ML Aggarwal Class 8 Chapter 1 Rational Numbers Exercise 1.1 ~ LearnsCBSE

Exercise 1.1

Question 1

(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

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

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

(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

(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

(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

(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

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

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

Exercise 1.2

(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 $9,18=72$

$-\frac{293}{72}$

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

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

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

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

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

$\frac{71}{45}$

3. 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.

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

4. 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⇒$\frac{-11}{10}$

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

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

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

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

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

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

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

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. givew statement is False

(ii)true

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

Exercise 1.3

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

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

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

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

=$\frac{5}{6}$

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

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

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 assoiative 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 disreibutive law of multiplication over addition

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

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

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

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

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

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

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

⇒$\frac{64+75-144}{90}$

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

⇒ $\frac{-1}{18}$

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

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

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

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

$\frac{1}{5}$

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

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.

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

(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 detin-ed

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

8. 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 $\frac{4}{5}$

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

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 $)$

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

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

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

10. Additive inverse of $9=-9$

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

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

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

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

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

$=-1$

Exercise 1.4

Q1.

i) $-\frac{3}{7} \div 4$

=$-\frac{3}{7} \div \frac{4}{1}$

=$\frac{-3}{7} \times \frac{1}{4}$

=$\frac{-3 \times 1}{7 \times 4}$

=$\frac{-3}{28}$

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

=$\frac{37}{8} \div\left(\frac{-4}{9}\right)$

=$\frac{37}{8} \times\left(\frac{9}{-4}\right)$

=$\frac{37 \times 9}{8 \times(-4)}$

=$-\frac{333}{32}$

=$-10 \frac{13}{32}$

iii) $\frac{8}{9} \div \frac{-3}{5}$

=$\frac{-8}{9} \times \frac{5}{-3}$

=$\frac{-8 \times 5}{9 x-3}$

=$\frac{-40}{-27}$

=$\frac{40}{27}=1 \frac{13}{27}$

Q2.

i) true

ii)false

iii)false

iv)true

v)ture

vi)false

Q3. let unknown number be $x$

⇒$x \times 2 \frac{4}{9}=\frac{-11}{12}$

⇒$x \times \frac{22}{9}=\frac{-11}{12}$

$x=\frac{-11}{12}$ x $\frac{22}{9}$

⇒$\frac{-11}{12} \times \frac{9}{22}$

$x=\frac{-3}{8}$

Other number $=\frac{-3}{8}$

4. Let unknown number be x

$x \times\left(\frac{-7}{12}\right)=\frac{5}{14}$

$x=\frac{5}{14} \div\left(\frac{-7}{12}\right)$

$=\frac{5}{14} \times \frac{12}{-7}$

$=\frac{5 \times 12}{14 \times(-7)}$

$=-\frac{60}{98}$

$=\frac{-30}{49}$

5. Let unkonw number be x

$\frac{-3}{x}=\frac{-9}{13}$

$x=-\frac{3}{1} \div\left(\frac{-9}{13}\right)$

$=\frac{-3}{1} \times \frac{13}{-9}$

$x=\frac{13}{3}=4 \frac{1}{3}$

Sum of number = $\frac{-13}{8}+\frac{5}{12}$

$=\frac{(-13 \times 3)+(5 \times 2)}{24} \quad$ LCM of $812=24$

⇒$\frac{-39+10}{24}$

⇒$\frac{-29}{24}$

sum of numbers = $\frac{-29}{24}$

difference of numbers= $\frac{(-13 \times 3)-(5 \times 2)}{24}$

$\frac{-39-10}{24}$

Difference of numbers $=\frac{-49}{24}$

Requried product = sum of numbers ÷ difference of numbers

$\begin{aligned} &=\frac{-29}{24} \div\left(-\frac{49}{24}\right) \\=& \frac{-29}{24} \times \frac{24}{-49} \\ &=\frac{29}{49} \end{aligned}$

$\begin{aligned} 7 . \quad \text { Sum of two numbers } &=\frac{8}{3}+\frac{4}{7} \\ &=\frac{(8 \times 7)+(3 \times 4)}{21} \text { LCM Of } 3,7=21 \\ &=\frac{56+12}{21} \\ \text { Sum of two numbers } &=\frac{68}{21} \end{aligned}$

$\begin{aligned} \text { Product of given numbers } &=\frac{-3}{7} \times \frac{14}{9} \\ &=\frac{-2}{3} \end{aligned}$

$\begin{aligned} \text { Required product } &=\frac{\text { Sum of } \frac{8}{3} \text { and } \frac{4}{7}}{\text { product of }\frac{3}{1} \text { and } \frac{14}{9}} \\ &=\frac{68}{21} \div\left(\frac{-2}{3}\right) \\ &=\frac{68}{21} \times \frac{3}{-2} \\ &=\frac{-34}{7} \end{aligned}$

