ML AGGARWAL CLASS 8 CHAPTER 9 Direct and Inverse Variation Exercise 9.2

 Exercise 9.2

Question 1

Sol :

(i) more speed , lets time taken by train to corer a fixed distance 

Speed and time or inversely proportional 


(ii) more people at work less time taken to finish work 

less people at work , more time taken to finish work 

i .e This is a inverse variation


Question 2

Sol :

(i) $90 \times 10-900 ; 15 \times 60=900 ; 45 \times 20=900 ; 30 \times 30 \cdot 900 ;$
$20 \times 45=900$

i.e $90 \times 10=15 \times 60=20 \times 45=30 \times 30=45 \times 20=900=$ Constant

ie x y= constant

$\therefore x$ and y are in inverse variation


(ii) $75 \times 10=750: 45 \times 30=1350 ; 30 \times 25=750 ; 20 \times 35.750$

$10 \times 65=650$

ie $75 \times 10 \neq 45 \times 30 \neq 30 \times 25=20 \times 35 \neq 10 \times 15 \neq$ Constant

ஃ x and y are not in inverse variation

Question 3

Sol :
Given volume inversely proportional to pressure 

$V \alpha \frac{1}{p}$

i.e PV= constant

Volume of a mass = 630 $c m^{3}=V_{1}$

pressure of mercury $=360 \mathrm{~mm}=p_{1}$

if volume is $720 \mathrm{~cm}^{3}$, pressur $9 p_{2}=9$

$p_{1} v_{1}=p_{2} v_{2}$

p = 315 mm

∴ pressure of mercury = 315mm

Question 4

Sol :
No of children are 20 

each received 4 sweets 

∴ Total no of sweets are 20 $\times$ 4 = 80 sweets 

Given number of children reduced by 4 i .e 20 - 4 = 16 

Now 16 children were present

$\therefore$ if is a case of inverse Variation

$\therefore$ each children get $\frac{80}{16}$ sweets i.e 5 sweets


Question 5

Sol :
Pooja has money to buy 36 oranges at the rate

of 24.5 per orange i.e

She has money $36 \times 4.5$

=₹ 162

Now the price of orange is increased by 90 paise

i.e New price of orange is $4.50+0.90={Rs } 5.40$

Now she takes only $\frac{162}{5.4}$ oranges

i.e 30 oranges only

Question 6

Sol :
In 8 days, number of men required to construct a

wall =12

In 6 days, how many. were required ？

ie No of men required $=x$.

$x=\frac{8 \times 12}{6}$

x=16

$\therefore 16$ men required to construct a wall in 6 days

Question 7

Sol :
Total no of taps = 8 

these takes 27 minutes to fill a tank 

 out of 8 , 2 taps go out of order i .e 

the remainnig no of taps are 6 

Let the 5 taps taken x minutes 

Note that lesser the number of pipes, more will be the 

time required to fill the tank '

So , it is a case of inverse variation

$\begin{aligned} 8 \times 27 &=x \times 6 \\ x &=\frac{8 \times 27}{6} \end{aligned}$

x = 36 minutes

∴ Time taken to fill the tank by 6 taps is 36 minutes 

Question 8

Sol :
No. of person contact to complete a part of stadiums 

in  9 months = 560 persons

Let the no of persons to complete a part of stadium in 5 months = x 
more month , less person are required
i.e It is a case of inverse variation 

$9 \times 560=8 \times x$
$x=9 \times 70$
x=630

Now , no of persons extra required is 630- 560 = 70 persons 

Question 9

Sol :
A batch of bottles were packed in 30 boxes with 10 bottles in each box 

Now 12 bottles in each box can be filled or packed in 

how many boxes , Let it be x 

More boxes ,less bottles were packed 

i.e It is a case of inverse variation

$30 \times 10=12 \times x$

$n=\frac{30 \times 10}{12}$

x=25

∴ 25 boxes with 12 bottles in each box

Question 10

Sol :
Vandana takes 24 minutes to reach school wilt a

Speed of 5 km /h

Now, how much speed is required to reach school with in 20 minutes

More time less speed is required 

i.e it is case of inverse variation

∴ Let the speed b 'x' 

$\begin{aligned} 5 \times 24=& x \times 20 \\ x &=\frac{5 \times 24}{20} \\ & x=6 \mathrm{~km} / \mathrm{h} \end{aligned}$

∴ Speed required = 6 km/h

Question 11

Sol :
After 15 days , the food for 80 soldiers for (60 - 15)days 

i.e 45 days 

So the number of soldiers in ford 80 + 20= 100

Let the food last for x days , when these are 100 soldiers in food 

note that more the number of soldiers in ford , the sooner the food exhaust

$\begin{aligned} 80 \times 45=& 100 \times x \\ x=& \frac{80 \times 45}{100} \\ x=36 \end{aligned}$

∴ The food will last for 36 days 

Question 12

Sol :
1200 Soldiers in a fort had enough food for 28 days 

after 4 days some soldier were sent be x 

$\therefore$ The food lasted for 32 more days

If is a case of inverse variation

$\begin{aligned} 1200-x &=\frac{1200 \times 24}{32} \\ 1200-x &=900 \\ x &=1200-900=300 \end{aligned}$

$\therefore$ No of Soldier were left = 300. Soldiers

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