ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.1

 Exercise 1.1

Question 1

Sol :

(i) $\frac{4}{7}+\frac{5}{7}$

⇒$\frac{4+5}{7}$

⇒$\frac{9}{7}$


(ii) $\frac{7}{-13}$ and $\frac{4}{-13}$

⇒$\frac{7}{-13}+\frac{4}{-13}$

⇒$\frac{7\times (-1)}{-13 \times (-1)}+\frac{4\times (-1)}{-13\times (-1)}$ Make denominator as +ve number

⇒$\frac{-7}{13}+\frac{-4}{13}$

⇒$\frac{-7+(-4)}{13}=\frac{-11}{13}$


Question 2

Sol :

To verify $\frac{5}{11}+4 \frac{3}{9}$

$\frac{5}{11}+\frac{39}{9}$

$\frac{5 \times 9+39 \times 11}{99}$ LCM of 11, 9 = 99

⇒$\frac{45+429}{99}$

⇒$\frac{474}{99}=\frac{158}{33}$


(ii) $\frac{-4}{9}+2 \frac{12}{13}$

⇒$\frac{-4 \times 13+38 \times 9}{117}$ L.C.M of 9 , 13=117

⇒$\frac{-52+342}{17}$

⇒$\frac{290}{117}$


Question 3

Sol :

To verify the commutative property of addition, we have to 

Show

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

L.H.S⇒$\frac{-4}{3}+\frac{3}{7}$

⇒$\frac{-4 \times 7+3 \times 3}{21}$  LCM OF 3,7 = 21

⇒$\frac{-28+9}{21}$

⇒LCM =  $\frac{-19}{21}$

⇒RHS =   $\frac{3}{7}+\left(\frac{-4}{3}\right)$

⇒$\frac{3 \times 3+(-4) \times 7}{21} \quad \mathrm{LCM}$ of $3,7=21$

⇒RHS =  $\frac{9-28}{21}$

 ∴ LHS = RHS

⇒$\frac{-4}{3}+\frac{3}{7} a=\frac{3}{7}+\left(\frac{-4}{3}\right)$


(ii) To verify commutative law of addition, we have  

to show  $\left(\frac{-2}{-5}\right)+\frac{1}{3}=\frac{1}{3}+\left(\frac{-2}{-5}\right)$  

LHS ⇒$\frac{-2}{-5}+\frac{1}{3}$

⇒$\frac{-2 x(-1)}{-5 x(-1)}+\frac{1}{3} \quad$ Make denominator +ve number

⇒$\frac{2}{5}+\frac{1}{3}$

⇒$\frac{2 \times 3+1 \times 5}{15} \quad$ Lcm OF $5,3=15$

⇒$\frac{6+5}{15}$

⇒LCM $=\frac{11}{15}$


RHS ⇒  

 ⇒ $=\frac{1}{3}+\frac{2}{5}$

$=\frac{1 \times 5+2 \times 3}{15} \cdot$ LCM of $3,5=15$

$=\frac{5+6}{15}$

R.H.S $=\frac{11}{15}$

LHS = RHS 

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

∴Commutative law of addition i verified 


(iii) $\frac{9}{11}$ and $\frac{2}{13}$

To verity the commutative show of addition, we have to

show $\quad \frac{9}{11}+\frac{2}{13}=\frac{2}{13}+\frac{9}{11}$

⇒L.H.S $=\frac{9}{11}+\frac{2}{13}$

⇒$=\frac{(9 \times 13)+(2 \times 11)}{143} \quad$ LCM OF  $11,13=143$

⇒$=\frac{117+22}{143}$

⇒LCM  $=\frac{139}{143}$


⇒R.H.S $=\frac{2}{13}+\frac{9}{11}$

$=\frac{(2 \times 10)+(9 \times 13)}{143} \mathrm{LCM}$ of $13,11=143$

⇒$=\frac{22+117}{143}$

RHS  $=\frac{139}{143}$

⇒L.H.S $=$ R.H.S

$\frac{9}{11}+\frac{2}{13}=\frac{2}{13}+\frac{9}{11}$

∴ Commutative law of addition is verified


 Question 4

Sol :
(i) The additive inverse of $\frac{2}{-3}=-\left(\frac{2}{-3}\right)$
$=-\left(\frac{2 x-1}{-3 x-1}\right)$
$=-\left(\frac{-2}{3}\right)$
=$\frac{2}{3}$

(ii) The additive inverse of $\frac{-7}{-12}=-\left(\frac{-7}{-12}\right)$
$=-\left(\frac{-7 x(-1)}{-12 x(-1)}\right)$
$=\left(\frac{-7}{12}\right)$
$=\frac{-7}{12}$

Question 5

Sol :
(i) $x=\frac{10}{13}$
$-x=-\frac{10}{13}$
$-(-x)=-\left(-\frac{16}{3}\right)$
$-(-x)=\frac{-1 x(-10)}{3}$
-(-x)=x

