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A certain pipe can fill a swimming pool in $2$ hours; another pipe can fill it in $5$ hours; a third pipe can empty the pool in $6$ hours. With all three pipes turned on exactly at the same time, and starting with an empty pool, how long will it take to fill the pool?

My try:

Let $m$ be the time required, then:

$m = \dfrac{1}{\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{1}{6}} = 1.875$

Question:

Is it correct?

Mathsource
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Mathslover
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1 Answers1

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Yes this is correct, let $P/h$ denotes the "Pool per Hour" unit.

  • pipe $1$ volumetric flow rate is $\frac{1}{2} P/h$ (a half pool per hour $\equiv$ one pool in two hours)
  • pipe $2$ volumetric flow rate is $\frac{1}{5} P/h$
  • pipe $3$ volumetric flow rate is $-\frac{1}{6} P/h$

It follows that the total debit is $$\frac{1}{2}+\frac{1}{5}-\frac{1}{6} = \frac{8}{15} P/h,$$ and thus the $3$ pipes fill $8$ pools per $15$ hours, which is equivalent to $1$ pool per $15/8 = 1.875$ hours.

Surb
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