Aptitude - Pipes and Cistern - Discussion

Let , the time taken by waste pipe to empty the tank is x.

Then part feed in one min.
(1/20+1/24) - 1/X = 1/15
11/120 -1/X = 1/15,
11X-120/120X = 1/15,
165X -1800= 120 X,
By solving we get x = 40 min.
So 40 min is the time which is taken by waste pipe to empty the entire tank

In problem it is given that waste pipe discharge 3 gallons per min.
40*3= 120 gallons is a capacity of tank.

I have the best solution here.
take LCM of 20,24 = 120

Pipe c can empty 3 gallon/minute.
120/3 = 40 minute.
-----------------------------|
minute | efficiency |
-------------|---------------|
a = 20 | 6 |
b = 24 | 5 |
c = 40 | 3 |
******************************
(a+b-c) --> 11-3 = (+8)

120/8 = 15 minute.

120-gallon tank will fill in 15 minutes,
A can fill in 24 minutes or 5 gallons/minute,
B can fill in 20 minutes or 6 gallons/minute.

C can empty in 40 minutes or 3 gallons/minute <---question says.

A = 20 min,
B = 24 min,
Let, C = x min.

Then, capacity = lcm of A,B,C =120x.
So, efficiency of A = capacity / time taken by A,
= 120x/20.
= (+) 6x [since it's used for filling].

Similarly, efficiency of B = 120x /24 = (+) 5x,
Efficiency of C = 120x / x = (-) 120. [since it's used for emptying]
Therefore, total efficiency = 6x + 5x -120.

So, time = (capasity / total efficiency) unit.
= (120x / (6x + 5x -120)) min.

Now, (120x / (6x + 5x - 120)) = 15.
Or, 24 / (11x - 120) = 3,
Or, 33x - 360 = 24,
Or, 9x = 360,
Or, x = 40.

Therefore, capacity = lcm of 20, 24 and 40,
= 120.

So, the capacity of the tank is 120 gallon.

Thanks for the explanation @Bhakta B Monger.

How did -1/40? Please explain it.

Thank you @Swapnil.

Thank you @Swapnil.

How 3 gallons =1/40?

We can directly determine the total capacity of tank by taking LCM of 20 24 15 as 120.

Let x = no.of minutes to take empty the gallon by leakage.

(1/20+1/24-1/x)=1/15.

x will get = 40 min.

40*3 = 120 gallon.