S99-16

Official Solution:

A bathtub has two faucets, P and Q, and one drain. Faucet P alone can fill the whole tub in ten minutes, and faucet Q alone can fill the whole tub four minutes faster than the drain can empty the whole tub. With faucets P and Q both running and the drain unstopped, the tub fills in six minutes. How long would the drain take to empty the whole tub?

A. 5 and \(\frac{5}{11}\) minutes
B. 6 minutes
C. 10 minutes
D. 19 minutes
E. 30 minutes


Let \(t\) stand for the desired time, so that the drain can empty the tub in \(t\) minutes and faucet Q can fill the tub in \((t - 4)\) minutes. Also, the drain does negative work as it empties the tub. The rates for the pipes and the drain are thus

P: 1 tub in \(10\) min \( = \frac{1}{10}\) tub per min

Q: 1 tub in \((t - 4)\) min \( = \frac{1}{t - 4}\) tub per min

Drain: -1 tub in \(t\) min \(= -\frac{1}{t}\) tub per min

Using the fact that all three fixtures together take 6 minutes to fill 1 tub, set up an RTW chart, and use the chart to calculate the total work quantities in the last column.
Rate (tub/min) \(\times\) Time (min) \(=\) Total Work (tubs) P \(\frac{1}{10}\) \(\times\) 6 \(=\) \(\frac{3}{5}\) Q \(\frac{1}{t - 4}\) \(\times\) 6 \(=\) \(\frac{6}{t - 4}\) Drain \(\frac{-1}{t}\) \(\times\) 6 \(=\) \(\frac{-6}{t}\) Total n/a n/a 1
Set up an equation summing up the work:
\(\frac{3}{5} + \frac{6}{t - 4} - \frac{6}{t} = 1\)

Multiply by the common denominator, \(5t(t - 4)\):
\(3t(t - 4) + 6(5t) - 6(5)(t - 4) = 5t(t - 4)\)
\(3t^2 - 12t + 30t - 30t + 120 = 5t^2 - 20t\)
\(0 = 2t^2 - 8t - 120\)
\(0 = t^2 - 4t - 60\)
\(0 = (t - 10)(t + 6)\)

\(t = 10\) or \(t = -6\)

The negative value is absurd, so \(t = 10\); the drain can empty the tub in 10 minutes.


Answer: C