Exercise 1 A
Q1 | Ex-1A | Class-8 | Composite mathematics | Schand Solution | myhelper
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
⇒$\dfrac{-3}{4}\times \dfrac{6}{6}=\dfrac{-18}{24}$
(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
Q2 | Ex-1A | Cl-8 | Composite mathematics Schand Solution | myhelper
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
Q3 | Ex-1A | Cl-8 | Composite mathematics Schand Solution | myhelper
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)
Q4 | Ex-1A | Cl-8 | Composite mathematics Schand Solution | myhelper
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)
⇒$\dfrac{26}{39} > \dfrac{-27}{39}$
⇒$\dfrac{-22}{-33} > \dfrac{45}{-65}$
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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 $\dfrac{p}{q}<\dfrac{r}{s}$
(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
⇒⇒$\dfrac{-1}{101}<\dfrac{-1}{103}$
⇒$\dfrac{-11}{1111} < \dfrac{1}{-103}$
Q6 | Ex-1A | Cl-8 | Composite mathematics Schand Solution | myhelper
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]
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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
⇒$\dfrac{1}{4}<\dfrac{2}{7}<\dfrac{3}{8}$
(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
⇒$\dfrac{-2}{5}<\dfrac{-1}{3}<\dfrac{3}{-10}$
Q8 | Ex-1A | Cl-8 | Composite mathematics Schand Solution | myhelper
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
⇒$\dfrac{8}{10}>\dfrac{-3}{5}>\dfrac{-7}{10}>\dfrac{-17}{15}$
Help full for mee because i am going to class 8
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