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

Exercise 4E

Page-75

Question 1:

Multiply:

(i) $\frac{3}{4} \mathrm{by} \frac{5}{7}$
(ii)  $\frac{9}{8} \mathrm{by} \frac{32}{3}$
(iii) $\frac{7}{6} \mathrm{by} 24$
(iv) $\frac{-2}{3} \mathrm{by} \frac{6}{7}$
(v) $\frac{-12}{5} \mathrm{by} \frac{10}{-3}$
(vi) $\frac{25}{-9} \mathrm{by} \frac{3}{-10}$
(vii) $\frac{-7}{10} \mathrm{by} \frac{-40}{21}$
(viii) $\frac{-36}{5} \mathrm{by} \frac{20}{-3}$
(ix) $\frac{-13}{15} \mathrm{by} \frac{-25}{26}$

Answer 1:

$\begin{array}{l}\\ \left(\mathrm{i}\right)\frac{3}{4}\times \frac{5}{7}=\frac{(3\times 5)}{(4\times 7)}=\frac{15}{28}\\ \left(\mathrm{ii}\right)\frac{{\cancel{9}}^3}{{\cancel{8}}_1}\times \frac{\sout{3}{\sout{2}}^4}{{\sout{3}}_1}=\frac{(3\times 4)}{(1\times 1)}=12\\ \left(\mathrm{iii}\right)\frac{7}{{\sout{6}}_1}\times \frac{\sout{2}{\sout{4}}^4}{1}=7\times 4=28\\ \left(\mathrm{iv}\right)\frac{-2}{{\sout{3}}_1}\times \frac{{\sout{6}}^2}{7}=\frac{(-2\times 2)}{7}=\frac{-4}{7}\\ \text{(v) We need a positive denominator.}\\ \text{ ∴ }\frac{10}{-3}\times \frac{-1}{-1}=\frac{-10}{3}\\ \text{=}\frac{-\sout{1}{\sout{2}}^4}{{\sout{5}}_1}\times \frac{-\sout{1}{\sout{0}}^2}{{\sout{3}}_1}\\ =\left(-4\right)\times \left(-2\right)\\ =8\\ \end{array}$
$\begin{array}{l}\left(\mathrm{vi}\right)\frac{\sout{2}{\sout{5}}^5}{-{\sout{9}}_3}\times \frac{{\sout{3}}^1}{-\sout{1}{\sout{0}}_2}=\frac{5}{3}\times \frac{1}{2}=\frac{5}{6}\\ \left(\mathrm{vii}\right)\frac{-{\sout{7}}^1}{\sout{1}{\sout{0}}_1}\times \frac{-\sout{4}{\sout{0}}^4}{\sout{2}{\sout{1}}_3}=\frac{4}{3}\\ \left(\mathrm{viii}\right)\frac{-\sout{3}{\sout{6}}^{12}}{{\sout{5}}_1}\times \frac{\sout{2}{\sout{0}}^4}{-{\sout{3}}_1}=12\times 4=48\\ \left(\mathrm{ix}\right)\frac{-\sout{1}{\sout{3}}^1}{\sout{1}{\sout{5}}_3}\times \frac{-\sout{2}{\sout{5}}^5}{\sout{2}{\sout{6}}_2}=\frac{-1}{3}\times \frac{-5}{2}=\frac{5}{6}\end{array}$

Question 2:

Simplify:

(i) $\frac{3}{20}\times \frac{4}{5}$
(ii) $\frac{-7}{30}\times \frac{5}{14}$
(iii) $\frac{5}{-18}\times \frac{-9}{20}$
(iv) $\frac{-9}{8}\times \frac{-16}{3}$
(v) $-32\times \frac{-7}{36}$
(vi) $\frac{16}{-21}\times \frac{-14}{5}$

Answer 2:

