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

Exercise 4B

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Question 1:

Represent each of the following rational numbers on the number line:

(i) $\frac{1}{3}$
(ii) $\frac{2}{7}$
(iii) $\frac{7}{3}$
(iv) $\frac{22}{7}$
(v) $\frac{37}{8}$
(vi) $\frac{-1}{3}$
(vii) $\frac{-3}{4}$
(viii) $\frac{-12}{7}$
(ix) $\frac{36}{-5}$
(x) $\frac{-43}{9}$

Answer 1:

(i)


(ii)
 


(iii) (7/3)=2+(1/3)



(iv) 

$\frac{22}{7}$ can be written as $3 + \frac{1}{7}$. So, we need to move to the right of point 3. Then, we need to move $\frac{1}{7}$ distance more to the right.  




(v) $\frac{37}{8}$ can be written as 4+$\frac{5}{8}$.  So, we need to move to the right of point 4. Then, we need to move $\frac{5}{8}$ distance more to the right.




(vi) 



(vii) 


(viii)

$\frac{-12}{7}$ can be written as $-1-\frac{5}{7}$. So, we need to move to the left of point -1. Then, we need to move $\frac{5}{7}$ distance more to the left. 


(ix) 

$\frac{36}{-5}$ can be written as $-7-\frac{1}{5}$. So, we need to move to the left of point -7. Then, we need to move $\frac{1}{5}$ distance more to the left.  






(x) $\frac{-43}{9}$ can be written as $-4-\frac{7}{9}$. So, we need to move to the left of point -4. Then, we need to move $\frac{7}{9}$ distance more to the left. 

Question 2:

Which of the two rational numbers is greater in each of the following pairs?

(i) $\frac{5}{6} \mathrm{or} 0$
(ii) $\frac{-3}{5} \mathrm{or} 0$
(iii) $\frac{5}{8}\mathrm{or} \frac{3}{8}$
(iv) $\frac{7}{9} \mathrm{or} \frac{-5}{9}$
(v) $\frac{-6}{11} \mathrm{or} \frac{5}{-11}$
(vi) $\frac{-15}{4} \mathrm{or} \frac{-17}{4}$

Answer 2:

$\begin{array}{l}\\ \left(\text{i}\right)\frac{5}{6}. \mathrm{This} \mathrm{is} \mathrm{because} 0\text{ can be written as }\frac{0}{6}\text{ and }\frac{0}{6}<\frac{5}{6}.\\ \left(\text{ii}\right)\frac{-3}{5} <0. \mathrm{This} \mathrm{is} \mathrm{because}0\text{ can be written as }\frac{0}{5}\text{ and }-3<0.\\ \left(\text{iii}\right)\frac{5}{8} >\frac{3}{8}. \mathrm{This} \mathrm{is} \mathrm{because} \text{ 5 }>\text{3}.\\ \left(\text{iv}\right)\frac{7}{9} >\frac{5}{9}. \mathrm{This} \mathrm{is} \mathrm{because} \text{7} > \text{5.}\\ \left(\text{v}\right)\frac{-6}{11} <\frac{-5}{11}. \mathrm{This} \mathrm{is} \mathrm{because} -\text{6} < -\text{5.}\\ \left(\text{vi}\right)\frac{-15}{4} > \frac{-17}{4},-\text{15} > -\text{17}\end{array}$

Question 3:

Which of the two rational numbers is greater in each of the following pairs?

(i) $\frac{5}{9} \mathrm{or} \frac{-3}{-8}$
(ii) $\frac{4}{-3} \mathrm{or} \frac{-8}{7}$
(iii) $\frac{-12}{5} \mathrm{or} -3$
(iv) $\frac{7}{-9} \mathrm{or} \frac{-5}{8}$
(v) $\frac{4}{-5} \mathrm{or} \frac{-7}{8}$
(vi) $\frac{9}{-13} \mathrm{or} \frac{7}{-12}$

Answer 3:

