RD Sharma 2019 2020 Solutions for Class 7 Maths Chapter 5 - Operations On Rational Numbers

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Page No 5.10:

Question 1:

Multiply:
(i) $\frac{7}{11} by \frac{5}{4}$
(ii) $\frac{5}{7} by (\frac{- 3}{4})$
(iii) $\frac{(- 2)}{9} by \frac{5}{11}$
(iv) $\frac{- 3}{17} by \frac{- 5}{- 4}$

Answer:

$(i) \frac{7}{11} \times \frac{5}{4} = \frac{7 \times 5}{11 \times 4} = \frac{35}{44} (ii) \frac{5}{7} \times \frac{- 3}{4} = \frac{5 \times - 3}{7 \times 4} = \frac{- 15}{28} (iii) \frac{- 2}{9} \times \frac{5}{11} = \frac{- 2 \times 5}{9 \times 11} = \frac{- 10}{99}$

$(iv) \frac{- 3}{17} \times \frac{- 5}{- 4} = \frac{- 3 \times 5}{17 \times - 4} = \frac{- 15}{68}$

Page No 5.10:

Question 2:

Multiply:
(i) $\frac{- 5}{17} by \frac{51}{- 60}$
(ii) $\frac{- 6}{11} by \frac{- 55}{36}$
(iii) $\frac{- 8}{25} by \frac{- 5}{16}$
(iv) $\frac{6}{7} by \frac{- 49}{36}$

Answer:

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

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

$(iii) \frac{- 8}{25} \times \frac{- 5}{16} = \frac{- 8}{5 \times 5} \times \frac{- 5}{8 \times 2} = \frac{1}{10}$

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

Page No 5.10:

Question 3:

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

Answer:

$(i) \frac{- 16}{21} \times \frac{14}{5} = \frac{- 16 \times 14}{21 \times 5} = \frac{- 32}{15}$
$(ii) \frac{7}{6} \times \frac{- 3}{28} = \frac{7 \times - 3}{6 \times 28} = \frac{- 1}{8} (iii) \frac{- 19}{36} \times 16 = \frac{- 19 \times 16}{36} = \frac{- 76}{9}$

$(iv) \frac{- 13}{9} \times \frac{27}{- 26} = \frac{- 13 \times 27}{9 \times - 26} = \frac{3}{2}$

Page No 5.10:

Question 4:

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

Answer:

$(i) (- 5 \times \frac{2}{15}) - (- 6 \times \frac{2}{9}) = (- 5 \times \frac{2}{3 \times 5}) - (- 6 \times \frac{2}{3 \times 3}) = (\frac{- 2}{3}) - (\frac{- 4}{3}) = \frac{- 2 + 4}{3} = \frac{2}{3} (ii) (\frac{- 9}{4} \times \frac{5}{3}) + (\frac{13}{2} \times \frac{5}{6}) = (\frac{- 3 \times 3}{4} \times \frac{5}{3}) + (\frac{65}{12}) = (\frac{- 15}{4}) + (\frac{65}{12}) = \frac{- 15 \times 3}{4 \times 3} + \frac{65}{12} = \frac{- 45 + 65}{12} = \frac{20}{12} = \frac{5 \times 4}{3 \times 4} = \frac{5}{3}$

Page No 5.10:

Question 5:

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

Answer:

$(i) (\frac{13}{9} \times \frac{- 15}{2}) + (\frac{7}{3} \times \frac{8}{5}) + (\frac{3}{5} \times \frac{1}{2}) = (\frac{13}{3 \times 3} \times \frac{- 3 \times 5}{2}) + (\frac{56}{15}) + (\frac{3}{10}) = \frac{- 65}{6} + \frac{56}{15} + \frac{3}{10} = \frac{- 65 \times 5}{6 \times 5} + \frac{56 \times 2}{15 \times 2} + \frac{3 \times 3}{10 \times 3} = \frac{- 325}{30} + \frac{112}{30} + \frac{9}{30} = \frac{- 325 + 112 + 9}{30} = \frac{- 204}{30} = \frac{- 34}{5} (ii) (\frac{3}{11} \times \frac{5}{6}) - (\frac{9}{12} \times \frac{4}{3}) + (\frac{5}{13} \times \frac{6}{15}) = (\frac{3}{11} \times \frac{5}{2 \times 3}) - (\frac{3 \times 3}{4 \times 3} \times \frac{4}{3}) + (\frac{5}{13} \times \frac{3 \times 2}{5 \times 3}) = \frac{5}{22} - \frac{1}{1} + \frac{2}{13} = \frac{5 \times 13}{22 \times 13} - \frac{1 \times 286}{1 \times 286} + \frac{2 \times 22}{13 \times 22} = \frac{65}{286} - \frac{286}{286} + \frac{44}{286} = \frac{65 - 286 + 44}{286} = \frac{- 177}{286}$