8. Givew $\quad P=\frac{-3}{2}, \quad q=\frac{4}{5}, \quad \ r=\frac{-7}{12}$

$(p \div q) \div r=p \div(q \div \ r)$

$\begin{aligned} \text { L.H.S } &=(p \div q) \div r \\ &=\left(\frac{-3}{2} \div \frac{4}{5}\right) \div\left(\frac{-7}{12}\right) \\ &=\left(\frac{-3}{2} \times \frac{5}{4}\right) \div\left(\frac{-7}{12}\right) \\ &=\left(\frac{-15}{8}\right) \div\left(-\frac{7}{12}\right) \\ &=\frac{-15}{8} \times \frac{12}{-7} \\ &=\frac{-15}{8} \times \frac{12}{-7} \end{aligned}$

$\begin{aligned} \text { L.H.S } &=+\frac{45}{7} \\ \text { R.H.S } &=p \div(q \div \ r) \\ &=\frac{-3}{2} \div\left(\frac{4}{5} \div\left(\frac{-7}{12}\right)\right) \\ &=\frac{-3}{2} \div\left(\frac{4}{5} \times \frac{12}{-7}\right) \\ &=\frac{-3}{2} \div\left(-\frac{48}{35}\right) \\ &=\frac{-3}{2} \times-\frac{35}{48} \\ R \cdot H S &=\frac{35}{32} \\ L \cdot H \cdot S \neq R \cdot H \cdot S \\(p \div q) \div r \neq P \div(q \div \tau) \end{aligned}$

Exercise 1.5

i) $\frac{11}{4}=2.75$ <diagram to be added>

ii) $4 \frac{3}{5}=\frac{23}{5}=4.6$

iii $\}\frac{-9}{7}=-1 \frac{2}{7}$

iv) $\frac{-2}{-5}=\frac{2}{5}=0.4$

i) $A=\frac{3}{7}$

$B=\frac{7}{7}=1$

$C=\frac{8}{7}$

$D=\frac{12}{7}$

$E=\frac{13}{7}$

ii) $\quad P=\frac{-3}{8}$

$Q=\frac{-4}{8}=-\frac{1}{2}$

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

$S=\frac{-11}{8}$

$T=\frac{12}{8}=\frac{-3}{2}$

3. $\frac{-3}{7} \quad \frac{-3}{10}$ $\frac{-5} {20}\frac{-2}{10}$ $\frac{-3}{20} \frac{-1}{10}\frac{-1}{20}$ $0 \frac{1}{20} \frac{1}{10} \frac{3}{20} \frac{2}{10} \frac{5}{20} \frac{3}{10}$

$\begin{array}{ccccccc}\frac{7}{20} & \frac{4}{10} & \frac{9}{20} & \frac{5}{10} & \frac{11}{20} & \frac{6}{10} & \frac{2}{3}\end{array}$

4. <DIAGRAM TO BE ADDED>

5 .<DIAGRAM TO BE ADDED>

6) $\quad 1,2,3,4,5,6,7,8,9,10$

7) $\quad-5,-6,-7,-8,-9$

8) $-\frac{7}{3}$ (Numeritor is greater then denominator (numerically)

$-\frac{7}{3}<-1 ;-\frac{5}{11}, \frac{-1}{2}, \frac{-4}{9}>-1 .$

$\therefore \quad \frac{-7}{3}$ is difrerert among all rational numbers

Exercise 1.6

$\begin{aligned} \text { Total fruits } &=20 \mathrm{~kg} \\ \text { oranges } &=7 \frac{1}{6} \mathrm{~kg} \\ \text { Apples } &=8 \frac{2}{3} \mathrm{~kg} \end{aligned}$

let grapes $=x$

$7 \frac{1}{6}+8 \frac{2}{3}+x=20$

$\frac{43}{6}+\frac{26}{3}+x=20$

$\frac{(43 \times 1)+(26 \times 2)}{6}+x=20$

$\frac{43+52}{6}+x=20$

$\begin{aligned} \frac{95}{6}+x &=20 \\ x &=\frac{20}{1}-\frac{95}{6} \\ x &=\frac{120-95}{6} \\ x &=\frac{25}{6} \mathrm{~kg} \end{aligned}$

$\therefore$ Beg contain $4 \frac{1}{6}$ kg of grapes.

