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

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more