$\begin{array}{l}\left(i\right)\\ \frac{3}{\sout{2}{\sout{0}}_5}\times \frac{{\sout{4}}^1}{5}\\ =\frac{3\times 1}{5\times 5}\\ =\frac{3}{25}\\ \\ \left(ii\right)\\ \frac{-{\sout{7}}^1}{\sout{3}{\sout{0}}_6}\times \frac{{\sout{5}}^1}{\sout{1}{\sout{4}}_2}\\ =\frac{-1\times 1}{6\times 2}\\ =\frac{-1}{12}\\ \\ \left(iii\right)\\ \frac{{\sout{5}}^1}{-\sout{1}{\sout{8}}_2}\times \frac{-{\sout{9}}^1}{\sout{2}{\sout{0}}_4}\\ =\frac{1\times (-1)}{-2\times 4}\\ =\frac{-1}{-8}=\frac{1}{8}\\ \\ \left(iv\right)\\ \frac{-{\sout{9}}^3}{{\sout{8}}_1}\times \frac{-\sout{1}{\sout{6}}^2}{{\sout{3}}_1}\\ =(-3)\times (-2)\\ =6\\ \\ \left(v\right)\\ \frac{-32}{1}\times \frac{-7}{36}\\ =\frac{-\sout{3}{\sout{2}}^8\times (-7)}{1\times \sout{3}{\sout{6}}_9}\\ =\frac{-8\times (-7)}{9}\\ =\frac{56}{9}\\ \\ \\ \text{(vi)}\\ \text{We need a positive denominator.}\\ \therefore \frac{16}{-21}\times \frac{-1}{-1}=\frac{-16}{21}\\ \mathrm{Now}, \frac{-16}{\sout{2}{\sout{1}}_3}\times \frac{-\sout{1}{\sout{4}}^2}{5}\\ \\ =\frac{(-16)\times (-2)}{3\times 5}\\ =\frac{32}{15}\end{array}$

Question 3:

Simplify:

(i) $\frac{7}{24}\times -48$
(ii) $\frac{-19}{36}\times 16$
(iii) $\frac{-3}{4}\times \frac{4}{3}$
(iv) $-13\times \frac{17}{26}$
(v) $\frac{-13}{5}\times -10$
(vi) $\frac{-9}{16}\times \frac{-64}{27}$

Answer 3:

$\begin{array}{l}\text{(i)}\\ \frac{7}{\sout{2}{\sout{4}}_1}\times (-\sout{4}{\sout{8}}^2)\\ =7\times (-2)\\ =-14\\ \left(ii\right)\\ \frac{-19}{\sout{3}{\sout{6}}_9}\times \sout{1}{\sout{6}}^4\\ =\frac{-19}{9}\times 4\\ =\frac{-76}{9}\\ \\ \left(iii\right)\\ \frac{-{\sout{3}}^1}{{\sout{4}}_1}\times \frac{{\sout{4}}^1}{{\sout{3}}_1}\\ =-1\\ \\ \left(iv\right)\\ -13\times \frac{17}{26}\\ =\frac{-\sout{1}{\sout{3}}^1\times 17}{\sout{2}{\sout{6}}^2}\\ =\frac{-17}{2}\\ \\ \left(v\right)\\ \frac{-13}{{\sout{5}}_1}\times (-\sout{1}{\sout{0}}^2)\\ =26\\ \\ \left(vi\right)\\ \frac{(-{\sout{9}}^1)}{\sout{1}{\sout{6}}_1}\times \frac{(-\sout{6}{\sout{4}}^4)}{\sout{2}{\sout{7}}_3}\\ =\frac{4}{3}\\ \end{array}$

Question 4:

Simplify:

(i) $\left(\frac{13}{8}\times \frac{12}{13}\right)+\left(\frac{-4}{9}\times \frac{3}{-2}\right)$
(ii) $\left(\frac{16}{15}\times \frac{-25}{8}\right)+\left(\frac{-14}{27}\times \frac{6}{7}\right)$
(iii) $\left(\frac{6}{55}\times \frac{-22}{9}\right)-\left(\frac{26}{125}\times \frac{-10}{39}\right)$
(iv) $\left(\frac{-12}{7}\times \frac{-14}{27}\right)-\left(\frac{-8}{45}\times \frac{9}{16}\right)$

Answer 4:

