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A large water tank is serviced by 3 inlet pipes and one drain pipe. Pipe A fills the tank in 6 hours alone at full force; pipe B fills the tank in 9 hours alone at full force; pipe C fills the tank in 15 hours alone at full force. The drain pipe empties a full tank in five hours if no inlet pipe is running.
Suppose the tank is empty. The drain pipe is open. Pipes A and B are turned on at full force. When the tank is one-third full, pipe C is turned on at full force. When the tank is three-quarters full, the drain pipe is closed. To the nearest minute, how long would it take to completely fill the tank in this scenario?
1 Answer
- chevalier rougeLv 41 decade agoFavorite Answer
there are 3 steps: A+B-D,A+B+C-D, A+B+C
the speed of A = 1V/6, one volume per 6 hr or 1/6 V/hr
B= 1/9 V/hr
C= 1/15 V/hr
D= 1/5 V/hr
the first stage A+B-D = 1/6 +1/9 -1/5 = (15+10-18)/90=7/90
the 2nd stage opens C :=7/90+1/15=13/90
the 3rd stage closes D :=13/90-1/15=31/90
so you need T=t1+t2+t3
where 1/3=t1(7/90)
3/4-1/3=t2(13/90)
and 1-3/4=t3(31/90)
finally the number that you get is T is the number of hours needed. remember to change the decimals into minutes
:-) cool problem!!
