# RD Sharma Solutions Class 7 Operations On Rational Numbers Exercise 5.2

## RD Sharma Solutions Class 7 Chapter 5 Exercise 5.2

#### Exercise 5.2

Q1. Subtract the first rational number from the second in each of the following:

(i) $\frac{3}{8},\frac{5}{8}$

$\frac{5}{8}-\frac{3}{8}=\frac{5-3}{8}\\ =\frac{2}{8}\\$

(ii) $\frac{-7}{9},\frac{4}{9}$

$\frac{-7}{9}+\frac{4}{9}=\frac{4}{9}-\frac{-7}{9}\\=\frac{4+7}{9}\\ =\frac{11}{9}\\$

(iii) $\frac{-2}{11},\frac{-9}{11}$

$\frac{-2}{11}+\frac{-9}{11}=\frac{-9}{11}+\frac{2}{11}\\ =\frac{-9+2}{11}\\ =\frac{-7}{11}$

(iv) $\frac{11}{13},\frac{-4}{13}$

$\frac{-4}{13}-\frac{11}{13}=\frac{-4-11}{13}\\ =\frac{-15}{13}\\$

Q2. Evaluate each of the following:

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

LCM of 3 and 5 is 15

$\frac{2}{3}=\frac{2\times5}{3\times5}=\frac{10}{15}\\ \frac{3}{5}=\frac{3\times3}{5\times3}=\frac{9}{15}\\ \frac{2}{3}-\frac{3}{5}=\frac{10}{15}-\frac{9}{15}\\ =\frac{1}{15}$

(ii) $-\frac{4}{7}-\frac{2}{-3}$

LCM of 3 and 7 is 21

$\frac{-4}{7}=\frac{-4\times3}{7\times3}=\frac{-12}{21}\\ \frac{2}{-3}=\frac{2\times7}{-3\times7}=\frac{14}{21}\\ \frac{-4}{7}-\frac{2}{-3}=\frac{-12}{21}-\frac{-14}{21}\\ =\frac{14}{21}-\frac{12}{21}\\ =\frac{14-12}{21}\\ =\frac{2}{21}$

(iii) $\frac{4}{7}-\frac{-5}{-7}$

$\frac{4}{7}-\frac{-5}{-7}\\ \frac{4}{7}-\frac{5}{7}=\frac{4-5}{7}\\ =\frac{-1}{7}$

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

$\frac{2}{1}-\frac{-5}{-9}\\ =\frac{-2\times9-5\times1}{9\times1}\\ =\frac{-18-5}{9}\\ =\frac{-23}{9}$

Q3. The sum of the two numbers is $\frac{5}{9}$. If one of the numbers is $\frac{1}{3}$, find the other.

Required number= $\frac{5}{9}-\frac{1}{3}\\$

LCM of 3 and 9 is 9

$\frac{1}{3}=\frac{1\times3}{3\times3}=\frac{3}{9}$

Therefore required number= $\frac{5}{9}-\frac{3}{9}$

=  $\frac{2}{9}$

Q4. The sum of two numbers is $\frac{-1}{3}$. If one of the numbers is $\frac{-12}{3}$, find the other.

Let the required number be x

$\frac{-12}{3}+x=\frac{-1}{3}\\ x=\frac{-1}{3}-\frac{-12}{3}\\ x=\frac{-1+12}{3}\\ x=\frac{11}{3}$

The required number is $\frac{11}{3}$

Q5. The sum of two numbers is $\frac{-4}{3}$. If one of the numbers is -5, find the other.

Let the required number be x

$-5+x=\frac{-4}{3}\\ x=\frac{-4}{3}+5\\ x=\frac{-4}{3}+\frac{5\times3}{1\times3}\\ x=\frac{-4}{3}+\frac{15}{3}\\ x=\frac{-4+15}{3}\\ x=\frac{11}{3}$

The required number is $\frac{11}{3}$

Q6. The sum of two rational numbers is -8. If one of the numbers is $\frac{-15}{7}$, find the other.

Let the required number be x

$\frac{-15}{7}+x=-8\\ x=-8-\frac{-15}{7}\\ x=-8+\frac{15}{7}\\ x=\frac{8\times7}{1\times7}+\frac{15}{7}\\ x=\frac{15-56}{7}\\ x=\frac{-41}{7}\\$

The required number is $\frac{-41}{7}$

Q7. What should be added to $\frac{-7}{8}$ so as to get $\frac{5}{9}$?

