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1. Three pipes A, B and C can fill a tank in 6 hours, 9 hours and 12 hours respectively. B and C are opened for half an hour, then A is also opened. The time taken by the three pipes together to fill the remaining part of the tank is -

A. 3 hours

B. 2 hours

C. $2 \frac{1}{2}$ hours

D. $3 \frac{1}{2}$ hours

Solution:

In half an hour (B + C) must have filled

= \frac{4}{2} + \frac{3}{2} = \frac{7}{2} units

Capacity left

=  36 - \frac{7}{2} = \frac{65}{2} units

Now all pipes will fill the remaining tank

\begin{aligned} = \frac{65}{2 \times (6 + 4 + 3)} \\ = \frac{65}{2 \times 13} \\ = \frac{5}{2} \\ = 2 \frac{1}{2} hours \end{aligned}

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