$\begin{array}{l}\left(\text{i}\right)\frac{5}{9},\frac{-3}{-8}\\ \frac{(-3)\times (-1)}{(-8)\times (-1)}=\frac{3}{8}\\ \text{L}.\text{C}.\text{M. of 9 and 8 }is \text{72.}\\ \frac{5\times 8}{9\times 8}=\frac{40}{72}\\ \frac{3\times 9}{8\times 9}=\frac{27}{72}\\ 27<40\\ \frac{-3}{-8}<\frac{5}{9}\\ \text{So, }\frac{5}{9}\text{ is greater}\text{. }\end{array}$                                                                                   
$\begin{array}{l}\left(\text{ii}\right)\\ \frac{4}{-3} ,\frac{-8}{7}\\ \mathrm{We} \mathrm{will} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}. \\ \frac{4\times -1}{-3\times -1}=\frac{-4}{3}\\ \\ \begin{array}{l}\text{L}.\text{C}.\text{M. of 3 and 7 is 21.}\\ \\ \frac{-4\times \left(7\right)}{\left(3\right)\times \left(7\right)}=\frac{-28}{21}\\ \frac{(-8)\times 3}{7\times 3}=\frac{-24}{21}\\ (-\text{24})>(-\text{28})\\ \frac{-8}{7}>\frac{4}{(-3)}\\ \text{So, }\frac{-8}{7}\text{ is greater}\text{.}\end{array}\end{array}$


$\begin{array}{l}\\ \left(\text{iii}\right)\\ \frac{-12}{5},-3\\ \begin{array}{l} \text{L}.\text{C}.\text{M. of 5 and 1 is 5.}\\ \\ \frac{-12\times 1}{5\times 1}=\frac{-12}{5}\\ \frac{-3\times 5}{1\times 5}=\frac{-15}{5}\\ -\text{12}>-15\\ \frac{-12}{5}>-3\\ \frac{-12}{5}\text{}\text{is greater. }\end{array}\\ \end{array}$
$\begin{array}{l}\left(\text{iv}\right)\\ \frac{-7}{9},\frac{-5}{8}\\ \text{L}.\text{C}.\text{M. of 9 and 8 is 72.}\\ \\ \frac{-7\times 8}{9\times 8}=\frac{-56}{72}\\ \frac{-5\times 9}{8\times 9}=\frac{-45}{72}\\ -\text{56}<-\text{45}\\ \frac{-7}{9}<\frac{-5}{8}\end{array}$                                                                          

$\begin{array}{l}\left(\text{v}\right)\frac{4}{-5},\frac{-7}{8}\\ \mathrm{We} \mathrm{will} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}. \\ \begin{array}{l}\frac{4\times -1}{-5\times -1}=\frac{-4}{5}\\ \begin{array}{l}\text{L}.\text{C}.\text{M. of 5 and 8 is 4}0.\\ \\ \frac{-4\times 8}{5\times 8}=\frac{-32}{40}\\ \frac{-7\times 5}{8\times 5}=\frac{-35}{40}\\ -\text{32}>-\text{35}\end{array}\\ \frac{-4}{5}>\frac{-7}{8}\\ \end{array}\end{array}$                            


$\begin{array}{l}\left(\text{vi}\right)\frac{9}{-13},\frac{7}{-12}\\ \mathrm{We} \mathrm{will} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}. \\ \frac{9\times -1}{-13\times -1}=\frac{-9}{13}\\ \frac{7\times -1}{-12\times -1}=\frac{-7}{12}\\ \text{ L}.\text{C}.\text{M. of 13 and 12 is }156.\\ \\ \frac{-9\times 12}{13\times (-12)}=\frac{-108}{156}\\ \frac{-7\times 13}{(-12)\times 13}=\frac{-91}{156}\\ -\text{1}0\text{8}<-\text{91}\\ \frac{9}{-13}<\frac{7}{-12}\end{array}$

Question 4:

Fill in the blanks with the correct symbol out of >, = and <:

