}\phantom{\rule{0ex}{0ex}}=\frac{2×9+7×1}{9}\phantom{\rule{0ex}{0ex}}=\frac{18+7}{9}\phantom{\rule{0ex}{0ex}}=\frac{25}{9}$

Hence, the correct answer is option (d).

#### Question 2:

Mark the correct alternative in each of the following:

What should be subtracted from $\frac{-2}{3}$ to get $\frac{4}{5}$?

(a) $\frac{22}{15}$                                 (b) $\frac{-22}{15}$                                 (c) $\frac{15}{22}$                                 (d) $\frac{-15}{22}$

Difference of the given number and required number = $\frac{4}{5}$

Given number = $\frac{-2}{3}$

∴ Required number = Given number − Difference of the numbers

$=\frac{-2}{3}-\frac{4}{5}\phantom{\rule{0ex}{0ex}}=\frac{-2}{3}+\left(\frac{-4}{5}\right)\phantom{\rule{0ex}{0ex}}=\frac{-2×5+\left(-4\right)×3}{15}\phantom{\rule{0ex}{0ex}}=\frac{-10+\left(-12\right)}{15}\phantom{\rule{0ex}{0ex}}=\frac{-22}{15}$

Hence, the correct answer is option (b).

#### Question 3:

Mark the correct alternative in each of the following:

Reciprocal of $\frac{-3}{4}$ is

(a) $\frac{3}{4}$                                 (b) $\frac{4}{3}$                                 (c) $\frac{-4}{3}$                                 (d) None of these

We know that the reciprocal of the rational number $\frac{a}{b}$ is ${\left(\frac{a}{b}\right)}^{-1}=\frac{b}{a}$.

∴ Reciprocal of $\frac{-3}{4}$

$={\left(\frac{-3}{4}\right)}^{-1}\phantom{\rule{0ex}{0ex}}=\frac{4}{-3}\phantom{\rule{0ex}{0ex}}=\frac{4×\left(-1\right)}{-3×\left(-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{-4}{3}$

Hence, the correct answer is option (c).

#### Question 4:

Mark the correct alternative in each of the following:

The multiplicative inverse of $\frac{4}{-5}$ is

(a) $-\frac{4}{5}$                                 (b) $\frac{5}{4}$                                 (c) $\frac{5}{-4}$                                 (d) $\frac{-5}{-4}$

We know that the multiplicative inverse of the rational number $\frac{a}{b}$ is $\frac{b}{a}$.

∴ Multiplicative inverse of $\frac{4}{-5}$$=\frac{-5}{4}=\frac{5}{-4}$

Hence, the correct answer is option (c).

#### Question 5:

Mark the correct alternative in each of the following:

$1÷\frac{-5}{7}=$

(a) $\frac{2}{7}$                                 (b) $\frac{5}{7}$                                 (c) $-\frac{2}{7}$                                 (d) $\frac{-7}{5}$

Hence, the correct answer is option (d).

#### Question 6:

Mark the correct alternative in each of the following:

$\frac{-5}{13}+?=-1$

(a) $\frac{8}{13}$                                 (b) $\frac{-8}{13}$                                 (c) $\frac{-18}{13}$                                 (d) $\frac{18}{13}$

$⇒?=\frac{-13+5}{13}\phantom{\rule{0ex}{0ex}}⇒?=\frac{-8}{13}$

Hence, the correct answer is option (b).

#### Question 7:

Mark the correct alternative in each of the following:

$0÷\frac{3}{5}=$

(a) 0                                   (b) $\frac{5}{3}$                                   (c) $\frac{3}{5}$                                   (d) $-\frac{3}{5}$

We know that 0 divided by any non-zero rational number is always 0.

Hence, the correct answer is option (a).

#### Question 8:

Mark the correct alternative in each of the following:

$-2\frac{3}{7}+4=?$

(a) $\frac{-11}{7}$                                   (b) $\frac{11}{7}$                                   (c) $\frac{-45}{7}$                                   (d) $\frac{45}{7}$

$?=-2\frac{3}{7}+4\phantom{\rule{0ex}{0ex}}⇒?=-\frac{17}{7}+\frac{4}{1}\phantom{\rule{0ex}{0ex}}⇒?=\frac{-17×1+4×7}{7}\phantom{\rule{0ex}{0ex}}⇒?=\frac{-17+28}{7}$
$⇒?=\frac{11}{7}$

Hence, the correct answer is option (b).

#### Question 9:

Mark the correct alternative in each of the following:

If the product of two non-zero rational numbers is 1, then they are

(a) additve inverse of each other                                 (b) multiplicative inverse of each other

(c) reciprocal of each other                                         (d) both (b) and (c)

For every non-zero rational number $\frac{a}{b}$ there exists a rational number $\frac{b}{a}$ such that

$\frac{a}{b}×\frac{b}{a}=1$

Here, the rational number $\frac{b}{a}$ is called the multiplicative inverse or reciprocal of $\frac{a}{b}$.

Thus, if the product of two non-zero rational numbers is 1, then they are multiplicative inverse or reciprocal of each other.

Hence, the correct answer is option (d).

#### Question 10:

Mark the correct alternative in each of the following:

The product $3\frac{1}{7}×1\frac{5}{6}×1\frac{2}{5}×1\frac{1}{11}$ is equal to

(a) $5\frac{8}{5}$                                 (b) $5\frac{4}{5}$                                   (c) $8\frac{4}{5}$                                   (d) $7\frac{4}{5}$

Hence, the correct answer is option (c).

#### Question 11:

Mark the correct alternative in each of the following:

$\frac{-7}{13}-\left(\frac{-8}{15}\right)=$

(a) $-\frac{239}{195}$                                   (b) $\frac{29}{195}$                                   (c) $\frac{-29}{195}$                                   (d) None of these

Hence, the correct answer is option (d).

#### Question 12:

Mark the correct alternative in each of the following:

$1÷\frac{1}{3}=$

(a) $\frac{1}{3}$                                  (b) 3                                   (c) $1\frac{1}{3}$                                   (d) $3\frac{1}{3}$

Hence, the correct answer is option (b).

#### Question 13:

Mark the correct alternative in each of the following:

$\left(-2\right)÷\left(-\frac{5}{3}\right)=$

(a) $\frac{5}{6}$                                  (b) $-\frac{5}{6}$                                   (c) $\frac{6}{5}$                                   (d) $\frac{-6}{5}$

Hence, the correct answer is option (c).

#### Question 14:

Mark the correct alternative in each of the following:

If $\frac{x}{2}+\frac{1}{3}=1$, then x =

(a) $\frac{3}{4}$                                   (b) $\frac{4}{3}$                                   (c) $-\frac{3}{4}$                                   (d) $\frac{-4}{3}$

$\frac{x}{2}+\frac{1}{3}=1\phantom{\rule{0ex}{0ex}}⇒\frac{x}{2}=1-\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{2}=\frac{3×1-1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{2}=\frac{3-1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{x}{2}=\frac{2}{3}$

Hence, the correct answer is option (b).

#### Question 15:

Mark the correct alternative in each of the following:

$\frac{5}{4}-\frac{7}{6}-\frac{-2}{3}=$

(a) $\frac{3}{4}$                                   (b) $-\frac{3}{4}$                                   (c) $\frac{-7}{12}$                                   (d) $\frac{7}{12}$

Hence, the correct answer is option (a).

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