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1. Three taps A, B and C can fill a tank in 12, 15 and 20 hours respectively. If A is open all the time and B and C are open for one hour each alternately, the tank will be full in:

A. 6 hours

B. $6 \frac{2}{3}$ hours

C. 7 hours

D. $7 \frac{1}{2}$ hours

Solution:

\begin{aligned} (A + B)'s 1 hour work \\ = \frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20} \\ (A + C)'s 1 hour work \\ = \frac{1}{12} + \frac{1}{20} = \frac{8}{60} = \frac{2}{15} \\ Part filled in 2 hrs \\ = \frac{3}{20} + \frac{2}{15} = \frac{17}{60} \\ Part filled in 6 hrs \\ = 3 \times \frac{17}{60} = \frac{17}{20} \\ Remaining part \\ = 1 - \frac{17}{20} = \frac{3}{20} \end{aligned}

Now it is the turn of A and B and

\frac{3}{20}

part is filled by A and B in 1 hour
∴ Total time taken to fill tank
= (6 + 1) hrs
= 7 hrs

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