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(D) Let the total capacity of the tank be the $LCM$ of $4, 8, 12,$ and $10$,which is $120$ units.Efficiency of $P = 120/4 = 30$ units/hr.Efficiency of

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$P, Q$ and $S$ are opened

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(D) Let the total capacity of the tank be the $LCM$ of $4, 8, 12,$ and $10$,which is $120$ units.Efficiency of $P = 120/4 = 30$ units/hr.Efficiency of $Q = 120/8 = 15$ units/hr.Efficiency of $R = 120/12 = 10$ units/hr.Efficiency of $S = -120/10 = -12$ units/hr (negative because it empties).$(i)$ $Q$ opened alone: Time $= 120/15 = 8$ hours.(ii) $P$ and $S$ opened: Net efficiency $= 30 - 12 = 18$ units/hr. Time $= 120/18 = 6.67$ hours.(iii) $P, R$ and $S$ opened: Net efficiency $= 30 + 10 - 12 = 28$ units/hr. Time $= 120/28 = 4.28$ hours.(iv) $P, Q$ and $S$ opened: Net efficiency $= 30 + 15 - 12 = 33$ units/hr. Time $= 120/33 = 3.64$ hours.Comparing the times: $3.64 < 4.28 < 6.67 < 8$. Therefore,option $(d)$ takes the least time.

Three taps $P, Q$ and $R$ separately can fill a tank completely in $4, 8$ and $12$ hours respectively. Another tap $S$ can empty the filled tank in $10$ hours. Which of the following combinations will fill the empty tank in the least amount of time?

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