Mathematics Solutions Solutions for Class 7 Math Chapter 5 - Operations On Rational Numbers

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Page No 36:

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) $2 \frac{3}{11} + 1 \frac{3}{77}$

Answer:

$(i) \frac{5}{36} + \frac{6}{42}$
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
$\frac{5}{36} + \frac{6}{42} = \frac{5 \times 7}{36 \times 7} + \frac{6 \times 6}{42 \times 6} = \frac{35}{252} + \frac{36}{252} = \frac{35 + 36}{252} = \frac{71}{252}$
$(ii) 1 \frac{2}{3} + 2 \frac{4}{5} = \frac{1 \times 3 + 2}{3} + \frac{2 \times 5 + 4}{5} = \frac{5}{3} + \frac{14}{5}$
Now, LCM of 3 and 5 is 15.
$\frac{5}{3} + \frac{14}{5} = \frac{5 \times 5}{3 \times 5} + \frac{14 \times 3}{5 \times 3} = \frac{25}{15} + \frac{42}{15} = \frac{67}{15} = 4 \frac{7}{15}$
$(iii) \frac{11}{17} + \frac{13}{19}$
Now, LCM of 17 and 19 is 323.
$\frac{11}{17} + \frac{13}{19} = \frac{11 \times 19}{17 \times 19} + \frac{13 \times 17}{19 \times 17} = \frac{209}{323} + \frac{221}{323} = \frac{430}{323}$
$(iv) 2 \frac{3}{11} + 1 \frac{3}{77} = \frac{2 \times 11 + 3}{11} + \frac{1 \times 77 + 3}{77} = \frac{25}{11} + \frac{80}{77}$
Now, LCM of 11 and 77 is 77.
$\frac{25}{11} + \frac{80}{77} = \frac{25 \times 7}{11 \times 7} + \frac{80 \times 1}{77 \times 1} = \frac{175}{77} + \frac{80}{77} = \frac{255}{77} = 3 \frac{24}{77}$

Page No 36:

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) $4 \frac{1}{2} - 3 \frac{1}{3}$

Answer:

$(i) \frac{7}{11} - \frac{3}{7}$
Now, LCM of 11 and 7 is 77.
$\frac{7}{11} - \frac{3}{7} = \frac{7 \times 7}{11 \times 7} - \frac{3 \times 11}{7 \times 11} = \frac{49}{77} - \frac{33}{77} = \frac{49 - 33}{77} = \frac{16}{77}$
$(ii) \frac{13}{36} - \frac{2}{40}$
Now, LCM of 36 and 40 is 360.
$\frac{13}{36} - \frac{2}{40} = \frac{13 \times 10}{36 \times 10} - \frac{2 \times 9}{40 \times 9} = \frac{130}{360} - \frac{18}{360} = \frac{130 - 18}{360} = \frac{112}{360} = \frac{112 \div 8}{360 \div 8} (since, HCF of 112 and 360 is 8) = \frac{14}{45}$
$(iii) 1 \frac{2}{3} - 3 \frac{5}{6} = \frac{1 \times 3 + 2}{3} - \frac{3 \times 6 + 5}{6} = \frac{5}{3} - \frac{23}{6}$
Now, LCM of 3 and 6 is 6.
$\frac{5}{3} - \frac{23}{6} = \frac{5 \times 2}{3 \times 2} - \frac{23 \times 1}{6 \times 1} = \frac{10}{6} - \frac{23}{6} = \frac{10 - 23}{6} = - \frac{13}{6}$
$(iv) 4 \frac{1}{2} - 3 \frac{1}{3} = \frac{4 \times 2 + 1}{2} - \frac{3 \times 3 + 1}{3} = \frac{9}{2} - \frac{10}{3}$
Now, LCM of 2 and 3 is 6.
$\frac{9}{2} - \frac{10}{3} = \frac{9 \times 3}{2 \times 3} - \frac{10 \times 2}{3 \times 2} = \frac{18}{6} - \frac{20}{6} = \frac{18 - 20}{6} = - \frac{2}{6} = - \frac{1}{3}$

Page No 36:

Question 3:

Multiply the following rational numbers.

(i) $\frac{3}{11} \times \frac{2}{5}$

(ii) $\frac{12}{5} \times \frac{4}{15}$

(iii) $\frac{(- 8)}{9} \times \frac{3}{4}$

(iv) $\frac{0}{6} \times \frac{3}{4}$

Answer:

$(i) \frac{3}{11} \times \frac{2}{5} = \frac{3 \times 2}{11 \times 5} = \frac{6}{55}$
$(ii) \frac{12}{5} \times \frac{4}{15} = \frac{12 \times 4}{5 \times 15} = \frac{48}{75} = \frac{48 \div 3}{75 \div 3} (Since, HCF of 48 and 75 is 3) = \frac{16}{25}$
$(iii) \frac{(- 8)}{9} \times \frac{3}{4} = \frac{(- 8) \times 3}{9 \times 4} = \frac{- 24}{36} = \frac{- 24 \div 12}{36 \div 12} (Since, HCF of 24 and 36 is 12) = - \frac{2}{3}$
$(iv) \frac{0}{6} \times \frac{3}{4} = \frac{0 \times 3}{6 \times 4} = \frac{0}{24} = 0$

Page No 36:

Question 4:

Write the multiplicative inverse.

