The time (t) required to empty a tank varies inversely as the rate (r) of pumping. A pump can empty a tank in 90 minutes at the rate of 1200 L/min. How long will the pump take to empty the tank at 3000 L/min?

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The time (t) required to empty a tank varies inversely as the rate (r) of pumping. A pump can empty a tank in 90 minutes at the rate of 1200 L/min. How long will the pump take to empty the tank at 3000 L/min?

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Answer 1 · 2016-05-03T12:33:26

$t = 36 minutes$ $t=36" minutes"$

Explanation:

$From first principles$ $color(brown)("From first principles")$

90 minutes at 1200 L/min means that the tank holds $90 \times 1200 L$ $90xx1200 L$

To empty the tank at a rate of 3000 L/m will take the time of

$\frac{90 \times 1200}{3000} = \frac{108000}{3000} = 36 minutes$ $(90xx1200)/3000 = (108000)/3000 = 36" minutes"$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Using the method implied in the question$ $color(brown)("Using the method implied in the question")$

$t α \frac{1}{r} \Rightarrow t = \frac{k}{r}$ $t" "alpha" "1/r" "=>" "t=k/r" "$ where k is the constant of variation

Known condition: $t = 90; r = 1200$ $t=90" ; "r=1200$

$\Rightarrow 90 = \frac{k}{1200} \Rightarrow k = 90 \times 1200$ $=>90=k/1200=> k=90xx1200$

So $t = \frac{90 \times 1200}{r}$ $t=(90xx1200)/r$

Thus at $r = 3000$ $r=3000$ we have

$t = \frac{90 \times 1200}{3000}$ $t=(90xx1200)/(3000)$

Observe that this is exactly the same as in first principles.

$t = 36 minutes$ $t=36" minutes"$

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The time (t) required to empty a tank varies inversely as the rate (r) of pumping. A pump can empty a tank in 90 minutes at the rate of 1200 L/min. How long will the pump take to empty the tank at 3000 L/min?

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