Aptitude - Pipes and Cistern - Discussion

Very good explanation, Thank you @Ishan.

Question is asked in terms of time, not volume (work). So, 30 mins is the answer. LCM method assumes half volume not half time.

Thanks @Viz.

Thanks @Sanjay Pahade.

A = 60 min.
B = 40 min.

Total work= 120.

The capacity of A and B is 2 and 3 respectively.

In Q, is clear that B is used for half the time, so we will take 20 min.

So, in 20 min B will fill = 60 work.
Remaining work = 60.
Now, both A and B is used for other half means is clear that for other half work that is 60.

So the capacity of A+B=5.

Time = 60/5 = 12 min.
Total time = 20 + 12 = 32 min.

Why not 32?

Total = 240 units.
Efficiency of A = 4 unit.
Efficiency of B = 6.

So required time = (240 *1/2) /6+(120/10) = 20+12 = 32.
Why ans is 30?

LCM(60,40) = 120.
Suppose the capacity of the tanker is 120litre.
Then, the quantity filled by pipe A in 1 min=120/60= 2 litres.
The quantity filled by pipe B in 1 min=120/40=3 litre.

The quantity filled by pipe A and B together in 1 min= 2 + 3 = 5 litre.

Suppose the tanker is filled in x minutes.

Then, to fill the tanker from the empty state, B is used for x/2 minutes and A &B together is used for the rest x/2 minutes.

3x/2 + 5x/2 = 120;
x/2(3+5) = 120;
8x/2 = 120
X = 30.

Required time =30 minutes.

I agree The right answer is 32.

Cross-check:

TA = 60 min.
TB = 40 min.
As per the given statement, A and B run together for 15 min (=Answer/2=30/2), and B runs alone for the next 15 mins (=Answer/2=30/2).
P1 = Part filled by first 15 min = 15*(1/TA+1/TB) = 15*(1/60+1/40) = 15*5/120=5/8.
P2 = Part filled by next 15 min = 15*(1/TB) = 15*(1/40) = 3/8.
Full tank = summation of both parts = P1+P2 = 5/8+3/8 = 1.
Hence it is proved that the given explanation is correct.