Page No 5.13:

Question 1:

Divide:
(i) $1 by \frac{1}{2}$
(ii) $5 by \frac{- 5}{7}$
(iii) $\frac{- 3}{4} by \frac{9}{- 16}$
(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) $\frac{2}{3} by \frac{- 7}{12}$

Answer:

$(i) 1 \div \frac{1}{2} = 1 \times \frac{2}{1} = 2 (ii) 5 \div \frac{- 5}{7} = 5 \times \frac{7}{- 5} = - 7 (iii) \frac{- 3}{4} \div \frac{9}{- 16} = \frac{- 3}{4} \times \frac{- 16}{9} = \frac{- 3}{4} \times \frac{- 4 \times 4}{3 \times 3} = \frac{4}{3} (iv) \frac{- 7}{8} \div \frac{- 21}{16} = \frac{- 7}{8} \times \frac{- 16}{21} = \frac{- 7}{8} \times \frac{- 8 \times 2}{7 \times 3} = \frac{2}{3}$

$(v) \frac{7}{- 4} \div \frac{63}{64} = \frac{7}{- 4} \times \frac{64}{63} = \frac{- 7}{4} \times \frac{4 \times 16}{7 \times 9} = \frac{- 16}{9} (vi) 0 \div \frac{- 7}{5} = 0 \times \frac{- 7}{5} = 0 (vii) \frac{- 3}{4} \div - 6 = \frac{- 3}{4} \times \frac{- 1}{6} = \frac{- 3}{4} \times \frac{- 1}{2 \times 3} = \frac{1}{8} (viii) \frac{2}{3} \div \frac{- 7}{12} = \frac{2}{3} \times \frac{- 12}{7} = \frac{2}{3} \times \frac{- 4 \times 3}{7} = \frac{- 8}{7}$

Page No 5.13:

Question 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) $\frac{40}{98} \div (- 20)$

Answer:

Page No 5.13:

Question 3:

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

Answer:

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

Then, we have
$x \times - 10 = 15 \Rightarrow x = 15 \times \frac{1}{- 10} = 5 \times 3 \times \frac{- 1}{2 \times 5} = \frac{- 3}{2}$

Page No 5.13:

Question 4:

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

Answer:

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

Then, we have
$x \times \frac{- 4}{15} = \frac{- 8}{9} \Rightarrow x = \frac{- 8}{9} \times \frac{- 15}{4} = \frac{- 2 \times 4}{3 \times 3} \times \frac{- 3 \times 5}{4} = \frac{10}{3}$

Page No 5.13:

Question 5:

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

Answer:

Let x be the number by which we should multiply $\frac{- 1}{6} to get \frac{- 23}{9} .$
Then, according to the question, we have
$\frac{- 1}{6} \times x = \frac{- 23}{9} \Rightarrow x = \frac{- 23}{9} \times (- 6) = \frac{46}{3}$

Page No 5.13:

Question 6:

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

Answer:

Let x be the number by which we multiply $\frac{- 15}{28} to get the product \frac{- 5}{7} .$
Then, we have

$x \times \frac{- 15}{28} = \frac{- 5}{7} \Rightarrow x = \frac{- 5}{7} \times \frac{28}{- 15} = \frac{- 5}{7} \times \frac{7 \times 4}{- 5 \times 3} = \frac{4}{3}$

Page No 5.14:

Question 7:

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

Answer:

Let x be the number required. Then, we have
$x \times \frac{- 8}{13} = 24 \Rightarrow x = 24 \times \frac{- 13}{8} = - 13 \times 3 = - 39$

Page No 5.14:

Question 8:

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

Answer:

Let x be the number by which we should multiply $\frac{- 3}{4} to get \frac{2}{3} .$
Then, we have
$\frac{- 3}{4} \times x = \frac{2}{3} \Rightarrow x = \frac{2}{3} \times \frac{4}{- 3} = \frac{- 8}{9}$

Page No 5.14:

Question 9:

Find (x + y) ÷ (x + y), if
(i) $x = \frac{2}{3}, y = \frac{3}{2}$
(ii) $x = \frac{2}{5}, y = \frac{1}{2}$
(iii) $x = \frac{5}{4}, y = \frac{- 1}{3}$

Answer:

(i) x = $\frac{2}{3}, y = \frac{3}{2}$
Then, (x+y) = $\frac{2}{3} + \frac{3}{2} = \frac{2 \times 2}{3 \times 2} + \frac{3 \times 3}{2 \times 3} = \frac{4}{6} + \frac{9}{6} = \frac{13}{6}$
$(x - y) = \frac{2}{3} - \frac{3}{2} = \frac{4}{6} - \frac{9}{6} = \frac{- 5}{6}$
Then, $(x + y) \div (x - y) = \frac{13}{6} \div \frac{- 5}{6} = \frac{13}{6} \times \frac{6}{- 5} = \frac{- 13}{5}$ .