(i) $\frac{-3}{7}......\frac{6}{-13}$
(ii) $\frac{5}{-13}......\frac{-35}{91}$
(iii) $-2......\frac{-13}{5}$
(iv) $\frac{-2}{3}......\frac{5}{-8}$
(v) $0......\frac{-3}{-5}$
(vi) $\frac{-8}{9}......\frac{-9}{10}$

Answer 4:

$\begin{array}{l}\left(\text{i}\right)\frac{-3}{7}>\frac{-6}{13}\\ \text{L}.\text{C}.\text{M. of 7 and} \text{13 is 91.}\\ \frac{-3\times 13}{7\times 13}=\frac{-39}{91}\\ \frac{-6\times 7}{13\times 7}=\frac{-42}{91}\\ \frac{-39}{91}>\frac{-42}{91}\\ \left(\text{ii}\right)\frac{5}{-13}=\frac{-35}{91}\\ \text{L}.\text{C}.\text{M. of 13 and 91 is 91.}\\ \frac{5\times (-7)}{-13\times (-7)}=\frac{-35}{91}\\ \left(\text{iii}\right)-\text{2}>\frac{-13}{5}\\ \text{L}.\text{C}.\text{M of 1 and 5 is 5.}\\ \frac{-2\times 5}{1\times 5}=\frac{-10}{5}\\ \frac{-13\times 1}{5\times 1}=\frac{-13}{5}\\ \frac{-10}{5}>\frac{-13}{5}\end{array}$                                                                                    
$\begin{array}{l}\left(\text{iv}\right) \frac{-2}{3}<\frac{-5}{8}\\ \text{L}.\text{C}.\text{M. of 8 and 3 is 24.}\\ \frac{-2\times 8}{3\times 8}=\frac{-16}{24}\\ \frac{-5\times 3}{8\times 3}=\frac{-15}{24}\\ \frac{-16}{24}<\frac{-15}{24}\end{array}$                                                

$\begin{array}{l}\left(\text{v}\right) 0<\frac{3}{5}\\ \text{L}.\text{C}.\text{M. of 1 and 5 is 5.}\\ \frac{0\times 1}{1\times 5}=\frac{0}{5}\\ \frac{3\times 1}{5\times 1}=\frac{3}{5}\\ \frac{0}{5}<\frac{3}{5}\end{array}$                                            

$\begin{array}{l}\left(\text{vi}\right) \frac{-8}{9}>\frac{-9}{10}\\ \text{L}.\text{C}.\text{M. of 9 and 1}0is \text{9}0.\\ \frac{-8\times 10}{9\times 10}=\frac{-80}{90}\\ \frac{-9\times 9}{10\times 9}=\frac{-81}{90}\\ \frac{-80}{90}>\frac{-81}{90}\end{array}$                                               

Question 5:

Arrange the following rational numbers in ascending order:

(i) $\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}$
(ii) $\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}$
(iii) $\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}$
(iv) $\frac{2}{3},\frac{3}{4},\frac{5}{-6},\frac{-7}{12}$

Answer 5:

$\begin{array}{l}\left(\text{i}\right)\frac{2}{5},\frac{7}{10},\frac{8}{15},\frac{13}{30}\\ \text{L}.\text{C}.\text{M. of 5}, \text{1}0, \text{15 and 3}0 \mathrm{is} \text{3}0\\ \end{array}$                                       



$\begin{array}{l}\frac{2\times 6}{5\times 6}=\frac{12}{30}\\ \frac{7\times 3}{10\times 3}=\frac{21}{30}\\ \frac{8\times 2}{15\times 2}=\frac{16}{30}\\ \frac{13\times 1}{30\times 1}=\frac{13}{30}\\ \text{Required order: }\frac{2}{5}<\frac{13}{30}<\frac{8}{15}<\frac{7}{10}\end{array}$

$\begin{array}{l}\left(ii\right)\frac{-3}{4},\frac{5}{-12},\frac{-7}{16},\frac{9}{-24}\\ \mathrm{First}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}.\\ \frac{-3}{4},\frac{5\times -1}{-12\times -1},\frac{-7}{16},\frac{9\times -1}{-24\times -1}\\ \begin{array}{l}\frac{-3}{4},\frac{-5}{12},\frac{-7}{16},\frac{-9}{24}\\ \begin{array}{l}\\ \\ \\ \\ \\ \end{array}\end{array}\end{array}$


