RS Aggarwal 2019,2020 solution class 7 chapter 4 Rational Numbers Exercise 4D

Exercise 4D

Page-72

Question 1:

Find the additive inverse of:

(i) 5
(ii) −9
(iii) $\frac{3}{14}$
(iv) $\frac{-11}{15}$
(v) $\frac{15}{-4}$
(vi) $\frac{-18}{-13}$
(vii) 0
(viii) $\frac{1}{-6}$

Answer 1:

(i) Additive inverse of 5 is −5.

(ii) Additive inverse of −9 is 9.

(iii) Additive inverse of $\begin{array}{l}\frac{3}{14} is \frac{-3}{14}\\ \end{array}$.

(iv) Additive inverse of $\frac{-11}{15} is \frac{11}{15}$.

(v) Additive inverse of $\frac{15}{-4}=\frac{15\times (-1)}{(-4)\times (-1)}$
                                         
                                          $$\begin{gathered} = \frac{-15}{4} \\ =\frac{15}{4} \end{gathered}$$

(vi) Additive inverse of $\frac{-18}{-13}=\frac{-18\times (-1)}{(-13)\times (-1)}$
                                              $$\begin{gathered} = \frac{18}{13} \\ =\frac{-18}{13} \end{gathered}$$

(vii) Additive inverse of​ 0 is 0.

(viii) Additive inverse of   $\frac{1}{-6}=\frac{1\times (-1)}{(-6)\times (-1)}$
                                                $$\begin{gathered} = \frac{-1}{6} \\ =\frac{1}{6} \end{gathered}$$

Question 2:

Subtract:

(i) $\frac{3}{4} \mathrm{from} \frac{1}{3}$
(ii) $\frac{-5}{6} \mathrm{from} \frac{1}{3}$
(iii) $\frac{-8}{9} \mathrm{from} \frac{-3}{5}$
(iv) $\frac{-9}{7} \mathrm{from} -1$
(v) $\frac{-18}{11} \mathrm{from} 1$
(vi) $\frac{-13}{9} \mathrm{from} 0$
(vii) $\frac{-32}{13} \mathrm{from} \frac{-6}{5}$
(viii) $-7 \mathrm{from} \frac{-4}{7}$
(ix) $\frac{5}{9} \mathrm{from} \frac{-2}{3}$
(x) $5 \mathrm{from} \frac{-3}{5}$

Answer 2:

$\begin{array}{l}\left(i\right)\\ \frac{1}{3}-\frac{3}{4}=\frac{1}{3}+ (\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{3}{4})\\ \text{ L.C.M.} \text{of} \text{3 and 4 is 12.}\\ \frac{1}{3}+\left(-\frac{3}{4}\right)=\frac{4+(-9)}{12}=\frac{-5}{12}\end{array}$


$\begin{array}{l}\left(ii\right) \\ \\ \frac{1}{3}-\frac{(-5)}{6}=\frac{1}{3}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-5}{6}\right)\\ =\frac{1}{3}+\frac{5}{6}\\ \mathrm{L.C.M}.\text{ of 3 and 6 is 6.}\\ \begin{array}{l}=\frac{2+5}{6}\\ =\frac{7}{6}\end{array}\end{array}$

$\begin{array}{l}\left(iii\right) \\ \\ \frac{(-3)}{5}-\left(\frac{-8}{9}\right)=\frac{(-3)}{5}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-8}{9}\right)\\ =\frac{(-3)}{5}+\frac{8}{9}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 5 \mathrm{and} 9 \mathrm{is} 45.\\ \begin{array}{l}=\frac{-27+40}{45}\\ =\frac{13}{45}\end{array}\end{array}$

$\begin{array}{l}\left(iv\right)\\ \\ \frac{(-1)}{1}-\frac{(-9)}{7}=\frac{(-1)}{1}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-9}{7}\right)\\ =\frac{(-1)}{1}+\frac{9}{7}\\ L.C.M. of 7 and 1 is 7.\\ \begin{array}{l}=\frac{-7+9}{7}\\ =\frac{2}{7}\end{array}\end{array}$

$\begin{array}{l}\left(v\right)\\ \\ \frac{1}{1}-\frac{(-18)}{11}=\frac{1}{1}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-18}{11}\right)\\ =\frac{\left(1\right)}{1}+\frac{18}{11} \\ =\frac{11+18}{11} \\ =\frac{29}{11}\end{array}$

(vi)
$$\begin{gathered} 0-\left(\frac{-13}{9}\right)=0+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-13}{9}\right) \\ =0+\frac{13}{9} \\ =\frac{13}{9} \end{gathered}$$

