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

Exercise 1 B


Question 1

Answer True (T) or False (F)
(i) Subtraction is commutative for rational numbers. F

(ii) To subtract \dfrac{c}{d}\text{ from } \dfrac{a}{b} , we add the additive inverse of \dfrac{a}{b}\text{ to } \dfrac{c}{d} . F

(iii) 0 is its own additive inverse . T

(iv) The additive inverse of \dfrac{-21}{-30}\text{ is }\dfrac{-21}{\phantom{-}30} . T

(v) While subtracting three or more rational numbers , they can be grouped in any order . F


Question 2

Add the following rational numbers
(i) \dfrac{4}{9}\text{ and }\dfrac{2}{9}

Sol :

\dfrac{4}{9}+\dfrac{2}{9}

\dfrac{6}{9}=\dfrac{2}{3}


(ii) \dfrac{14}{23}\text{ and }\dfrac{9}{23}

Sol :

\dfrac{-14}{23}+\dfrac{9}{23}

\dfrac{-14+9}{23}=\dfrac{-5}{23}


(iii) \dfrac{-9}{10}\text{ and }\dfrac{3}{4}

Sol :

L.C.M of 4 and 10 is = 2×2×5=20

\begin{tabular}{c|c}  2 & 4,10 \\  \hline 2 & 2,5 \\  \hline 5 & 1,5 \\  \hline & 1,1  \end{tabular}

\dfrac{-18 + 15}{20}\dfrac{-9\times 2 + 3\times 5}{20}

\dfrac{-3}{20}


(iv) \dfrac{-5}{12}\text{ and }\dfrac{-7}{9}

Sol :

\dfrac{-5}{12}+\dfrac{(-7)}{9}

L.C.M of 12 and 9 is = 2×2×3×3=36

\begin{tabular}{l|l} 2 & 12,9 \\ \hline 2 & 6,9 \\ \hline 3 & 3,9 \\ \hline 3 & 1,3 \\ \hline & 1,1 \end{tabular}

\dfrac{-5\times 3+(-7)\times 4}{36}

\dfrac{-15-28}{36}

\dfrac{-43}{36}


(v) \dfrac{7}{10}\text{ and }\dfrac{-8}{15}

Sol :

\dfrac{7}{10}+\left(\dfrac{-8}{15}\right)

L.C.M of 10 and 15 is = 2×3×5=30

\begin{tabular}{c|c}  2 & 10,15 \\  \hline 3 & 5,15 \\  \hline 5 & 5,5 \\  \hline &1,1  \end{tabular}

\dfrac{21-16}{30}\dfrac{7\times 3-8\times 2}{30}

\dfrac{5}{30}=\dfrac{1}{6}


(vi) -7\dfrac{3}{11}\text{ and }(-8)

Sol :

-\dfrac{7\times 11+3}{11}+\dfrac{-8}{1}

-\dfrac{80}{11}-\dfrac{8}{1}

\dfrac{-80-8\times 11}{11}

\dfrac{-80-88}{11}

\dfrac{-168}{11}=-15\dfrac{3}{11}


(vii) 4\dfrac{1}{3}\text{ and }2\dfrac{5}{7}

Sol :

\dfrac{13}{3}+\dfrac{19}{7}

L.C.M of 3 and 7 is 21

\dfrac{13\times 7+19\times 3}{21}

\dfrac{91+57}{21}

\dfrac{148}{21}=7\dfrac{1}{21}


(viii) 3\dfrac{5}{6}\text{ and }\dfrac{-5}{-12}

Sol :

\dfrac{23}{6}+\dfrac{5}{12}

L.C.M of 6 and 12 is =2×2×3=12

\begin{tabular}{c|c}  2 & 6,12 \\  \hline 2 & 3,6 \\  \hline 3 & 3,3 \\  \hline & 1,1  \end{tabular}

\dfrac{46+5}{12}\dfrac{23\times 2+5\times 1}{12}

\dfrac{51}{12}=\dfrac{17}{4}=4\dfrac{1}{4}


Question 3

Simplify:
(i) \dfrac{1}{4}+\dfrac{-6}{7}+\dfrac{3}{14}

Sol :

\dfrac{1}{4}-\dfrac{6}{7}+\dfrac{3}{14}

L.C.M of 4 , 7 and 14 is =2×2×3 =28

 

\begin{tabular}{c|c}  2 & 4,7,14 \\ \hline  2 & 2,7,7 \\ \hline  7 & 1,7,7 \\ \hline  &1,1,1  \end{tabular}

