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

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