Question : Pipes A and B can empty a full tank in 18 hours and 24 hours, respectively. Pipe C alone can fill the tank in 36 hours. If the tank is $\frac{5}{6}$ full and all the three pipes are opened together, then in how many hours will the tank be emptied?
Option 1: $10 \frac{1}{2}$
Option 2: $12 \frac{1}{2}$
Option 3: $10$
Option 4: $12$
Correct Answer: $12$
Solution :
Time taken by pipe A to empty a full tank = 18 hr
Part of tank emptied by pipe A in an hour = $\frac{1}{18}$
Time taken by pipe B to empty a full tank = 24 hr
Part of tank emptied by pipe B in an hour = $\frac{1}{24}$
Time taken by pipe C to fill a full tank = 36 hr
Part of tank filled by pipe C in an hour = $\frac{1}{36}$
Part of the tank emptied by pipes A, B, and C in an hour = $\frac{1}{18}+\frac{1}{24}-\frac{1}{36}$
= $\frac{4+3-2}{72}$
= $\frac{5}{72}$
Time taken by pipes A, B and C to empty the tank = $\frac{72}{5}$ hr
So, the time taken by pipes A, B, and C to empty $\frac{5}{6}$th of the tank = $\frac{5}{6}×\frac{72}{5}$ = $\frac{72}{6}$ = 12 hours
Hence, the correct answer is $12$.
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