\dfrac{7-24+6}{28}\dfrac{7\times 1-6\times 4+3\times 2}{28}

\dfrac{13-24}{28}

\dfrac{-11}{28}


(ii) \dfrac{7}{15}+\dfrac{-9}{25}+\dfrac{-3}{10}

Sol :

\dfrac{7}{15}-\dfrac{9}{25}+\dfrac{-3}{10}

L.C.M of 15 , 25 and 10 is =2×3×5×5 =150

 

\begin{tabular}{c|c}  2 & 15,25,10 \\ \hline  3 & 15,25,5 \\\hline  5 & 5,25,5 \\\hline  5 & 1,5,1 \\  \hline &1,1,1  \end{tabular}

\dfrac{70-54-45}{150}\dfrac{7\times 10-9\times 6-3\times 15}{150}

\dfrac{70-99}{150}

\dfrac{29}{150}


(iii) 1\dfrac{1}{3}+\dfrac{2}{-9}+\dfrac{-5}{6}

Sol :

Denominator can not be negative

\left(\dfrac{3\times 1+1}{3}\right)+\left(\dfrac{2}{-9}\times \dfrac{-1}{-1}\right)-\dfrac{5}{6}

\dfrac{4}{3}-\dfrac{2}{9}-\dfrac{5}{6}

L.C.M of 3 , 9 and 6 is =2×3×3 =18

\begin{tabular}{l|l}  2 & 3,9,6 \\  \hline 3 & 3,9,3 \\  \hline 3 & 1,3,1 \\  \hline&1,1,1  \end{tabular}

\dfrac{24-4-15}{18}\dfrac{4\times 6-2\times 2-5\times 3}{18}

\dfrac{24-19}{18}

\dfrac{5}{18}


(iv) 4\dfrac{1}{5}+\left(-5\dfrac{3}{10}\right)+1\dfrac{1}{2}

Sol :

\dfrac{21}{5}-\dfrac{53}{10}+\dfrac{3}{2}

L.C.M of 5 , 10 and 2 is =2×5 =10

\begin{tabular}{l|l}  2 & 5,10,2 \\  \hline 5 & 5,5,1 \\  \hline&1,1,1  \end{tabular}

\dfrac{42-53+15}{10}\dfrac{21\times 2-53\times 1+3\times 5}{10}

\dfrac{57-53}{10}

\dfrac{4}{10}=\dfrac{2}{5}


Question 4

Verify the following.
(i) \dfrac{3}{8}+\dfrac{-2}{3}=\dfrac{-2}{3}+\dfrac{3}{8}

Sol :

They are equal to  each other by commutative property

ALTERNATE METHOD

L.C.M of 8 and 3 is =2×2×2×3 =24

  \begin{tabular}{l|l}  2 & 8,3 \\  2 & 4,3 \\  2 & 2,3 \\  \hline 3 & 1,3 \\  \hline & 1,1  \end{tabular}

\dfrac{9-16}{24}=\dfrac{-16+9}{24}\dfrac{3\times 3-2\times 8}{24}=\dfrac{-2\times 8+3\times 3}{24}

\dfrac{-7}{24}=\dfrac{-7}{24}


(ii) \dfrac{-7}{11}+\dfrac{15}{-22}=\dfrac{15}{-22}+\dfrac{-7}{11}

Sol :

They are equal to  each other by commutative property

\frac{a}{b}+\frac{c}{d}=\frac{c}{d}+\frac{a}{b} \quad             (by commutative property)

                                                                              ALTERNATE METHOD

-\dfrac{7}{11}-\dfrac{15}{22}=-\dfrac{15}{22}-\dfrac{7}{11}

[denominator cannot be negative]

LCM of 11 and 22 is 22.

