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

Exercise 4C

Page-69

Question 1:

Add the following rational numbers:

(i) $\frac{12}{7 } \mathrm{and} \frac{3}{7}$
(ii) $\frac{-2}{5 } \mathrm{and} \frac{1}{5}$
(iii) $\frac{3}{-8 } \mathrm{and} \frac{1}{8}$
(iv) $\frac{-5}{11} \mathrm{and} \frac{7}{-11}$
(v) $\frac{9}{-13 } \mathrm{and} \frac{-11}{-13}$
(vi) $\frac{-2}{9} \mathrm{and} \frac{-5}{9}$
(vii) $\frac{-17}{9} \mathrm{and} \frac{-11}{9}$
(viii) $\frac{-3}{7 } \mathrm{and} \frac{5}{-7}$

Answer 1:

(i)
$\frac{12}{7}+\frac{3}{7}=\frac{12+3}{7}=\frac{15}{7}$

(ii)
$\frac{-2}{5}+\frac{1}{5}=\frac{-2+1}{5}=\frac{-1}{5}$

(iii)

$\frac{3}{-8}\times \frac{-1}{-1}=\frac{-3}{8}$

$\frac{-3}{8}+\frac{1}{8}=\frac{-3+1}{8}=\frac{-2}{8}$

(iv)

$$\begin{gathered} \frac{7}{-11}\times \frac{-1}{-1}=\frac{-7}{11} \\ \frac{-5}{11}+\frac{-7}{11}=\frac{-5+(-7)}{11}=\frac{-5-7}{11}=\frac{-12}{11} \end{gathered}$$


(v)

$\frac{-11}{-13}\times \frac{-1}{-1}=\frac{11}{13}$

=$\frac{-9}{13}+\frac{11}{13}=\frac{-9+11}{13}=\frac{2}{13}$

(vi)

$\frac{-2}{9}+\frac{-5}{9}=\frac{-2-5}{9}=\frac{-7}{9}$

(vii)

$\frac{(-17)}{9}+\frac{(-11)}{9}=\frac{-17-11}{9}=\frac{-28}{9}$

(viii)
$\frac{5}{-7}\times \frac{-1}{-1}=\frac{-5}{7}$


$\frac{-3}{7}+\frac{(-5)}{7}=\frac{-3-5}{7}=\frac{-8}{7}$

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Question 2:

Add the following rational numbers:

(i) $\frac{-2}{5 } \mathrm{and} \frac{3}{4}$
(ii) $\frac{-5}{9} \mathrm{and} \frac{2}{3}$
(iii) $-4 \mathrm{and} \frac{1}{2}$
(iv) $\frac{-7}{27 } \mathrm{and} \frac{5}{18}$
(v) $\frac{-5}{36} \mathrm{and} \frac{-7}{12}$
(vi) $\frac{1}{-9 } \mathrm{and} \frac{4}{-27}$
(vii) $\frac{-9}{24} \mathrm{and} \frac{-1}{8}$
(viii) $\frac{27}{-4} \mathrm{and} \frac{-15}{8}$

Answer 2:

(i)$\frac{-2}{5}+\frac{3}{4}$

$\begin{array}{l}\text{The denominators of the given rational numbers are 5 and 4}.\\ L.C.M. of \text{5 and 4 is} \text{2}0.\\ \frac{-2}{5}=\frac{(-2)\times 4}{5\times 4}=\frac{-8}{20}\\ \frac{3}{4}=\frac{3\times 5}{4\times 5}=\frac{15}{20}\\ \\ \mathrm{Now},\frac{(-8)}{20}+\frac{15}{20}=\frac{-8+15}{20}=\frac{7}{20}\end{array}$

(ii)$\frac{-5}{9}+\frac{2}{3}$

$\begin{array}{l}\text{The denominators of the given rational numbers are 9 and 3. } \end{array}$



