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

Exercise 1 B

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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

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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

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

⇒$\dfrac{-9\times 2 + 3\times 5}{20}$  

⇒$\dfrac{-18 + 15}{20}$

⇒$\dfrac{-3}{20}$

(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

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

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

⇒$\dfrac{-15-28}{36}$

⇒$\dfrac{-43}{36}$

(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

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

⇒$\dfrac{7\times 3-8\times 2}{30}$

⇒$\dfrac{21-16}{30}$

⇒$\dfrac{5}{30}=\dfrac{1}{6}$

(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{array}{c|c}2 & 6,12 \\\hline 2 & 3,6 \\\hline 3 & 3,3 \\\hline & 1,1\end{array} $

⇒$\dfrac{23\times 2+5\times 1}{12}$  

⇒$\dfrac{46+5}{12}$

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

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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{array}{c|c}2 & 4,7,14 \\ \hline 2 & 2,7,7 \\ \hline 7 & 1,7,7 \\ \hline &1,1,1\end{array} $


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

⇒$\dfrac{7-24+6}{28}$

⇒$\dfrac{13-24}{28}$

⇒$\dfrac{-11}{28}$

(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

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

⇒$\dfrac{7\times 10-9\times 6-3\times 15}{150}$

⇒$\dfrac{70-54-45}{150}$

⇒$\dfrac{70-99}{150}$

⇒$\dfrac{-29}{150}$

(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{array}{l|l}

2 & 3,9,6 \\

\hline 3 & 3,9,3 \\

\hline 3 & 1,3,1 \\

\hline&1,1,1

\end{array} $

⇒$\dfrac{4\times 6-2\times 2-5\times 3}{18}$ 

⇒$\dfrac{24-4-15}{18}$

⇒$\dfrac{24-19}{18}$

⇒$\dfrac{5}{18}$

(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{array}{l|l}

2 & 5,10,2 \\

\hline 5 & 5,5,1 \\

\hline&1,1,1

\end{array} $


⇒$\dfrac{21\times 2-53\times 1+3\times 5}{10}$

⇒$\dfrac{42-53+15}{10}$

⇒$\dfrac{57-53}{10}$

⇒$\dfrac{4}{10}=\dfrac{2}{5}$

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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{array}{l|l}

2 & 8,3 \\ \hline

2 & 4,3 \\ \hline

2 & 2,3 \\

\hline 3 & 1,3 \\

\hline & 1,1

\end{array} $

⇒$\dfrac{3\times 3-2\times 8}{24}=\dfrac{-2\times 8+3\times 3}{24}$

⇒$\dfrac{9-16}{24}=\dfrac{-16+9}{24}$

⇒$\dfrac{-7}{24}=\dfrac{-7}{24}$

(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

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

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

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

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

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

 

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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 negative)

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

[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]

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

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

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

 

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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 :

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

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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{array}{c|c}

2 & 6,15 \\ \hline

3 & 3,15 \\\hline

5 & 1,5 \\\hline

& 1,1

\end{array}$

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

$=\frac{25-34}{30}$

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

 

(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

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

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

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

 

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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) $\frac{-8}{9}$

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

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

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

 

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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)

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

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

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

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

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

 

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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

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

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

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

$=\frac{-31}{18}$

 

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

Sol:

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

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



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

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

$=\frac{3+16}{22}$

$=\frac{19}{22}$


(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

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

$=\frac{16+15}{20}$

$=\frac{31}{20}$

 

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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 $\frac{4}{15}$

 

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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

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

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

$=\frac{-27}{30}$

$or-\frac{9}{10}$

 

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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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