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

Then, $(x + y) \div (x - y) = \frac{9}{10} \div \frac{- 1}{10} = - 9$

(iii) x = $\frac{5}{4}, y = \frac{- 1}{3}$
Then, (x+y) = $\frac{5}{4} + \frac{- 1}{3} = \frac{5 \times 3}{4 \times 3} + \frac{- 1 \times 4}{3 \times 4} = \frac{15}{12} + \frac{- 4}{12} = \frac{11}{12}$
$(x - y) = \frac{5}{4} - \frac{- 1}{3} = \frac{5}{4} + \frac{1}{3} = \frac{19}{12}$
Then, $(x + y) \div (x - y) = \frac{11}{12} \div \frac{19}{12} = \frac{11}{12} \times \frac{12}{19} = \frac{11}{19} .$

Page No 5.14:

Question 10:

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

Answer:

The cost of $7 \frac{2}{3} = \frac{23}{3} met res$ of rope = Rs. $12 \frac{3}{4} = \frac{51}{4}$ .
Then, the cost of 1 metre of rope = Rs. $\frac{51}{4} \div \frac{23}{3} = \frac{51}{4} \times \frac{3}{23} = \frac{153}{92} = Rs . 1 \frac{61}{92} .$

Page No 5.14:

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.

Answer:

The cost of $2 \frac{1}{3} = \frac{7}{3} met res of cloth$ = $Rs . 75 \frac{1}{4} = \frac{301}{4}$ .
The cost of 1 metre of cloth = Rs. $\frac{301}{4} \div \frac{7}{3} = \frac{301}{4} \times \frac{3}{7} = \frac{43 \times 7}{4} \times \frac{3}{7} = \frac{129}{4} = 32 \frac{1}{4} .$

Page No 5.14:

Question 12:

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

Answer:

Let x be the number required.

Then, we have

$\frac{- 33}{16} \div x = \frac{- 11}{4} \Rightarrow \frac{- 33}{16} \times \frac{1}{x} = \frac{- 11}{4} \Rightarrow \frac{- 33}{16} \times \frac{4}{- 11} = x x = \frac{- 3 \times 11}{4 \times 4} \times \frac{4}{- 11} = \frac{3}{4}$

Page No 5.14:

Question 13:

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

Answer:

$The sum of \frac{- 13}{5} and \frac{12}{7} is \frac{- 13}{5} + \frac{12}{7} = \frac{- 13 \times 7}{5 \times 7} + \frac{12 \times 5}{7 \times 5} = \frac{- 91}{35} + \frac{60}{35} = \frac{- 91 + 60}{35} = \frac{- 31}{35}$

$The product of \frac{- 31}{7} and \frac{- 1}{2} is \frac{- 31}{7} \times \frac{- 1}{2} = \frac{31}{14}$

Then, according to the question, we have

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

Page No 5.14:

Question 14:

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

Answer:

$The sum of \frac{65}{12} and \frac{8}{3} is \frac{65}{12} + \frac{8}{3} = \frac{65}{12} + \frac{8 \times 4}{3 \times 4} = \frac{65}{12} + \frac{32}{12} = \frac{65 + 32}{12} = \frac{97}{12} The difference of \frac{65}{12} and \frac{8}{3} is \frac{65}{12} - \frac{8}{3} = \frac{65}{12} - \frac{8 \times 4}{3 \times 4} = \frac{65}{12} - \frac{32}{12} = \frac{65 - 32}{12} = \frac{33}{12}$

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

$\frac{97}{12} \div \frac{33}{12} = \frac{97}{12} \times \frac{12}{33} = \frac{97}{33}$

Page No 5.14:

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?

Answer:

Total cloth given = 54 metres
Total number of pairs of trousers made = 24
Length of cloth required for each pair of trousers = $\frac{54}{24} = \frac{9 \times 6}{4 \times 6} = \frac{9}{4} metres .$

Page No 5.16:

Question 1:

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

Answer:

$Since - 4 < - 3 < - 2 < 1 < 0 < 1 < 2 < 3, six rational numbers between \frac{- 4}{8} and \frac{3}{8} are \frac{- 3}{8}, \frac{- 2}{8}, \frac{- 1}{8}, \frac{0}{8}, \frac{1}{8}, \frac{2}{8}$

Page No 5.16:

Question 2:

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

Answer:

Since

$- 4 < - 3 < - 2 < - 1 < 0 < 1 < 2 < 3 < 4 < 5 < 6 < 7, 10 rational numbers between \frac{7}{13} and \frac{- 4}{13} are, \frac{- 3}{13}, \frac{- 2}{13}, \frac{- 1}{13}, \frac{0}{13}, \frac{1}{13}, \frac{2}{13}, \frac{3}{13} \frac{4}{13}, \frac{5}{13}, \frac{6}{13}$

Page No 5.16:

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.