$\begin{array}{l} \text{L.C.M. of 9 and 3 is} \text{9.}\\ \frac{-5}{9}=\frac{(-5)\times 1}{9\times 1}=\frac{-5}{9}\\ \\ \frac{2}{3}=\frac{2\times 3}{3\times 3}=\frac{6}{9}\\ \\ \mathrm{Now}, \frac{(-5)}{9}+\frac{6}{9}\\ =\frac{-5+6}{9}\\ =\frac{1}{9}\end{array}$
(iii)$-4+\frac{1}{2}$


$\begin{array}{l}\text{The denominators of the given rational numbers are 1 and 2. } \\ \text{L.C.M. of 1 and 2 is 2}.\\ \frac{-4}{1}=\frac{(-4)\times 2}{1\times 2}=\frac{-8}{2}\\ \\ \frac{1}{2}=\frac{1\times 1}{2\times 1}=\frac{1}{2}\\ \\ \mathrm{Now}, \frac{(-8)}{2}+\frac{1}{2}\\ =\frac{-8+1}{2}\\ =\frac{-7}{2}\end{array}$

(iv)
$\frac{-7}{27}+\frac{5}{18}$


The denominators of the given rational numbers are 27 and 18.

L.C.M. of 27 and 18 is 54.

$\begin{array}{l}\\ \frac{-7}{27}=\frac{(-7)\times 2}{27\times 2}=\frac{-14}{54}\\ \frac{5}{18}=\frac{5\times 3}{18\times 3}=\frac{15}{54}\\ \\ \mathrm{Now}, \frac{(-14)}{54}+\frac{15}{54}=\frac{-14+15}{54}\\ =\frac{1}{54}\\ \end{array}$

(v)$\frac{-5}{36}+\left(\frac{-7}{12}\right)$


The denominators of the given rational numbers are 36 and 12.
L.C.M. of 36 and 12 is 36.

$\begin{array}{l}\frac{-5}{36}=\frac{(-5)\times 1}{36\times 1}=\frac{-5}{36}\\ \frac{-7}{12}=\frac{-7\times 3}{12\times 3}=\frac{-21}{36}\\ \\ \mathrm{Now}, \frac{(-5)}{36}+\frac{(-21)}{36}=\frac{-5-21}{36}\\ \frac{-26}{36}=\frac{-13}{18} (\text{26 and 36 are divided by 2 .)}\\ \end{array}$

(vi)
$\frac{1}{-9}+\left(\frac{4}{-27}\right)$

$\begin{array}{l}\text{We need a positive denominator}.\\ \frac{1}{-9}\times \frac{-1}{-1}=\frac{-1}{9}\text{ and }\frac{4}{-27}\times \frac{-1}{-1}=\frac{-4}{27}\\ \\ \text{The denominators of the given rational numbers are 9} \text{and} \text{27}\end{array}$

$\begin{array}{l}\text{L.C.M. of 9 and 27 is} \text{27}\\ \frac{-1}{9}=\frac{(-1)\times 3}{9\times 3}=\frac{-3}{27}\\ \frac{-4}{27}=\frac{-4\times 1}{27\times 1}=\frac{-4}{27}\\ \\ \frac{(-3)}{27}+\frac{(-4)}{27}=\frac{-3-4}{27}\\ \\ =\frac{-7}{27}\end{array}$


(vii)
$$\begin{gathered} \frac{-9}{24}+\left(\frac{-1}{18}\right) \\ \mathrm{The} \mathrm{denominators} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{numbers} \mathrm{are} 24 \mathrm{and} 18 \end{gathered}$$

$$\begin{gathered} \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 24 \mathrm{and} 18 \mathrm{is} 72. \\ \therefore \frac{-9}{24}=\frac{-9\times 3}{24\times 3}=\frac{-27}{72} \\ \frac{-1}{18}=\frac{-1\times 4}{18\times 4}=\frac{-4}{72} \\ \mathrm{Now}, \frac{-27}{72}+\left(\frac{-4}{72}\right) \\ =\frac{-27+(-4)}{72} \\ = \frac{-27-4}{72} \\ =\frac{-31}{72} \end{gathered}$$


