ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.3

 Exercise 1.3

Question 1

Sol :

(i)⇒$\frac{6}{-7} \times \frac{14}{30}$

=$\frac{6 \times 14}{-7 \times 30}$

=$-\frac{84}{210}$

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


(ii)  $6 \frac{2}{3} \times 1 \frac{2}{7}$

⇒$\frac{20}{3} \times \frac{9}{7}$

⇒$\frac{20 \times 9}{3 \times 7}$

=$\frac{180}{21}$

=$\frac{60}{7}$


(iii)  $\frac{25}{-9} \times \frac{-3}{10}$

=$\frac{25 \times(-3)}{-9 \times 10}$

=$\frac{-75}{-90}$

=$\frac{-5}{-6}$

=$\frac{5}{6}$


Question 2

Sol :

(i) To verify commutative property of multiplication, we have 

to show    $\frac{4}{5} \times \frac{-7}{8}=\frac{-7}{8} \times \frac{4}{5}$

 L.H.S ⇒ $\frac{4 \times-7}{5 \times 8}$

⇒ $\frac{-28}{40}$

L.H.S ⇒ $\frac{-7}{10}$

R.H.S⇒ $\frac{-7}{8} \times \frac{4}{5}$

⇒$\frac{-7 \times 4}{8 \times 5}$

⇒$\frac{-28}{40}$

R.H.S⇒ $\frac{-7}{10}$

∴ L.H.S =R.H.S

∴ Commutative property of multiplication is verified

Question 3

Sol :

$\quad \frac{3}{5} \times\left(\frac{-4}{7} \times \frac{-8}{9}\right)=\left(\frac{3}{5} \times \frac{-4}{7}\right) \times \frac{-8}{9}$

L.H.S $=\frac{3}{5} \times\left(\frac{-4}{7} \times \frac{-8}{9}\right)$

$=\frac{3}{5} \times\left(\frac{-4 \times(-8)}{7 \times 9}\right)$

$=\frac{3}{5} \times\left(\frac{32}{63}\right)$

$=\frac{3 \times 32}{5 \times 63}$

$=\frac{96}{315}$

L.H.S⇒ $\frac{32}{165}$

$\begin{aligned} \text { R.H.S } &=\left(\frac{3}{5} \times \frac{-4}{7}\right) \times\left(\frac{-8}{9}\right) \\ &=\left(\frac{-3 \times(-4)}{5 \times 7}\right) \times\left(\frac{-8}{9}\right) \\ &=\frac{-12}{35} \times \frac{-8}{9} \\ &=\frac{-12 \times(-8)}{35 \times 9} \\ &=\frac{96}{315} \\  R.H.S &=\frac{32}{105} \end{aligned}$ 

L.H.S=R.H.S, HENCE PROVED

∴This law is called associative property of multiplication

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


L.H.S $=\frac{5}{9} \times\left(\frac{-3}{2}+\frac{7}{5}\right)$

$=\frac{5}{9} \times\left(\frac{(-3 \times 5)+(7 \times 2)}{10}\right) \quad$ LCM of $2,5=10$

$=\frac{5}{9} \times\left(\frac{-15+14}{10}\right)$

$=\frac{5}{9} \times\left(\frac{-1}{10}\right)$

L.H.S= $\frac{-1}{18}$

R.H.S= $\left(\frac{5}{9} \times \frac{-3}{2}\right)+\left(\frac{5}{9} \times \frac{7}{5}\right)$

$\left(\frac{5 \times(-3)}{18}\right]+\left[\frac{5 \times 7}{9 \times 5}\right]$

$=\left[\frac{5 \times(-3)}{18}\right]+\left[\frac{5 \times 7}{9 \times 5}\right]$

$=\frac{-15}{18}+\frac{35}{45}$

$=\frac{(-15 \times 5)+(35 \times 2)}{90}$

$=\frac{-75+70}{90}$

$=\frac{-5}{90}$

R.H.S $=\frac{-1}{18}$

L.H.S = R.H,S, Hence proved

This law is called distributive law of multiplication over addition

Question 4

Sol :

(i)  12 reciprocal of 12 =$\frac{1}{12}$

∴ $\frac{1}{12}$ is multiplicative  inverse of 12

$\begin{array}{ll}\text { ii\} } & \text { reciprocal of } \frac{2}{3}=\frac{3}{2} \\ \therefore & \frac{3}{2} \text { is multipl:cative inverse of } \frac{2}{3}\end{array}$


