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Bunuel
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

First pump time = 24 hours

Second pump time = (3/2)*24 = 36 hours (since speed is 2/3 so the time should be 3/2 times)

Let total time to fill the tank = T

the First pump works for T hours and second pump works for (T-8) Hours

i.e. T/24 + (T-8)/36 = 1

i.e. 3T + 2*(T-8) = 72

i.e. 5T = 56


i.e. T = 11 1/5 hour = 11 hours 12 mins

Answer: Option C

should be (D) ?
3T + 2*(T-8) = 72
5T = 88
T = 17.6 h = 17 h 36 min
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+1 for option D. The key here is to recognize that if the speed is 2/3 , the time taken becomes 3/2.
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Bunuel
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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First pump takes 24hrs.
Second pump works at 2/3 the speed of first, hence it takes 3/2 of the time, i.e.36hrs.
Work done per hour by first pump=1/24
Work done per hour by second pump=1/36
Overall work done per hour=1/24+1/36=5/72.
First pump in 8 hrs filled [8*(1/24)] of the tank, i.e. 1/3rd of the tank.
Time taken to fill remaining 2/3 of tank = (2/3)/(5/72) = 9hrs 36min
Therefore overall time taken = 8hrs + 9hrs36min = 17hours 36 min
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Let's take the total work as 72 units (LCM of 24 and 36)
A does 72/24 = 3 units of work and B does 72/36 = 2 units of work per hour

For 8 hours A does 8*3 = 24 units of work
Remaining work = 72 -24 = 48 units

A+B together complete 5 units of work

48/5 = 9.xx

Total time: 8+9.xx = 17.xx

Option D
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