<PRE><B>Question 1147308</B>
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Once again, tutor @josgarithmetic sets up the problem correctly for solving and then botches the arithmetic to come up with a nonsensical answer.  Clearly, if either tap alone can fill the tank faster than drain pipe can drain it, then with the two taps working together the tank will get filled at some point.&lt;br&gt;
And once again, tutor @maththerapy states that the first solution is clearly wrong and then shows the answer without saying or showing anything about how to get the answer, making the response useless to the student.&lt;br&gt;
And tutor @ikleyn, as usual, presents a clear and correct solution to the problem, using the standard algebraic approach.&lt;br&gt;
Here is an alternative to that approach that many students prefer because it avoids working with fractions.&lt;br&gt;
Consider the least common multiple of the three given times.  The LCM of 6, 9, and 15 is 90.&lt;br&gt;
Now suppose we have 90 minutes, and there are multiple tanks to be filled.&lt;br&gt;
Tap A can fill 90/6 = 15 tanks in 90 minutes; tap B can fill 90/9 = 10 tanks in 90 minutes.  And the drain pipe C can empty 90/15 = 6 tanks in 90 minutes.&lt;br&gt;
So in 90 minutes the number of tanks that could be filled is 15+10-6 = 19.&lt;br&gt;
And since in 90 minutes 19 tanks could be filled, the one tank can be filled in 90/19 minutes.</PRE>