ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.4

Exercise 1.4

Question 1

Sol :

(i) $-\frac{3}{7} \div 4$

=$-\frac{3}{7} \div \frac{4}{1}$

=$\frac{-3}{7} \times \frac{1}{4}$

=$\frac{-3 \times 1}{7 \times 4}$

=$\frac{-3}{28}$


(ii)  $4 \frac{5}{8} \div\left(\frac{-4}{9}\right)$

=$\frac{37}{8} \div\left(\frac{-4}{9}\right)$

=$\frac{37}{8} \times\left(\frac{9}{-4}\right)$

=$\frac{37 \times 9}{8 \times(-4)}$

=$-\frac{333}{32}$

=$-10 \frac{13}{32}$


(iii) $\frac{8}{9} \div \frac{-3}{5}$

=$\frac{-8}{9} \times \frac{5}{-3}$

=$\frac{-8 \times 5}{9 x-3}$

=$\frac{-40}{-27}$

=$\frac{40}{27}=1 \frac{13}{27}$
 

Question 2

Sol :
(i) true 

(ii) false 

(iii) false 

(iv) true 

(v) true 

(vi)false     


Question 3

Sol :
Let unknown number be x

⇒$x \times 2 \frac{4}{9}=\frac{-11}{12}$

⇒$x \times \frac{22}{9}=\frac{-11}{12}$

$x=\frac{-11}{12}$ x $\frac{22}{9}$

⇒$\frac{-11}{12} \times \frac{9}{22}$

$x=\frac{-3}{8}$

Other number $=\frac{-3}{8}$


Question 4

Sol :
Let unknown number be x

$x \times\left(\frac{-7}{12}\right)=\frac{5}{14}$

$x=\frac{5}{14} \div\left(\frac{-7}{12}\right)$

$=\frac{5}{14} \times \frac{12}{-7}$

$=\frac{5 \times 12}{14 \times(-7)}$

$=-\frac{60}{98}$

$=\frac{-30}{49}$


Question 5

Sol :

Let unknown number be x 

$\frac{-3}{x}=\frac{-9}{13}$

$x=-\frac{3}{1} \div\left(\frac{-9}{13}\right)$

$=\frac{-3}{1} \times \frac{13}{-9}$

$x=\frac{13}{3}=4 \frac{1}{3}$


Question 6

Sol :

Sum of number = $\frac{-13}{8}+\frac{5}{12}$

$=\frac{(-13 \times 3)+(5 \times 2)}{24} \quad$ LCM of $812=24$

⇒$\frac{-39+10}{24}$

⇒$\frac{-29}{24}$

sum of numbers = $\frac{-29}{24}$

difference of numbers= $\frac{(-13 \times 3)-(5 \times 2)}{24}$

$\frac{-39-10}{24}$

Difference of numbers $=\frac{-49}{24}$

Required product = sum of numbers ÷ difference of numbers 

$\begin{aligned} &=\frac{-29}{24} \div\left(-\frac{49}{24}\right) \\=& \frac{-29}{24} \times \frac{24}{-49} \\ &=\frac{29}{49} \end{aligned}$


Question 7

Sol :

$\begin{aligned} \text { Sum of two numbers } &=\frac{8}{3}+\frac{4}{7} \\ &=\frac{(8 \times 7)+(3 \times 4)}{21} \text { LCM Of } 3,7=21 \\ &=\frac{56+12}{21} \\ \text { Sum of two numbers } &=\frac{68}{21} \end{aligned}$

$\begin{aligned} \text { Product of given numbers } &=\frac{-3}{7} \times \frac{14}{9} \\ &=\frac{-2}{3} \end{aligned}$

$\begin{aligned} \text { Required product } &=\frac{\text { Sum of } \frac{8}{3} \text { and } \frac{4}{7}}{\text { product of }\frac{3}{1} \text { and } \frac{14}{9}} \\ &=\frac{68}{21} \div\left(\frac{-2}{3}\right) \\ &=\frac{68}{21} \times \frac{3}{-2} \\ &=\frac{-34}{7} \end{aligned}$


Question 8

Sol :

Given $\quad P=\frac{-3}{2}, \quad q=\frac{4}{5}, \quad \ r=\frac{-7}{12}$

$(p \div q) \div r=p \div(q \div \ r)$

$\begin{aligned} \text { L.H.S } &=(p \div q) \div r \\ &=\left(\frac{-3}{2} \div \frac{4}{5}\right) \div\left(\frac{-7}{12}\right) \\ &=\left(\frac{-3}{2} \times \frac{5}{4}\right) \div\left(\frac{-7}{12}\right) \\ &=\left(\frac{-15}{8}\right) \div\left(-\frac{7}{12}\right) \\ &=\frac{-15}{8} \times \frac{12}{-7} \\ &=\frac{-15}{8} \times \frac{12}{-7} \end{aligned}$

$\begin{aligned} \text { L.H.S } &=+\frac{45}{7} \\ \text { R.H.S } &=p \div(q \div \ r) \\ &=\frac{-3}{2} \div\left(\frac{4}{5} \div\left(\frac{-7}{12}\right)\right) \\ &=\frac{-3}{2} \div\left(\frac{4}{5} \times \frac{12}{-7}\right) \\ &=\frac{-3}{2} \div\left(-\frac{48}{35}\right) \\ &=\frac{-3}{2} \times-\frac{35}{48} \\ R \cdot H S &=\frac{35}{32} \\ L \cdot H \cdot S \neq R \cdot H \cdot S \\(p \div q) \div r \neq P \div(q \div \tau) \end{aligned}$

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