Mathematics Solutions Solutions for Class 7 Maths 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 Maths Operations On Rational Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 7 Maths Chapter 5 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 7 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Carry out the following additions of rational numbers.

(i)  $\frac{5}{36}+\frac{6}{42}$

(ii)  $1\frac{2}{3}+2\frac{4}{5}$

(iii)  $\frac{11}{17}+\frac{13}{19}$

(iv) $2\frac{3}{11}+1\frac{3}{77}$

At first, we will calculate the LCM of 36 and 42. The prime factorisation is 36 and 42 is,
36 = 2 × 2 × 3 × 3
42 = 2 × 3 × 7
Now, LCM of 36 and 42 = 2 × 2 × 3 × 3 × 7 = 252
$\frac{5}{36}+\frac{6}{42}\phantom{\rule{0ex}{0ex}}=\frac{5×7}{36×7}+\frac{6×6}{42×6}\phantom{\rule{0ex}{0ex}}=\frac{35}{252}+\frac{36}{252}\phantom{\rule{0ex}{0ex}}=\frac{35+36}{252}\phantom{\rule{0ex}{0ex}}=\frac{71}{252}$

Now, LCM of 3 and 5 is 15.
$\frac{5}{3}+\frac{14}{5}\phantom{\rule{0ex}{0ex}}=\frac{5×5}{3×5}+\frac{14×3}{5×3}\phantom{\rule{0ex}{0ex}}=\frac{25}{15}+\frac{42}{15}\phantom{\rule{0ex}{0ex}}=\frac{67}{15}\phantom{\rule{0ex}{0ex}}=4\frac{7}{15}$

Now, LCM of 17 and 19 is 323.
$\frac{11}{17}+\frac{13}{19}\phantom{\rule{0ex}{0ex}}=\frac{11×19}{17×19}+\frac{13×17}{19×17}\phantom{\rule{0ex}{0ex}}=\frac{209}{323}+\frac{221}{323}\phantom{\rule{0ex}{0ex}}=\frac{430}{323}$

Now, LCM of 11 and 77 is 77.
$\frac{25}{11}+\frac{80}{77}\phantom{\rule{0ex}{0ex}}=\frac{25×7}{11×7}+\frac{80×1}{77×1}\phantom{\rule{0ex}{0ex}}=\frac{175}{77}+\frac{80}{77}\phantom{\rule{0ex}{0ex}}=\frac{255}{77}=3\frac{24}{77}$

#### Question 2:

Carry out the following subtarctions involving rational numbers.

(i) $\frac{7}{11}-\frac{3}{7}$

(ii) $\frac{13}{36}-\frac{2}{40}$

(iii) $1\frac{2}{3}-3\frac{5}{6}$

(iv) $4\frac{1}{2}-3\frac{1}{3}$

Now, LCM of 11 and 7 is 77.
$\frac{7}{11}-\frac{3}{7}\phantom{\rule{0ex}{0ex}}=\frac{7×7}{11×7}-\frac{3×11}{7×11}\phantom{\rule{0ex}{0ex}}=\frac{49}{77}-\frac{33}{77}\phantom{\rule{0ex}{0ex}}=\frac{49-33}{77}\phantom{\rule{0ex}{0ex}}=\frac{16}{77}$

Now, LCM of 36 and 40 is 360.

Now, LCM of 3 and 6 is 6.
$\frac{5}{3}-\frac{23}{6}\phantom{\rule{0ex}{0ex}}=\frac{5×2}{3×2}-\frac{23×1}{6×1}\phantom{\rule{0ex}{0ex}}=\frac{10}{6}-\frac{23}{6}\phantom{\rule{0ex}{0ex}}=\frac{10-23}{6}\phantom{\rule{0ex}{0ex}}=-\frac{13}{6}$

Now, LCM of 2 and 3 is 6.
$\frac{9}{2}-\frac{10}{3}\phantom{\rule{0ex}{0ex}}=\frac{9×3}{2×3}-\frac{10×2}{3×2}\phantom{\rule{0ex}{0ex}}=\frac{18}{6}-\frac{20}{6}\phantom{\rule{0ex}{0ex}}=\frac{18-20}{6}\phantom{\rule{0ex}{0ex}}=-\frac{2}{6}\phantom{\rule{0ex}{0ex}}=-\frac{1}{3}$

#### Question 3:

Multiply the following rational numbers.

