S.chand books class 8 maths solution chapter 1 Rational Numbers exercise 1 A

Exercise 1 A

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Question 1

Fill in the blanks

(i) $\dfrac{\phantom{-}3}{-4}$ , expressed as a rational number with denominator 24

Sol :

Denominator can not be negative

⇒$\dfrac{\phantom{-}3}{-4}\times \dfrac{-1}{-1}$

On dividing and multiplying by 6

⇒$\dfrac{-3}{4}\times \dfrac{6}{6}=\dfrac{-18}{24}$

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(ii) $\dfrac{-4}{7}\dots \dfrac{4}{-11}$ (>,<,=)

Sol :

L.C.M of 7 and 11 is 77

⇒$\dfrac{-4}{7}\times \dfrac{11}{11}=\dfrac{-44}{77}$..(i)

Denominator can not be negative

⇒$\dfrac{4}{-11}\times \dfrac{-1}{-1}=\dfrac{-4}{11}$

⇒$\dfrac{-4}{11}\times \dfrac{7}{7}=\dfrac{-28}{77}$..(ii)

From (i) and (ii)

⇒$\dfrac{-44}{77}<\dfrac{-28}{77}$

⇒$\dfrac{-4}{7}<\dfrac{-4}{11}$

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(iii) The absolute value of $\dfrac{-21}{-29}=\dots$

Sol :

⇒$\dfrac{\not{-}21}{\not{-}29}=\dfrac{21}{29}$

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(iv) The rational numbers whose absolute value is $\dfrac{7}{8}$ are __ .

Sol :

⇒$\dfrac{-7}{\phantom{-}8}\text{ and }\dfrac{7}{8}$

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(v) The rational number which is neither positive nor negative is ___ .

Sol :

⇒0

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Question 2

Answer True (T) or False (F)

(i) Every rational number is a whole number. 

Sol : F

(ii) 0 is the smallest rational number .

Sol : F

(iii) Every fractional number is a rational number.

Sol : T

(iv) $\dfrac{4}{0}$ is a rational number .

Sol : F

(v) |x| = -x , if x<0

Sol : T

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Question 3

Write four rational numbers equivalent to each of the following rational numbers

NOTE: Equivalent rational numbers can be find by multiplying or dividing by same number to rational numbers

(i) $\dfrac{3}{7}$

Sol :

⇒$\dfrac{3}{7}\times \dfrac{2}{2}=\dfrac{6}{14}$..(i)

⇒$\dfrac{3}{7}\times \dfrac{3}{3}=\dfrac{9}{21}$..(ii)

⇒$\dfrac{3}{7}\times \dfrac{4}{4}=\dfrac{12}{28}$..(iii)

⇒$\dfrac{3}{7}\times \dfrac{5}{5}=\dfrac{15}{35}$..(iv)

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(ii) $\dfrac{-7}{9}$

Sol :

⇒$\dfrac{-7}{9}\times \dfrac{2}{2}=\dfrac{-14}{18}$..(i)

⇒$\dfrac{-7}{9}\times \dfrac{3}{3}=\dfrac{-21}{27}$..(ii)

⇒$\dfrac{-7}{9}\times \dfrac{4}{4}=\dfrac{-28}{36}$..(iii)

⇒$\dfrac{-7}{9}\times \dfrac{5}{5}=\dfrac{-35}{45}$..(iv)

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(iii) $\dfrac{5}{-12}$

Sol :

⇒$\dfrac{5}{-12}\times \dfrac{2}{2}=\dfrac{10}{-24}$..(i)

⇒$\dfrac{5}{-12}\times \dfrac{3}{3}=\dfrac{15}{-36}$..(ii)

⇒$\dfrac{5}{-12}\times \dfrac{4}{4}=\dfrac{20}{-48}$..(iii)

⇒$\dfrac{5}{-12}\times \dfrac{5}{5}=\dfrac{25}{-60}$..(iv)

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(iv) $\dfrac{-12}{13}$

Sol :

⇒$\dfrac{-12}{13}\times \dfrac{2}{2}=\dfrac{-24}{26}$..(i)

⇒$\dfrac{-12}{13}\times \dfrac{3}{3}=\dfrac{-36}{39}$..(ii)

⇒$\dfrac{-12}{13}\times \dfrac{4}{4}=\dfrac{-48}{52}$..(iii)

⇒$\dfrac{-12}{13}\times \dfrac{5}{5}=\dfrac{-60}{65}$..(iv)

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Question 4

Which of the two rational numbers is greater?

