Bunuel wrote:
Two different pumps are used to fill an empty water tank. The first pump can fill the tank in 24 hours and the second pump works at 2/3 the speed of the first pump. The first pump is used alone for 8 hours and then the second pump is added until the tank is full. How long does it take in total for the empty tank to be filled?
A. 9 hours and 36 minutes
B. 10 hours
C. 11 hours and 12 minutes
D. 17 hours and 36 minutes
E. 18 hours
If you happen not to see the inverse proportion shortcut, the "long way" is not bad at all. One hitch: at the end add time taken in
both stages
Use Stage 1 to find amount of work finished by Pump A working alone. Use Stage 2 to find time for both pumps to finish.
Stage 1: Pump A works alone. Work finished?
Pump A's rate = \(\frac{1}{24}\)
Pump A's time = 8 hours
Work finished by Pump A alone:
\(W=(r*t)=
(\frac{1}{24}*8)=\frac{8}{24}=\frac{1}{3}\)
of work is finished
So \(\frac{2}{3}\) remains.
Stage 2: Pumps A and B work together
Pump A's rate =\(\frac{1}{24}\)
Pump B's rate is \(\frac{2}{3}\) of A's rate
Pump B's rate: \((\frac{2}{3}*\frac{1}{24})=\frac{1}{36}\)
Combined rate of A and B:
\((\frac{1}{24}+\frac{1}{36})=(\frac{3}{72}+\frac{2}{72})=\frac{5}{72}\)
Work, W, remaining = \(\frac{2}{3}\)
Time taken by A and B working together?
Time\(=\frac{W}{r}=\frac{\frac{2}{3}}{\frac{5}{72}}=(\frac{2}{3}*\frac{72}{5})=\)
\(\frac{48}{5}\) hours
Total time taken to fill the
entire tank?
(Stage 1 + Stage 2) = (8 hrs + \(\frac{48}{5}\) hrs) =
\((\frac{40}{5}+\frac{48}{5})=\frac{88}{5}=17\frac{3}{5}\) hours
ANY fraction of an hour * 60 = # of minutes
(\(\frac{3}{5}\) * 60) = 36 minutes
17 hours, 36 minutes
Answer D
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