question with word problems

Question

a pool has an inlet pipe to fill it and an outlet pipe to empty it. It takes 2 hours longer to empty the pool than it does to fill. The inlet pipe is turned on to fill the pool, but the outlet pipe is accidentally left open. Despite this the pool fills in 8 hours. How long would it have taken the inlet pipe to fill the pool if the outlet pipe had been turned off? Round to the nearest tenth of an hour

Answer 1

The discharge rate of the outlet pipe is X and the input rate is Y. If the volume of the pool is V then X=V/T1 where T1 is the time it takes to empty the pool with volume V. To fill the same volume T2=V/Y. T1=T2+2.

When both pipes are open and working, the effective rate of fill=Y-X and T3=8 hours. Y-X=V/T3=V/8.

Therefore, since V=YT2=XT1=X(T2+2) we have Y=V/T2 and X=V/(T2+2), so Y-X=V(1/T2-1/(T2+2))=V/8.

Cancel out V: 1/T2-1/(T2+2)=1/8. Multiply through by 8T2(T2+2): 8T2+16-8T2=T2^2+2T2.

T2^2+2T2-16=0. Complete the square: T2^2+2T2+1=17; (T2+1)^2=17, so T2+1=√17 (T2 has to be positive), making T2=√17-1=3.12 hrs. T1=5.12 hrs, 2 hours longer. 

CHECK

1/3.12-1/5.12=1/8 (approx). To the nearest tenth of an hour the time to fill the pool would be T2=3.1 hours.