(i) $\dfrac{-8}{13} \text{ or } \dfrac{6}{13}$

Sol :

⇒$\dfrac{-8}{13} < \dfrac{6}{13}$

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(ii) $\dfrac{-5}{16} \text{ or } \dfrac{-6}{8}$

Sol :

L.C.M of 16 and 8 is 16

⇒$\dfrac{-5}{16}\times \dfrac{1}{1}=\dfrac{-5}{16}$..(i)

⇒$\dfrac{-6}{8}\times \dfrac{2}{2}=\dfrac{-12}{16}$..(ii)

From (i) and (ii)

⇒$\dfrac{-5}{16} > \dfrac{-12}{16}$

⇒$\dfrac{-5}{16} > \dfrac{-6}{8}$

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(iii) $\dfrac{-11}{25} \text{ or } 0$

Sol :

⇒$\dfrac{-11}{25} < 0$

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(iv) $\dfrac{-22}{-33} \text{ or } \dfrac{45}{-65}$

Sol :

⇒$\dfrac{-22}{-33}\div \dfrac{11}{11}=\dfrac{2}{3}$

Denominator can't be negative

⇒$\dfrac{45}{-65}\times \dfrac{-1}{-1}$

⇒$\dfrac{-45}{65}\div \dfrac{5}{5}=\dfrac{-9}{13}$

L.C.M of 3 and 13 is 39

⇒$\dfrac{2}{3}\times \dfrac{13}{13}=\dfrac{26}{39}$..(i)

⇒$\dfrac{-9}{13}\times \dfrac{3}{3}=\dfrac{-27}{39}$..(ii)

From (i) and (ii)

⇒$\dfrac{26}{39} > \dfrac{-27}{39}$

⇒$\dfrac{-22}{-33} > \dfrac{45}{-65}$

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Question 5

Which of the two rational numbers is smaller?

NOTE: To compare two rational numbers $\dfrac{p}{q}\text{ and }\dfrac{r}{s}$ , we compare their cross product

⇒$\dfrac{p}{q}\text{ and }\dfrac{r}{s}$

⇒p×s and r×q

(i) if p×s > r×q , then $\dfrac{p}{q}>\dfrac{r}{s}$

(ii) if p×s < r×q, then $\dfrac{p}{q}<\dfrac{r}{s}$

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(i) $\dfrac{-11}{8} \text{ or } \dfrac{17}{-8}$

Sol :

⇒$\dfrac{-11}{8} > \dfrac{-17}{8}$

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(ii) $\dfrac{12}{19} \text{ or } \dfrac{-9}{-19}$

Sol :

⇒$\dfrac{12}{19} > \dfrac{-9}{-19}$

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(iii) $\dfrac{21}{-5} \text{ or } 1$

Sol :

⇒$\dfrac{21}{-5} < 1$

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(iv) $\dfrac{-11}{1111} \text{ or } \dfrac{1}{-103}$

Sol :

⇒$\dfrac{-11}{1111}\div \dfrac{11}{11}=\dfrac{-1}{101}$..(i)

Denominator can not be negative

⇒$\dfrac{1}{-101}\times \dfrac{-1}{-1}=\dfrac{-1}{103}$..(ii)

From (i) and (ii)

⇒$\dfrac{-1}{101} \text{ and } \dfrac{-1}{103}$

Cross multiplication

⇒-103 < -101

⇒⇒$\dfrac{-1}{101}<\dfrac{-1}{103}$

⇒$\dfrac{-11}{1111} < \dfrac{1}{-103}$

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Question 6

Which of the symbols = , <  or > should replace the blank space ?

(i) $\left(\dfrac{10}{3}\right)\dots \left(1\right)$

Sol : >

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(ii) $\left(-\dfrac{1}{2}\right)\dots \left(-\dfrac{3}{5}\right)$

Sol : >

L.C.M of 2 and 5 is 10

⇒$\left(-\dfrac{1}{2}\right)\times \dfrac{5}{5}=-\dfrac{5}{10}$..(i)

⇒$\left(-\dfrac{3}{5}\right)\times \dfrac{2}{2}=-\dfrac{6}{10}$..(ii)

From (i) and (ii)

⇒$-\dfrac{5}{10}>-\dfrac{6}{10}$

⇒$\left(-\dfrac{1}{2}\right) > \left(-\dfrac{3}{5}\right)$

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(iii) $\left(-\dfrac{3}{4}\right)\dots \left(-\dfrac{2}{3}\right)$

