Pipes A and B can fill a tank in 16 hours and 24 hours, respectively, and pipe C alone can empty the full tank in x hours. All the pipes were opened together at 10:30 AM, but C was closed at 2:30 PM. If the tank was full at 8:30 PM on the same day, then what is the value of x?
96
Step 1 — Phase durations:
Phase 1 (A, B, C open): 10:30 AM to 2:30 PM = 4 hours. Phase 2 (A, B only): 2:30 PM to 8:30 PM = 6 hours.
Step 2 — Combined rate of A and B:
\(\dfrac{1}{16}+\dfrac{1}{24}=\dfrac{3+2}{48}=\dfrac{5}{48}\) tank/hour.
Step 3 — Equate total work to 1 tank:
\[4\left(\dfrac{5}{48}-\dfrac{1}{x}\right) + 6\cdot\dfrac{5}{48} = 1\]
Step 4 — Solve for x:
\(\dfrac{20}{48}+\dfrac{30}{48}-\dfrac{4}{x}=1 \Rightarrow \dfrac{50}{48}-1=\dfrac{4}{x} \Rightarrow \dfrac{1}{24}=\dfrac{4}{x}\).
Therefore \(x = 96\) hours.
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