$\begin{array}{l}\left(i\right)\\ (\frac{\sout{1}{\sout{3}}^1}{{\sout{8}}_2} \times \frac{\sout{1}{\sout{2}}^3}{\sout{1}{\sout{3}}_1}) + (\frac{-{\sout{4}}^2}{{\sout{9}}_3} \times \frac{{\sout{3}}^1}{-{\sout{2}}_1})\\ =\frac{3}{2} + \frac{2}{3}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 2 \mathrm{and} 3 \mathrm{is} 6.\\ \begin{array}{l}=\frac{9 + 4}{6}\\ \frac{13}{6}\\ \left(ii\right)\\ (\frac{16}{15}\times \frac{-25}{8}) + (\frac{-14}{27}\times \frac{6}{7})\end{array}\\ =(\frac{\sout{1}{\sout{6}}^2}{\sout{1}{\sout{5}}_3}\times \frac{-\sout{2}{\sout{5}}^5}{{\sout{8}}_1}) + (\frac{-\sout{1}{\sout{4}}^2}{27}\times \frac{6}{{\sout{7}}_1})\\ =[\frac{2}{3}\times \frac{(-5)}{1}] + [\frac{(-2)}{27}\times \frac{6}{1}]\\ =\frac{(-10)}{3}+\frac{(-\sout{1}{\sout{2}}^{{}^4})}{\sout{2}{\sout{7}}^9}\\ =\frac{-10}{3}+\frac{-4}{9}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 3 \mathrm{and} 9 \mathrm{is} 9.\\ \begin{array}{l}=\frac{-30-4}{9}\\ =\frac{-34}{9}\\ \\ \left(iii\right)\\ (\frac{6}{55}\times \frac{-22}{9})-(\frac{26}{125}\times \frac{-10}{39})\end{array}\\ =(\frac{{\sout{6}}^2}{\sout{5}{\sout{5}}_5}\times \frac{-\sout{2}{\sout{2}}^2}{{\sout{9}}_3})-(\frac{\sout{2}{\sout{6}}^2}{\sout{1}\sout{2}{\sout{5}}_{25}}\times \frac{-\sout{1}{\sout{0}}^2}{\sout{3}{\sout{9}}_3})\\ =[\frac{(-4)}{15}-\frac{(-4)}{75}]\\ =\frac{-4}{15}+\frac{4}{75}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 15 \mathrm{and} 75 \mathrm{is} 75.\\ \begin{array}{l}=\frac{-20+4}{75}\\ =\frac{-16}{75}\\ \left(iv\right)\\ (\frac{-\sout{1}{\sout{2}}^4}{{\sout{7}}_1}\times \frac{-\sout{1}{\sout{4}}^2}{\sout{2}{\sout{7}}_9})-(\frac{-{\sout{8}}^1}{\sout{4}{\sout{5}}_5}\times \frac{{\sout{9}}^1}{{\sout{16}}_2})\end{array}\\ =[\frac{(-4)}{1}\times \frac{(-2)}{9}]-[\frac{-1}{5}\times \frac{1}{2}]\\ =\frac{8}{9}+\frac{1}{10}\\ \mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 9 \mathrm{and} 10 \mathrm{is} 90.\\ \begin{array}{l}=\frac{80+9}{90}\\ =\frac{89}{90}\end{array}\end{array}$

Question 5:

Find the cost of $3\frac{1}{3}$ metres of cloth at Rs $40\frac{1}{2}$ per metre.

Answer 5:

$\begin{array}{l}\text{Cost of 1 meter cloth }=\text{Rs }40\frac{1}{2}\\ \text{Cost of }3\frac{1}{2}\text{ meter cloth }=\text{ Rs }(40\frac{1}{2}\times 3\frac{1}{2})\\ =\mathrm{Rs} (\frac{81}{2}\times \frac{7}{2})\\ =\mathrm{Rs} \frac{567}{4}\\ =\mathrm{Rs}141.75\end{array}$

Question 6:

A bus is moving at an average speed of $46\frac{2}{3}$ km/h. How much distance will it cover in $2\frac{2}{5}$ huors?

Answer 6:

$\begin{array}{l}\text{Distance covered in 1 hour }= 46\frac{2}{3} \text{km}\\ \text{Distance coverd in }2\frac{2}{5}\text{ hours = }(46\frac{2}{3}\times 2\frac{2}{5})\\ =(\frac{\sout{1}\sout{4}{\sout{0}}^{28}}{\sout{3}}\times \frac{\sout{1}{\sout{2}}^4}{{\sout{5}}^1})\\ =(28\times 4)\\ =112\text{ km}\\ \text{Hence, the required distance is 112 km.}\end{array}$