Class 7 Operations on Rational Numbers Exercise 5.2

\begin{array}{l} \text {Q1. Subtract the first rational number from the second in each of the following: } \\ \text {(i) } \quad \frac{3}{8}, \frac{5}{8} \quad \text {(ii) } \quad \frac{-7}{9}, \frac{4}{9} \\ \text {(iii) } \quad \frac{-2}{11} , \frac{-9}{11} \quad \text {(iv) } \quad \frac{11}{13}, \frac{-4}{13} \end{array}\begin{array}{l} \text {Sol. } \\\text {(i) } \quad \frac{5}{8}-\frac{3}{8} \\ =\frac{5-3}{8} \quad \quad [\text {Denominators are same so add the numerator}] \\ =\frac{2}{8} \\ \\\text {(ii) } \quad \frac{4}{9} – \frac{-7}{9} \\ =\frac{4+7}{9} \quad \quad [\text {Denominators are same so add the numerator}] \\ =\frac{11}{9} \\ \\\text {(iii) } \quad \frac{-9}{11} – \frac{-2}{11} \\ =\frac{-9+2}{11} \quad \quad [\text {Denominators are same so add the numerator}] \\ =\frac{-7}{11} \\ \\\text {(iv) } \quad \frac{-4}{13} – \frac{11}{13} \\ =\frac{-4-11}{13} \quad \quad [\text {Denominators are same so add the numerator}] \\ =\frac{-15}{13} \\ \end{array}\begin{array}{l} \text {Q2. Evaluate each of the following: } \\ \text {(i) } \quad \frac{2}{3}-\frac{3}{5} \quad \text {(ii) } \quad -\frac{4}{7}-\frac{2}{-3} \quad \text {(iii) } \quad \frac{4}{7}-\frac{-5}{-7} \quad \text {(iv) } \quad -2-\frac{5}{9} \end{array}\begin{array}{l} \text {Sol. } \\\text {(i) If } \frac {p}{q} \text { and } \frac {r}{s} \text { are two rational numbers such that q and s } \\ \text { do not have a common factor other than 1, then } \\ \frac {p}{q} + \frac {r}{s} = \frac {p \times s + r \times q}{q \times s} \\ \therefore \quad \frac{2}{3}-\frac{3}{5} =\frac{2 \times 5-3 \times 3}{15}=\frac{10-9}{15} \\ =\frac{1}{15} \\ \\\text {(ii) } \quad -\frac{4}{7}-\frac{2}{-3} =\frac{-4 \times 3+2 \times 7}{21} \\ =\frac{-12+14}{21}=\frac{2}{21} \\ \\\text {(iii) } \quad \frac{4}{7}-\frac{-5}{-7} \\ =\frac{4-5}{7} \quad \quad [\text {Denominators are same so add the numerator}] \\ =\frac{-1}{7} \\ \\\text {(iv) } \quad -2-\frac{5}{9} =\frac{-2 \times 9-5}{9} \\ =\frac{-18-5}{9}=\frac{-23}{9} \\ \end{array}\begin{array}{l} \text {Q3. The sum of the two numbers is } \frac{5}{9} \text {. If one of the numbers is } \frac{1}{3} \text {, find the other. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{1}{3}+x=\frac{5}{9} \\ \Rightarrow \quad x =\frac{5}{9}-\frac{1}{3} \\ \Rightarrow \quad x =\frac{5 – 1 \times 3}{9} \\ \Rightarrow \quad x =\frac{5-3}{9} \\ \Rightarrow \quad x =\frac{2}{9} \\\therefore \text {The other number is } \frac{2}{9}\end{array}\begin{array}{l} \text {Q4. The sum of two numbers is } \frac{-1}{3} \text {. If one of the numbers is } \frac{-12}{3}\text {, find the other.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{-12}{3}+x =\frac{-1}{3} \\ \Rightarrow \quad x =\frac{-1}{3}-\frac{-12}{3} \\ \Rightarrow \quad x =\frac{-1+12}{3} \\ \Rightarrow \quad x =\frac{11}{3} \\\therefore \text {The other number is } \frac{11}{3}\end{array}\begin{array}{l} \text {Q5. The sum of two numbers is } \frac{-4}{3} \text {. If one of the numbers is -5, find the other. } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ -5+x =\frac{-4}{3} \\ \Rightarrow \quad x =\frac{-4}{3}-\frac{-5}{1} \\ \Rightarrow \quad x =\frac{-4+5 \times 3}{3} \\ \Rightarrow \quad x =\frac{-4+15}{3} \\ \Rightarrow \quad x =\frac{11}{3} \\\therefore \text {The other number is } \frac{11}{3}\end{array}\begin{array}{l} \text {Q6. The sum of two rational numbers is -8. If one of the numbers is } \frac{-15}{7} \text {, find the other.