Let the required number be x

$\frac{-7}{8}+x=\frac{5}{9}\\ x=\frac{5}{9}-\frac{-7}{8}\\ x=\frac{5\times8}{9\times8}-\frac{-7\times9}{8\times9}\\ x=\frac{40}{72}-\frac{-63}{72}\\ x=\frac{40+63}{72}\\ x=\frac{103}{72}\\$

The required number is $\frac{103}{72}$

Q8. What number should be added to $\frac{-5}{11}$ so as to get $\frac{26}{33}$?

Let the required number be x

$\frac{-5}{11}+x=\frac{26}{33}\\ x=\frac{26}{33}-\frac{-5}{11}\\ x=\frac{26}{33}-\frac{-5\times3}{11\times3}\\ x=\frac{26}{33}-\frac{-15}{33}\\ x=\frac{26+15}{33}\\ x=\frac{41}{33}\\$

The required number is $\frac{41}{33}$

Q9. What number should be added to $\frac{-5}{7}$ to get $\frac{-2}{3}$?

Let the required number be x

$\frac{-5}{7}+x=\frac{-2}{3}\\ x=\frac{-2}{3}-\frac{-5}{7}\\ x=\frac{-2\times7}{3\times7}-\frac{-5\times3}{7\times3}\\ x=\frac{-14}{21}-\frac{-15}{21}\\ x=\frac{-14+15}{21}\\ x=\frac{1}{21}\\$

The required number is $\frac{1}{21}$

Q10. What number should be subtracted from $\frac{-5}{3}$ to get $\frac{5}{6}$?

Let the required number be x

$\frac{-5}{3}-x=\frac{5}{6}\\ -x=\frac{5}{6}-\frac{-5}{3}\\ -x=\frac{5}{6}-\frac{-5\times2}{3\times2}\\ -x=\frac{5}{6}-\frac{-10}{6}\\ -x=\frac{5+10}{6}\\ -x=\frac{15}{6}\\ x=-\frac{15}{6}\\$

The required number is $\frac{15}{6}$

Q11. What number should be subtracted from $\frac{3}{7}$ to get $\frac{5}{4}$?

Let the required number be x

$\frac{3}{7}-x=\frac{5}{4}\\ -x=\frac{5}{4}-\frac{3}{7}\\ -x=\frac{5\times7}{4\times7}-\frac{3\times4}{7\times4}\\ -x=\frac{35}{28}-\frac{12}{28}\\ -x=\frac{35-12}{28}\\ -x=\frac{23}{28}\\ x=-\frac{23}{28}\\$

The required number is $\frac{23}{28}$

Q12. What should be added to $\left(\frac{2}{3}+\frac{3}{5} \right )$ to get $\frac{-2}{15}$?

Let the required number be x

$\left(\frac{2}{3}+\frac{3}{5} \right )+x=\frac{-2}{15}\\ \left(\frac{2\times5}{3\times5}+\frac{3\times3}{5\times3} \right )+x=\frac{-2}{15}\\ \left(\frac{10}{15}+\frac{9}{15} \right )+x=\frac{-2}{15}\\ \frac{19}{15}+x=\frac{-2}{15}\\ x=\frac{-2}{15}-\frac{19}{15}\\ x=\frac{-2-19}{15}\\ x=\frac{-21}{15}\\ x=\frac{-7}{5}$

The required number is $\frac{-7}{5}$

Q13. What should be added to $\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5} \right )$ to get 3?

Let the required number be x

$\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{5} \right )+x=3\\ \left(\frac{1\times15}{2\times15}+\frac{1\times10}{3\times10}+\frac{1\times6}{5\times6} \right )+x=3\\ \left(\frac{15+10+6}{30} \right )+x=3\\ \frac{31}{30}+x=3\\ x=3-\frac{31}{30}\\ x=\frac{3\times30}{1\times30}-\frac{31}{30}\\ x=\frac{90}{30}-\frac{31}{30}\\ x=\frac{59}{30}\\$

The required number is $\frac{59}{30}$

Q14. What should be subtracted from $\left(\frac{3}{4}-\frac{2}{3} \right )$ to get $\frac{-1}{6}$

Let the required number be x

$\left(\frac{3}{4}-\frac{2}{3} \right )-x=\frac{-1}{6}\\ \left(\frac{3\times3}{4\times3}-\frac{2\times4}{3\times4} \right )-x=\frac{-1}{6}\\ \left(\frac{9}{12}-\frac{8}{12} \right )-x=\frac{-1}{6}\\ \frac{1}{12}-x=\frac{-1}{6}\\ -x=\frac{-1}{6}-\frac{1}{12}\\ -x=\frac{-1\times2}{6\times2}-\frac{1}{12}\\$