$\begin{array}{l}\text{L}.\text{C}.\text{M. of 4}, \text{12}, \text{16 and} \text{24 is 48.}\\ \frac{-3\times 12}{4\times 12}=\frac{-36}{48}\\ \frac{-5\times 4}{12\times 4}=\frac{-20}{48}\\ \frac{-7\times 3}{16\times 3}=\frac{-21}{48}\\ \frac{-9\times 2}{24\times 2}=\frac{-18}{48}\\ \text{Required order: }\frac{-3}{4}<\frac{-7}{16}<\frac{-5}{12}<\frac{-9}{24}\end{array}$
$\begin{array}{l}\left(\text{iii}\right)\frac{-3}{10},\frac{7}{-15},\frac{-11}{20},\frac{17}{-30}\\ \text{First, we need to convert the negative denominators to make them positive.}\\ \frac{-3}{10},\frac{7\times -1}{-15\times -1},\frac{-11}{20},\frac{17\times -1}{-30\times -1}\\ \frac{-3}{10},\frac{-7}{15},\frac{-11}{20},\frac{-17}{30}\end{array}$



$\begin{array}{l}\begin{array}{l}\text{L}.\text{C}.\text{M of 1}0,\text{15},\text{2}0,\text{3}0=60\\ \frac{-3\times 6}{10\times 6}=\frac{-18}{60}\\ \frac{-7\times 4}{15\times 4}=\frac{-28}{60}\\ \frac{-11\times 3}{20\times 3}=\frac{-33}{60}\\ \frac{-17\times 2}{30\times 2}=\frac{-34}{60}\\ \text{Therefore, }\frac{-34}{60}<\frac{-33}{60}<\frac{-28}{60}<\frac{-18}{60}\end{array}\\ \text{i}\text{.e}\text{. }\frac{-17}{30}<\frac{-11}{20}<\frac{-7}{15}<\frac{-3}{10}\\ \left(\text{iv}\right)\frac{2}{3},\frac{3}{4},\frac{5}{-6},\frac{-7}{12}\\ \text{First, we need to convert the negative denominators to positive ones.}\\ \frac{2}{3},\frac{3}{4},\frac{5\times -1}{-6\times -1},\frac{-7}{12}\\ \frac{2}{3},\frac{3}{4},\frac{-5}{6},\frac{-7}{12}\end{array}$



$\begin{array}{l}\begin{array}{l}\text{L}.\text{C}.\text{M of 3},\text{4},\text{6},\text{12}=\text{12}\\ \frac{2\times 4}{3\times 4}=\frac{8}{12}\\ \frac{3\times 3}{4\times 3}=\frac{9}{12}\\ \frac{-5\times 2}{6\times 2}=\frac{-10}{12}\\ \frac{-7\times 1}{12\times 1}=\frac{-7}{12}\end{array}\\ \\ \text{Therefore, the correct order is }\frac{-5}{6}<\frac{-7}{12}<\frac{2}{3}<\frac{3}{4}.\end{array}$

Question 6:

Arrange the following rational numbers in descending order:

(i) $\frac{-2}{5},\frac{7}{-10},\frac{-11}{15},\frac{19}{-30}$
(ii) $-2,\frac{-13}{6},\frac{8}{-3},\frac{1}{3}$
(iii) $\frac{-4}{9},\frac{5}{-12},\frac{-7}{18},\frac{2}{-3}$
(iv) $\frac{17}{-30},\frac{11}{-15},\frac{-7}{10},\frac{3}{5}$

Answer 6:

