ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.2

Exercise 1.2

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Q1 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 1

Subtract :

(i) $2\frac{3}{5}\text{ from } \frac{-3}{7}$

Sol :

$=\frac{-3}{7}-\left(\frac{13}{5}\right)$

$=\frac{-15-91}{35}=\frac{-106}{35}=-3\frac{1}{35}$

(ii) $\frac{-4}{9} \text{ from } 3\frac{5}{8}$

Sol :

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

$=\frac{29}{8}+\frac{4}{9}$

$=\frac{261+32}{72}$ (LCM of 8,9=72)

$=\frac{293}{72}=4\frac{5}{72}$

(iii) $-3\frac{1}{5} \text{ from } -4\frac{7}{9}$

Sol :

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

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

$=\frac{-215+144}{45}=\frac{-71}{45}$

$=-1\frac{26}{45}$

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Q2 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 2

Sum of two rational numbers is $\frac{3}{5}$. If one of them is , $\frac{-2}{7}$, find the other.

Sol :

Given

Sum of two rational numbers is $\frac{3}{5}$

One of the number is $-\frac{2}{7}$

Hence, the other number is calculated as follows :

Other number$=\frac{3}{5}-\left(\frac{-2}{7}\right)$

$=\frac{3}{5}+\frac{2}{7}$

Taking LCM , we get

$=\frac{21+10}{35}=\frac{31}{35}$

Therefore , the other number is $\frac{31}{35}$

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Q3 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 3

What rational number should be added to $\frac{-5}{11}$ to get $\frac{-7}{8}$ ?

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 $11,8=88$

$=\frac{-77+40}{88}$

$x=\frac{-37}{88}$

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Q4 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 4

What rational number should be subtracted from $-4\frac{3}{5}$ to get $-3\frac{1}{2}$

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⇒$\frac{-11}{10}$

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Q5 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 5

Subtract the sum of $\frac{-5}{7} \text { and }\frac{-8}{3}$ from the sum of $\frac{5}{2} \text{ and } \frac{-11}{12}$

Sol :

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

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

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

$=\left(\frac{30- 11}{12}\right)-\left(\frac{-15-56}{21}\right)$

$=\left(\frac{19}{12}\right)-\left(\frac{-71}{21}\right)=\frac{19}{12}+\frac{71}{21}$

$=\frac{19 \times 7+ 71 \times 4}{84}=\frac{133+284}{84}$

$=\frac{417}{84}=\frac{139}{28}=4\frac{27}{28}$

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Q6 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 6

If $x=\frac{-4}{7}$ and $y=\frac{2}{5}$ verify that

(i) x-y≠y-x

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

Sol :

(i)

$x=\frac{-4}{7}$ and $y=\frac{2}{5}$

Now, 

$x-y=\frac{-4}{7}-\left(\frac{2}{5}\right)$

Taking LCM

$=\frac{-4}{7}-\frac{-2}{5}=\frac{-20-14}{35}=\frac{-34}{35}$

And $y-x=\frac{2}{5}-\left(\frac{-4}{7}\right)$

$=\frac{2}{5}+\frac{4}{7}=\frac{14+20}{35}=\frac{34}{35}$

ஃx-y ≠ y-x

(ii)

$x=\frac{-4}{7}$ and $y=\frac{2}{5}$

Now,

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

LHS=-(x+y)

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

$=-\left(\frac{-4\times 5 + 2 \times 7 }{35}\right)$

$=- \times \frac{-20+14}{35}=- \times \frac{-6}{35}=\frac{6}{35}$

RHS=-x+(-y)

$=-\left(\frac{-4}{7}\right)+\left(- \times \frac{2}{5} \right)$

$=\frac{+4}{7}+\frac{-2}{5}=\frac{+4}{7}-\frac{2}{5}$

$=\frac{4\times 5-2\times 7}{35}=\frac{20-14}{35}$

$=\frac{6}{35}$

ஃ-(x+y)=-x+(-y)

Sol :

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

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

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Q7 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 7

If $x=\frac{4}{9} ;  \quad y=\frac{-7}{12} ; \quad z=\frac{-2}{3}$ , then verify that x – (y – z) ≠ (x – y)– z

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

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

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

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

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

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

⇒$\frac{37+24}{36}$

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

∴ $x-(y-z) \neq(x-y)-z$

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Q8 | Ex-1.2 | Class 8 | ML Aggarwal | Rational Numbers | myhelper

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Question 8

Which of the following statement is true / false ?

(i) $\frac{2}{3}-\frac{4}{5}$ is not a rational number.

Sol :

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

Taking LCM

$-\frac{10-12}{15}=\frac{-2}{15}$

Is a rational number

Hence, the given statement is false

(ii) $\frac{-5}{7}$ is the additive inverse of $\frac{5}{7}$

Sol :

The given statement is true.

(iii) 0 is the additive inverse of its own.

Sol :

The given statement is true.

(iv) Commutative property holds for subtraction of rational numbers.

Sol :

Let us take,

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

We know that,

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

$\frac{2}{4}\neq \frac{-2}{4}$

Therefore, the given statement is true.

(v) Associative property does not hold for subtraction of rational numbers.

Sol :

The given statement is true.

(vi) 0 is the identity element for subtraction of rational numbers.

Sol :

Let us take,

$\frac{7}{8}-0=\frac{7}{8}$

But $0-\frac{7}{8}=\frac{-7}{8}$

$\frac{7}{8} \neq \frac{-7}{8}$

Therefore, the given statement is false.