RD Sharma solution class 7 chapter 4 Rational numbers Exercise 4.6

Exercise 4.6

Page-4.26

Question 1:

Draw the number line and represent the following rational numbers on it:
(i) $\frac{2}{3}$
(ii) $\frac{3}{4}$
(iii) $\frac{3}{8}$
(vi) $\frac{-5}{8}$
(v) $\frac{-3}{16}$
(vi) $\frac{-7}{3}$
(vii) $\frac{22}{-7}$
(viii) $\frac{-31}{3}$

Answer 1:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

Question 2:

Which of the two rational numbers in each of the following pairs of rational numbers is greater?
(i) $\frac{-3}{8}, 0$
(ii) $\frac{5}{2}, 0$
(iii) $\frac{-4}{11}, \frac{3}{11}$
(iv) $\frac{-7}{12}, \frac{5}{-8}$
(v) $\frac{4}{-9}, \frac{-3}{-7}$
(vi) $\frac{-5}{8}, \frac{3}{-4}$
(vii) $\frac{5}{9}, \frac{-3}{-8}$
(viii) $\frac{5}{-8}, \frac{-7}{12}$

Answer 2:

(i) We know that every positive rational number is greater than zero and every negative rational number is smaller than zero. Thus,

$\frac{-3}{8}>0$
(ii) $\frac{5}{2}>0.$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iii) $\frac{-4}{8}<\frac{3}{11}.$ Because every positive rational number is greater than zero and every negative rational number is smaller than zero.
(iv)

$$\begin{gathered} \frac{-7}{12}=\frac{-7\times 2}{12\times 2}=\frac{-14}{24} \mathrm{and} \frac{5}{-8}=\frac{-5\times 3}{8\times 3}=\frac{-15}{24} \\ \mathrm{Therefore}, \frac{-7}{12}>\frac{5}{-8} \end{gathered}$$

(v)

$$\begin{gathered} \frac{4}{-9}=\frac{-4\times 7}{9\times 7}=\frac{-28}{63} \mathrm{and} \frac{-3}{-7}=\frac{3\times 7}{7\times 9}=\frac{21}{63} \\ \mathrm{Therefore}, \frac{4}{-9}<\frac{-3}{-7} \end{gathered}$$
(vi)

$$\begin{gathered} \frac{-5}{8} \mathrm{and} \frac{3}{-4}=\frac{-3\times 2}{4\times 2}=\frac{-6}{8} \\ \mathrm{Therefore}, \frac{-5}{8}>\frac{3}{-4} \end{gathered}$$

(vii)

$$\begin{gathered} \frac{5}{9}=\frac{5\times 8}{9\times 8}=\frac{40}{72} \mathrm{and} \frac{-3}{-8}=\frac{3\times 9}{8\times 9}=\frac{27}{72} \\ \mathrm{Therefore},\frac{5}{9}>\frac{-3}{-8} \end{gathered}$$

(viii)
$$\begin{gathered} \frac{-7}{12}=\frac{-7\times 2}{12\times 2}=\frac{-14}{24} \mathrm{and} \frac{5}{-8}=\frac{-5\times 3}{8\times 3}=\frac{-15}{24} \\ \mathrm{Therefore}, \frac{-7}{12}>\frac{5}{-8} \end{gathered}$$

Question 3:

Which of the two rational numbers in each of the following pairs of rational numbers is smaller?
(i) $\frac{-6}{-13}, \frac{7}{13}$
(ii) $\frac{16}{-5}, 3$
(iii) $\frac{-4}{3}, \frac{8}{-7}$
(iv) $\frac{-12}{5}, -3$

Answer 3:

(i) $\frac{-6}{-13}=\frac{6}{13}<\frac{7}{13}$
(ii) $\frac{16}{-5}<3$
(iii)
$$\begin{gathered} \frac{-4}{3}=\frac{-4\times 7}{3\times 7}=\frac{-28}{21} \mathrm{and} \frac{8}{-7}=\frac{-8\times 3}{7\times 3}=\frac{-24}{21} \\ \mathrm{Therefore}, \frac{-4}{3}<\frac{8}{-7} \end{gathered}$$
(iv)
$$\begin{gathered} \frac{-12}{5 }\mathrm{and} -3 = \frac{-3\times 5}{1\times 5}=\frac{-15}{5} \\ \mathrm{Therefore} \frac{-12}{5}>-3 \end{gathered}$$

