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) 90×10−900;15×60=900;45×20=900;30×30⋅900;
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 75×10≠45×30≠30×25=20×35≠10×15≠ 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
1200−x=1200×24321200−x=900x=1200−900=300
∴ No of Soldier were left = 300. Soldiers
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