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Two water pumps, working simultaneously at their respective constant

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Re: Two water pumps, working simultaneously at their respective constant  [#permalink]

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New post 13 Nov 2019, 05:19
\(5R=\frac{1}{4} \implies 3R=\frac{20}{3}\) (10 sec)
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Re: Two water pumps, working simultaneously at their respective constant  [#permalink]

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New post 24 Jan 2020, 02:53
Rates Pool/Hour; Both of their rate is 1/4 (Since they fill 1 pool in 4 hours)

Fast - X;
Slow - X/1.5 = 10X/15 = 2x/3

X + 2X/3 = 1/4 => X= 3/20 Pool per Hour

Since we marked faster one with the X if we want it's time we just need it's reciprocal 20/3 H.
(Work=timexRate => Time => Work/Time => Time = 1/(3/20) = 20/3)
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Re: Two water pumps, working simultaneously at their respective constant   [#permalink] 24 Jan 2020, 02:53

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Two water pumps, working simultaneously at their respective constant

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