Answer:

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

Page No 5.16:

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) $\frac{25}{9}$

Answer:

Sum of the given number and the required number = 2

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

∴ Required number = Sum of the numbers − Given number

$= 2 - (\frac{- 7}{9}) = \frac{2}{1} + \frac{7}{9} = \frac{2 \times 9 + 7 \times 1}{9} = \frac{18 + 7}{9} = \frac{25}{9}$

Hence, the correct answer is option (d).

Page No 5.16:

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) $\frac{- 15}{22}$

Answer:

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

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

∴ Required number = Given number − Difference of the numbers

$= \frac{- 2}{3} - \frac{4}{5} = \frac{- 2}{3} + (\frac{- 4}{5}) = \frac{- 2 \times 5 + (- 4) \times 3}{15} = \frac{- 10 + (- 12)}{15} = \frac{- 22}{15}$

Hence, the correct answer is option (b).

Page No 5.16:

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

Answer:

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

∴ Reciprocal of $\frac{- 3}{4}$

$= {(\frac{- 3}{4})}^{- 1} = \frac{4}{- 3} = \frac{4 \times (- 1)}{- 3 \times (- 1)} = \frac{- 4}{3}$

Hence, the correct answer is option (c).

Page No 5.16:

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) $\frac{- 5}{- 4}$

Answer:

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).

Page No 5.16:

Question 5:

Mark the correct alternative in each of the following:

$1 \div \frac{- 5}{7} =$

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

Answer:

$1 \div \frac{- 5}{7} = 1 \times \frac{7}{- 5} (x \div y = x \times \frac{1}{y}) = \frac{7}{- 5} = \frac{7 \times (- 1)}{- 5 \times (- 1)} = \frac{- 7}{5}$

Hence, the correct answer is option (d).

Page No 5.17:

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) $\frac{18}{13}$

Answer:

$\frac{- 5}{13} + ? = - 1 \Rightarrow ? = - 1 - (\frac{- 5}{13}) \Rightarrow ? = - 1 + \frac{5}{13} [- (\frac{- 5}{13}) = \frac{5}{13}] \Rightarrow ? = \frac{- 1 \times 13 + 5}{13}$
$\Rightarrow ? = \frac{- 13 + 5}{13} \Rightarrow ? = \frac{- 8}{13}$

Hence, the correct answer is option (b).

Page No 5.17:

Question 7:

Mark the correct alternative in each of the following:

$0 \div \frac{3}{5} =$

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

Answer:

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

$∴ 0 \div \frac{3}{5} = 0 (0 \div \frac{a}{b} = 0)$

Hence, the correct answer is option (a).

Page No 5.17:

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) $\frac{45}{7}$

Answer:

$? = - 2 \frac{3}{7} + 4 \Rightarrow ? = - \frac{17}{7} + \frac{4}{1} \Rightarrow ? = \frac{- 17 \times 1 + 4 \times 7}{7} \Rightarrow ? = \frac{- 17 + 28}{7}$
$\Rightarrow ? = \frac{11}{7}$

Hence, the correct answer is option (b).

Page No 5.17:

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)

Answer:

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

$\frac{a}{b} \times \frac{b}{a} = 1$

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).

Page No 5.17:

Question 10:

Mark the correct alternative in each of the following:

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

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

Answer:

$3 \frac{1}{7} \times 1 \frac{5}{6} \times 1 \frac{2}{5} \times 1 \frac{1}{11} = \frac{22}{7} \times \frac{11}{6} \times \frac{7}{5} \times \frac{12}{11} = \frac{22 \times 11 \times 7 \times 12}{7 \times 6 \times 5 \times 11} (\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}) = \frac{44}{5} = \frac{8 \times 5 + 4}{5} = 8 \frac{4}{5}$

Hence, the correct answer is option (c).

Page No 5.17:

Question 11:

Mark the correct alternative in each of the following:

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

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

Answer:

$\frac{- 7}{13} - (\frac{- 8}{15}) = \frac{- 7}{13} + \frac{8}{15} [- (\frac{- 8}{15}) = \frac{8}{15}] = \frac{- 7 \times 15 + 8 \times 13}{195} (LCM of 13 and 15 = 195) = \frac{- 105 + 104}{195} = \frac{- 1}{195}$

Hence, the correct answer is option (d).

Page No 5.17:

Question 12:

Mark the correct alternative in each of the following:

$1 \div \frac{1}{3} =$

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

Answer:

$1 \div \frac{1}{3} = 1 \times 3 (x \div y = x \times \frac{1}{y}) = 3$

Hence, the correct answer is option (b).