(viii)$\frac{27}{-4}+\left(\frac{-15}{8}\right)$

$\begin{array}{l}\text{We need a positive denominator. }\\ \frac{27}{-4}\times \frac{-1}{-1}=\frac{-27}{4}\\ \text{The denominators of the given rational numbers are 4 and 8. } \end{array}$

$\begin{array}{l}\text{L.C.M. of 4 and 8 is 8.}\\ \frac{-27}{4}=\frac{-27\times 2}{4\times 2}=\frac{-54}{8}\\ \frac{(-15)}{8}=\frac{(-15)\times 1}{8\times 1}=\frac{-15}{8}\\ \mathrm{Now}, \frac{-54}{8}+\frac{(-15)}{8}\\ =\frac{-54-15}{8}\\ =\frac{-69}{8}\end{array}$

Question 3:

Evaluate:

(i) $\frac{-3}{5}+\frac{7}{5}+\frac{-1}{5}$
(ii) $\frac{-12}{7}+\frac{3}{7}+\frac{-2}{7}$
(iii) $\frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}$
(iv) $\frac{-16}{9}+\frac{-5}{12}+\frac{7}{18}$
(v) $-3+\frac{1}{8}+\frac{-2}{5}$
(vi) $\frac{-13}{8}+\frac{5}{16}+\frac{-1}{4}$

Answer 3:

$\begin{array}{l}\left(i\right)\\ \frac{-3}{5}+\frac{7}{5}+\frac{-1}{5}\\ \text{L.C.M. of the given rational number is 5}.\\ \\ \frac{(-3)}{5}+\frac{7}{5}+\frac{(-1)}{5}\\ =\frac{-3+7-1}{5}\\ =\frac{-4+7}{5}\\ =\frac{3}{5}\end{array}$

(ii)
$\begin{array}{l}\frac{\begin{array}{l}\\ -12\end{array}}{7}+\frac{3}{7}+\frac{-2}{7}\\ =\frac{(-12)}{7}+\frac{3}{7}+\frac{(-2)}{7}\\ =\frac{-12+3-2}{7}\\ =\frac{-14+3}{7}\\ =\frac{-11}{7}\end{array}$


$\begin{array}{l}\left(\mathrm{iii}\right) \frac{11}{-12}+\frac{3}{-8}+\frac{1}{4}\\ \\ \text{ We need a positive denominator.}\\ \\ \frac{11}{-12}\times \frac{-1}{-1}=\frac{-11}{12}\mathrm{and}\frac{3}{-8}\times \frac{-1}{-1}=\frac{-3}{8}\\ \text{L.C.M. of the denominators 12}, \text{8 and} \text{4 is 24. }\\ \therefore \frac{-11\times 2}{12\times 2}=\frac{-22}{24}\\ \\ \frac{-3\times 3}{8\times 3}=\frac{-9}{24}\\ \\ \frac{1\times 6}{4\times 6}=\frac{6}{24}\\ \mathrm{Now},\frac{(-22)}{24}+\frac{(-9)}{24}+\frac{6}{24}\\ =\frac{-22-9+6}{24}\\ \text{=}\frac{-31+6}{24}\\ =\frac{-25}{24}\end{array}$                                     
$\begin{array}{l}\left(\mathrm{iv}\right)\frac{-16}{9}+\frac{-5}{12}+\frac{7}{18}\\ \\ \text{ L.C.M. of the denominators } \text{9}, \text{12 and} \text{18 is 36.}\\ \frac{-16\times 4}{9\times 4}=\frac{-64}{36}\\ \frac{-5\times 3}{12\times 3}=\frac{-15}{36}\\ \frac{7\times 2}{18\times 2}=\frac{14}{36}\\ \mathrm{Now}, \frac{(-64)}{36}+\frac{(-15)}{36}+\frac{14}{36}\\ =\frac{-64-15+14}{36}\\ \frac{-79+14}{36}\\ =\frac{-65}{36}\end{array}$                                     

