Two water taps running together can fill a tank in 9(3/8) hrs. The tap of larger diameter takes 10 hrs less than the smaller one to fill the tank separately . Find the time in which each one can separately fill the tank.
Answers
Sol :
Let the tap of the larger diameter fills the tank alone in (x – 10) hours.
In 1 hour, the tap of the smaller diameter can fill 1/x part of the tank.
In 1 hour, the tap of the larger diameter can fill 1/(x – 10) part of the tank.
Two water taps together can fii a tank in 75 / 8 hours.
But in 1 hour the taps fill 8/75 part of the tank.
1 / x + 1 / (x – 10) = 8 / 75.
( x – 10 + x ) / x ( x – 10) = 8 / 75.
2( x – 5) / ( x2 – 10 x) = 8 / 75.
4x2 – 40x = 75x – 375.
4x2 – 115x + 375 = 0
4x2 – 100x – 15x + 375 = 0
4x ( x – 25) – 15( x – 25) = 0
( 4x -15)( x – 25) = 0.
x = 25, 15/ 4.
But x = 15 / 4 then x – 10 = -25 /4 which is not possible since time
But x = 25 then x – 10 = 15.
Larger diameter of the tap can the tank 15 hours and smaller diameter of the tank can fill
the tank in 25 hours.