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

Exercise 1 A


Question 1

Fill in the blanks

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

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}


(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}


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

Sol :

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


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

Sol :

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


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

Sol :

0


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


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)


(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)


(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)


(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)


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}


(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}


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

Sol :

\dfrac{-11}{25} < 0


(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}


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}


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

Sol :

\dfrac{-11}{8} > \dfrac{-17}{8}


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

Sol :

\dfrac{12}{19} > \dfrac{-9}{-19}


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

Sol :

\dfrac{21}{-5} < 1


(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}


Question 6

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

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

Sol : >


(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)


(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)


(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)


(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}


(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}


(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}


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

Sol : <

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


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}


(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}


(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}


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}


(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}


(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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