$\begin{array}{l}\left(v\right)-3+\frac{1}{8}=\frac{-2}{5}\\ \text{L.C.M. of the denominators} \text{1}, \text{8 and 5 is 4}0.\\ \\ \frac{-3\times 40}{1\times 40}=\frac{-120}{40}\\ \\ \frac{1\times 5}{8\times 5}=\frac{5}{40}\\ \\ \frac{-2\times 8}{5\times 8}=\frac{-16}{40}\\ \mathrm{Now}, \frac{(-120)}{40}+\frac{5}{40}+\frac{(-16)}{40}\\ =\frac{-120+5-16}{40}\\ = \frac{-136+5}{40}=\frac{-131}{40} \end{array}$                                

$\begin{array}{l}\left(vi\right)\frac{-13}{8}+\frac{5}{16}+\frac{-1}{4}\\ \text{L.C.M. of the denominator 8}, \text{16 and 4 is 16.}\\ \frac{-13\times 2}{8\times 2}=\frac{-26}{16}\\ \\ \frac{5\times 1}{16\times 1}=\frac{5}{16}\\ \\ \frac{-1\times 4}{4\times 4}=\frac{-4}{16}\\ \mathrm{Now},\frac{(-26)}{16}+\frac{5}{16}+\frac{(-4)}{16}\\ =\frac{-26+5-4}{16}\\ \\ \mathrm{Now},\frac{-30+5}{16}=\frac{-25}{16}\end{array}$

Question 4:

Simplify:

(i) $\frac{-8}{15}+\frac{2}{-3}$
(ii) $\frac{-7}{10}+\frac{13}{-15}+\frac{27}{20}$
(iii) $-1+\frac{7}{-9}+\frac{11}{12}$
(iv) $\frac{-11}{39}+\frac{5}{26}+2$
(v) $2+\frac{-1}{2}+\frac{-3}{4}$
(vi) $\frac{-9}{11}+\frac{2}{3}+\frac{-3}{4}$

Answer 4:

$\begin{array}{l}\left(i\right)\\ \frac{-8}{15}+\frac{2}{-3}\\ \\ \text{ We need a positive denominator. }\\ \therefore \frac{2}{-3}\times \frac{-1}{-1}=\frac{-2}{3}\\ \text{ Now, L.C.M. of 15 and 3 is 15.}\\ \frac{-8}{15}=\frac{-8\times 1}{15\times 1}=\frac{-8}{15}\\ \\ \frac{-2}{3}=\frac{-2\times 5}{3\times 5}=\frac{-10}{15}\\ \\ \mathrm{Now}, \frac{-8}{15}+\frac{-10}{15}\\ =\frac{-8-10}{15}\\ \\ =\frac{-18}{15}\\ =\frac{-6}{5}\end{array}$                                           

$\begin{array}{l}\left(ii\right)\\ \frac{-7}{10}+\frac{13}{-15}+\frac{27}{20}\\ \\ \text{We need a} \text{positive denominator.}\\ \frac{13}{-15}\times \frac{-1}{-1}=\frac{-13}{15}\\ \text{Now}, \text{L.C.M. of 1}0, \text{15 and 2}0\text{ is 6}0.\\ \therefore \frac{-7}{10}=\frac{-7\times 6}{10\times 6}=\frac{-42}{60}\\ \\ \frac{-13}{15}=\frac{-13\times 4}{15\times 4}=\frac{-52}{60}\\ \\ \frac{27}{20}=\frac{27\times 3}{20\times 3}=\frac{81}{60}\\ \\ \\ \mathrm{Now}, \frac{-42}{60}+\frac{-52}{60}+\frac{81}{60}\\ =\frac{(-42)+(-52)+\left(81\right)}{60}\\ =\frac{-94+81}{60}\\ =\frac{-13}{60}\end{array}$                                            