(i) $\frac{3}{11}×\frac{2}{5}$

(ii) $\frac{12}{5}×\frac{4}{15}$

(iii) $\frac{\left(-8\right)}{9}×\frac{3}{4}$

(iv) $\frac{0}{6}×\frac{3}{4}$

#### Question 4:

Write the multiplicative inverse.

(i)

(ii)

(iii)

(iv) 7

(v) $-7\frac{1}{3}$

It is known that, the multiplicative inverse of any rational number a is the reciprocal of the rational number i.e., $\frac{1}{a}$.
(i) Multiplicative inverse of $\frac{2}{5}=\frac{1}{\left(\frac{2}{5}\right)}=\frac{5}{2}$
(ii) Multuplicative inverse of $-\frac{3}{8}=\frac{1}{-\frac{3}{8}}=-\frac{8}{3}$
(iii) Multiplicative inverse of $-\frac{17}{39}=\frac{1}{-\frac{17}{39}}=-\frac{39}{17}$
(iv) Multiplicative inverse of 7 = $\frac{1}{7}$
(v) The given number is $-7\frac{1}{3}$.

Multiplicative inverse of $-\frac{22}{3}=\frac{1}{-\frac{22}{3}}=-\frac{3}{22}$

#### Question 5:

Carry out the divisions of rational numbers.

(i) $\frac{40}{12}÷\frac{10}{4}$

(ii) $\frac{-10}{11}÷\frac{-11}{10}$

(iii) $\frac{-7}{8}÷\frac{-3}{6}$

(iv) $\frac{2}{3}÷\left(-4\right)$

(v) $2\frac{1}{5}÷5\frac{3}{6}$

(vi) $\frac{-5}{13}÷\frac{7}{26}$

(vii) $\frac{9}{11}÷\left(-8\right)$

(viii) $5÷\frac{2}{5}$

#### Question 1:

Write three rational numbers that lie between the two given numbers.

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i) The given numbers are $\frac{2}{7}$ and $\frac{6}{7}$.
We know that,
2 < 3 < 4 < 5 < 6

Hence, 3 rational numbers between $\frac{2}{7}$ and $\frac{6}{7}$ are :
$\frac{3}{7},\frac{4}{7}$ and $\frac{5}{7}$.

(ii) The given numbers are $\frac{4}{5}$ and $\frac{2}{3}$.
Let us convert these numbers into fractions with equal denominators.
$\frac{4}{5}=\frac{4×6}{5×6}=\frac{24}{30}\phantom{\rule{0ex}{0ex}}\frac{2}{3}=\frac{2×10}{3×10}=\frac{20}{30}$
We know that,
20 < 21 < 22 < 23 < 24

Hence, 3 rational numbers between $\frac{2}{3}$ and $\frac{4}{5}$ are :
$\frac{21}{30},\frac{22}{30}$ and $\frac{23}{30}$.

(iii) The given numbers are $-\frac{2}{3}$ and $\frac{4}{5}$.
Let us convert each of given numbers into fractions with equal denominators.
$-\frac{2}{3}=\frac{-2×5}{3×5}=-\frac{10}{15}\phantom{\rule{0ex}{0ex}}\frac{4}{5}=\frac{4×3}{5×3}=\frac{12}{15}$
We know that,
−10 < −9 < −8 < −7 <...........< 1 < 2 < 3 < 4 <..........< 12

Hence, 3 rational numbers between $-\frac{2}{3}$ and $\frac{4}{5}$ are:
$-\frac{9}{15},-\frac{7}{15}$ and $\frac{4}{15}.$

(iv) The given numbers are $\frac{7}{9}$ and $-\frac{5}{9}.$
We know that,
−5 < −4 < −3 < −2 < −1 < 0 <.....< 6 < 7

Hence, 3 rational numbers between $-\frac{5}{9}$ and $\frac{7}{9}$ are:
and $\frac{6}{9}.$

(v) The given numbers are $-\frac{3}{4}$ and $\frac{5}{4}$.
We know that,
−3 < −2 < −1 < 0 < 1 < 2 < 3 < 4 < 5

Hence, 3 rational numbers between $-\frac{3}{4}$ and $\frac{5}{4}$ are:
and $\frac{3}{4}.$

