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) 90×10900;15×60=900;45×20=900;30×30900;
20×45=900

i.e 90×10=15×60=20×45=30×30=45×20=900= Constant

ie x y= constant

x and y are in inverse variation


(ii) 75×10=750:45×30=1350;30×25=750;20×35.750

10×65=650

ie 75×1045×3030×25=20×3510×15 Constant

ஃ x and y are not in inverse variation

Question 3

Sol :
Given volume inversely proportional to pressure 

Vα1p

i.e PV= constant

Volume of a mass = 630 cm3=V1

pressure of mercury =360 mm=p1

if volume is 720 cm3, pressur 9p2=9

p1v1=p2v2

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 × 4 = 80 sweets 

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

Now 16 children were present

if is a case of inverse Variation

each children get 8016 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 36×4.5

=₹ 162

Now the price of orange is increased by 90 paise

i.e New price of orange is 4.50+0.90=Rs5.40

Now she takes only 1625.4 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=8×126

x=16

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

8×27=x×6x=8×276

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 

9×560=8×x
x=9×70
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

30×10=12×x

n=30×1012

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' 

5×24=x×20x=5×2420x=6 km/h

∴ 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

80×45=100×xx=80×45100x=36

∴ 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 

The food lasted for 32 more days

If is a case of inverse variation

1200x=1200×24321200x=900x=1200900=300

No of Soldier were left = 300. Soldiers

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