$\begin{array}{l}\\ \left(\text{i}\right)\frac{-2}{5},\frac{7}{-10},\frac{-11}{15},\frac{19}{-30}\\ \mathrm{First}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}.\\ \frac{-2}{5},\frac{7\times -1}{-10\times -1},\frac{-11}{15},\frac{19\times -1}{-30\times -1}\\ \frac{-2}{5},\frac{-7}{10},\frac{-11}{15},\frac{-19}{30}\\ \end{array}$


$\begin{array}{l}\begin{array}{l}\text{L}.\text{C}.\text{M. of 5}, \text{1}0, 1\text{5 and 3}0 \mathrm{is} \text{3}0.\\ \frac{-2\times 6}{5\times 6}=\frac{-12}{30},\\ \frac{-7\times 3}{10\times 3}=\frac{-21}{30},\\ \frac{-11\times 2}{15\times 2}=\frac{-22}{30},\\ \frac{-19\times 1}{30\times 1}=\frac{-19}{30},\end{array}\\ \\ \text{Correct order:}\frac{-2}{5}>\frac{19}{-30}>\frac{7}{-10}>\frac{-11}{15}\\ \left(\text{ii}\right)-2,\frac{-13}{6},\frac{8}{-3},\frac{1}{3}\\ \mathrm{First}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}.\\ -2,\frac{-13}{6},\frac{8\times -1}{-3\times -1},\frac{1}{3}\\ -2,\frac{-13}{6},\frac{-8}{3},\frac{1}{3}\end{array}$



$\begin{array}{l}\begin{array}{l}\text{L}.\text{C}.\text{M. of 6}, \text{3 and 3 is 6.}\\ \frac{-2\times 6}{1\times 6}=\frac{-12}{6},\\ \frac{-13\times 1}{6\times 1}=\frac{-13}{6},\\ \frac{-8\times 2}{3\times 2}=\frac{-16}{6},\end{array}\\ \frac{1\times 2}{3\times 2}=\frac{2}{6},\\ \text{Correct order: }\frac{1}{3}>-2>\frac{-13}{6}>\frac{-8}{3}\end{array}$
$\begin{array}{l}\left(\text{iii}\right)\frac{-4}{9},\frac{5}{-12},\frac{-7}{18},\frac{2}{-3}\\ \mathrm{First}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}.\\ \frac{-4}{9},\frac{5\times -1}{-12\times -1},\frac{-7}{18},\frac{2\times -1}{-3\times -1}\\ \frac{-4}{9},\frac{-5}{12},\frac{-7}{18},\frac{-2}{3}\end{array}$


$\begin{array}{l}\begin{array}{l}\text{L}.\text{C}.\text{M. of 9}, \text{12}, \text{18 and 3 is }36.\\ \frac{-4\times 4}{9\times 4}=\frac{-16}{36},\\ \frac{-5\times 3}{12\times 3}=\frac{-15}{36},\\ \frac{-7\times 2}{18\times 2}=\frac{-14}{36},\\ \frac{-2\times 12}{3\times 12}=\frac{-24}{36}\end{array}\\ \\ \text{Correct order: }\frac{-7}{18}>\frac{-5}{12}>\frac{-4}{9}>\frac{-2}{3}\\ \left(\text{iv}\right)\frac{17}{-30},\frac{11}{-15},\frac{-7}{10},\frac{3}{5}\\ \mathrm{First}, \mathrm{we} \mathrm{need} \mathrm{to} \mathrm{convert} \mathrm{each} \mathrm{negative} \mathrm{denominator} \mathrm{into} \mathrm{positive}.\\ \frac{17\times -1}{-30\times -1},\frac{11\times -1}{-15\times -1},\frac{-7}{10},\frac{3}{5}\\ \frac{-17}{30},\frac{-11}{15},\frac{-7}{10},\frac{3}{5}\\ \begin{array}{l}\\ \\ \\ \\ \\ \end{array}\end{array}$


$\begin{array}{l}\text{L}.\text{C}.\text{M. of 3}0, \text{15}, \text{1}0 \mathrm{and} \text{5 is 3}0.\\ \frac{-17\times 1}{30\times 1}=\frac{-17}{30},\\ \frac{-11\times 2}{15\times 2}=\frac{-22}{30},\\ \frac{-7\times 3}{10\times 3}=\frac{-21}{30},\\ \frac{3\times 6}{5\times 6}=\frac{18}{30},\\ \text{Correct order: }\frac{3}{5}>\frac{17}{-30}>\frac{-7}{10}>\frac{11}{-15}\end{array}$

Question 7:

Which of the following statements are true?

