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Pipe A can fill a tank in 24 hours and pipe B in 32 hours. If both the pipes are opened in an empty tank at the same time, then the time taken to fill the tank is:

Solution

✅ Correct Option: 3

Pipe A alone takes 24 hours to fill the tank.

Pipe B alone takes 32 hours to fill the tank.

When pipes work together, their rates of work add up.


Pipe A's work in 1 hour = 124\frac{1}{24} of the tank

Pipe B's work in 1 hour = 132\frac{1}{32} of the tank


Combined work in 1 hour = 124+132\frac{1}{24} + \frac{1}{32}

To add these fractions, find the LCM of 24 and 32, which is 96.

124=496\frac{1}{24} = \frac{4}{96}

132=396\frac{1}{32} = \frac{3}{96}

Combined work = 496+396\frac{4}{96} + \frac{3}{96}

Combined work = 796\frac{7}{96} of the tank in 1 hour


Time to fill the complete tank:

Time = 1÷7961 \div \frac{7}{96}

Time = 1×9671 \times \frac{96}{7}

Time = 967\frac{96}{7} hours


Converting to mixed number:

967=1357\frac{96}{7} = 13\frac{5}{7} hours

Therefore, the time taken to fill the tank is 135713\frac{5}{7} hours.

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