(i) $\frac{2}{5}$

(ii) $\frac{- 3}{8}$

(iii) $\frac{- 17}{39}$

(iv) 7

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

Answer:

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 $\frac{2}{5} = \frac{1}{(\frac{2}{5})} = \frac{5}{2}$
(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 $- 7 \frac{1}{3}$ .
$Now, - 7 \frac{1}{3} = - (7 + \frac{1}{3}) = - (\frac{21 + 1}{3}) = - \frac{22}{3}$
Multiplicative inverse of $- \frac{22}{3} = \frac{1}{- \frac{22}{3}} = - \frac{3}{22}$

Page No 36:

Question 5:

Carry out the divisions of rational numbers.

(i) $\frac{40}{12} \div \frac{10}{4}$

(ii) $\frac{- 10}{11} \div \frac{- 11}{10}$

(iii) $\frac{- 7}{8} \div \frac{- 3}{6}$

(iv) $\frac{2}{3} \div (- 4)$

(v) $2 \frac{1}{5} \div 5 \frac{3}{6}$

(vi) $\frac{- 5}{13} \div \frac{7}{26}$

(vii) $\frac{9}{11} \div (- 8)$

(viii) $5 \div \frac{2}{5}$

Answer:

$(i) \frac{40}{12} \div \frac{10}{4} = \frac{40}{12} \times \frac{4}{10} = \frac{40 \times 4}{12 \times 10} = \frac{160}{120} = \frac{160 \div 40}{120 \div 40} (Since, HCF of 160 and 120 is 40) = \frac{40}{3}$
$(ii) \frac{- 10}{11} \div \frac{- 11}{10} = (- \frac{10}{11}) \times (- \frac{10}{11}) = \frac{(- 10) \times (- 10)}{11 \times 11} = \frac{100}{121}$
$(iii) \frac{- 7}{8} \div \frac{- 3}{6} = (- \frac{7}{8}) \times (- \frac{6}{3}) = \frac{(- 7) \times (- 6)}{8 \times 3} = \frac{42}{24} = \frac{42 \div 6}{24 \div 6} (Since, HCF of 42 and 24 is 6) = \frac{7}{4}$
$(iv) \frac{2}{3} \div (- 4) = \frac{2}{3} \times (- \frac{1}{4}) = \frac{2 \times (- 1)}{3 \times 4} = \frac{- 2}{12} = \frac{- 2 \div 2}{12 \div 2} (Since, HCF of 2 and 12 is 2) = \frac{- 1}{6}$
$(v) 2 \frac{1}{5} \div 5 \frac{3}{6} = \frac{2 \times 5 + 1}{5} \div \frac{5 \times 6 + 3}{6} = \frac{11}{5} \div \frac{33}{6} = \frac{11}{5} \times \frac{6}{33} = \frac{11 \times 6}{5 \times 33} = \frac{66}{165} = \frac{66 \div 33}{165 \div 33} (Since, HCF of 66 and 165 is 33) = \frac{2}{5}$

Page No 38:

Question 1:

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

(i) $\frac{2}{7}, \frac{6}{7}$

(ii) $\frac{4}{5}, \frac{2}{3}$

(iii) $- \frac{2}{3}, \frac{4}{5}$

(iv) $\frac{7}{9}, - \frac{5}{9}$

(v) $\frac{- 3}{4}, \frac{+ 5}{4}$

(vi) $\frac{7}{8}, \frac{- 5}{3}$

(vii) $\frac{5}{7}, \frac{11}{7}$

(viii) $0, \frac{- 3}{4}$

Answer:

