RD Sharma solution class 7 chapter 4 Rational numbers Exercise 4.3

Exercise 4.3

Page-4.12

Question 1:

Determine whether the following rational numbers are in the lowest form or not:

(i) $\frac{65}{84}$
(ii) $\frac{-15}{32}$
(iii) $\frac{24}{128}$
(iv) $\frac{-56}{-32}$

Answer 1:

(i) We observe that 65 and 84 have no common factor i..e., their HCF is 1.

Thus, $\frac{65}{84}$ is in its lowest form.

(ii) We observe that $-$15 and 32 have no common factor i..e., their HCF is 1.

Thus, $\frac{-15}{32}$is  in  its lowest form.

(iii) HCF of 24 and 128 is not 1.

Thus, given rational number is not in its simplest form.

(iv) HCF of 56 and 32 is 8.

Thus, given rational number is not in its simplest form.

Question 2:

Express each of the following rational numbers to the lowest form:
(i) $\frac{4}{22}$
(ii) $\frac{-36}{180}$
(iii) $\frac{132}{-428}$
(iv) $\frac{-32}{-56}$

Answer 2:

Lowest form of:
$$\begin{gathered} \left(i\right) \frac{4}{22} \mathrm{is}: \\ 4 =2\times 2 \\ 22 = 2\times 11 \\ \mathrm{HCF} \mathrm{of} 4 \mathrm{and} 22 \mathrm{is} 2. \\ \text{D}\mathrm{ividing} \mathrm{the} \mathrm{fraction} \mathrm{by} 2 , \mathrm{we} \mathrm{get} \frac{2}{11}. \\ \left(\mathrm{ii}\right) \frac{-36}{180} \mathrm{is}: \\ 36 = 3\times 3\times 2\times 2 \\ 180 = 5\times 3\times 3\times 2\times 2 \\ \mathrm{HCF} \mathrm{of} 36 \mathrm{and} 180 \mathrm{is} 36. \\ \text{D}\mathrm{ividing} \mathrm{the} \mathrm{fraction} \mathrm{by} 36 , \mathrm{we} \mathrm{get} \frac{-1}{5} \end{gathered}$$
$$\begin{gathered} \left(\mathrm{iii}\right) \frac{132}{-428} \mathrm{is}: \\ 132 = 2\times 3\times 2\times 11 \\ 428 = 2\times 2\times 107 \\ \mathrm{HCF} \mathrm{of} 132 \mathrm{and} 428 \mathrm{is} 4. \\ \text{D}\mathrm{ividing} \mathrm{the} \mathrm{fraction} \mathrm{by} 4 , \mathrm{we} \mathrm{get} \frac{33}{-107} \\ \left(\mathrm{iv}\right) \frac{-32}{-56} \mathrm{is}: \\ 32 =2\times 2\times 2\times 2\times 2 \\ 56 = 2\times 2\times 2\times 7 \\ \mathrm{HCF} \mathrm{of} 32 \mathrm{and} 56 \mathrm{is} 8. \\ \text{D}\mathrm{ividing} \mathrm{the} \mathrm{fraction} \mathrm{by} 8, \mathrm{we} \mathrm{get}\frac{4}{7} \end{gathered}$$

Question 3:

Fill in the blanks:
(i) $\frac{-5}{7}=\frac{\dots }{35}=\frac{\dots }{49}$
(ii) $\frac{-4}{-9}=\frac{\dots }{18}=\frac{12}{\dots }$
(iii) $\frac{6}{-13}=\frac{-12}{\dots }=\frac{24}{\dots }$
(iv) $\frac{-6}{\dots }=\frac{3}{11}=\frac{\dots }{-55}$

Answer 3:

$$\begin{gathered} \text{(i) } \\ \mathrm{Here}, \frac{-5\times 5}{7\times 5}= \frac{-25}{35} \\ \text{A}\mathrm{lso} , \frac{-5\times 7}{7\times 7} = \frac{-35}{49}. \\ \mathrm{Therefore}, \frac{-5}{7}=\frac{-25}{35}=\frac{-35}{49} \\ \left(\mathrm{ii}\right) \mathrm{Here}, \frac{-4\times -2}{-9\times -2} = \frac{8}{18} \\ \text{A}\mathrm{lso}, \frac{-4\times -3}{-9\times -3} = \frac{12}{27} \\ \mathrm{Therefore}, \frac{-4}{-9}=\frac{8}{18}=\frac{12}{27} \\ \left(\mathrm{iii}\right) \mathrm{Here}, \frac{6\times -2}{-13\times -2}= \frac{-12}{26} \\ \mathrm{Also}, \frac{6\times 4}{-13\times 4} = \frac{24}{-52} \\ \mathrm{Therefore}, \frac{6}{-13}=\frac{-12}{26}=\frac{24}{-52} \\ \left(\mathrm{iv}\right) \mathrm{Here}, \frac{-6/-2}{-22/-2} = \frac{3}{11} \\ \mathrm{Also}, \frac{-6}{-22} = \frac{3\times -5}{11\times -5} =\frac{-15}{-55} \\ \mathrm{Therefore}, \frac{-6}{-22}=\frac{3}{11}=\frac{-15}{-55} \end{gathered}$$