Sol : <

L.C.M of 4 and 3 is 12

⇒$\left(-\dfrac{3}{4}\right)\times \dfrac{3}{3}=-\dfrac{9}{12}$..(i)

⇒$\left(-\dfrac{2}{3}\right)\times \dfrac{4}{4}=-\dfrac{8}{12}$..(ii)

From (i) and (ii)

⇒$-\dfrac{9}{12}>-\dfrac{8}{12}$

⇒$\left(-\dfrac{3}{4}\right) > \left(-\dfrac{2}{3}\right)$

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(iv) $\left(-2\dfrac{3}{7}\right)\dots \left(-2\dfrac{3}{5}\right)$

Sol : >

⇒$\left(-\dfrac{17}{7}\right)\dots \left(-\dfrac{13}{5}\right)$

L.C.M of 7 and 5 is 35

⇒$-\dfrac{17}{7} \times \dfrac{5}{5}=-\dfrac{58}{35}$..(i)

⇒$-\dfrac{13}{5} \times \dfrac{7}{7}=-\dfrac{91}{35}$..(ii)

From (i) and (ii)

⇒$\dfrac{-58}{35}>\dfrac{-91}{35}$

⇒$\left(-\dfrac{17}{7}\right)>\left(-\dfrac{13}{5}\right)$

⇒$\left(-2\dfrac{3}{7}\right) > \left(-2\dfrac{3}{5}\right)$

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(v) $\dfrac{8}{7}\dots \dfrac{12}{9}$

Sol : <

L.C.M of 7 and 9 is 63

⇒$\dfrac{8}{7}\times \dfrac{9}{9} =\dfrac{72}{63}$..(i)

⇒$\dfrac{12}{9}\times \dfrac{7}{7} =\dfrac{84}{63}$..(ii)

From (i) and (ii)

⇒$\dfrac{72}{63} < \dfrac{84}{63}$

⇒$\dfrac{8}{7} < \dfrac{12}{9}$

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(vi) $\dfrac{-2}{3}\dots \dfrac{4}{-7}$

Sol : <

L.C.M of 3 and 7 is 21

⇒$\dfrac{-2}{3} \times \dfrac{7}{7}=\dfrac{-14}{21}$..(i)

Denominator can not be negative

⇒$\dfrac{4}{-7} \times \dfrac{-1}{-1}=\dfrac{-4}{7}$

⇒$\dfrac{-4}{7} \times \dfrac{3}{3}=\dfrac{-12}{21}$..(ii)

From (i) and (ii)

⇒$\dfrac{-14}{21} < \dfrac{-12}{21}$

⇒$\dfrac{-2}{3} < \dfrac{4}{-7}$

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(vii) $\dfrac{13}{-15}\dots \dfrac{10}{-11}$

Sol : >

L.C.M of 15 and 11 is 165

Denominator can not be negative

⇒$\dfrac{13}{-15} \times \dfrac{-1}{-1}=\dfrac{13}{-15}$

⇒$\dfrac{-13}{15} \times \dfrac{11}{11}=\dfrac{-143}{165}$..(i)

And

⇒$\dfrac{10}{-11} \times \dfrac{-1}{-1}=\dfrac{-10}{11}$

⇒$\dfrac{-10}{11} \times \dfrac{15}{15}=\dfrac{-150}{165}$..(ii)

From(i) and (ii)

⇒$\dfrac{-143}{165} > \dfrac{-150}{165}$

⇒$\dfrac{13}{-15} > \dfrac{10}{-11}$

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(viii) $0\dots \dfrac{-5}{-9}$

Sol : <

$0 < \dfrac{5}{9}$ [negative sign cancelled out]

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Question 7

Arrange in order from least to greatest

(i) $\dfrac{1}{4},\dfrac{3}{8},\dfrac{2}{7}$

Sol :

L.C.M of 4 , 8 and 7 is 56

⇒$\dfrac{1}{4}\times \dfrac{14}{14}=\dfrac{14}{56}$..(i)

⇒$\dfrac{3}{8}\times \dfrac{7}{7}=\dfrac{21}{56}$..(ii)

⇒$\dfrac{2}{7}\times \dfrac{8}{8}=\dfrac{16}{56}$..(iii)

From (i) , (ii) and (iii)

⇒$\dfrac{14}{56}<\dfrac{16}{56}<\dfrac{21}{56}$

or

⇒$\dfrac{1}{4}<\dfrac{2}{7}<\dfrac{3}{8}$

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(ii) $\dfrac{-8}{5},\dfrac{-3}{2},\dfrac{-19}{11}$

