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- anonymous

A tank has two taps that can fill it
within 24 hours and 30 hours respectively, and a drain that
empties in 16 hours. If we open simultaneously
the whole set (two faucets and a drain), how
time the tank will be filled?

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Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
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- anonymous

- schrodinger

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- JamesJ

The way to think of this and all problems like this is in terms of rates.
The first tap fills 1/24th of a tank per hour; the second tap at a rate of 1/30.
Therefore their combined rate is \( 1/24 + 1/30 \).
The drain drains the tank at a rate of 1/16.
Therefore the combined rate of all three is
\[ R = \frac{1}{24} + \frac{1}{30} - \frac{1}{16} \]
Then the time required to fill the tank is \(1/R\).

- anonymous

I have found 10,43 hours. Am i right?

- JamesJ

What value do you have for R? 10.43 hours isn't right.

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- anonymous

What is the right value?

- JamesJ

As R = 1/24 + 1/30 - 1/16, just do the math. What value do you get for R?

- JamesJ

put it in your calculator if you like.

- JamesJ

I get R = 0.0125 = 1/80 . Hence the number of hours 1/R = ...

- JamesJ

For the record, if there was no drain, then the two taps would fill the sink at a combined rate of 1/24 + 1/30 = 0.075
and therefore the sink would fill in 1/0.075 = 13.3 hours.
Whatever answer you get with the drain, it had obviously better be longer than 13.3 hours.

- anonymous

I get 0,1249 for R.

- anonymous

So the righ answer is 80 hours.

- JamesJ

R = 0.0125 and hence 1/R = 80 hours, yes.

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