Page No 5.17:

Question 13:

Mark the correct alternative in each of the following:

$(- 2) \div (- \frac{5}{3}) =$

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

Answer:

$(- 2) \div (- \frac{5}{3}) = - 2 \times (- \frac{3}{5}) (x \div y = x \times \frac{1}{y}) = \frac{- 2}{1} \times \frac{(- 3)}{5} = \frac{- 2 \times (- 3)}{1 \times 5} (\frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d}) = \frac{6}{5}$

Hence, the correct answer is option (c).

Page No 5.17:

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

Answer:

$\frac{x}{2} + \frac{1}{3} = 1 \Rightarrow \frac{x}{2} = 1 - \frac{1}{3} \Rightarrow \frac{x}{2} = \frac{3 \times 1 - 1}{3} \Rightarrow \frac{x}{2} = \frac{3 - 1}{3} \Rightarrow \frac{x}{2} = \frac{2}{3}$
$\Rightarrow \frac{2 x}{2} = \frac{2 \times 2}{3} (Multiplying both sides by 2) \Rightarrow x = \frac{4}{3}$

Hence, the correct answer is option (b).

Page No 5.17:

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) $\frac{7}{12}$

Answer:

$\frac{5}{4} - \frac{7}{6} - \frac{- 2}{3} = \frac{5}{4} + (\frac{- 7}{6}) + \frac{2}{3} [- (\frac{- 2}{3}) = \frac{2}{3}] = \frac{5 \times 3 + (- 7) \times 2 + 2 \times 4}{12} (LCM of 3, 4 and 6 = 12) = \frac{15 - 14 + 8}{12} = \frac{9}{12}$

$= \frac{9 \div 3}{12 \div 3} (Dividing numerator and denominator by 3) = \frac{3}{4}$

Hence, the correct answer is option (a).

Page No 5.4:

Question 1:

Add the following rational numbers:
(i) $\frac{- 5}{7} and \frac{3}{7}$
(ii) $\frac{- 15}{4} and \frac{7}{4}$
(iii) $\frac{- 8}{11} and \frac{- 4}{11}$
(iv) $\frac{6}{13} and \frac{- 9}{13}$

Answer:

$(i) \frac{- 5}{7} + \frac{3}{7} = \frac{- 5 + 3}{7} = \frac{- 2}{7} (ii) \frac{- 15}{4} + \frac{7}{4} = \frac{- 15 + 7}{4} = \frac{- 8}{4} = - 2 (iii) \frac{- 8}{11} + \frac{- 4}{11} = \frac{- 8 - 4}{11} = \frac{- 12}{11} (iv) \frac{6}{13} + \frac{- 9}{13} = \frac{6 - 9}{13} = \frac{- 3}{13}$

Page No 5.4:

Question 2:

(i) $\frac{3}{4} and \frac{- 3}{5}$
(ii) $- 3 and \frac{3}{5}$
(iii) $\frac{- 7}{27} and \frac{11}{18}$
(iv) $\frac{31}{- 4} and \frac{- 5}{8}$

Answer:

(i)

$\frac{3}{4} + \frac{- 3}{5} LCM of the denominators 4 and 5 is 20 . Now, we express \frac{3}{4} and \frac{- 3}{5} into forms in which both of them have the same denominator 20 . \frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20} \frac{- 3}{5} = \frac{- 3 \times 4}{5 \times 4} = \frac{- 12}{20} Therefore, \frac{3}{4} + \frac{- 3}{5} = \frac{15}{20} + \frac{- 12}{20} = \frac{15 - 12}{20} = \frac{3}{20}$

(ii)

$- 3 + \frac{3}{5} LCM of the denominators 1 and 5 is 5 . Now, we express - 3 and \frac{3}{5} into forms in which both of them have the same denominator 5 . \frac{- 3}{1} = \frac{3 \times 5}{1 \times 5} = \frac{- 15}{5} \frac{3}{5} = \frac{3 \times 1}{5 \times 1} = \frac{3}{5} Therefore, - 3 + \frac{3}{5} = \frac{- 15}{5} + \frac{3}{5} = \frac{- 12}{5}$

(iii)

$\frac{- 7}{27} + \frac{11}{18} LCM of the denominators 27 and 18 is 54 . Now, we express \frac{- 7}{27} and \frac{11}{18} into forms in which both of them have the same denomiator 54 . \frac{- 7}{27} = \frac{- 7 \times 2}{27 \times 2} = \frac{- 14}{54} \frac{11}{18} = \frac{11 \times 3}{18 \times 3} = \frac{33}{54} Therefore, \frac{- 7}{27} + \frac{11}{18} = \frac{- 14}{54} + \frac{33}{54} = \frac{- 14 + 33}{54} = \frac{19}{54}$

(iv)