Page-4.27

Question 4:

Fill in the blanks by the correct symbol out of >, =, or <:
(i) $\frac{-6}{7}.....\frac{7}{13}$
(ii) $\frac{-3}{5}.....\frac{-5}{6}$
(iii) $\frac{-2}{3}.....\frac{5}{-8}$
(iv) $0 .....\frac{-2}{5}$

Answer 4:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Because} \mathrm{every} \mathrm{positive} \mathrm{number} \mathrm{is} \mathrm{greater} \mathrm{than} \mathrm{a} \mathrm{negative} \mathrm{number}, \frac{-6}{7}<\frac{7}{13}. \\ \left(\mathrm{ii}\right) \mathrm{On} \mathrm{multiplying} \frac{-3}{5} \mathrm{by} \frac{6}{6}, \mathrm{we} \mathrm{get} \frac{-18}{30}. \\ \mathrm{On} \mathrm{multiplying} \frac{-5}{6} \mathrm{by} \frac{5}{5}, \mathrm{we} \mathrm{get} \frac{-25}{30}. \\ \mathrm{Because} -18> -25, \frac{-3}{5}>\frac{-5}{6}. \\ \left(\mathrm{iii}\right) \mathrm{On} \mathrm{multiplying} \frac{-2}{3} \mathrm{by} \frac{8}{8}, \mathrm{we} \mathrm{get} \frac{-16}{24}. \\ \mathrm{On} \mathrm{multiplying} \frac{5}{-8} \mathrm{by} \frac{3}{3}, \mathrm{we} \mathrm{get} \frac{15}{-24} = \frac{-15}{24}. \\ \mathrm{Because} -15 >-16, \frac{-2}{3}<\frac{5}{-8}. \\ \left(\mathrm{iv}\right) \mathrm{Because} \mathrm{every} \mathrm{positive} \mathrm{number} \mathrm{is} \mathrm{greater} \mathrm{than} \mathrm{a} \mathrm{negative} \mathrm{number}, 0>\frac{-2}{5}. \end{gathered}$$

Question 5:

Arrange the following rational numbers in ascending order:
(i) $\frac{3}{5}, \frac{-17}{-30}, \frac{8}{-15}, \frac{-7}{10}$
(ii) $\frac{-4}{9}, \frac{5}{-12}, \frac{7}{-18}, \frac{2}{-3}$

Answer 5:

(i) Ascending order:

$$\begin{gathered} \mathrm{Since}, \mathrm{LCM}\hspace{0.17em}\mathrm{of} 5, -30, -15, 10 \mathrm{is} 30. \\ \mathrm{Multiplying} \mathrm{the} \mathrm{numerators} \mathrm{and} \mathrm{denominators} \mathrm{to} \mathrm{get} \mathrm{the} \mathrm{denominator} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{LCM}, \\ \frac{3}{5}=\frac{3\times 6}{5\times 6}=\frac{18}{30}, \\ \frac{17}{30}= \frac{17\times 1}{30\times 1} = \frac{17}{30}, \\ \frac{8}{-15}=\frac{-8\times 2}{15\times 2}=\frac{-16}{30}, \\ \frac{-7}{10}=\frac{-7\times 3}{10\times 3}=\frac{-21}{30}. \\ \mathrm{Order} \mathrm{is} -21 <-16<17<8. \\ \mathrm{Order} \mathrm{is} \frac{-7}{10}<\frac{8}{-15}<\frac{17}{30}<\frac{3}{5}. \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{Since}, \mathrm{LCM}\hspace{0.17em}\mathrm{of} 9, -12, -18, 3 \mathrm{is} 36. \\ \mathrm{Multiplying} \mathrm{the} \mathrm{numerators} \mathrm{and} \mathrm{denominators} \mathrm{to} \mathrm{get} \mathrm{the} \mathrm{denominator} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{LCM}, \\ \frac{-4}{9}=\frac{-4\times 4}{9\times 4}=\frac{-16}{36} \\ \frac{5}{-12}=\frac{-5\times 3}{12\times 3}=\frac{-15}{36} \\ \frac{7}{-18}=\frac{-7\times 2}{18\times 2}=\frac{-14}{36} \\ \frac{2}{-3}=\frac{-2\times 12}{3\times 12}=\frac{-24}{36}. \\ \mathrm{Order} \mathrm{is} -24 <-16<-15<-14. \\ \mathrm{Order} \mathrm{is} \frac{2}{-3}<\frac{-4}{9}<\frac{5}{-12}<\frac{7}{-18}. \end{gathered}$$

