We are given that the rate of the first pump is 1/24. Since the second pump has a rate that is 2/3 of the first pump, the rate of the second pump is (1/24) x (2/3) = 2/72 = 1/36.

If the first pump is used for 8 hours, 1/24 x 8 = 8/24 = 1/3 of the tank is filled, leaving 2/3 left to be filled.

Since the second pump is then added to complete the job, the combined rate is 1/24 + 1/36 = 3/72 + 2/72 = 5/72. We can let t = the time it takes to complete the job:

(5/72)t = 2/3

5t/72 = 2/3

15t = 144

t = 144/15 = 48/5 = 9 ⅗ hours = 9 hours and 36 minutes

Thus, the tank is filled in 8 hours + 9 hours and 36 minutes = 17 hours and 36 minutes.

Answer: D
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should be (D) ?
3T + 2*(T-8) = 72
5T = 88
T = 17.6 h = 17 h 36 min

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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