} \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\\frac{-15}{7}+x = -8 \\ \Rightarrow \quad x=\frac{-8}{1}-\frac{-15}{7} \\ \Rightarrow \quad x=\frac{-8 \times 7+15}{7} \\ \Rightarrow \quad x=\frac{-56+15}{7} \\ \Rightarrow \quad x=\frac{-41}{7} \\\therefore \text {The other number is } \frac{-41}{7} \end{array}\begin{array}{l} \text {Q7. What should be added to } \frac{-7}{8} \text { so as to get } \frac{5}{9} ? \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{-7}{8}+x = \frac{5}{9} \\ \Rightarrow \quad x=\frac{5}{9} – \frac{-7}{8} \\ \Rightarrow \quad x=\frac{5 \times 8 + 7 \times 9}{9 \times 8} \\ \Rightarrow \quad x=\frac{40+63}{72} \\ \Rightarrow \quad x=\frac{103}{72} \\\therefore \text {The required number is } \frac{103}{72} \end{array}\begin{array}{l} \text {Q8. What number should be added to } \frac{-5}{11} \text { so as to get } \frac{26}{33} ? \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\\frac{-5}{11} + x = \frac{26}{33} \\ \Rightarrow \quad x= \frac{26}{33}-\frac{-5}{11} \\ \Rightarrow \quad x=\frac{26+5 \times 3}{33} \\ \Rightarrow \quad x=\frac{26+15}{33} \\ \Rightarrow \quad x=\frac{41}{33} \\\therefore \text {The required number is } \frac{41}{33} \end{array}\begin{array}{l} \text {Q9. What number should be added to } \frac{-5}{7} \text { to get } \frac{-2}{3} ? \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{-5}{7} +x =\frac{-2}{3} \\ \Rightarrow \quad x=\frac{-2}{3}-\frac{-5}{7} \\ \Rightarrow \quad x=\frac{-2 \times 7+5 \times 3}{21} \\ \Rightarrow \quad x=\frac{-14+15}{21} \\ \Rightarrow \quad x=\frac{1}{21} \\\therefore \text {The required number is } \frac{1}{21} \end{array}\begin{array}{l} \text {Q10. What number should be subtracted from } \frac{-5}{3} \text { to get } \frac{5}{6} ? \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{-5}{3}-x =\frac{5}{6} \\ \Rightarrow \quad x=\frac{-5}{3}-\frac{5}{6} \\ \Rightarrow \quad x=\frac{-5 \times 2}{3 \times 2}-\frac{5}{6} \\ \Rightarrow \quad x=\frac{-10-5}{6} =\frac{-15}{6} =\frac{-5}{2} \\\therefore \text {The required number is } \frac{-5}{2} \end{array}\begin{array}{l} \text {Q11. What number should be subtracted from } \frac{3}{7} \text { to get } \frac{5}{4} ? \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{3}{7}-x =\frac{5}{4} \\ \Rightarrow \quad x=\frac{3}{7}-\frac{5}{4} \\ \Rightarrow \quad x=\frac{3 \times 4-5 \times 7}{28} \\ \Rightarrow \quad x=\frac{12-35}{28} =\frac{-23}{28} \\\therefore \text {The required number is } \frac{-23}{28} \end{array}\begin{array}{l} \text {Q12. What should be added to } (\frac{2}{3}+\frac{3}{5}) \text { to get } \frac{-2}{15} ? \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{2}{3}+\frac{3}{5}+x=\frac{-2}{15}$ Rightarrow \quad \frac{2 \times 5 + 3 \times 3}{3 \times 5}+x=\frac{-2}{15} \\ \Rightarrow \quad \frac{19}{15}+x=\frac{-2}{15} \\ \Rightarrow \quad x=\frac{-2}{15}-\frac{19}{15} \\ \Rightarrow \quad x=\frac{-21}{15} \\ \Rightarrow \quad x=\frac{-7 \times 3}{5 \times 3} \\ \Rightarrow \quad x=\frac{-7}{5} \\\therefore \text {The required number is } \frac{-7}{5} \end{array}\begin{array}{l} \text {Q13. What should be added to } (\frac{1}{2}+\frac{1}{3}+\frac{1}{5}) \text { to get 3 ? } \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{1}{2}+\frac{1}{3}+\frac{1}{5}+x =3 \\ \Rightarrow \quad \frac{1 \times 15 + 1 \times 