.$\begin{array}{l}\left(\mathrm{vii}\right)\\ \\ \frac{(-6)}{5}-\frac{(-32)}{13}=\frac{(-6)}{5}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-32}{13}\right)\\ =\frac{-6}{5}+\frac{32}{13}\\ L.C.M. of 5 \mathrm{and} 13 \mathrm{is} 65\\ \begin{array}{l}=\frac{-78+160}{65}\\ =\frac{82}{65}\end{array}\end{array}$


$$\begin{gathered} \begin{array}{l}\left(\mathrm{viii}\right) \\ \\ \\ \begin{array}{l}\begin{array}{l}\frac{-4}{7}-\frac{(-7)}{1}=\frac{-4}{7}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-7}{1}\right)\\ =\frac{(-4)}{7}+\frac{7}{1}\\ \text{L.C.M. of 7 and 1 is 7.}\end{array}\\ =\frac{-4+49}{7}\end{array}\end{array} \\ =\frac{45}{7} \end{gathered}$$
$\begin{array}{l}\left(\mathrm{ix}\right)\\ \\ \\ \frac{-2}{3}-\frac{5}{9}=\frac{-2}{3}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{5}{9}\right)\\ =\frac{-2}{3}-\frac{5}{9}\\ \text{ L.C.M. of} \mathrm{3 and 9 is 9}.\\ \begin{array}{l}=\frac{-6-5}{9}\\ =\frac{-11}{9}\end{array}\end{array}$

$\begin{array}{l}\text{(x)}\\ \\ \frac{-3}{5}-\frac{5}{1}=\frac{-3}{5}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{5}{1}\right)\\ =\frac{-3}{5}-\frac{5}{1}\\ \text{ L.C.M. of} \mathrm{5 and 1 is 5}. \\ \begin{array}{l}=\frac{-3-25}{5}\\ =\frac{-28}{5} \end{array}\end{array}$

Question 3:

Evaluate:

(i) $\frac{3}{4}-\frac{4}{5}$
(ii) $-3-\frac{4}{7}$
(iii) $\frac{7}{24}-\frac{19}{36}$
(iv) $\frac{14}{15}-\frac{13}{20}$
(v) $\frac{4}{9}-\frac{2}{-3}$
(vi) $\frac{7}{11}-\frac{-4}{-11}$
(vii) $\frac{-5}{14}-\frac{-2}{7}$
(viii) $\frac{-5}{-8}-\frac{-3}{4}$

Answer 3:

$\begin{array}{l}\text{( i)}\\ \\ \frac{3}{4}-\frac{4}{5}\\ =\frac{3}{4}-\frac{4}{5}=\frac{3}{4}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{4}{5}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \text{4 and 5 is 20.} \\ \begin{array}{l}=\frac{15-16}{20}\\ =\frac{-1}{20}\end{array}\\ \end{array}$                                        


$\begin{array}{l}\left(ii\right)\\ \\ \frac{-3}{1}-\frac{4}{7}=\frac{-3}{1}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{4}{7}\right)\\ =\frac{-3}{1}-\frac{4}{7} \\ =\frac{-21-4}{7} \\ =\frac{-25}{7}\end{array}$

$\begin{array}{l}\left(\mathrm{iii}\right)\\ \frac{7}{24}-\frac{19}{36}= \frac{7}{24}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{19}{36}\right)\\ = \frac{7}{24}-\frac{19}{36}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 24 \mathrm{and} 36 \mathrm{is} 72.\\ \begin{array}{l}=\frac{21-38}{72}\\ =\frac{-17}{72}\end{array}\end{array}$                                

$\begin{array}{l}\left(iv\right)\\ \\ \frac{14}{15}-\frac{13}{20}=\frac{14}{15}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{13}{20}\right)\\ \\ =\frac{14}{15}-\frac{13}{20}\\ \\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 15 \mathrm{and} 20 \mathrm{is} 60.\\ \begin{array}{l}=\frac{56-39}{60}\\ \\ =\frac{17}{60}\end{array}\\ \end{array}$

$\begin{array}{l}\left(\mathrm{v}\right)\\ \\ \frac{4}{9}-\frac{2}{(-3)}\\ \text{We need a positive denominator.}\\ \frac{2}{-3}\times \frac{(-1)}{(-1)}=\frac{-2}{3}\\ \\ \frac{4}{9}-\frac{(-2)}{3}\text{ =}\frac{4}{9}\text{+ }\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-2}{3}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 3 \mathrm{and} 9 \mathrm{is} 9.\\ \begin{array}{l}=\frac{4+6}{9}\\ =\frac{10}{9}\end{array}\end{array}$