Sol :

L.C.M of 5 , 2 and 11 is 110

⇒$\dfrac{-8}{5}\times \dfrac{22}{22}=\dfrac{-176}{110}$..(i)

⇒$\dfrac{-3}{2}\times \dfrac{55}{55}=\dfrac{-165}{110}$..(ii)

⇒$\dfrac{-19}{11}\times \dfrac{10}{10}=\dfrac{-190}{110}$..(iii)

From (i) , (ii) and (iii)

⇒$\dfrac{-190}{110}<\dfrac{-176}{110}<\dfrac{-165}{110}$

or

⇒$\dfrac{-19}{11}<\dfrac{-8}{5}<\dfrac{-3}{2}$

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(iii) $\dfrac{-2}{5},\dfrac{3}{-10},\dfrac{-1}{3}$

Sol :

L.C.M of 5 , 10 and 3 is 30

⇒$\dfrac{-2}{5}\times \dfrac{6}{6}=\dfrac{-12}{30}$..(i)

⇒$\dfrac{-3}{10}\times \dfrac{3}{3}=\dfrac{-9}{30}$..(ii)

⇒$\dfrac{-1}{3}\times \dfrac{10}{10}=\dfrac{-10}{30}$..(iii)

From (i) , (ii) and (iii)

⇒$\dfrac{-12}{30}<\dfrac{-10}{30}<\dfrac{-9}{30}$

or

⇒$\dfrac{-2}{5}<\dfrac{-1}{3}<\dfrac{3}{-10}$

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Question 8

Arrange in descending order .

(i) $\dfrac{3}{4},\dfrac{6}{7},\dfrac{9}{14},\dfrac{7}{8}$

Sol :

L.C.M of 4 , 7 , 14 and 8 is 56

⇒$\dfrac{3}{4}\times \dfrac{14}{14}=\dfrac{42}{56}$..(i)

⇒$\dfrac{6}{7}\times \dfrac{8}{8}=\dfrac{48}{56}$..(ii)

⇒$\dfrac{9}{14}\times \dfrac{4}{4}=\dfrac{36}{56}$..(iii)

⇒$\dfrac{7}{8}\times \dfrac{7}{7}=\dfrac{49}{56}$..(iv)

From (i) , (ii) , (iii) and (iv)

⇒$\dfrac{49}{56}>\dfrac{48}{56}>\dfrac{42}{56}>\dfrac{36}{56}$

or

⇒$\dfrac{7}{8}>\dfrac{6}{7}>\dfrac{3}{4}>\dfrac{9}{14}$

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(ii) $\dfrac{1}{11},\dfrac{-1}{3},\dfrac{-3}{12}$

Sol :

L.C.M of 11 , 3 and 12 is 132

⇒$\dfrac{1}{11}\times \dfrac{12}{12}=\dfrac{12}{132}$..(i)

⇒$\dfrac{-1}{3}\times \dfrac{44}{44}=\dfrac{-44}{132}$..(ii)

⇒$\dfrac{-3}{12}\times \dfrac{11}{11}=\dfrac{-33}{132}$..(iii)

From (i) , (ii) and (iii)

⇒$\dfrac{12}{132}>\dfrac{-33}{132}>\dfrac{-44}{132}$

or

⇒$\dfrac{1}{11}>\dfrac{-3}{12}>\dfrac{-1}{3}$

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(iii) $\dfrac{3}{-5},\dfrac{-17}{15},\dfrac{8}{10},\dfrac{-7}{10}$

Sol :

L.C.M of 5 , 15 ,10 and 10 is 30

⇒$\dfrac{-3}{5}\times \dfrac{6}{6}=\dfrac{-18}{30}$..(i)

⇒$\dfrac{-17}{15}\times \dfrac{2}{2}=\dfrac{-34}{30}$..(ii)

⇒$\dfrac{8}{10}\times \dfrac{3}{3}=\dfrac{24}{30}$..(iii)

⇒$\dfrac{-7}{10} \times \dfrac{3}{3}=\dfrac{-21}{30}$..(iv)

From (i) , (ii) , (iii) and (iv)

⇒$\dfrac{24}{30}>\dfrac{-18}{30}>\dfrac{-21}{30}>\dfrac{-34}{30}$

or

⇒$\dfrac{8}{10}>\dfrac{-3}{5}>\dfrac{-7}{10}>\dfrac{-17}{15}$

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