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Two pipes A and B can fill a tank in 5 hours and 6 hours respectively. Pipe C can empty it in 12 hours. If all three pipes are opened together, then the time taken to fill the tank is:

Solution

✅ Correct Option: 4

Pipe A fills the tank in 5 hours, so in 1 hour it fills 15\frac{1}{5} of the tank.

Pipe B fills the tank in 6 hours, so in 1 hour it fills 16\frac{1}{6} of the tank.

Pipe C empties the tank in 12 hours, so in 1 hour it empties 112\frac{1}{12} of the tank.


When all three pipes are opened together, the combined work in 1 hour is:

15+16−112\frac{1}{5} + \frac{1}{6} - \frac{1}{12}

The LCM of 5, 6, and 12 is 60.

Converting each fraction:

15=1260\frac{1}{5} = \frac{12}{60}

16=1060\frac{1}{6} = \frac{10}{60}

112=560\frac{1}{12} = \frac{5}{60}

Combined rate =1260+1060−560= \frac{12}{60} + \frac{10}{60} - \frac{5}{60}

Combined rate =1760= \frac{17}{60} of tank per hour


If 1760\frac{17}{60} of the tank is filled in 1 hour, then the time to fill the complete tank is:

Time =1÷1760= 1 \div \frac{17}{60}

Time =1×6017= 1 \times \frac{60}{17}

Time =6017= \frac{60}{17} hours


Converting to mixed fraction:

60÷17=360 \div 17 = 3 with remainder 99

6017=3917\frac{60}{17} = 3\frac{9}{17} hours

Therefore, the time taken to fill the tank is 39173\frac{9}{17} hours.

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