$\begin{array}{l}\\ \text{(vi)}\\ \\ \frac{7}{11}-\frac{(-4)}{(-11)}\\ \text{We need a positive denominator. }\\ \frac{-4}{-11}\times \frac{(-1)}{(-1)}=\frac{4}{11}\\ \\ \frac{7}{11}-\frac{4}{11}=\frac{7}{11}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of}\frac{4}{11} \right)\\ =\frac{7-4}{11}\\ =\frac{3}{11}\\ \end{array}$
$$\begin{gathered} \left(\mathrm{vii}\right) \frac{-5}{14}-\frac{-2}{7}=\frac{-5}{14}+\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-2}{7}\right) \\ \frac{-5}{14}+\frac{2}{7} \\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 14 \mathrm{and} 7 \mathrm{is} 14. \\ \frac{-5+4}{14} \\ =\frac{-1}{14} \end{gathered}$$




$\begin{array}{l}\left(viii\right)\\ \\ \left(\frac{-5}{-8}\right)-\left(\frac{-3}{4}\right)\\ \\ \text{ We need a positive denominator. }\\ \\ \frac{-5}{-8}\times \frac{-1}{-1}=\frac{5}{8}\\ \\ \frac{5}{8}-\frac{(-3)}{4}\text{ =}\frac{5}{8}\text{+ }\left(\mathrm{additive} \mathrm{inverse} \mathrm{of} \frac{-3}{4}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 8 \mathrm{and} 4 \mathrm{is} 8.\\ =\frac{5+6}{8}\\ \\ =\frac{11}{8}\end{array}$    [L.C.M. of 8 and 4 is 8.]

Question 4:

Subtract the sum of $\frac{-36}{11}$ and $\frac{49}{22}$ from the sum of $\frac{33}{8}$ and $\frac{-19}{4}$.

Answer 4:

$\begin{array}{l}\\ \text{First we will find the sum of }\frac{-36}{11} \text{and }\frac{49}{22}\text{.} \\ \\ \frac{-36}{11}+\frac{49}{22}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 11 \mathrm{and} 22 \mathrm{is} 22.\\ =\frac{-72+49}{22}\\ \\ =\frac{-23}{22}\\ \\ \text{Now. we have to find the sum of }\frac{33}{8}\mathrm{and}\frac{-19}{4}.\\ \frac{33}{8}+\frac{(-19)}{4}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 8 \mathrm{and} 4 \mathrm{is} 8.\\ \begin{array}{l}=\frac{33-38}{8}\\ =\frac{-5}{8}\end{array}\\ \\ \mathrm{Now}, \frac{(-5)}{8}-\frac{(-23)}{22}\\ =\frac{-5}{8}+\frac{23}{22}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 8 \mathrm{and} 22 \mathrm{is} 88.\\ \begin{array}{l}=\frac{-55+92}{88}\\ =\frac{37}{88}\end{array}\end{array}$

Question 5:

The sum of two rational numbers is $\frac{4}{21}$. If one of them is $\frac{5}{7}$, find the other.

Answer 5:

$\begin{array}{l}\\ \text{Let the other number that be }\mathit{\text{x.}}\\ \\ \frac{5}{7}+x=\frac{4}{21}\\ =>x=\frac{4}{21}-\frac{5}{7}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 21 \mathrm{and} 7 \mathrm{is} 21.\\ \begin{array}{l}=>x=\frac{4-15}{21}\\ =>x=\frac{-11}{21}\end{array}\\ \text{Hence, the required number is }\frac{-11}{21}\text{.}\end{array}$

Question 6:

The sum of two rational numbers is $\frac{-3}{8}$. If one of them is $\frac{3}{16}$, find the other.

Answer 6:

$\begin{array}{l}\\ \text{Let the other number be}\mathit{\text{ x. }}\\ \\ \frac{3}{16}+x=\frac{-3}{8}\\ \\ =>x=\frac{-3}{8}-\frac{3}{16}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 8 \mathrm{and} 16 \mathrm{is} 16.\\ =>x=\frac{-6-3}{16}\\ \\ =>x=\frac{-9}{16}\\ \end{array}$

Question 7:

The sum of two rational numbers is −3. If one of them is $\frac{-15}{7}$, find the other.

Answer 7:

$\begin{array}{l}\\ \text{ Let the other number be }\mathit{\text{x. }}\\ \frac{-15}{7}+x=-3\\ =>x=-3-\frac{(-15)}{7}\\ \\ \begin{array}{l}=>x=\frac{-21+15}{7}\\ =>x = \frac{-6}{7}\\ \text{Hence, the required number is }\frac{-6}{7}\end{array}\\ \end{array}$

Question 8:

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

Answer 8:

$\begin{array}{l}\text{ Let the required number be }\mathit{\text{x}}\text{. }\\ \\ -5+x=\frac{-4}{3}\\ =>x=\frac{-4}{3}+5\\ =>x=\frac{-4+15}{3}\\ =>x=\frac{11}{3}\\ \\ \text{Hence, the requied number is }\frac{11}{3}\text{.}\end{array}$

Question 9:

What should be added to $\frac{-3}{8}$ to get $\frac{5}{12}$?

