Question

A tank has two taps ,p and q fitted to it. tap p fills the tank, while tap q empties the tank. the time taken by tap p to fill the tank is 25% less than the time taken by tap q to empty the tank . when both the taps were opened simultaneously , it was found that they took 50% more time working together to fill the tank than expected . the reason for this was found to be a leak that was present at the bottom of the tank. if the taken by tap p alone to fill the tank is 3 hours ,find the time taken by the leak alone to empty a full tank? 1)12hours 2)18hours 3)24hours 4)36hours

Answer 1

Time taken by P to fill = 25% less time than the time taken by Q to empty.
Let Q empties the tank in X hours
Hence P fills in: X - 25% of X = X - 25X/100 = 75X/100 = 3X/4 hours
Given, P fills in 3 hours
3 = 3X/4 hours
X = 4 hours.
Hence, Q takes 4 hours to empty the tank.

Now, P in 1 hour fills: 1/3 part of the tank
Q in 1 hour empties: 1/4 part of the tank
In 1 hour both taps do:
1/3 - 1/4 = 4-3/12 = 1/12 part filled... ['-' was used coz, Q empties]
! hour=== 1/12 part filled
X hours=== 1 part
X = 12 hours
So, in normal condition the tank has to be filled in 12 hours.
But due to a leak it took 50% more time, that is, 12 +50% of 12 = 18 hours.

So, with leak, in 1 hour only 1/18 part is filled.
Let the leak empties the tank in 'L' hours.
In 1 hour, it empties: 1/L part
1/18 = 1/3 - 1/4 - 1/L
1/18 = 4-3/12 - 1/L = 1/12 - 1/L
1/L = 1/12 - 1/18 = 3-2/36 = 1/36
L = 36 hours.
Hence the time taken by leak alone to empty the full tank is 36 hours.
Hope it helps.