Question

Pipe A and Pipe B can fill a tank into 20 hours and 12 hours respectively. Pipe C can empty it in 10 hours. If all three pipes are open alternatively for 1 hour but start is with A, then the whole tank will fill in how many hours?

1. 120 hours
2. 80 hours 
3. 100 hours and 30 minutes
4. 90 hours and 48 minutes
5. None of these

Answer 1

Correct Answer - Option 2 : 80 hours 

Given:

Pipe A can fill a tank in = 20 hours

Pipe B can fill a tank in = 12 hours

Pipe C can empty a tank in = 10 hours

Concept Used:

If a pipe takes x hours to fill a tank, the part of the tank filled by the pipe in 1 hour = 1/x

Calculation:

Let the total capacity of tank be LCM (20, 12, 10) = 60 litres

Now,

Tank filled by A in 1 hour = 60/20 = 3 litre/hour

Tank filled by B in 1 hour = 60/12 = 5 litre/hour

Tank empty by C in 1 hour = 60/10 = (– 6) litre/hour

When A, B and C work alternatively,

Tank filled by (A + B + C) in 3 hours = (3 + 5 – 6) = 2 litres

We conclude that, 

In (26 × 3) = 78 hours, the part of the tank filled = 2 × 26 = 52 litres

So, volume left = 60 - 52 = 8 litres

After the 79th hour, tank filled = 3 litres

Now, remaining volume to be filled  = 8 - 3 = 5 litre

Now, pipe B will take 1 hours to fill 5 litres 

So, the total time taken = 79 + 1 = 80 hours

∴ The whole tank will be fill in 80 hours 

When the tank full in 80 hours not need to empty it. So, in such cases take min time.