# RD Sharma Solutions Class 7 Operations On Rational Numbers Exercise 5.1

## RD Sharma Solutions Class 7 Chapter 5 Exercise 5.1

#### Exercise 5.1

Q1. Add the following rational numbers:

(i) $\frac{-5}{7}\;\;and\;\;\frac{3}{7}$

We have,

$\frac{-5}{7}+\frac{3}{7}$

=$\frac{-5+3}{7}$

=$\frac{-2}{7}$

(ii) $\frac{15}{4}\;\;and\;\;\frac{7}{4}$

We have,

$\frac{-15}{4}+\frac{7}{4}$

=$\frac{-15+7}{4}$

=$\frac{-8}{4}$

=-2

(iii) $\frac{-8}{11}\;\;and\;\;\frac{-4}{11}$

We have,

$\frac{-8}{11}+\frac{-4}{11}$

=$\frac{-8-4}{11}$

=$\frac{-12}{11}$

(iv) $\frac{6}{13}\;\;and\;\;\frac{-9}{13}$

We have,

$\frac{6}{13}+\frac{-9}{13}$

$\frac{6}{13}-\frac{9}{13}$

=$\frac{6-9}{13}$

=$\frac{-3}{13}$

Q2. Add the following rational numbers:

(i) $\frac{3}{4}\;\;and\;\;\frac{-3}{5}$

If  $\frac{p}{q}$ and $\frac{r}{s}$ are two rational numbers such that q and s do not have a common factor

$\frac{p}{q}+\frac{r}{s}=\frac{p\times s+r\times q}{q\times s}$

$\frac{3}{4}+\frac{-3}{5}=\frac{3\times\left(5 \right )+\left(-3 \right )\times 4}{4\times 5}\\ =\frac{15-12}{20}\\ =\frac{3}{20}$

(ii) $\frac{-3}{1}\;\;and\;\;\frac{3}{5}$

If  $\frac{p}{q}$ and $\frac{r}{s}$ are two rational numbers such that q and s do not have a common factor

$\frac{p}{q}+\frac{r}{s}=\frac{p\times s+r\times q}{q\times s}$

$\frac{3}{1}+\frac{3}{5}=\frac{5\times\left(-3 \right )+\left(3 \right )\times 1}{5}\\ =\frac{-15+3}{5}\\ =\frac{-12}{5}$

(iii) $\frac{-3}{1}\;\;and\;\;\frac{3}{5}$

LCM of 27 and 18 is 54

$\frac{-7}{27}=\frac{-7\times 2}{27\times2}=\frac{-14}{54}\\ \frac{11}{18}=\frac{11\times3}{18\times3}=\frac{33}{54}\\ \frac{-7}{27}+\frac{11}{18}=\frac{-14}{54}+\frac{33}{54}\\ =\frac{33-14}{54}\\ =\frac{19}{54}$

(iv) $\frac{31}{-4}\;\;and\;\;\frac{-5}{8}$

LCM of 4 and 8 is 4

$\frac{31}{-4}=\frac{31\times 2}{-4\times2}=\frac{62}{-8}\\ \frac{31}{-4}+\frac{-5}{8}=\frac{62}{-8}+\frac{-5}{8}\\ =\frac{-62-5}{8}\\ =\frac{-67}{8}$

Q3. Simplify

(i) $\frac{8}{9}+\frac{-11}{6}$

$\frac{8}{9}-\frac{11}{6}$

LCM of 9 and 6 is 18

$\frac{8}{9}=\frac{8\times 2}{9\times2}=\frac{16}{18}\\ \frac{11}{6}=\frac{11\times 3}{6\times3}=\frac{33}{18}\\ \frac{8}{9}+\frac{-11}{6}=\frac{16}{18}-\frac{33}{18}\\ =\frac{16-33}{18}\\ =\frac{-17}{18}$

(ii) $\frac{-5}{16}+\frac{7}{24}$

LCM of 16 and 24 is 48

$\frac{-5}{16}=\frac{-5\times 3}{16\times3}=\frac{-15}{48}\\ \frac{7}{24}=\frac{7\times 2}{24\times2}=\frac{14}{48}\\ \frac{-5}{16}+\frac{7}{24}=\frac{-15}{48}+\frac{14}{48}\\ =\frac{14-15}{48}\\ =\frac{-1}{48}$

(iii) $\frac{1}{-12}+\frac{2}{-15}$

$\frac{-1}{12}-\frac{2}{15}$

LCM of 12 and 15 is 60

$\frac{-1}{12}=\frac{-1\times 5}{12\times5}=\frac{-5}{60}\\ \frac{-2}{15}=\frac{-2\times 4}{15\times4}=\frac{-8}{60}\\ \frac{-1}{12}-\frac{2}{15}=\frac{-5}{60}+\frac{-8}{60}\\ =\frac{-5-8}{60}\\ =\frac{-13}{60}$

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

LCM of 19 and 57 is 57

$\frac{-8}{19}=\frac{-8\times 3}{19\times3}=\frac{-24}{57}\\ \frac{-8}{19}+\frac{-4}{57}=\frac{-24}{57}+\frac{-4}{57}\\ =\frac{-24-4}{57}\\ =\frac{-28}{57}$

Q4. Add and express the sum as a mixed fraction:

(i) $\frac{-12}{5}+\frac{43}{10}$

LCM of 5 and 10 is 10

$\frac{-12}{5}=\frac{-12\times 2}{5\times2}=\frac{-24}{10}\\ \frac{-12}{5}+\frac{43}{10}=\frac{-24}{10}+\frac{43}{10}\\ =\frac{-24+43}{10}\\ =\frac{19}{10}\\ =1\frac{9}{10}$

(ii) $\frac{24}{7}+\frac{-11}{4}$

LCM of 7 and 4 is 28

$\frac{24}{7}=\frac{24\times 4}{7\times4}=\frac{96}{28}\\ \frac{-11}{4}=\frac{-11\times 7}{4\times7}=\frac{-77}{28}\\ \frac{24}{7}+\frac{-11}{4}=\frac{96}{28}+\frac{-77}{28}\\ =\frac{96}{28}-\frac{77}{28}\\ =\frac{96-77}{28}\\ =\frac{19}{28}\\$

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

LCM of 6 and 8 is 24

$\frac{-31}{6}=\frac{-31\times 4}{6\times4}=\frac{-124}{24}\\ \frac{-27}{8}=\frac{-27\times 3}{8\times3}=\frac{-81}{24}\\ \frac{-31}{6}+\frac{-27}{8}=\frac{-124}{24}+\frac{-81}{24}\\ =\frac{-124}{24}-\frac{81}{24}\\ =\frac{-124-81}{24}\\ =\frac{-205}{24}\\ =-8\frac{13}{24}$