ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.2

Exercise 1.2

Question 1

Sol :
(i) $\quad 2 \frac{2}{3}=\frac{2 \times 3+2}{3}=\frac{8}{3}$

=$2 \frac{2}{3}-\left(-\frac{3}{7}\right)$

=$\frac{8}{3}-\left(-\frac{3}{7}\right)$

=$\frac{8}{3}+\frac{3}{7}$

=$\frac{(8 \times 7)+(3 \times 3)}{21} \quad$  LCM of $3,7=21$

=$\frac{56+9}{21}$

=$\frac{65}{21}$

$\frac{-4}{9}-\left(3 \frac{5}{8}\right)$

$\frac{-4}{9}-\left(\frac{29}{8}\right)$

$-\frac{4}{9}-\frac{29}{8}$

$\frac{-4 \times 8-(29 \times 9)}{72} \quad$ LCM of $9,18=72$

$-\frac{293}{72}$


(ii) $-3 \frac{1}{5}-\left(-4 \frac{7}{9}\right)$

$-\frac{16}{5}-\left(\frac{-43}{9}\right)$

$\frac{-16}{5}+\frac{43}{9}$

$\frac{(-16 \times 9)+(43 \times 5)}{45}$ LCM OF 9,15=45

$\frac{-144+215}{45}$

$\frac{71}{45}$


Question 3

Sol :
Let the Unknown number as x.

 $\begin{aligned}-\frac{5}{11}+x &=\frac{-7}{8} \\ x &=\frac{-7}{8}-\left(-\frac{5}{11}\right) \\ &=\frac{-7}{8}+\frac{5}{11} \end{aligned}$    

$=\frac{(-7 \times 11)+(8 \times 5)}{88} \quad$ LCM Of $11,8=88$

$=\frac{-77+45}{88}$

$x=\frac{-32}{88}$

$x=\frac{-4}{11}$


(ii)   let the unknown number as x.  

$\begin{aligned}-\frac{2}{7}+x &=\frac{3}{5} \\ x &=\frac{3}{5}-\left(-\frac{2}{7}\right) \\ x &=\frac{3}{5}+\frac{2}{7} \\ &=\frac{(3 \times 7)+(2 \times 5)}{35} \\ &=\frac{21+10}{35} \\ x &=\frac{31}{35} \end{aligned}$


Question 4

Sol :
Let the Unknown number be x

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

$-\frac{23}{5}-x=-\frac{7}{2}$

$\begin{aligned}-\frac{23}{5} &=-\frac{7}{2}+x \\ x &=-\frac{23}{5}+\frac{7}{2} \\ &=\frac{(-23 \times 2)+(7 \times 5)}{10} \end{aligned}$      LCM OF 5,2=10 

⇒$\frac{-46+35}{10}$

x⇒$\frac{-11}{10}$


Question 5

Sol :

$\left[\frac{-5}{7}+\left(\frac{-8}{3}\right)\right]-\left[\frac{5}{2}+\left(\frac{-11}{12}\right)\right]$

$\left[\frac{(-5 \times 3)+(-8 \times 7)}{21}\right]-\left[\frac{5 \times 6+(-11 \times 1)}{12}\right]$

$\frac{-284-133}{84}-\frac{417}{84}=\frac{-139}{28}=-\frac{139}{28} \|$


Question 6

Sol :

$\quad x=\frac{4}{9} ; \quad y=\frac{-7}{12}$

Consider

x-y =  $\frac{4}{9}-\left(-\frac{7}{12}\right)$

$=\frac{4}{9}+\frac{7}{12}$

⇒$\frac{(4 \times 4)+(7 \times 3)}{36}$               (∴LCM OF 9,12= 36)

⇒$\frac{16+21}{36}$

⇒$x-y=\frac{37}{36}$

Consider     

y-x = $\frac{-7}{12}-\left(\frac{4}{9}\right)$

    

⇒$\frac{-7}{12}-\frac{4}{9}$

⇒$\frac{(-7 \times 3)-(4 \times 4)}{36}$         LCM of  9, 12=36

⇒$\frac{-21-16}{36}$

y-x ⇒$\frac{-37}{36}$

ஃx-y ≠ y-x


Question 7

Sol :

$x=\frac{4}{9} ;  \quad y=\frac{-7}{12} ; \quad z=\frac{-2}{3}$

Consider

$x-(y-z)=\frac{4}{9}-\left(\frac{-7}{12}-\left(\frac{-2}{3}\right)\right)$

$=\frac{4}{9}-\left(\frac{-7}{12}+\frac{2}{3}\right)$

⇒$\frac{4}{9}-\left(\frac{(-7 \times 1)+(2 \times 4)}{12}\right)$

⇒$\frac{4}{9}-\left(\frac{-7+8}{12}\right)$

⇒$\frac{4}{9}-\frac{1}{12}$

⇒$\frac{(4 \times 4)-(1 \times 3)}{36}$

=$\frac{16-3}{36}$

$x-(y-z)$  =$\frac{13}{36}$ 

       Consider       ( x-y)-z  =$\left[\frac{4}{9}-\left(\frac{-7}{12}\right)\right]-\left(\frac{-2}{3}\right)$

$=\left[\frac{4}{9}+\frac{7}{12}\right]+\frac{2}{3}$

$=\left[\frac{(4 \times 4)+(-7 \times 3)}{36}\right]+\frac{2}{3}$

$=\frac{16+21}{36}+\frac{2}{3}$

$=\frac{37}{36}+\frac{2}{3}$

$=\frac{(37 \times 1)+(2 \times 12)}{36}$

⇒$\frac{37+24}{36}$

$(x-y)-z=\frac{61}{36}$

∴ $x-(y-z) \neq(x-y)-z$


Question 8

Sol :

(i) ⇒$\frac{2}{3}-\frac{4}{5}$

$\frac{(2 \times 5)-(4 \times 3)}{15}$     LCM = 3,5 = 15

⇒$\frac{10-12}{15}$

⇒$\frac{-2}{5}$   It is a Rational Number

So. given statement is False


(ii)true

$\frac{-5}{7}+\frac{5}{7}=0$                              

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