Question 1117269
A joint-work problem.   Some tutors have lessons here covering how to solve this type of problem.  You should investigate and learn from them if you want to master it. <br>

The approach is to see how much of the tank fills in a unit of time.  A good choice for the unit of time is one minute, because the fill and empty times are given in minutes.  We of course assume the tank is empty at the start.<br>

A fills 1/6 of the tank per minute.
B fills 1/9 of the tank per minute.
C "fills" -1/15 of the tank per minute.   Here the minus sign indicates it is actually emptying the tank. <br>  


Fill rate = {{{ (1/6) + (1/9) - (1/15) =  (15/90) + (10/90) - (6/90) = 19/90 }}} tanks per minute

The time needed to fill one tank is therefore  {{{ 1/((19/90)) = 90/19 }}} or about {{{ highlight(4.7368)}}} minutes (4 min, 44.2 sec).  <br>

Check:  This is more of a qualitative check.  Does our answer make sense?
  Say the pipe emptying the tank was turned off.   The two taps would fill the tank at a rate  (1/6)+(1/9) = 5/18 tank/min.   This corresponds to a fill time of  18/5 = 3.6 minutes.   In the problem scenario the pipe is emptying the tank at 1/15 tank/min which is comparatively slower than the fill rate,  so we'd expect the fill time to be extended, but only moderately.  Our answer of 4.7368 minutes is moderately longer than 3.6 minutes so at least the answer seems about right.