Discussion
A tank has three taps P, Q and R. Taps P and Q can fill the tank in 1.5 and 2 hours, respectively. Tap R can empty the completely filled tank in just half hour. Tap P is opened at 8 am, tap Q is opened at 9am and tap R is opened at 10 am. At what exact time will the tank be empty?
Answer: C
Work in rates: a pipe that fills a tank in \( n \) units of time fills \( \tfrac{1}{n} \) of it per unit, and rates add (an emptying pipe counting negative).
Per hour: \( P = \tfrac{2}{3} \), \( Q = \tfrac{1}{2} \), \( R = -2 \).
By 10 a.m. P has run 2 hours and Q one hour:
\[ 2 \times \frac{2}{3} + 1 \times \frac{1}{2} = \frac{4}{3} + \frac{1}{2} = \frac{11}{6} \]
From 10 a.m. the net rate turns negative:
\[ \frac{2}{3} + \frac{1}{2} - 2 = -\frac{5}{6} \text{ per hour} \]
\[ t = \frac{11/6}{5/6} = 2.2 \text{ hours} = 2 \text{ h } 12 \text{ min} \]
\[ 10{:}00 + 2{:}12 = \textbf{12{:}12 p.m.} \]
The calculation treats the inflow as accumulating without an overflow, which is the standard convention for these problems.
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