=\dfrac{-7 \times 2-15 \times 1}{22}

=\dfrac{-15 \times 1-7 \times 2}{22}

\dfrac{-14-15}{22}=\dfrac{-15-14}{22}

\dfrac{-29}{22}=-\dfrac{29}{22}

 


(iii) -8+\dfrac{-11}{-12}=\dfrac{-11}{-12}+(-8)

Sol :

They are equal to  each other by commutative property

ALTERNATE METHOD

-8\dfrac{-11}{-12}=\dfrac{-11}{-12}-8

-8+\frac{11}{12}=\frac{11}{12}-8

\frac{-8 \times 12+11 \times 1}{12}=\frac{11 \times 1-8 \times 12}{12}

-\frac{96+11}{12}=-\frac{96+11}{12}

\dfrac{-107}{12}=\dfrac{-107}{12}

 


Question 5

Verify that:
(i) \left(\dfrac{-5}{8}+\dfrac{9}{8}\right)+\dfrac{13}{8}=\dfrac{-5}{8}+\left(\dfrac{9}{8}+\dfrac{13}{8}\right)

Sol :

They are equal to  each other by associative property

ALTERNATE METHOD

\left(\dfrac{-5+9}{8}\right)+\dfrac{13}{8}=\dfrac{-5}{8}+\left(\dfrac{9+13}{8}\right)

\dfrac{4}{8}+\dfrac{13}{8}=\dfrac{-5}{8}+\dfrac{22}{8}

\frac{17}{8}=\frac{17}{8}

 


(ii) -20+\left(\dfrac{3}{-5}+\dfrac{-7}{-10}\right)=\left(-20+\dfrac{3}{-5}\right)+\dfrac{-7}{-10}

Sol :

They are equal to  each other by associative property

ALTERNATE METHOD

(denominator can not be begative)

-20+\left(-\frac{3}{5}+\frac{7}{10}\right)=\left(-20-\frac{3}{5}\right)+\frac{7}{10}

[L.C.M of 5 and 10 is 10]

-20+\left(\frac{-3 \times2+7 \times 1}{10}\right)=\left(\frac{-20 \times 5-3 \times 1}{5}\right)+\frac{7}{10}

-20+\left(\frac{-6+7}{10}\right)=\left(\frac{-100-3}{5}\right)+\frac{7}{10}

-20+\frac{1}{10}=-\frac{103}{5}+\frac{7}{10}

[L.C.M of 5 and 10 is 10]

\left(\frac{-20 \times 10+1 \times 1}{10}\right)=\left(\frac{-103 \times 2+7 \times 1}{10}\right)

\frac{-200+1}{10}=\frac{-208+7}{10}

-\frac{199}{10}=-\frac{199}{10}

 


Question 6

Find the additive inverse of each of the following.
(i) \dfrac{10}{11}

Sol :

-\left(\dfrac{10}{11}\right)


(ii) 0

Sol :

0=0


(iii) \dfrac{21}{8}

Sol :

-\left(\dfrac{21}{8}\right)


(iv) \dfrac{-11}{-8}

Sol :

-\left(\dfrac{\not{-}11}{\not{-}8}\right)

-\left(\dfrac{11}{8}\right)=\dfrac{-11}{8}


(v) \dfrac{-26}{13}

Sol :

-\left(\dfrac{-26}{13}\right)=\dfrac{26}{13}


(vi) \dfrac{18}{-39}

Sol :

-\left(\dfrac{18}{-39}\right)=\dfrac{18}{39}


Question 7

Arrange and simplify:
(i) \dfrac{1}{2}+\dfrac{-3}{5}+\dfrac{3}{2}

Sol :

\frac{1}{2}+\left(-\frac{3}{5}\right)+\frac{3}{2}

=\frac{1}{2}+\frac{3}{2}-\frac{3}{5}

=\frac{1+3}{2}-\frac{3}{5}

=\frac{4}{2}-\frac{3}{5}

[L.C.M of 2 and 5 is 10]

=\frac{4 \times 5-3 \times 2}{10}

=\frac{20-6}{10}

=\dfrac{14}{10} = \dfrac{7}{5}=1 \dfrac{2}{5}

 


(ii) \dfrac{28}{17}+\dfrac{35}{17}+\dfrac{-16}{17}+\dfrac{-23}{17}

Sol :

=\frac{28+35-16-23}{17}

=\frac{63-39}{17}

=\dfrac{24}{17}=1\dfrac{7}{17}

 


(iii) \dfrac{2}{3}+\dfrac{-3}{5}+\dfrac{1}{6}+\dfrac{-8}{15}

Sol :

\frac{2}{3}+\left(\frac{-3}{5}\right)+\frac{1}{6}+\left(\frac{-8}{15}\right)

