Two water taps together can fill a tank in 938 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time(in hrs.) in which tap of smaller diameter can separately fill the tank.
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 1x part of the tank in 1 hour. Similarly, the tap with larger diameter can fill 1x−10 part of the tank in 1 hour.
It is given that the tank is filled in 758 hours that is, the taps fill 875 part of the tank in 1 hour. Then,
1x+1x−10=875
x−10+xx(x−10)=875
2x−10x2−10x=875
75(2x−10)=8(x2−10x)
150x−750=8x2−80x
8x2−230x+750=0
4x2−115x+375=0
4x2−100x−15x+375=0
4x(x−25)−15(x−25)=0
(4x−15)(x−25)=0
4x−15=0
x=154
Or,
x−25=0
x=25
When x=154, then, x−10=154−10
=15−404
=−254
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.