Three taps $P,$ $Q$ and $R$ can fill a drum in $10, 20$ and $30$ hours respectively. If $P$ is open all the time and $Q$ and $R$ are open for one hour each alternately, then the drum will be full in

6

hours

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6.5

hours

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7

hours

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7.5

hours

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Solution

The correct option is C $7$ hours
Drum filled by both $P$ and $Q$ in $1$ hour $= \frac{1}{10} + \frac{1}{20} = \frac{3}{20}$
Drum filled by both $P$ and $R$ in $1$ hour $= \frac{1}{10} + \frac{1}{30} = \frac{2}{15}$
Drum filled in first $2$ hours $= \frac{3}{20} + \frac{2}{15} = \frac{17}{60}$
So, drum filled in $6$ hours $= 3 \times \frac{17}{60} = \frac{17}{20}$
$∴$ Remaining part $= 1 - \frac{17}{20} = \frac{3}{20}$
Now, it is the turn of $P$ and $Q$ and $\frac{3}{20}$ part is filled by $P$ and $Q$ in $1$ hour.
Therefore, total time taken to fill the drum $= (6 + 1) = 7$ hours

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