$\frac{- 31}{4} + \frac{- 5}{8} LCM of the denominators 4 and 8 is 8 . Now, we express \frac{- 31}{4} and \frac{- 5}{8} into forms in which both of them have the same denomiator 8 . \frac{- 31}{4} = \frac{- 31 \times 2}{4 \times 2} = \frac{- 62}{8} \frac{- 5}{8} = \frac{- 5 \times 1}{8 \times 1} = \frac{- 5}{8} Therefore, \frac{- 31}{4} + \frac{- 5}{8} = \frac{- 62}{8} + \frac{- 5}{8} = \frac{- 62 - 5}{8} = \frac{- 67}{8}$

Page No 5.4:

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) $\frac{- 8}{19} + \frac{- 4}{57}$

Answer:

$(i) \frac{8}{9} + \frac{- 11}{6} LCM of the denominators 9 and 6 is 18 . Now, we express \frac{8}{9} and \frac{- 11}{6} into forms in which both of them have the same denominator as LCM . \frac{8}{9} = \frac{8 \times 2}{9 \times 2} \frac{- 11}{6} = \frac{- 11 \times 3}{6 \times 3} Therefore, \frac{8 \times 2}{9 \times 2} + \frac{- 11 \times 3}{6 \times 3} = \frac{16 - 33}{18} = \frac{- 17}{18}$

$(ii) \frac{- 5}{16} + \frac{7}{24} LCM of the denominators 16 and 24 is 48 . Now, we express \frac{- 5}{16} and \frac{7}{24} into forms in which both of them have the same denominator as LCM . \frac{- 5}{16} = \frac{- 5 \times 3}{16 \times 3} \frac{7}{24} = \frac{7 \times 2}{24 \times 2} Therefore, \frac{- 5 \times 3}{16 \times 3} + \frac{7 \times 2}{24 \times 2} = \frac{- 15 + 14}{48} = \frac{- 1}{48} (iii) \frac{1}{- 12} + \frac{2}{- 15} LCM of the denominators 12 and 15 is 60 . Now, we express \frac{- 1}{12} and \frac{- 2}{15} into forms in which both of them have the same denominator as LCM . \frac{- 1}{12} = \frac{- 1 \times 5}{12 \times 5} \frac{- 2}{15} = \frac{- 2 \times 4}{15 \times 4} Therefore, \frac{- 1 \times 5}{12 \times 5} + \frac{- 2 \times 4}{15 \times 4} = \frac{- 5 - 8}{60} = \frac{- 13}{60}$

$(iv) \frac{- 8}{19} + \frac{- 4}{57} LCM of the denominators 19 and 57 is 57 . Now, we express \frac{- 8}{19} and \frac{- 4}{57} into forms in which both of them have the same denominator as LCM . \frac{8}{9} = \frac{- 8 \times 3}{19 \times 3} \frac{- 4}{57} = \frac{- 4 \times 1}{57 \times 1} Therefore, \frac{- 8 \times 3}{19 \times 3} + \frac{- 4}{57} = \frac{- 24 - 4}{57} = \frac{- 28}{57}$

Page No 5.4:

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) $\frac{- 31}{6} + \frac{- 27}{8}$

Answer:

$(i) \frac{- 12}{5} + \frac{43}{10} LCM of the denominators 5 and 10 is 10 . Now, we express \frac{- 12}{5} and \frac{43}{10} into forms in which both of them have the same denominator as LCM . \frac{- 12}{5} = \frac{12 \times 2}{5 \times 2} \frac{43}{10} = \frac{43 \times 1}{10 \times 1} Therefore, \frac{- 12 \times 2}{5 \times 2} + \frac{43}{10} = \frac{- 24 + 43}{10} = \frac{19}{10} = 1 \frac{9}{10} (ii) \frac{24}{7} + \frac{- 11}{4} LCM of the denominators 7 and 4 is 28 . Now, we express \frac{24}{7} and \frac{- 11}{4} into forms in which both of them have the same denominator as LCM . \frac{24}{7} = \frac{24 \times 4}{7 \times 4} \frac{- 11}{4} = \frac{- 11 \times 7}{4 \times 7} Therefore, \frac{24 \times 4}{7 \times 4} + \frac{- 11 \times 7}{4 \times 7} = \frac{96 - 77}{28} = \frac{19}{28}$

$(iii) \frac{- 31}{6} + \frac{- 27}{8} LCM of the denominators 6 and 8 is 24 . Now, we express \frac{- 31}{6} and \frac{- 27}{8} into forms in which both of them have the same denominator as LCM . \frac{- 31}{6} = \frac{- 31 \times 4}{6 \times 4} \frac{- 27}{8} = \frac{- 27 \times 3}{8 \times 3} Therefore, \frac{- 31 \times 4}{6 \times 4} + \frac{- 27 \times 3}{8 \times 3} = \frac{- 124 - 81}{24} = \frac{- 205}{24}$

