RD Sharma solution class 7 chapter 5 Operations On Rational numbers Exercise 5.1

Exercise 5.1

Page-5.4

Question 1:

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

Answer 1:

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

Question 2:

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

Answer 2:

(i)

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

(ii)

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

(iii)

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

(iv)

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

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 3:

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

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


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

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 4:

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


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