10 + 1 \times 6}{2 \times 3 \times 5}+x =3 \\ \Rightarrow \quad \frac{31}{30}+x =3 \\ \Rightarrow \quad x=3-\frac{31}{30} \\ \Rightarrow \quad x=\frac{3 \times 30}{1 \times 30}-\frac{31}{30} \\ \Rightarrow \quad x=\frac{90}{30}-\frac{31}{30} \\ \Rightarrow \quad x=\frac{59}{30} \\\therefore \text {The required number is } \frac{59}{30} \end{array}\begin{array}{l} \text {Q14. What should be subtracted from } (\frac{3}{4}-\frac{2}{3}) \text { to get } \frac{-1}{6} ? \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let the required number be x. } \\ \text {According to question, we have } \\ \frac{3}{4}-\frac{2}{3}-x =\frac{-1}{6} \\ \Rightarrow \quad \frac{3 \times 3 – 2 \times 4}{4 \times 3} -x =\frac{-1}{6} \\ \Rightarrow \quad \frac{1}{12}-x =\frac{-1}{6} \\ \Rightarrow \quad x=\frac{1}{12}-\frac{-1}{6} \\ \Rightarrow \quad x=\frac{1}{12}-\frac{-1 \times 2}{6 \times 2} \\ \Rightarrow \quad x=\frac{1}{12}-\frac{-2}{12} \\ \Rightarrow \quad x=\frac{3}{12} =\frac{1}{4} \\\therefore \text {The required number is } \frac{1}{4} \end{array}\begin{array}{l} \text {Q15. Simplify: } \\ \text {(i) } \quad \frac{-3}{2}+\frac{5}{4}-\frac{7}{4} \quad \text {(ii) } \quad \frac{5}{3}-\frac{7}{6}+\frac{-2}{3} \\ \text {(iii) } \quad \frac{5}{4}-\frac{7}{6}-\frac{-2}{3} \quad \text {(iv) } \quad \frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7} \end{array}\begin{array}{l} \text {Sol. } \\ \text {(i) LCM of denominators 2 and 4 is 4.} \\\frac{-3}{2}+\frac{5}{4}-\frac{7}{4} \\ =\frac{-3 \times 2}{2 \times 2}+\frac{5}{4}-\frac{7}{4} \\ =\frac{-6+5-7}{4} \\ =\frac{-8}{4}=-2 \\ \\\text {(ii) LCM of denominators 3 and 6 is 6.} \\ \frac{5}{3}-\frac{7}{6}+\frac{-2}{3} \\ =\frac{5 \times 2}{3 \times 2}-\frac{7}{6}+\frac{-2 \times 2}{3 \times 2} \\ =\frac{10-7-4}{6}=\frac{-1}{6} \\ \\\text {(iii) LCM of denominators 4, 6 and 3 is 12.} \\ \frac{5}{4}-\frac{7}{6}-\frac{-2}{3} \\ =\frac{5 \times 3}{4 \times 3}-\frac{7 \times 2}{6 \times 2}-\frac{-2 \times 4}{3 \times 4} \\ =\frac{15-14+8}{12} \\ =\frac{9}{12}=\frac{3}{4} \\ \\\text {(iv) LCM of denominators 3 and 6 is 6.} \\ \frac{-2}{5}-\frac{-3}{10}-\frac{-4}{7} \\ =\frac{-2 \times 14}{5 \times 14}-\frac{-3 \times 7}{10 \times 7}-\frac{-4 \times 10}{7 \times 10} \\ =\frac{-28+21+40}{70}=\frac{33}{70} \\\end{array}\begin{array}{l} \text {Q16. Fill in the blanks: } \\ \text {(i) } \quad \frac{-4}{13}-\frac{-3}{26}=\ldots \quad \text {(ii) } \quad \frac{-9}{14}+\ldots =-1 \\ \text {(iii) } \quad \frac{-7}{9}+\ldots =3 \quad \text {(iv) } \quad \ldots +\frac{15}{23}=4 \end{array}\begin{array}{l} \text {Sol. } \\ \text {Let assume that the required number be x. } \\\text {(i) } \quad \frac{-4}{13}-\frac{-3}{26}=x \\ \Rightarrow \quad x = \frac{-4 \times 2}{13 \times 2}-\frac{-3}{26} \\ \Rightarrow \quad x=\frac{-8+3}{26} \\ \Rightarrow \quad x=\frac{-5}{26} \\ \\\text {(ii) } \quad \frac{-9}{14}+x =-1 \\ \Rightarrow \quad x=-1-\frac{-9}{14} \\ \Rightarrow \quad x=\frac{-14+9}{14} \\ \Rightarrow \quad x=\frac{-5}{14} \\ \\\text {(iii) } \quad \frac{-7}{9}+x =3 \\ \Rightarrow \quad x=\frac{3 \times 9}{1 \times 9}-\frac{-7}{9} \\ \Rightarrow \quad x=\frac{27+7}{9} \\ \Rightarrow \quad x=\frac{34}{9} \\ \\\text {(iv) } \quad x +\frac{15}{23}=4 \\ \Rightarrow \quad x=4-\frac{15}{23} \\ \Rightarrow \quad x=\frac{4 \times 23}{1 \times 23}-\frac{15}{23} \\ \Rightarrow \quad x=\frac{92-15}{23} \\ \Rightarrow \quad x=\frac{77}{23} \\\end{array}