=\frac{2}{3}+\frac{1}{6}-\frac{3}{5}-\frac{8}{15}

L.C.M of 3 and 6 is 6

L.C.M of 5 and 15 is 15

=\frac{2 \times 2+1 \times 1}{6}+\frac{-3 \times 3-8 \times 1}{15}

=\frac{4+1}{6}+\frac{-9-8}{15}

L.C.M of 15 and 6 is 30

=\frac{5}{6}-\frac{17}{15}

\begin{tabular}{c|c}  2 & 6,15 \\ \hline  3 & 3,15 \\\hline  5 & 1,15 \\\hline  & 1,1  \end{tabular}

=\frac{5 \times 5-17 \times 2}{30}

=\frac{25-34}{30}

=-\frac{9}{30} \text{ or}-\frac{3}{10}

 


(iv) \dfrac{3}{5}+\dfrac{5}{3}+\dfrac{-11}{5}+\dfrac{-2}{3}

Sol :

\frac{3}{5}+\frac{5}{3}+\frac{-11}{5}+\frac{-2}{3}

=\frac{3}{5}-\frac{11}{5}+\frac{5}{3}-\frac{2}{3}

=\frac{3-11}{5}+\frac{5-2}{3}

=\frac{-8}{5}+\frac{3}{3}

L.C.M of 5 and 3 is 15

=\frac{-8 \times 3+3 \times 5}{15}

=\frac{-24+15}{15}

=\frac{-9}{15} \text{ or}-\frac{3}{5}

 


(v) \dfrac{4}{3}+\dfrac{3}{5}+\dfrac{-2}{3}+\dfrac{-11}{5}

Sol :

\frac{4}{3}+\frac{3}{5}+\frac{-2}{3}+\frac{-11}{5}

=\frac{4}{3}-\frac{2}{3}+\frac{3}{5}-\frac{11}{5}

=\frac{4-2}{3}+\frac{3-11}{5}

=\frac{2}{3}+\frac{-8}{5}

L.C.M of 5 and 3 is 15

=\frac{2 \times 5-8 \times 3}{15}

=\frac{-14}{15}

 

(vi) \dfrac{4}{9}+\dfrac{5}{3}+\dfrac{-4}{5}+\dfrac{7}{9}+\dfrac{-2}{3}+\dfrac{9}{5}

Sol :

\frac{4}{9}+\frac{5}{3}+\frac{-4}{5}+\frac{7}{9}+\frac{-2}{3}+\frac{9}{5}

=\frac{4}{9}+\frac{7}{9}+ \frac{5}{3}+\frac{-2}{3}-\frac{4}{5}+\frac{9}{5}

=\frac{4+7}{9}+\frac{5-2}{3}+\frac{-4+9}{5}

=\frac{11}{9}+\frac{3}{3}+\frac{5}{5}

L.C.M of 9,3,5 is 45

=\frac{11 \times 5+3 \times 15+5 \times 9}{45}

=\frac{55+45+45}{45}

=\frac{145}{45} \text{ or} \frac{29}{9}=3 \dfrac{2}{9}

 


Question 8

Verify that -(-x)=x , when x= (i)\dfrac{7}{6}

(ii)\dfrac{-8}{9}

Sol :

(i) \frac{7}{6}

\Rightarrow-\left(-\frac{7}{6}\right)=\frac{7}{6}

\Rightarrow \frac{7}{6}=\frac{7}{6}

 

(ii) \frac{-8}{9}

\Rightarrow-\left[-\left(\frac{-8}{9}\right)\right]=-\frac{8}{9}

\Rightarrow \quad-\left[+\frac{8}{9}\right]=\frac{-8}{9}

\Rightarrow-\frac{8}{9}=-\frac{8}{9}

 


Question 9

Verify that -(x+y)=(-x)+(-y) , when (i)x=\dfrac{3}{4},y=\dfrac{6}{7} (ii)x=\dfrac{-3}{4},y=\dfrac{-6}{7}

Sol :

(i) x=\frac{3}{4}, y=\frac{8}{7}

-\left(\frac{3}{4}+\frac{6}{7}\right)=\left(-\frac{3}{4}\right)+\left(-\frac{6}{7}\right)

-\left(\frac{3 \times 7+6\times 4}{28}\right)=-\frac{3}{4}-\frac{6}{7}

-\left(\frac{21+24}{28}\right)=\left(\frac{-3 \times 7-6 \times 4}{28}\right)

\frac{-45}{28}=\frac{-21-24}{28}

-\frac{45}{28}=-\frac{45}{28}

 

