Time taken to empty the tank The problem is: 
An electric pump takes 3 hrs to fill a tank, but due to a leak in the tank now it takes 3 and 1/2 hrs to fill the tank. If the tank is full, how much time(in hrs) will the leak require to empty that tank?
A)21
B)20
C)23
D)22
I faced this question in an aptitude test for an interview. I couldn't find the solution. Hence I require your help.
 A: Some definitions


*

*The volume of the tank - $V$

*The time to fill the tank if there were no leak - $T_0$

*The time to fill the tank with the leak $T_l$


We know then that the rate at which the pump can fill the tank is $R_0 = V/T_0$. Assuming the leak leaks at a constant rate $R_l$ then the rate with which the tank is actually filled is $R_0 - R_l$. 
The time to fill volume $V$ at this modified rate is just $T_l = V/(R_0-R_l)$. Substituting in $R_0$ and rearranging gives:
$$\frac{T_l}{T_l/T_0-1} = \frac{V}{R_l}$$
Once the tank is full we assume the leak still just removes water at constant rate $R_l$ thus the time to empty the tank is just $V/R_l$ which we have shown is $\frac{T_l}{T_l/T_0-1}$, i.e. $3.5/(3.5/3 - 1) = 21$.
A: I would say the answer is 21 (but I'm not a fully math person):


*

*Assume the tank is 15 liter.

*Then the pump fills with speed 15 l/hr / 3 hr = 5 l/hr

*Due to the leak it takes 3,5 hr. Meaning the tank filling is 5 l * 3,5 = 17,5 l.

*The difference is 17,5 l - 15 l = 2,5 l ... during 3,5 hours 

*This means 2,5 / 3,5 = 0,714 l/hr to be spoilt.

*The tank filling is 15 l: 15 l / 0,87 = 21 
So answer A is correct.

A: Let the Volume of the tank be V.
So without leakage the tank is V/3 full in 1 hour.
But with leakage the tank is on V/(3.5) i.e 2V/7 full in 1 hour.
So the leakage contribution is V/3-(2V/7) i.e. 6V/21 - 7V/21  i.e  -V/21.
So the full tank will get emptied in 21 hours when the pump is turned off. 
Hence the answer is option (A).
A: Lets assume that total volume is v and hole rate or empty volume is y hour so as given total time taken to fill is 7/2 hours
V/7/2=V/3-V/y
1/y=1/3-2/7
y=21hours
A: Input = (1/3) tank per hour
Input - output = (2/7)tank per hour
Output = 1/3 - 2/7 = 1/21 tank per hour, 
so takes 21
Regards - Ian H