$\begin{array}{l}\left(iii\right)\\ -1+\frac{7}{-9}+\frac{11}{12}\\ \text{We need a} \text{positive denominator.}\\ \frac{7}{-9}\times \frac{-1}{-1}=\frac{-7}{9}\\ \text{Now}, \text{L.C.M. of 1}, \text{9 and 12 is 36.}\\ \frac{-1}{1}=\frac{-1\times 36}{1\times 36}=\frac{-36}{36}\\ \\ \frac{-7}{9}=\frac{-7\times 4}{9\times 4}=\frac{-28}{36}\\ \\ \frac{11}{12}=\frac{11\times 3}{12\times 3}=\frac{33}{36}\\ \\ \\ \frac{-36}{36}+\frac{-28}{36}+\frac{33}{36}\\ =\frac{-36-28+33}{36}\\ =\frac{-64+33}{36}\\ =\frac{-31}{36}\\ =\frac{-5}{4}\end{array}$                                                 


$\begin{array}{l}\left(iv\right)\\ \frac{-11}{39}+\frac{5}{26}+\frac{2}{1}\\ \text{ L.C.M. of 39},\text{ 26 and 1 is 78.}\\ \\ \frac{-11}{39}=\frac{-11\times 2}{39\times 2}=\frac{-22}{78}\\ \\ \frac{5}{26}=\frac{5\times 3}{26\times 3}=\frac{15}{78}\\ \\ \frac{2}{1}=\frac{2\times 78}{1\times 78}=\frac{156}{78}\\ \text{Now },\frac{-22}{78}+\frac{15}{78}+\frac{156}{78}\\ =\frac{-22+171}{78}\\ =\frac{149}{78}\end{array}$                                                                      
 
   $$\begin{gathered} \left(v\right) \\ 2+\frac{-1}{2}+\frac{-3}{4} \end{gathered}$$2+−12+−3

$\begin{array}{l}\text{L.C.M. of 2 and 4 is 4.}\\ 2=\frac{2\times 4}{1\times 4}=\frac{8}{4}\\ \frac{-1}{2}=\frac{-1\times 2}{2\times 2}=\frac{-2}{4}\\ \frac{-3}{4}=\frac{-3\times 1}{4\times 1}=\frac{-3}{4}\\ \frac{8}{4}+\frac{(-2)}{4}+\frac{(-3)}{4}\\ =\frac{8-2-3}{4}\\ =\frac{3}{4}\end{array}$                                                           

$\begin{array}{l}\left(vi\right)\frac{-9}{11}+\frac{2}{3}+\frac{-3}{4}\\ \text{L.C.M. of 11}, \text{3 and 4 is 132.}\\ \frac{-9}{11}=\frac{-9\times 12}{11\times 12}=\frac{-108}{132}\\ \frac{2}{3}=\frac{2\times 44}{3\times 44}=\frac{88}{132}\\ \frac{-3}{4}=\frac{-3\times 33}{4\times 33}=\frac{-99}{132}\\ \frac{-108}{132}+\frac{88}{132}+\frac{(-99)}{132}\\ =\frac{-108+88-99}{132}\\ =\frac{-207+88}{132}=\frac{-119}{132}\end{array}$                                                     

Question 5:

Express each of the following rational numbers as the sum of an integer and a rational number:

(i) $\frac{12}{5}$
(ii) $\frac{-11}{7}$
(iii) $\frac{-25}{9}$
(iv) $\frac{-103}{20}$

Answer 5:

$\begin{array}{l}\left(i\right) \frac{12}{5}=2\frac{2}{5}=2+\frac{2}{5}\\ \left(ii\right) \frac{-11}{7}=\left(-\frac{11}{7}\right)=\left(-1\frac{4}{7}\right)=-1+\left(\frac{-4}{7}\right)\\ \left(iii\right) \frac{-25}{9}=\left(-\frac{25}{9}\right)=-\left(2\frac{7}{9}\right)=-2+\left(\frac{-7}{9}\right)\\ \left(iv\right) \frac{-103}{20}=-\left(\frac{103}{20}\right)=-\left(5\frac{3}{20}\right)=-5+\left(\frac{-3}{20}\right)\end{array}$