(ii) x=-\frac{3}{4} \quad, y=-\frac{6}{7}

-(x+y)=(-x)+(-y)

-\left[-\frac{3}{4}+-\frac{6}{7}\right]=\left[-\left(-\frac{3}{4}\right)\right]+\left[-\left(-\frac{6}{7}\right)\right]

-\left[\frac{-3 \times 7 -6 \times 4}{28}\right]=\frac{3}{4}+\frac{6}{7}

-\left[\frac{-21+24}{28}\right]=\frac{3 \times 7+6 \times 4}{28}

-\left[\dfrac{-45}{28}\right]=\dfrac{21+24}{28}

\frac{45}{28}=\frac{45}{28}

 


Question 10

Subtract:
(i) \dfrac{2}{9}\text{ from }\dfrac{7}{9}

Sol :

=\frac{7}{9}-\frac{2}{9}

=\frac{7-2}{9}

=\frac{5}{9}

(ii) \dfrac{8}{9}\text{ from }\dfrac{-5}{6}

Sol :

=-\frac{5}{6}-\frac{8}{9}

LCM of 6,9 is 18

\begin{tabular}{c|c}
2 & 12,9 \\
\hline 2 & 6,9 \\
\hline 3 & 3,9 \\
\hline 3 & 1,3 \\
\hline & 1,1
\end{tabular}

=\frac{-5 \times 3-8 \times 2}{18}

=\frac{-15-16}{18}

=\frac{-31}{18}

 

(iii) \dfrac{-8}{11}\text{ from }\dfrac{3}{22}

Sol:

=\frac{3}{22}-\frac{-8}{11}

=\frac{3}{22}+\frac{8}{11}

\begin{tabular}{c|c}  2 & 22,11 \\ \hline 11 & 11,11  \\  \hline & 1,1  \end{tabular}

\frac{3 \times 1 +8\times 2}{22}

=\frac{3+16}{22}

=\frac{19}{22}

(iv) \dfrac{3}{-4}\text{ from }\dfrac{4}{5}

Sol :

=\frac{4}{5}-\left(\frac{3}{-4}\times\frac{-1}{-1}\right)

=\frac{4}{5}-\frac{-3}{4}

=\frac{4}{5}+\frac{3}{4}

LCM of 4,5 is 20

=\frac{4 \times{4}+3 \times 5}{20}

=\frac{16+15}{20}

=\frac{31}{20}

 


Question 11

The sum of two rational numbers is \dfrac{-19}{60} . If one of the numbers is \dfrac{-7}{12} , find the other .

Sol :

Sum of two rational numbers is \frac{-19}{60}

One of the number is -\frac{7}{12} and Let the other be x

\frac{-7}{12}+x=-\frac{19}{60}

\left.x=-\frac{19}{60}+\frac{7}{12} \text { (on transposing }\right)

LCM of 60 and 12 is 60

=\frac{-19 \times 1+7 \times 5}{60}

=\frac{-19+35}{60}=\frac{16}{60} or \frac{4}{15}

 


Question 12

What number should be subtracted from \dfrac{-14}{15} to get \dfrac{-1}{30} ?

Sol :

let x be subtstaded from -\frac{14}{15} to get -\frac{1}{30}

-\frac{14}{15}-x=\frac{-1}{30}

-\frac{14}{15}+\frac{1}{30}=x

or

x=\frac{-14}{15}+\frac{1}{30}

LCM of 15 and 30 is 30

x=\frac{-14 \times 2+1 \times 1}{30}

x=\frac{-28+1}{30}

=\frac{-27}{30}

or-\frac{9}{10}

 


Question 13

What should be subtracted from \left(\dfrac{3}{4}+\dfrac{1}{3}+\dfrac{2}{5}\right) to get \dfrac{1}{2} ?

Sol :

\left[\frac{3}{4}+\frac{1}{3}+\frac{2}{5}\right]-x=\frac{1}{2}

\left[\frac{3}{4}+\frac{1}{3}+\frac{2}{5}\right]-\frac{1}{2}=x

x=\frac{3}{4}+\frac{1}{3}+\frac{2}{5}-\frac{1}{2}

LCM of 4,3,5,2 is 60

x=\frac{3 \times 15+1 \times 20+2 \times 12-1 \times 30}{60}

x=\frac{45+20+24-30}{60}

=\frac{89-30}{60}

=\frac{59}{60}

 


 

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