SOLUTION: two taps can fill a tank in 15 and 12 minutes respectively.a third tap can empty it in 20 minutes.if all the taps are opened at the same time then in how much time will the tank b

Algebra ->  Rate-of-work-word-problems -> SOLUTION: two taps can fill a tank in 15 and 12 minutes respectively.a third tap can empty it in 20 minutes.if all the taps are opened at the same time then in how much time will the tank b      Log On


   

Question 622900: two taps can fill a tank in 15 and 12 minutes respectively.a third tap can empty it in 20 minutes.if all the taps are opened at the same time then in how much time will the tank be filled??
Found 2 solutions by josmiceli, ankor@dixie-net.com:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Add the rates of filling and subtract the rate of
emptying to get the net rate of filling
Tap 1: ( 1 tank filled ) / ( 15 min )
Tap 2: ( 1 tank filled ) / ( 12 min )
Tap 3: ( 1 tank emptied ) / ( 20 min )
Let +t+ = the time needed to fill the tank
------------
+1%2F15+%2B+1%2F12+-+1%2F20+=+1%2Ft+
Multiply both sides by +60t+
+4t+%2B+5t+-+3t+=+60+
+6t+=+60+
+t+=+10+
It takes 10 minutes to fill the tank

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
two taps can fill a tank in 15 and 12 minutes respectively.
a third tap can empty it in 20 minutes.
if all the taps are opened at the same time then in how much time will the tank be filled?
:
Let t = time required when all three taps are open.
Let the completed job = 1; (a full tank)
:
A typical shared work equation.
:
Fill + Fill - Empty = a full tank
t%2F15 + t%2F12 - t%2F20 = 1
multiply by 60 to clear the denominators, results
4t + 5t - 3t = 60
6t = 60
t = 60/6
t = 10 minutes with all taps open