Page No 5.7:

Question 1:

Subtract the first rational number from the second in each of the following:
(i) $\frac{3}{8}, \frac{5}{8}$
(ii) $\frac{- 7}{9}, \frac{4}{9}$
(iii) $\frac{- 2}{11}, \frac{- 9}{11}$
(iv) $\frac{11}{13}, \frac{- 4}{13}$

Answer:

$(i) \frac{5}{8} - \frac{3}{8} = \frac{2}{8} (ii) \frac{4}{9} - \frac{- 7}{9} = \frac{11}{9}$
$(iii) \frac{- 9}{11} - \frac{- 2}{11} = \frac{- 9 + 2}{11} = \frac{- 7}{11} (iv) \frac{- 4}{13} - \frac{11}{13} = \frac{- 4 - 11}{13} = \frac{- 15}{13}$

Page No 5.7:

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) $- 2 - \frac{5}{9}$

Answer:

$(i) \frac{2}{3} - \frac{3}{5} = \frac{2 \times 5 - 3 \times 3}{15} = \frac{10 - 9}{15} = \frac{1}{15} (ii) - \frac{4}{7} - \frac{2}{- 3} = \frac{- 4 \times 3 + 2 \times 7}{21} = \frac{- 12 + 14}{21} = \frac{2}{21}$

$(iii) \frac{4}{7} - \frac{- 5}{- 7} = \frac{4 - 5}{7} = \frac{- 1}{7} (iv) - \frac{2}{1} - \frac{5}{9} = \frac{- 2 \times 9 - 5}{9} = \frac{= - 18 - 5}{9} = \frac{- 23}{9}$

Page No 5.7:

Question 3:

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

Answer:

First number = $\frac{1}{3}$
Let the second number = x.
According to the question, we have
$\frac{1}{3} + x = \frac{5}{9} \Rightarrow x = \frac{5}{9} - \frac{1}{3} = \frac{5 - 1 \times 3}{9} = \frac{5 - 3}{9} = \frac{2}{9}$

Page No 5.7:

Question 4:

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

Answer:

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

Page No 5.7:

Question 5:

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

Answer:

First number = $- 5$
Let the second number = x.
Then, according to the question, we have
$- 5 + x = \frac{- 4}{3} \Rightarrow x = \frac{- 4}{3} - \frac{- 5}{1} = \frac{- 4 + 5 \times 3}{3} = \frac{- 4 + 15}{3} = \frac{11}{3}$

Page No 5.7:

Question 6:

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

Answer:

First number = $\frac{- 15}{7}$
Let the second number = x.
Then, according to the question, we have
$\frac{- 15}{7} + x = - 8 \Rightarrow x = \frac{- 8}{1} - \frac{- 15}{7} = \frac{- 8 \times 7 + 15}{7} = \frac{- 56 + 15}{7} = \frac{- 41}{7}$

Page No 5.7:

Question 7:

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

Answer:

Let x be added to $\frac{- 7}{8} to get \frac{5}{9} .$
Then, according to the question, we have
$x + \frac{- 7}{8} = \frac{5}{9} = > x = \frac{5}{9} - \frac{- 7}{8} = \frac{5 \times 8 + 7 \times 9}{72} = \frac{40 + 63}{72} = \frac{103}{72}$

Page No 5.7:

Question 8:

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

Answer:

Let x be added to $\frac{- 5}{11} to get \frac{26}{33} .$
Then, according to the question, we have
$x + \frac{- 5}{11} = \frac{26}{33} = > x = \frac{26}{33} - \frac{- 5}{11} = \frac{26 + 5 \times 3}{33} = \frac{26 + 15}{33} = \frac{41}{33}$

Page No 5.7:

Question 9:

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

Answer:

Let x be added to $\frac{- 5}{7} to get \frac{- 2}{3} .$
Then, according to the question, we have
$x + \frac{- 5}{7} = \frac{- 2}{3} \Rightarrow x = \frac{- 2}{3} - \frac{- 5}{7} = \frac{- 2 \times 7 + 5 \times 3}{21} = \frac{- 14 + 15}{21} = \frac{1}{21}$

Page No 5.8:

Question 10:

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

Answer:

Let x be the number that should be subtracted from $\frac{- 5}{3} to get \frac{5}{6} .$
Then, according to the question, we have
$\frac{- 5}{3} - x = \frac{5}{6} \Rightarrow x = \frac{- 5}{3} - \frac{5}{6} = \frac{- 5 \times 2}{3 \times 2} - \frac{5}{6} = \frac{- 10 - 5}{6} = \frac{- 15}{6} = \frac{- 5}{2}$

Page No 5.8:

Question 11:

