Pipes and Cisterns – Two inlets and one outlet with net partial fill in one hour: Three pipes are connected to a tank. In one hour, pipe 1 can fill 1/2 of the tank, pipe 2 can fill 1/3 of the tank, and pipe 3 is an outlet (empties). When all three are opened together, 7/12 of the tank is filled in 1 hour. How long would the third pipe alone take to empty a full tank?

Introduction / Context:Net effect over one hour is known. Express the outlet’s unknown rate as x (per hour) and solve using the fact that the resulting net equals 7/12 tank per hour.

Given Data / Assumptions:

• Pipe 1 rate = 1/2 per h.
• Pipe 2 rate = 1/3 per h.
• Pipe 3 is an outlet with rate = x per h (to be subtracted).
• Net with all open = 7/12 per h.

Concept / Approach:Set up the rate equation: 1/2 + 1/3 − x = 7/12 and solve for x, then invert to get the emptying time.

Step-by-Step Solution:1/2 + 1/3 = 5/6 = 10/12.10/12 − x = 7/12 ⇒ x = 3/12 = 1/4 per h.Therefore, outlet alone empties in 1 / (1/4) = 4 h.

Verification / Alternative check:Recombine: 1/2 + 1/3 − 1/4 = 6/12 + 4/12 − 3/12 = 7/12 per h, matches the statement.

Why Other Options Are Wrong:2, 3, 5, and 6 h would not produce 7/12 per h net.

Common Pitfalls:Using 1/2 + 1/3 = 2/5 (incorrect) or forgetting to subtract the outlet.

Final Answer:4 h