(iii)  reciprocal of $\frac{-4}{7}=\frac{7}{-4}$ or  $-\frac{7}{4}$

$\therefore \frac{-7}{4}$ is multiplicative inverse of $\frac{-4}{7}$


(iv) $\frac{-3}{8} \times\left(\frac{-7}{13}\right)=\frac{-3 \times(-7)}{8 \times 13}=\frac{21}{104}$

reciprocal of $\frac{21}{104}$ in $\frac{104}{21}$

∴ $\frac{104}{21}$ in multiplicative inverse of $\frac{-3}{8} \times\left(\frac{-7}{13}\right)$


Question 5

Sol :

(i) $\frac{2}{5} \times \frac{-3}{7}-\frac{1}{14}=\frac{3}{7} \times \frac{3}{5}$

=$\frac{2 \times(-3)}{5 \times 7}-\frac{1}{14}-\frac{(3 \times 3)}{7 \times 5}$

=$\frac{-6}{35}-\frac{1}{14}-\frac{9}{35}$

⇒$\frac{(-6 \times 2)-(1 \times 5)-(9 \times 2)}{70} \quad$ LCM OF $35,14,35=70$

=$\frac{-12-5-18}{70}$

=$\frac{-35}{70}$

=$\frac{-1}{2}$


(ii) $\frac{8}{9} \times \frac{4}{5}+\frac{5}{6}-\frac{9}{5} \times \frac{8}{9}$

=$\frac{(8 \times 4)}{9 \times 5}+\frac{5}{6}-\frac{(9 \times 8)}{(5 \times 9)}$

=$\frac{32}{45}+\frac{5}{6}-\frac{72}{45}$

=$\frac{(32 \times 2)+(5 \times 15)-(72 \times 2)}{90}$    L.C.M of 45,6,45= 90

=$\frac{64+75-144}{90}$

⇒$\frac{64+75-144}{90}$

=$\frac{-5}{90}$

⇒ $\frac{-1}{18}$


(ii) $\frac{-3}{7} \times \frac{14}{15} \times \frac{7}{12} \times\left(\frac{-30}{35}\right)$

=$\frac{-3 \times 14 \times 7}{7 \times 15 \times 12} \times\left(\frac{-30}{35}\right)$

=$\frac{-294}{1260} \times\left(\frac{-30}{35}\right)$

=$\frac{-294 \times(-30)}{1260 \times 35}$

$\frac{1}{5}$ 


Question 6

Sol :

$P=\frac{-8}{27}, q=\frac{3}{4}, \quad \ r=\frac{-12}{15}$

(i) $P{}(q \times \ r)=(p \times q) \times \ r$

L.H.S⇒P x (q x r) 

$=\frac{-8}{27} \times\left(\frac{3}{4} \times\left(\frac{-12}{15}\right)\right)$

$=\frac{-8}{27} \times\left(\frac{3 \times(-12)}{4 \times 15}\right)$

$=\frac{-8}{27} \times\left(\frac{-36}{60}\right)$

$=\frac{-8 \times(-36)}{27 \times 60}$

$\begin{aligned} \ L \cdot H \cdot S &=\frac{8}{45} \\ R \cdot H \cdot S &=(P \times q) \times \ r \\ &=\left(\frac{-8}{27} \times \frac{3}{4}\right) \times\left(\frac{-12}{15}\right) \\ &=\left(\frac{-8 \times 3}{27 \times 4}\right) \times\left(\frac{-12}{15}\right) \\ &=\frac{-24}{108} \times\left(\frac{-12}{15}\right) \\ &=\frac{-24 \times-12}{108 \times 15} \end{aligned}$ 

R.H.S $=\frac{8}{45}$

∴ L.H.S = R.H.S, Hence verified


(ii)

 $\begin{aligned} P \times(q-\ r) &=P \times q-P \times r . \\ L . H \cdot S &=P \times(q-\ r) \\ &=\frac{-8}{27} \times\left(\frac{3}{4}-\left(\frac{-12}{15}\right)\right) \\ &=\frac{-8}{27} \times\left(\frac{3}{4}+\frac{12}{15}\right) \\=& \frac{-8}{27} \times\left(\frac{(3 \times 15)+(12 \times 4)}{60}\right) \\=& \frac{-8}{27} \times\left(\frac{45+48}{60}\right) \\=& \frac{-8}{27} \times \frac{93}{60} \\=& \frac{-8}{27} \times \frac{31}{20} \end{aligned}$ 