$-x=\frac{-2}{12}-\frac{1}{12}\\ -x=\frac{-2-1}{12}\\ -x=\frac{-3}{12}\\ x=\frac{3}{12}\\ x=\frac{1}{4}\\$

The required number is $\frac{1}{4}$

Q15. Simplify:

(i) $\left(\frac{-3}{2}+\frac{5}{4}-\frac{7}{4} \right )$

$\left(\frac{-3}{2}+\frac{5}{4}-\frac{7}{4} \right )\\ =\left(\frac{-3\times2}{2\times2}+\frac{5}{4}-\frac{7}{4} \right )\\ =\left(\frac{-6}{4}+\frac{5}{4}-\frac{7}{4} \right )\\ =\left(\frac{-6+5-7}{4} \right )\\ =\left(\frac{-13+5}{4} \right )\\ =\left(\frac{-8}{4} \right )\\ =-2$

(ii) $\left(\frac{5}{3}-\frac{7}{6}+\frac{-2}{3} \right )$

$\left(\frac{5}{3}-\frac{7}{6}+\frac{-2}{3} \right )\\ =\left(\frac{5\times2}{3\times2}-\frac{7}{6}+\frac{-2\times2}{3\times2} \right )\\ =\left(\frac{10}{6}-\frac{7}{6}+\frac{-4}{6} \right )\\ =\left(\frac{10-7-4}{6} \right )\\ =\left(\frac{10-11}{6} \right )\\ =\left(\frac{-1}{6} \right )\\$

(iii) $\left(\frac{5}{4}-\frac{7}{6}-\frac{-2}{3} \right )$

$\left(\frac{5}{4}-\frac{7}{6}-\frac{-2}{3} \right )\\ =\left(\frac{5\times3}{4\times3}-\frac{7\times2}{6\times2}-\frac{-2\times4}{3\times4} \right )\\ =\left(\frac{15}{12}-\frac{14}{12}-\frac{-8}{12} \right )\\ =\left(\frac{15-14+8}{12} \right )\\ =\left(\frac{9}{12} \right )\\ =\left(\frac{3}{4} \right )\\$

(iv) $\left(\frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7} \right )$

$\left(\frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7} \right )\\ \left(\frac{-2\times14}{5\times14}-\frac{-3\times7}{10\times7}-\frac{-4\times10}{7\times10} \right )\\ \left(\frac{-28}{70}-\frac{-21}{70}-\frac{-40}{70} \right )\\ \left(\frac{-28+21+40}{70} \right )\\ \left(\frac{33}{70} \right )\\$

Q16. Fill in the blanks:

(i) $\frac{-4}{13}-\frac{-3}{26}=….$

$\frac{-4}{13}-\frac{-3}{26}=\frac{-4\times2}{13\times2}-\frac{-3}{26}\\ =\frac{-8+3}{26}\\ =\frac{-5}{26}$

$\frac{-4}{13}-\frac{-3}{26}=\frac{-5}{26}$

(ii) $\frac{-9}{14}+….=-1$

$\frac{-9}{14}+x=-1\\ x=-1-\left(\frac{-9}{14} \right )\\ x=-\frac{1\times14}{1\times14}-\left(\frac{-9}{14} \right )\\ x=-\frac{14}{14}-\left(\frac{-9}{14} \right )\\ x=\frac{-14+9}{14}\\ x=\frac{-5}{14}\\$

$\frac{-9}{14}+ \frac{-5}{14}=-1$

(iii) $\frac{-7}{9}+….=3$

$\frac{-7}{9}+x=3\\ x=3-\frac{-7}{9}\\ x=\frac{3\times9}{1\times9}-\frac{-7}{9}\\ x=\frac{27}{9}-\frac{-7}{9}\\ x=\frac{27+7}{9}\\ x=\frac{34}{9}\\$

$\frac{-7}{9}+\frac{34}{9}=3$

(iv) $….+\frac{15}{23}=4$

$x+\frac{15}{23}=4\\ x=4-\frac{15}{23}\\ x=\frac{4\times23}{1\times23}-\frac{15}{23}\\ x=\frac{92}{23}-\frac{15}{23}\\ x=\frac{92-15}{23}\\ x=\frac{77}{23}\\$

$\frac{77}{23}+\frac{15}{23}=4$<