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"A large tank can be filled by 2 similar small pumps and 1 larger pump working together in 1 hour and 12 minutes. The larger pump takes one hour less than the smaller pump to fill up the tank alone. Find how long each pump takes, given that V/P = T where V = volume of the tank, P = Pump, T = Time."

These are the exact words from a mathematics exam question and I'm unsure how to interpret P (I assumed it was the rate). If someone could provide a solution, that would be great.

Mark Geha
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  • Please clarify if both small pumps running concurrently take 1 hour more than the larger pump or one of the small pumps running in isolation takes 1 hour more than the larger pump working alone. This makes a difference in the answer. – vvg Sep 23 '20 at 02:18
  • It should mean in isolation – Mark Geha Sep 23 '20 at 07:29
  • Not to nitpick, but when you say 'run in isolation' for baselining do you mean, (a) two small pumps concurrently (i.e., both of them) run without the large pump or (b) one small pump run in isolation without the large pump and the other small pump. @MatthewHolder's answer below is correct for scenario (a). – vvg Sep 23 '20 at 13:16
  • lol I thought I was correct but someone down voted me >:-O – Matthew H. Sep 23 '20 at 13:42
  • And @vvgiri, doesn't my answer correspond to scenario (b)? The time $t_S$ represent the amount of time is takes one small pump to fill the tank working all by itself. The expressing $1/t_S$ represents the rate at which the small pump fills the tank working all by itself. – Matthew H. Sep 23 '20 at 13:44
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    @MatthewHolder - I made a typo. Sorry, your response is for scenario (b). I am not sure why there is a downvote. If the downvoter could add a comment, it can help in providing an improvement or correction. – vvg Sep 23 '20 at 13:48

1 Answers1

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This reminds me of the painting a room word problem. Let's proceed as follows.

Suppose the large pump can fill the tank in $t_L$ hours working alone, and the small pump can fill the tank in $t_S$ hours working alone.

Then $t_L=t_S-1$ and $\frac{1}{t_S}+\frac{1}{t_S}+\frac{1}{t_L}=\frac{1}{1.2}$. Combining these equations and simplifying yields $\frac{2}{t_S}+\frac{1}{t_S-1}=\frac{1}{1.2}$ This implies $t_S=4$ and $t_L=3$. (Note: $t_S=3/5$ is extraneous).

Matthew H.
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  • $2/t_s + 1/(t_s - 1) = 1/1.2$ gives $t_s = 3/5$ or $4$ hours. (Quadratic equation in $t_s$ with two real roots).

    See https://www.wolframalpha.com/input/?i=solve+2%2Ft+%2B+1%2F%28t+-1%29+%3D+1%2F1.2+over+the+reals

     – vvg Sep 23 '20 at 02:41
  • whoops! thank you – Matthew H. Sep 23 '20 at 12:08