Pipes A and B can fill a tank in 15 hours and 20 hours, respectively, while Pipe C can empty a full tank in 25 hours. All three pipes are opened together. After 10 hours, Pipe C is closed and A and B continue. What is the total time (from the start) required to fill the tank completely?

Difficulty: Easy

12 hours
13 1/2 hours
16 hours
18 hours
None of these

Correct Answer: 12 hours

Explanation:

Introduction / Context:
This is a two-phase problem: a mixed phase with both inlets and the outlet, followed by an inlet-only phase. Compute the fraction completed in the first 10 hours, then finish with A and B only.

Given Data / Assumptions:

A = 1/15 job/hour.
B = 1/20 job/hour.
C = −1/25 job/hour.
First 10 hours: A + B + C. After that: A + B.

Concept / Approach:
Phase 1 work = 10 * (1/15 + 1/20 − 1/25). Phase 2 time = (remaining) / (1/15 + 1/20). Sum both phases for total time.

Step-by-Step Solution:

Phase 1 rate = LCM 300 ⇒ (20 + 15 − 12)/300 = 23/300 job/hour.Work after 10 h = 10 * 23/300 = 23/30.Remaining = 1 − 23/30 = 7/30.A + B rate = 1/15 + 1/20 = 7/60 job/hour.Phase 2 time = (7/30) / (7/60) = 2 hours.Total time = 10 + 2 = 12 hours.

Verification / Alternative check:
Back-substitution confirms totals reach exactly 1 job.

Why Other Options Are Wrong:
13 1/2 / 16 / 18 hours do not match the exact two-phase computation.

Common Pitfalls:
Dropping the outlet term during Phase 1 or miscomputing the remaining fraction.

Pipes A and B can fill a tank in 15 hours and 20 hours, respectively, while Pipe C can empty a full tank in 25 hours. All three pipes are opened together. After 10 hours, Pipe C is closed and A and B continue. What is the total time (from the start) required to fill the tank completely?

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