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

Answer:

Let x be the number that should be subtracted from $\frac{3}{7} to get \frac{5}{4} .$
Then, according to the question, we have
$\frac{3}{7} - x = \frac{5}{4} \Rightarrow x = \frac{3}{7} - \frac{5}{4} = \frac{3 \times 4 - 5 \times 7}{28} = \frac{12 - 35}{28} = \frac{- 23}{28}$

Page No 5.8:

Question 12:

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

Answer:

$\frac{2}{3} + \frac{3}{5} = \frac{2 \times 5}{3 \times 5} + \frac{3 \times 3}{5 \times 3} = \frac{10}{15} + \frac{9}{15} = \frac{19}{15}$
Let x be the number that should be added to $\frac{19}{15}$ to get $\frac{- 2}{15}$
Then, we have
$\frac{19}{15} + x = \frac{- 2}{15} = > x = \frac{- 2}{15} - \frac{19}{15} = \frac{- 21}{15} = \frac{- 7 \times 3}{5 \times 3} = \frac{- 7}{5}$

Page No 5.8:

Question 13:

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

Answer:

Let x be added to $(\frac{1}{2} + \frac{1}{3} + \frac{1}{5}) = (\frac{1 \times 15}{2 \times 15} + \frac{1 \times 10}{3 \times 10} + \frac{1 \times 6}{5 \times 6}) = (\frac{15}{30} + \frac{10}{30} + \frac{6}{30}) = \frac{31}{30} to get 3 .$
Then, we have
$\frac{31}{30} + x = 3 \Rightarrow x = 3 - \frac{31}{30} = \frac{3 \times 30}{1 \times 30} - \frac{31}{30} = \frac{90}{30} - \frac{31}{30} = \frac{59}{30}$

Page No 5.8:

Question 14:

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

Answer:

Let x be the number that should be subtracted from $\frac{3}{4} - \frac{2}{3} = \frac{3 \times 3}{4 \times 3} - \frac{2 \times 4}{3 \times 4} = \frac{9}{12} - \frac{8}{12} = \frac{1}{12} to get \frac{- 1}{6} .$

Then, we have
$\frac{1}{12} - x = \frac{- 1}{6} \Rightarrow x = \frac{1}{12} - \frac{- 1}{6} = \frac{1}{12} - \frac{- 1 \times 2}{6 \times 2} = \frac{1}{12} - \frac{- 2}{12} = \frac{3}{12} = \frac{1}{4}$

Page No 5.8:

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) $\frac{- 2}{5} - \frac{- 3}{10} - \frac{- 4}{7}$

Answer:

$(i) \frac{- 3}{2} + \frac{5}{4} - \frac{7}{4} = \frac{- 3 \times 2}{2 \times 2} + \frac{5}{4} - \frac{7}{4} = \frac{- 6 + 5 - 7}{4} = \frac{- 8}{4} = - 2 (ii) \frac{5}{3} - \frac{7}{6} + \frac{- 2}{3} = \frac{5 \times 2}{3 \times 2} - \frac{7}{6} + \frac{- 2 \times 2}{3 \times 2} = \frac{10 - 7 - 4}{6} = \frac{- 1}{6}$

$(iii) \frac{5}{4} - \frac{7}{6} - \frac{- 2}{3} = \frac{5 \times 3}{4 \times 3} - \frac{7 \times 2}{6 \times 2} - \frac{- 2 \times 4}{3 \times 4} = \frac{15 - 14 + 8}{12} = \frac{9}{12} = \frac{3}{4} (iv) \frac{- 2}{5} - \frac{- 3}{10} - \frac{- 4}{7} = \frac{- 2 \times 14}{5 \times 14} - \frac{- 3 \times 7}{10 \times 7} - \frac{- 4 \times 10}{7 \times 10} = \frac{- 28 + 21 + 40}{70} = \frac{33}{70}$

Page No 5.8:

Question 16:

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

Answer:

(i) $\frac{- 4}{13} - \frac{- 3}{26} = \frac{- 4 \times 2}{13 \times 2} - \frac{- 3}{26} = \frac{- 8 + 3}{26} = \frac{- 5}{26}$
$(ii) \frac{- 9}{14} + x = - 1 \Rightarrow x = - 1 - \frac{- 9}{14} = \frac{- 14 + 9}{14} = \frac{- 5}{14} (iii) \frac{- 7}{9} + x = 3 \Rightarrow x = \frac{3 \times 9}{1 \times 9} - \frac{- 7}{9} = \frac{27 + 7}{9} = \frac{34}{9}$

(iv)
$x + \frac{15}{23} = 4 \Rightarrow x = 4 - \frac{15}{23} = \frac{4 \times 23}{1 \times 23} - \frac{15}{23} = \frac{92 - 15}{23} = \frac{77}{23}$

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