(vi) The given numbers are $\frac{7}{8}$ and $-\frac{5}{3}$.
Let us convert each of the given numbers into fractions with equal denominators.
$\frac{7}{8}=\frac{7×3}{8×3}=\frac{21}{24}\phantom{\rule{0ex}{0ex}}-\frac{5}{3}=\frac{-5×8}{3×8}=-\frac{40}{24}\phantom{\rule{0ex}{0ex}}$
We know that,
−40 < −39 <....< −13 < −12 <......<11 < 12 <....17 <.... 21

Hence, 3 rational numbers between $-\frac{5}{3}$ and $\frac{7}{8}$ are:
and $\frac{17}{24}.$

(vii) The given numbers are $\frac{5}{7}$ and $\frac{11}{7}$.
We know that,
5 < 6 < 7 < 8 < 9 < 10 < 11
$\therefore \frac{5}{7}<\frac{6}{7}<\frac{7}{7}<\frac{8}{7}<\frac{9}{7}<\frac{10}{7}<\frac{11}{7}$
Hence, 3 rational numbers between $\frac{5}{7}$ and $\frac{11}{7}$ are :
$\frac{6}{7},\frac{8}{7}$ and $\frac{9}{7}$

(viii) The given numbers are 0 and $-\frac{3}{4}$.
Let us convert each of the given numbers into fractions with equal denominators.
$0=\frac{0×8}{1×8}=\frac{0}{8}\phantom{\rule{0ex}{0ex}}-\frac{3}{4}=\frac{-3×2}{4×2}=-\frac{6}{8}$
We know that,
−6 < −5 < −4 < −3 < −2 < −1 < 0

Hence, 3 rational numbers between $-\frac{6}{8}$ and 0 are:
and $-\frac{1}{8}.$

#### Question 1:

Write the following rational numbers in decimal form.

(i) $\frac{13}{4}$

(ii) $\frac{-7}{8}$

(iii) $7\frac{3}{5}$

(iv) $\frac{5}{12}$

(v) $\frac{22}{7}$

(vi) $\frac{4}{3}$

(vii) $\frac{7}{9}$

(i) The given number is $\frac{13}{4}$.

The decimal form of $\frac{13}{4}$ is 3.25.
(ii) The given number is $-\frac{7}{8}$.

The decimal form of $-\frac{7}{8}$ is − 0.875.
(iii) The given number is $7\frac{3}{5}$.
$7\frac{3}{5}=\frac{7×5+3}{5}=\frac{35+3}{5}=\frac{38}{5}$

The decimal form of $7\frac{3}{5}$ is 7.6.
(v) The given number is $\frac{22}{7}$.

The decimal form of $\frac{22}{7}$ is $3.\overline{)142857}$.
(vi) The given number is $\frac{4}{3}$.

The decimal form of $\frac{4}{3}$ is $1.\overline{)3}$.
(vii) The given number is $\frac{7}{9}$.

The decimal form of $\frac{7}{9}$ is $0.\overline{)7}$.

Simplify:
50 × 5 ÷ 2 + 24

50 × 5 ÷ 2 + 24
= 250 ÷ 2 + 24
= 125 + 24
= 149

#### Question 2:

Simplify:
(13 × 4) ÷ 2 – 26

(13 × 4) ÷ 2 – 26
= 52 ÷ 2 – 26
= 26 – 26
= 0

#### Question 3:

Simplify:
140 ÷ [( – 11) × ( – 3) – ( – 42) ÷ 14 – 1]

140 ÷ [( – 11) × ( – 3) – ( – 42) ÷ 14 – 1]
= 140 ÷ [33 – ( – 42) ÷ 14 – 1]
= 140 ÷ [33 + 42 ÷ 14 – 1]
= 140 ÷ [33 + 3 – 1]
= 140 ÷ [36 – 1]
= 140 ÷ 35
= 4

#### Question 4:

Simplify:
{(220 – 140) + [10 × 9 + ( – 2 × 5)]} – 100

{(220 – 140) + [10 × 9 + (– 2 × 5)]} – 100
= {80 + [10 × 9 + (– 10)]} – 100
= {80 + [10 × 9  –  10]} – 100
= {80 + [90 – 10]} –  100
= {80 + 80} –  100
= 160 – 100
= 60

#### Question 5:

Simplify:
$\frac{3}{5}+\frac{3}{8}÷\frac{6}{4}$