(i) $\frac{-3}{5}$ lies to the left of 0 on the number line.
(ii) $\frac{-12}{7}$ lies to the right of 0 on the number line.
(iii) $\frac{1}{3} \mathrm{and} \frac{-5}{2}$ lie on opposite sides of 0 on the number line.
(iv) $\frac{-18}{-13}$ lies to the left of 0 on the number line.
(v) $\frac{-5}{-8}$ lies on the right of $\frac{-5}{7}$ on the number line.

Answer 7:

$\begin{array}{l}\\ \left(\text{i}\right)\text{True}. This is because \text{negative rational numbers lie to the left of 0 on the number line}\text{.}\\ \left(\text{ii}\right)\text{False}. This is because \text{negative rational numbers do not lie to the right of 0}\text{.}\\ \left(\text{iii}\right)\text{True}. This is because \text{positive rational numbers lie to the right, while negative rational numbers to the left of 0}\text{.}\\ \left(\text{iv}\right)\text{False}. This is because \text{it is positive, which means it lies to the right of 0}\text{.}\\ \left(\text{v}\right)\text{True}. This is because \mathrm{a} \text{positive rational number lies to the right of a negative rational number}\text{.}\\ \end{array}$

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Question 8:

Find five rational numbers between −3 and −2.

Answer 8:

$\begin{array}{l}\mathrm{L}.\mathrm{C}.\mathrm{M}. \mathrm{of} 2 \mathrm{and} 3 \mathrm{is} 6. \\ -3=\frac{-3\times 6}{1\times 6}=\frac{-18}{6}\\ -2=\frac{-2\times 6}{1\times 6}=\frac{-12}{6}\\ \\ \mathrm{Therefore} ,\frac{-17}{6},\frac{-16}{6},\frac{-15}{6},\frac{-14}{6} \mathrm{and} \frac{-13}{6} \mathrm{are} \mathrm{the} \mathrm{five} \mathrm{rational} \mathrm{number} \mathrm{between} -3 \mathrm{and} -2.\end{array}$

Question 9:

Find five rational numbers between −1 and 1.

Answer 9:

$\begin{array}{l}-1=\frac{-1\times 5}{1\times 5}, 1=\frac{1\times 5}{1\times 5}\\ \frac{-5}{5}\mathrm{and}\frac{5}{5}\\ \begin{array}{l}\text{Hence, the five rational numbers between }-1\text{ and }1 \mathrm{are} \frac{-4}{5},\frac{-3}{5},\frac{-2}{5},\frac{-1}{5} \mathrm{and} \frac{1}{5}.\\ \end{array}\end{array}$

Question 10:

Find five rational numbers between $\frac{-3}{5}$ and $\frac{-1}{2}$.

Answer 10:

$\begin{array}{l}\\ \frac{-3}{5}\text{ and }\frac{-1}{2}\\ \text{L}\text{.C}\text{.M. of 5 and 2 is 10}\text{.}\\ \frac{-3\times 2}{5\times 2}=\frac{-6\times 4}{10\times 4}=\frac{-24\times 2}{40\times 2}=\frac{-48}{80},\\ \frac{-1\times 5}{2\times 5}=\frac{-5\times 4}{10\times 4}=\frac{-20\times 2}{40\times 2}=\frac{-40}{80},\\ \mathrm{Hence}, \mathrm{the} \mathrm{f}\text{ive rational numbers between }\frac{-3}{5}\text{ and }\frac{-1}{2}\text{ are }\frac{-45}{80},\frac{-44}{80},\frac{-43}{80},\frac{-42}{80} \mathrm{and} \frac{-41}{80}.\\ \end{array}$