Question : Pipes P and Q can fill a tank in 10 and 12 hours respectively and C can empty it in 6 hours. If all three are opened at 7 AM at what time will one-fourth of the tank be filled?
Option 1: 10 AM
Option 2: 10 PM
Option 3: 11 PM
Option 4: 11 AM
Correct Answer: 10 PM
Solution :
Pipe P's hourly work = $\frac{1}{10}$ unit
Pipe Q's hourly work = $\frac{1}{12}$ unit
Pipe C's hourly work = $-\frac{1}{6}$ unit
When all pipes are working together, let the tank be filled in $x$ hours.
According to the question,
$\frac{1}{10}+\frac{1}{12}-\frac{1}{6}=\frac{1}{x}$
$⇒\frac{1}{x}=\frac{6+5-10}{60}$
$⇒\frac{1}{x}=\frac{1}{60}$
$\therefore x =60$
Thus, $\frac{1}{4}$th of the tank is filled in $\frac{1}{4}×60 = 15$ hours
At 7:00 AM, all pipes start working, so the required time is (7:00 AM + 15 hours) = 10:00 PM
Hence, the correct answer is 10 PM.
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