Answer 9:

 Let the required number be x.

$\begin{array}{l}\frac{\begin{array}{l}\\ -3\end{array}}{8}+x=\frac{5}{12}\\ =>x=\frac{5}{12}-\frac{(-3)}{8}\\ =>x=\frac{10+9}{24}\\ =>x=\frac{19}{24}\\ \text{Hence, the required number is }\frac{19}{24}\text{.}\\ \end{array}$

Question 10:

What should be added to $\frac{-12}{5}$ to get 3?

Answer 10:

Let the number that is to be added be x.

$\begin{array}{l}\frac{\begin{array}{l}\\ -12\end{array}}{5}+x=3\\ =>x=3-\frac{(-12)}{5}\\ =>x=\frac{15+12}{5}\\ =>x=\frac{27}{5}\\ \text{Hence, the required number is }\frac{27}{5}\text{.}\\ \end{array}$

Question 11:

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

Answer 11:

Let the number that is to be added be x.

$\begin{array}{l}\frac{-5}{7}+x=\frac{-2}{3}\\ =>x=\frac{-2}{3}-\frac{(-5)}{7}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 3 \mathrm{and} 7 \mathrm{is} 21.\\ \begin{array}{l}=>x=\frac{-14+15}{21}\\ =>x=\frac{1}{21}\\ \text{Hence, the required number is }\frac{1}{21}\text{.}\end{array}\\ \\ \end{array}$

Question 12:

What should be added to $\frac{2}{9}$ to get −1?

Answer 12:

Let the number that is to be added be x.
$\begin{array}{l}\frac{2}{9}+x=-1\\ =>x=-1-\frac{2}{9}\\ =>x=\frac{-9-2}{9}\\ =>x=\frac{-11}{9}\\ \text{Hence, the required number is }\frac{-11}{9}\text{.}\\ \end{array}$

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Question 13:

What should be added to $\left(\frac{-13}{4}+\frac{-3}{8}\right)$ to get 1?

Answer 13:

$\begin{array}{l}\text{Let the required} \text{number that is to be added be} \mathit{\text{x. }}\\ \left(\frac{-13}{4}+\frac{-3}{8}\right)+x=1\\ =>x=1-\left(\frac{-13}{4}+\frac{(-3)}{8}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 4 \mathrm{and} 8 \mathrm{is} 8.\\ \begin{array}{l}=1-\left(\frac{-26-3}{8}\right)\\ =1-\left(\frac{-29}{8}\right)\\ =1+\frac{29}{8}\\ =\frac{8+29}{8}\end{array}\\ =\frac{37}{8}\\ \text{Hence, the required number is }\frac{37}{8}\text{.}\end{array}$

Question 14:

What should be subtracted from $\frac{-3}{4}$ to get $\frac{5}{6}$?

Answer 14:

$\begin{array}{l}\text{Let the required} \text{number that is to be subtracted} \mathrm{be} \text{x}.\\ \frac{-3}{4}-x=\frac{5}{6}\\ =>-x=\frac{5}{6}-\frac{(-3)}{4}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 6 \mathrm{and} 4 \mathrm{is} 12.\\ \begin{array}{l}=>-x=\frac{10+9}{12}\\ =>-x=\frac{19}{12}\\ =>-x\times \left(\text{ -1}\right)=\frac{19}{12}\times \left(-1\right)\end{array}\\ \\ =>x=-\frac{19}{12}\\ \text{Hence, the required number is }\frac{-19}{12}\text{.}\end{array}$

Question 15:

What should be subtracted from $\frac{-2}{3}$ to get $\frac{-5}{6}$?

Answer 15:

$\begin{array}{l}\text{Let the required number that is to be subtracted} \text{be} \text{x}.\\ \frac{-2}{3}-x=\frac{-5}{6}\\ =>-x=\frac{-5}{6}-\frac{(-2)}{3}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 6 \mathrm{and} 3 \mathrm{is} 6.\\ \begin{array}{l}=>-x=\frac{-5+4}{6}\\ =>-x=\frac{-1}{6}\\ =>x=\frac{1}{6}\end{array}\end{array}$

Question 16:

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

Answer 16:

$\begin{array}{l}\text{Let the required} \text{number that is to be subtracted} \mathrm{be} \mathit{\text{x}}.\\ \\ \frac{-3}{4}-x=1\\ =>-x=1-\frac{(-3)}{4}\\ \\ =>-x=\frac{4+3}{4}\\ =>-x=\frac{7}{4}\\ \\ =>x=-\frac{7}{4}\\ \end{array}$