Here is How I tried and solved this sum :
A + B + C = t min
Pipe A = t minutes
Pipe B = 10 minutes
Pipe C = (t-12) minutes
Let the total capacity of the tank be 1 unit, Since we know that all pipes filled equal share of tank we can assume that A B and C all three filled 1/3 unit each.
Since it is mentioned that A + B = t minutes and also that each pipe filled equal share of tank so in this case we can say that A and B both filled the tank in 1/2 time
we can say that,
B=1/2t ---- 1 equation
B= 10 min ---- 2 equation
By using equation 1 and 2
T= 20 min
Now, C fills 1/3 unit of tank in (t-12) min so for full tank it will take 3 * (t-12) min
(t-12) = 20-12 = 8
3*8= 24
C is the correct answer.
I hope I solved the question in the right way.
Bunuel
Pipes A, B and C together filled a tank in t minutes working in the following manner:
Pipe A was open for t minutes;
Pipe B was open for the first 10 minutes and then closed.
Two minutes after pipe B was closed, pipe C was opened and is kept open till the tank was full.
Each pipe filled an equal share of the tank. If pipes A and B, both opened, can fill the tank in t minutes. How long will it take pipe C alone to fill the tank ?
A. 36
B. 27
C. 24
D. 20
E. 18
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