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: 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}W\) finished

\(\frac{2}{3}W\) remains

Stage 2: Pumps A and B work together

Pump A's rate =

\(\frac{1}{24}\)Pump B's rate is

\(\frac{2}{3}\) that 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,

\(Time=\frac{W}{r}=\frac{\frac{2}{3}}{\frac{5}{72}}=(\frac{2}{3}*\frac{72}{5})=\frac{48}{5}\) hoursTotal time taken to fill the

entire tank?

(Stage 1 + Stage 2) = (8 hours +

\(\frac{48}{5}\) hours) =

\((\frac{40}{5}+\frac{48}{5})=\frac{88}{5}=17\frac{3}{5}\) hours

Only one answer has 17 hours: D

Answer D

OR, 17\(\frac{3}{5}\)hrs = 17 hrs + how many minutes?

(ANY fraction of an hour * 60) = # of minutes

(\(\frac{3}{5}\) * 60) = 36 minutes, hence

17 hours, 36 minutes

Answer D

_________________

The only thing more dangerous than ignorance is arrogance.

-- Albert Einstein