(ii) $x=-\frac{5}{17}$
$-x=-1 \times \frac{-5}{17}$
$=\frac{(-1) \times(-5)}{17}$
$-x=\frac{5}{17}$
$-(-x)=-\left(\frac{5}{17}\right)$
$-(-x)=\frac{-1 \times 5}{17}$
$-(-x)=\frac{-5}{17}$
-(-x)=x

Question 6 

Sol :

(i) $\frac{4}{5}+\frac{11}{7}+\left(\frac{-7}{5}\right)+\left(\frac{-2}{7}\right)$

⇒$\left[\frac{4}{5}+\left(\frac{-7}{5}\right)\right]+\left[\frac{11}{7}+\left(\frac{-2}{7}\right)\right]$

(Using Commutative and associativity of addition)

⇒$\left[\frac{4-7}{5}\right]+\left[\frac{11-2}{7}\right]$

⇒$\left[\frac{-3}{5}\right]+\left[\frac{9}{7}\right]$

⇒$\frac{(-3 \times 7)+(9 \times 5)}{35} \quad$ LCM of $5,7=35$

⇒$\frac{-21+45}{35}=\frac{24}{35}$


(ii) $\frac{3}{7}+\frac{4}{9}+\left(\frac{-5}{21}\right)+\frac{2}{3}$ 

⇒$\left[\frac{3}{7}+\left(\frac{-5}{21}\right)\right]+\left[\frac{4}{9}+\frac{2}{3}\right]$

⇒(By using the commutative and associativity of addition)

⇒$\left[\frac{(3 \times 3)+(-5 \times 1)}{21}\right]+\left[\frac{4 \times 1+2 \times 3}{9}\right] \quad \begin{array}{l}\text { LCM OF } 7,21=21 \\ \text { LCM OF } 9,3=9\end{array}$

⇒$\left[\frac{9-5}{21}\right]+\left[\frac{4+6}{9}\right]$

⇒$\frac{4}{21}+\frac{16}{9}$

⇒$\frac{(4 \times 3)+(10 \times 7)}{63} \quad$ LCM of $21,9=63$

⇒$\frac{12+70}{63} .$

⇒$\frac{82}{63}=$  $\frac{19}{63}$


Question 7

Sol :

(i)  $\left(\frac{-4}{9}\right)+\frac{2}{3}$ in a Rational number

(ii) $\frac{43}{89}+\left(\frac{-51}{47}\right)=\left(\frac{-51}{47}\right)+\frac{43}{89}$

(iii) $\frac{2}{7}+\frac{0}{7}=\frac{2}{7}-0+\frac{2}{7}$

(iv) $\frac{4}{11}+\left[\left(\frac{-7}{12}\right)+\frac{9}{10}\right]=\left[\frac{4}{11}+\left(\frac{-7}{12}\right)\right]+\frac{9}{10}$

(v)  $\frac{5}{9}+\left(\frac{-5}{9}\right)=0=\left(\frac{-5}{9}\right)+\frac{5}{9}$


Question 8

Sol :
Sol :
GIVEN
$a=\frac{-11}{27}$
$b=\frac{4}{9}$
$c=\frac{-5}{18}$

$\begin{aligned} \text { L.H.S }  a+(b+c) &=\frac{-11}{27}+\left(\frac{4}{9}+\left(\frac{-5}{18}\right)\right) \\ &=\frac{-11}{27}+\left[\frac{4 \times 2+(-5 \times 1)}{18}\right] \end{aligned}$ LCM OF $9,18=18$

$=\frac{-11}{27}+\left[\frac{8-5}{18}\right]$

$=\frac{-11}{27}+\frac{3}{18}$

$=\frac{(-11 \times 2)+(3 \times 3)}{54}$  LCM OF 27,18=54

$=\frac{-22+9}{54}$

LHS  $=\frac{-13}{54}$

R.H.S $\begin{aligned}(a+b)+c &=\left(\frac{-11}{27}+\frac{4}{9}\right)+\left(-\frac{5}{18}\right) \\ &=\frac{(-11 \times 1)+(4 \times 3)}{27}+\left(\frac{-5}{18}\right) \end{aligned}$ 
LCM OF 27, 9=27

$\begin{aligned} &=\frac{-22+9}{54} \\ \text {L.H.S } &=\frac{-13}{54} \end{aligned}$

$\begin{aligned} \text { R.H.S } \quad(a+b)+C &=\left(\frac{-11}{27}+\frac{4}{9}\right)+\left(\frac{-5}{18}\right) \\ &=\frac{(-11 \times 1)+(4 \times 3)}{27}+\left(\frac{-5}{18}\right) \end{aligned}$  

LCM OF +27,9=27

$=\frac{-11+12}{27}+\left(\frac{-5}{18}\right)$

$=\frac{1}{27}+\left(\frac{-5}{18}\right)$

$=\frac{(1 \times 2)+(-5 \times 3)}{54}$ LCM OF 27,18= 54

$\frac{2-15}{54}$

R. H.S $=\frac{-13}{54}$

∴LCH =RHS 

a+(b+c)=(a+b)+c

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