Question 6:

Arrange the following rational numbers in descending order:
(i) $\frac{7}{8}, \frac{64}{16}, \frac{36}{-12}, \frac{5}{-4}, \frac{140}{28}$
(ii) $\frac{-3}{10}, \frac{17}{-30}, \frac{7}{-15}, \frac{-11}{20}$

Answer 6:

We have to arrange them in descending order.

(i)
$$\begin{gathered} \mathrm{Since}, \mathrm{LCM}\hspace{0.17em}\mathrm{of} 8, 16, -12, -4, 28 \mathrm{is} 336. \\ \mathrm{Multiplying} \mathrm{the} \mathrm{numerators} \mathrm{and} \mathrm{denominators}, \mathrm{to} \mathrm{get} \mathrm{the} \mathrm{denominator} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{LCM}, \\ \frac{7}{8}=\frac{7\times 42}{8\times 42}=\frac{294}{336}, \\ \frac{64}{16}=\frac{64\times 21}{16\times 21}=\frac{1344}{336}, \\ \frac{36}{-12}=\frac{-36\times 28}{12\times 28}=\frac{-1008}{336}, \\ \frac{5}{-4}=\frac{-5\times 84}{4\times 84}=\frac{-420}{336}, \\ \frac{140}{28}=\frac{140\times 12}{28\times 12}=\frac{1680}{336}. \\ \mathrm{Order} \mathrm{is} 1680> 1344 >294 >-420>-1008. \\ \mathrm{Order} \mathrm{is} \frac{140}{28}>\frac{64}{16}>\frac{7}{8}>\frac{5}{-4}>\frac{36}{-12}. \end{gathered}$$
(ii)

$$\begin{gathered} \mathrm{Since}, \mathrm{LCM}\hspace{0.17em}\mathrm{of} 10, -30, -15, 20 \mathrm{is} 60. \\ \mathrm{Multiplying} \mathrm{the} \mathrm{numerators} \mathrm{and} \mathrm{denominators}, \mathrm{to} \mathrm{get} \mathrm{the} \mathrm{denominator} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{LCM}, \\ \frac{-3}{10}=\frac{-3\times 6}{10\times 6}=\frac{-18}{60}, \\ \frac{17}{-30}=\frac{-17\times 2}{30\times 2}=\frac{-34}{60}, \\ \frac{7}{-15}=\frac{-7\times 4}{15\times 4}=\frac{-28}{60}, \\ \frac{-11}{20}=\frac{-11\times 3}{20\times 3}=\frac{-33}{60}, \\ \mathrm{Order} \mathrm{is}, -18>-28>-33>-34. \\ \mathrm{Order} \mathrm{is} \frac{-3}{10}>\frac{7}{-15}>\frac{-11}{20}>\frac{17}{-30}. \end{gathered}$$

Question 7:

Which of the following statements are true:
(i) The rational number $\frac{29}{23}$ lies to the left of zero on the number line.
(ii) The rational number $\frac{-12}{-17}$ lies to the left of zero on the number line.
(iii) The rational number $\frac{3}{4}$ lies to the right of zero on the number line.
(iv) The rational numbers $\frac{-12}{-5} \mathrm{and} \frac{-7}{17}$ are on the opposite side of zero on the number line.
(v) The rational numbers $\frac{-21}{-5} \mathrm{and} \frac{7}{-31}$ are on the opposite side of zero on the number line.
(vi) The rational number $\frac{-3}{-5}$ is one the right of $\frac{-4}{7}$ on the number line.

Answer 7:

(i) False; it lies to the right of zero because it is a positive number.
(ii) False; it lies to the right of zero because it is a positive number.
(iii) True
(iv) True; they are of opposite signs.
(v) False; they both are of same signs.
(v) True; they both are of opposite signs and positive number is greater than the negative number. Thus, it is on the right of the negative number.