Two water taps together can fill a tank in 9
hours. The tap of larger diameter takes 10hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answers
Step-by-step explanation:
Consider that the tap with smaller diameter fills the tank in x hours.
Then, the tap with larger diameter fills the tank in x−10 hours.
This shows that the tap with a smaller diameter can fill
x
1
part of the tank in 1 hour. Similarly, the tap with larger diameter can fill
x−10
1
part of the tank in 1 hour.
It is given that the tank is filled in
8
75
hours that is, the taps fill
75
8
part of the tank in 1 hour. Then,
x
1
+
x−10
1
=
75
8
x(x−10)
x−10+x
=
75
8
x
2
−10x
2x−10
=
75
8
75(2x−10)=8(x
2
−10x)
150x−750=8x
2
−80x
8x
2
−230x+750=0
4x
2
−115x+375=0
4x
2
−100x−15x+375=0
4x(x−25)−15(x−25)=0
(4x−15)(x−25)=0
4x−15=0
x=
4
15
Or,
x−25=0
x=25
When x=
4
15
, then, x−10=
4
15
−10
=
4
15−40
=−
4
25
This cannot be possible because time can never be negative.
When x=25, then,
x−10=25−10
x=25
Therefore, the tap of smaller diameter can separately fill the tank in 25 hours.