(i) The given numbers are $\frac{2}{7}$ and $\frac{6}{7}$ .
We know that,
2 < 3 < 4 < 5 < 6
$∴ \frac{2}{7} < \frac{3}{7} < \frac{4}{7} < \frac{5}{7} < \frac{6}{7}$
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.
$\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30} \frac{2}{3} = \frac{2 \times 10}{3 \times 10} = \frac{20}{30}$
We know that,
20 < 21 < 22 < 23 < 24
$∴ \frac{20}{30} < \frac{21}{30} < \frac{22}{30} < \frac{23}{30} < \frac{24}{30} \Rightarrow \frac{2}{3} < \frac{21}{30} < \frac{22}{30} < \frac{23}{30} < \frac{4}{5}$
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.
$- \frac{2}{3} = \frac{- 2 \times 5}{3 \times 5} = - \frac{10}{15} \frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15}$
We know that,
−10 < −9 < −8 < −7 <...........< 1 < 2 < 3 < 4 <..........< 12
$\Rightarrow - \frac{2}{3} < - \frac{9}{15} < - \frac{8}{15} < - \frac{7}{15} < . . . . . < \frac{1}{15} < \frac{2}{15} < \frac{3}{15} < \frac{4}{15} < . . . . . < \frac{4}{15}$
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
$∴ - \frac{5}{9} < - \frac{4}{9} < - \frac{3}{9} < - \frac{2}{9} < - \frac{1}{9} < 0 < . . . . . < \frac{6}{9} < \frac{7}{9}$
Hence, 3 rational numbers between $- \frac{5}{9}$ and $\frac{7}{9}$ are:
$- \frac{4}{9}, 0$ 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
$∴ - \frac{3}{4} < - \frac{2}{4} < - \frac{1}{4} < 0 < \frac{1}{4} < \frac{2}{4} < \frac{3}{4} < \frac{4}{4} < \frac{5}{4}$
Hence, 3 rational numbers between $- \frac{3}{4}$ and $\frac{5}{4}$ are:
$- \frac{2}{4}, - \frac{1}{4}$ 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.
$\frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24} - \frac{5}{3} = \frac{- 5 \times 8}{3 \times 8} = - \frac{40}{24}$
We know that,
−40 < −39 <....< −13 < −12 <......<11 < 12 <....17 <.... 21
$∴ - \frac{40}{24} < - \frac{39}{24} < . . . . . < - \frac{13}{24} < - \frac{12}{24} < . . . . < \frac{11}{24} < \frac{12}{24} < . . . . < \frac{17}{24} < . . . . < \frac{21}{24} \Rightarrow - \frac{5}{3} < - \frac{39}{24} < . . . . . . < - \frac{13}{24} < - \frac{12}{24} < . . . . < \frac{11}{24} < \frac{12}{24} < . . . . < \frac{17}{24} < . . . . < \frac{7}{8}$
Hence, 3 rational numbers between $- \frac{5}{3}$ and $\frac{7}{8}$ are:
$- \frac{13}{24}, \frac{11}{24}$ 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
$∴ \frac{5}{7} < \frac{6}{7} < \frac{7}{7} < \frac{8}{7} < \frac{9}{7} < \frac{10}{7} < \frac{11}{7}$
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.
$0 = \frac{0 \times 8}{1 \times 8} = \frac{0}{8} - \frac{3}{4} = \frac{- 3 \times 2}{4 \times 2} = - \frac{6}{8}$
We know that,
−6 < −5 < −4 < −3 < −2 < −1 < 0
$∴ - \frac{6}{8} < - \frac{5}{8} < - \frac{4}{8} < - \frac{3}{8} < - \frac{2}{8} < - \frac{1}{8} < \frac{0}{8} \Rightarrow - \frac{3}{4} < - \frac{5}{8} < - \frac{4}{8} < - \frac{3}{8} < - \frac{2}{8} < - \frac{1}{8} < 0$
Hence, 3 rational numbers between $- \frac{6}{8}$ and 0 are:
$- \frac{5}{8}, - \frac{2}{8}$ and $- \frac{1}{8} .$

Page No 41:

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) $\frac{7}{9}$

Answer:

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

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

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

$∴ \frac{38}{5} = 7.6$
The decimal form of $7 \frac{3}{5}$ is 7.6.
(v) The given number is $\frac{22}{7}$ .

$∴ \frac{22}{7} = 3.142857142857 . . . . = 3.142857$
The decimal form of $\frac{22}{7}$ is $3.142857$ .
(vi) The given number is $\frac{4}{3}$ .

$∴ \frac{4}{3} = 1.33 . . . . = 1.3$
The decimal form of $\frac{4}{3}$ is $1.3$ .
(vii) The given number is $\frac{7}{9}$ .

$∴ \frac{7}{9} = 0.77 . . . . = 0.7$
The decimal form of $\frac{7}{9}$ is $0.7$ .

Page No 42:

Question 1:

Simplify:
50 × 5 ÷ 2 + 24

Answer:

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

Page No 42:

Question 2:

Simplify:
(13 × 4) ÷ 2 – 26

Answer:

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

Page No 42:

Question 3:

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

Answer:

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

Page No 42:

Question 4:

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

Answer:

{(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

Page No 42:

Question 5:

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

Answer:

$\frac{3}{5} + \frac{3}{8} \div \frac{6}{4} = \frac{3}{5} + \frac{3}{8} \times \frac{4}{6} = \frac{3}{5} + \frac{1}{4} = \frac{4 \times 3 + 5 \times 1}{20} (Since, LCM of 5 and 4 is 20) = \frac{12 + 5}{20} = \frac{17}{20}$

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