Question

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?

Answer 1

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 \(1/R\).

Answer 2

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

Answer 3

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

Answer 4

What is the right value?

Answer 5

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

Answer 6

put it in your calculator if you like.

Answer 7

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

Answer 8

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.

Answer 9

I get 0,1249 for R.

Answer 10

So the righ answer is 80 hours.

Answer 11

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