L.H.S $=\frac{-62}{135}$

   R.H.S $=-p \times q-P \times \ r$

⇒$\frac{-8}{27} \times \frac{3}{4}-\left(\frac{-8}{27} \times\left(\frac{-12}{15}\right)\right.$

⇒$\frac{-8 \times 3}{27 \times 4}-\left(\frac{-8 \times(-12)}{27 \times 15}\right)$

⇒$\frac{-24}{108}-\frac{96}{405}$

⇒$-\frac{2}{9}-\frac{32}{135}$

⇒$\frac{(-2 \times 15)-(32 \times 1)}{135}$

$=\frac{-30-32}{135}$

$R \cdot H \cdot S=\frac{-62}{135}$ 

 $\therefore  {LHS}=$ RHS; Hence veritied.


Question 7

Sol :

(i) $\frac{2}{3} \times \frac{-4}{5}$ is a Rational number

(ii) $\frac{54}{81} \times \frac{-63}{108}=\frac{-63}{108} \times \frac{54}{81}$

(iii) $\frac{4}{5} \times 1=\frac{4}{5}=1 \times \frac{4}{5}$

(iv) $\frac{5}{-12} \times \frac{-12}{5}=1=\frac{-12}{5} \times \frac{5}{-12}$

(v) $\frac{3}{7} \times\left(\frac{-2}{8} \times \frac{5}{9}\right)=\left(\frac{3}{7} \times \frac{-2}{8}\right) \times \frac{5}{9}$

(vi) $\frac{-8}{9} \times\left[\frac{4}{13}+\frac{5}{17}\right]=\left(\frac{-8}{9} \times \frac{4}{13}\right)+\left(\frac{-8}{9} \times \frac{5}{17}\right)$

(vii) $\quad \frac{-6}{13} \times\left[\frac{8}{9}-\frac{4}{7}\right]=\frac{-6}{13} \times \frac{8}{9}-\left(\frac{-6}{13} \times \frac{4}{7}\right)$

(viii) $\quad \frac{16}{25} \times 0=0$

(ix) Not defined

$\begin{array}{ll}\text { x) } & 1,-1 \\ \text { xi }\rangle & x^{2} \\ \text { xii) } & 1 \\ \text { xiii)} & \text { negative }\end{array}$


Question 8

Sol :

NO,   

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

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

=$-1 \neq 1$

             $\therefore-1 \frac{1}{4}$ is not multiplicative inverse of $\frac{4}{5}$

              $\therefore \quad$ multiplicative inverse of $\frac{4}{5}$ should be $\frac{5}{4}$

 

Question 9

Sol :

(i)  $\left\{\frac{7}{5} \times\left(\frac{-3}{12}\right)\right\}+\left\{\frac{7}{5}+\frac{5}{12}\right\}$

=$\frac{7}{5} \times\left\{\frac{-3}{12}+\frac{5}{12}\right\}$ (-i distributive property)

=$\frac{7}{5} \times\left\{\frac{-3+5}{12}\right\}$

=$\frac{7}{5} \times \frac{2}{12}$

=$\frac{7}{30}$


(ii) $\left\{\frac{9}{16} \times \frac{4}{12}\right\}+\left\{\frac{9}{16} \times\left(\frac{-3}{9}\right)\right\}$

$\frac{9}{16} \times\left\{\frac{4}{12}+\left(\frac{-3}{9}\right)\right\} \quad(\because$ distributive property $)$

$\frac{9}{16} \times\left\{\frac{1}{3}+\left(\frac{-1}{3}\right)\right\}$

$\frac{9}{16}\left\{\frac{1}{3}-\frac{1}{3}\right\}$

$\frac{9}{16} \times 0=0$


Question 10

Sol :

Additive inverse of 9=-9

Multiplicative inverse of $9=\frac{1}{9}$

$\begin{aligned} \text { Required Sum } &=-9+\frac{1}{9} \\ \text =-\frac{81+1}{9} \\ &=-\frac{80}{9} \\ {}  &=-8 \frac{8}{9} \end{aligned}$


Question 11

Sol :

Additive inverse of $\frac{-2}{7}=\frac{2}{7}$ 

Multiplicative inverse ot $\frac{-2}{7}=\frac{-7}{2}$

Required product $=\frac{2}{7} \times\frac{-7}{2}$

=-1

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