Question : Pipes A and B can fill a tank in 12 minutes and 15 minutes, respectively. The tank when full can be emptied by pipe C in $x$ minutes. When all three pipes are opened simultaneously, the tank is full in 10 minutes. The value of $x$ is:
Option 1: 18
Option 2: 15
Option 3: 20
Option 4: 24
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Correct Answer: 20
Solution :
Pipe A can do 1 unit work = $\frac{1}{12}$ minutes
Pipe B can do 1 unit work = $\frac{1}{15}$ minutes
Pipe C can do 1 unit work = $-\frac{1}{x}$ minutes
Together they can fill the tank = $\frac{1}{10}$ minutes
Total work = Efficiency × days
$⇒\frac{1}{\text{total days}} = \frac{1}{\text{A}} + \frac{1}{\text{B}} - \frac{1}{\text{C}}$
$⇒\frac{1}{10} = \frac{1}{12} + \frac{1}{15} - \frac{1}{x}$
$⇒\frac{1}{x} = \frac{1}{12} + \frac{1}{15} - \frac{1}{10}$
$⇒\frac{1}{x} = \frac{(5 + 4 - 6)}{60}$
$⇒\frac{1}{x} = \frac{3}{60}$
$\therefore x = 20$
Hence, the correct answer is 20.
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