A tank can be filled by Pipe1 in 7 hours and by Pipe2 in 5 hours. Ther

Given

• A tank can be filled by Pipe1 in 7 hours and by Pipe2 in 5 hours.
• There is a waste pipe that is kept open when Pipe2 is working; the tank then takes 8 hours 30 minutes to fill.

To Find

• The time taken to fill the tank if all the three pipes are working.


Approach and Working Out

• Pipe 1 takes 7 hours and pipe 2 takes 5 hours.

• Pipe 2 and waste pipe takes 8.5 hours.

• Let us take the total capacity as a common multiple of 5 and 8.5 which is 85.

o Pipe 2, 17 units/hour
o Pipe 2 and the waste pipe, 10 units/hour
o The waste pipe will take out 7 units/hour.

• Pipe 1 will fill \(\frac{85}{7}\) units per hour.

• Total input/hour = Pipe 1 + Pipe 2 – Waste pipe.

o = \(\frac{85}{7} \)+ 17 – 7
o = \(\frac{155}{7}\)

• Total time = \(\frac{85}{(155/7) }\)= \(\frac{595}{155}\) = 3.84

o = 3 hours 50 minutes (approx)

Correct Answer: Option B

I would attempt this with reasoning.

Pipe 1 fills in 7 hours. If there were two pipes like 1, they will fill the tank in 3.5 hours (3 hours 30 minute). Since the combination of Pipe 2 & Drain pipe fills at 8.5 hours, I would consider that the answer would have to be just a little bit over 3 hours 30 minutes.

This narrows the answer choices to (B), (C) and may be (D).

Similar logic the other way - If two pipes of the same strength as (pipe 2+ drain pipe) were filling the tank, the tank will be filled in half the time i.e. 4 hours 15 minutes. Since pipe A is faster than this combination, the answer has to be less than 4 hours 15 minutes (which is choice C)

Only (B) remains. I would mark (B) without calculation.