2 . Total population of cily $=6,63,432$

Adult male $=\frac{1}{2}$ of total population

Adult females $2 \frac{1}{3}$ of total population

Adult male + Adult female terms + children = Total city

$\frac{1}{2}(6,63,432)+\frac{1}{3}(663432)+$ Children $=6163432$

$\frac{5}{6}(663432)+$ children $=663432$

children $=663432\left(1-\frac{5}{6}\right)$

$=663432 \times \frac{1}{6}$

Children $=110572$

no. of children in city $=110,572$

3. Total votes $=30$

no. of votes for Mr. $x=\frac{2}{5}$ of $30=\frac{2}{5}(30)$

no.ot votes for $M r \cdot z=\frac{1}{3} (30)$

Let Mr.Y votes $=x$

=$\therefore \quad \frac{2}{5}(30)+x+\frac{1}{3}(30)=30$

=$12 \ +x+10=30$

=$x+22=30$

=$x=30-22$

$x=8$

NO. of votes for mr. y = 8

4. Total earnings $=₹ 100$

Rupees spent on food $=₹ 14 \frac{2}{7}$

Rupees spent on petrol $=₹ 30 \frac{2}{3}$

Let Savingson that day $=x$

$14 \frac{2}{7}+30 \frac{2}{3}+x=100$

$\frac{100}{7}+\frac{92}{3}+x=100$

$\begin{aligned}\left(\frac{100 \times 3)+(92 \times 7)}{21}\right.&+x=100 \\ x &=100-\frac{944}{21} \end{aligned}$

$x=\frac{1156}{21}=55 \frac{1}{21}$

$\therefore$ savings $=₹ 55 \frac{1}{21}$

5. $\quad$ Total students $=400$

no.of $\operatorname{girls} =130$

no. of boys appeared for exam $=400-130$$=270$

no.ot boys passed in exam $=\frac{2}{3}(270)$ $=180$

no. of boys failed in exam $=$ Total boys - pessed boys

=270-180

=90

$\therefore$ no. of boys failed in exam $=90$.

6.speed of $\mathrm{Car} \ =40 \frac{2}{3} \mathrm{Km} / \mathrm{h}$

time $=\frac{9}{10} \mathrm{hr}$

distance travelled car = speed ｘtime

$=40 \frac{2}{3} \times \frac{9}{10}$

$=\frac{122}{3} \times \frac{9}{10}$

$=\frac{183}{5}$

$=36 \frac{3}{5} \mathrm{~km}$

7. Side of square = $5 \cdot \frac{7}{9} m$

$S=\frac{52}{9} m$

$\begin{aligned} \text { Area of square } &=S^{2} \\ &=\frac{52}{9} \times \frac{52}{9} \\ &=\frac{2704}{81} \\ \text { Area of square } &=33 \frac{31}{81} \mathrm{~m}^{2} \end{aligned}$

8. perimeter of rectangle = $15 \frac{3}{7} m$

length of rectangle (l) = $4 \frac{2}{7} m$

perimeter = 2 (l+b)

=$15 \frac{3}{7}=2\left(4 \frac{2}{7}+b\right)$

=$\frac{108}{7}=2\left(\frac{30}{7}+b\right)$

=$\frac{3 b}{7}+b=\frac{108}{7 \times 2}$

$\frac{30 }{7}+b=\frac{108}{7 \times 2}$

$\frac{30}{7}+b=\frac{54}{7}$

$b=\frac{54}{7}-\frac{30}{7}$

$b=\frac{24}{7}$

$b=3 \frac{3}{7} m$

9. Total length of rope = $325 \frac{4}{5} m$

Rahul cutoff $150 \frac{3}{5} \mathrm{~m}$ of rope

Remaining length of rope = $325 \frac{4}{5}-150 \frac{3}{5}$

$=\frac{1629}{5}-\frac{753}{5}$

$=\frac{876}{5} \mathrm{~m}$

Rahul made remaining rope into 3 equal parts

$\begin{aligned} \therefore \text { length of each part } &=\frac{876}{5} \div \frac{3}{1} \\ &=\frac{876}{5} \times \frac{1}{3}=\frac{292}{5}=58 \frac{2}{5} \mathrm{~m} \\ \therefore \text { length of each part } & \text { part }=58 \frac{2}{3} \mathrm{~m} \end{aligned}$

10. $3 \frac{1}{2}$ liters of petrol cost= ₹ $270 . \frac{3}{8}$

cost for 1 liter of petrol $=\frac{270 \frac{3}{8}}{3 \frac{1}{2}}$

$=\frac{2163}{8} \times \frac{2}{7}$

Cost for 1 liter of petrol $=\frac{309}{4}$

cost for 4 liters of petrol $=\frac{309}{4} \times 4$

$\therefore$ cost for 4 liters of petrol $= ₹ 309$

11.

Ramesh Spends $\frac{3}{8}$ of income on food $=\frac{3}{8} \times 40000$

=15000

Remaining money $=40000-15000$

=25000

remaining spend $\frac{1}{5}$ of remaining on LIC

$=\frac{1}{5}(25000)$

$=5000$

Remainine money $=25000-195000$

$=20000$

other expenses are $\frac{1}{2} of$ remaining money

= $=\frac{1}{2} \times 20000$

= 10,000

$\begin{aligned} \text { Remaining money } &=20,000-10,000 \\ &=10,0001\end{aligned}$

12. Let total bill Amount as $x$

Amount paid A will be $=\frac{x}{2}$

Amount paid $B_{1} C_{1} D$ will be $=x-\frac{x}{2}$

$=\frac{x}{2}$

given bill is shared equally among three

let bill paid by each one = y

$y+y+y=\frac{x}{2}$

$3 y=\frac{x}{2}$

$y=\frac{x}{6}$

$\therefore$ Each paid $\frac{1}{6}$ th of total bill.

13.

Let total no. of student s = х

no. of studentsof school come by car $=\frac{2}{5} x$

no.ot Students of school come by bus e $\frac{1}{4} x$

no of students of school come by walk $=x-\left(\frac{2}{5} x+\frac{1}{4} x\right)$

= x- $\left(\frac{(2 \times 4)+(1 \times 5)}{20} \cdot x\right)$

$=x-\frac{13}{20} \cdot x$

no.ot students of school come by walk $=\frac{7}{20} \cdot x$

no.ot students of school come by walk on their own

\begin{aligned} &=\frac{1}{3} \ of \left(\frac{7}{20} x\right) \\ &=\frac{7}{60} \cdot x \end{aligned}

$\begin{aligned} \therefore \quad \frac{7}{60} \cdot x &=224 \\ x &=\frac{224}{1} \times \frac{60}{7} \\ x &=1920 \\ \text { Total Students in school }=1920 \end{aligned}$

14. Total cost of Room $= ₹ 60,000$

let Mother's contribution $= ₹ x$

Elder son contribution $= ₹ \frac{3}{8} x$

younger son contribution $= ₹ \frac{1}{2} x$

$\begin{aligned} \therefore \quad x+\frac{3}{8} x+\frac{1}{2} x &=60,000 \\ \frac{( 8)+(3)+(4)}{8} \cdot x &=60,000 \\ \frac{15}{8} \cdot x &=60000 \\ x &=60000 \times \frac{8}{15} \\ x &=32000 \end{aligned}$

; Mother's contribution 2232,000

; Elder son's contribution e $\frac{3}{8} \times 32000=₹ 12,000$

; Younger son's Contribution = $\frac{1}{2} \times 32000=₹ 16,000$

15. Total students $=56$

let no.of girls $=x$

no.ot boys $=\frac{2}{5} x$

$\therefore \quad x+\frac{2}{5} x=56$

$\frac{7}{5} x=56$

$x=\frac{56}{1} \times \frac{5}{7}$

$x=40$

$\therefore$ no.ot girls $=40$

$\therefore$ no. ot boys $=56-40=16$

16. Let money posses by man $2 x$

$\frac{1}{10}$th money donated to school $=\frac{x}{10}$

Remaining money $=x-\frac{x}{10}=\frac{9 x}{10}$

$\frac{1}{6}$th remaining money $=\left(\frac{9 x}{10}\right) \times \frac{1}{6}$

Remaining money $=\frac{9 x}{10}-\frac{9 x}{10 \times 6}$

$=\frac{45 x}{60}$

Now, man distributed This money equally to hin three sons and each one gets $=₹50,000$

$\frac{45 x}{60} \div 3=50,000$

$\frac{45 x}{60} \times \frac{1}{3}=50000$

$\frac{3}{4} x \times \frac{1}{8}=50000$

$x=50000 \times 4$

$x=2,00,000$

17. Let a number be 'x'

$\frac{1}{4}$ of a number is added to $\frac{1}{3}$ of number

$\frac{x}{4}+\frac{x}{3}$ is 15 greater thaw half of number

$\begin{aligned} \frac{x}{4}+\frac{x}{3}=15 &+\frac{x}{2} \\ \frac{7 x}{12}=15 &+\frac{x}{2} \\ \frac{7 x}{12}-\frac{x}{2} &=15 \\ \frac{x}{12} &=15 \\ x &=15 \times 12 \\ x &=180 \end{aligned}$

18. Let the number be ' $x$ '

$\frac{x}{1} \div \frac{4}{5}=36+\left(x+\frac{4}{5}\right)$

$\frac{5 x}{4}=36+\frac{4 x}{5} .$

$\frac{5 x}{4}-\frac{4 x}{5}=36$

$\frac{(5 \times 5)-(4 \times 4)}{20} \cdot x=36$

$\frac{25-16}{20} \cdot x=36$

$\frac{9}{20} \cdot x=36$

$x=\frac{36}{9} \times 20$

$x=80$

$\therefore$ The given number is 80

⇒

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

x²

x₂

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Monday, March 15, 2021