SOLUTION: Pipes A and B can fill a tank in 45 minutes when used together. Alone, pipe A can fill the tank in one hour. How long would it take to fill the tank if only pipe B is used?

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Pipes A and B can fill a tank in 45 minutes when used together. Alone, pipe A can fill the tank in one hour. How long would it take to fill the tank if only pipe B is used?      Log On


   

Question 309889: Pipes A and B can fill a tank in 45 minutes when used together. Alone, pipe A can fill the tank in one hour. How long would it take to fill the tank if only pipe B is used?
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!
You must add their rates of pumping to get the rate
of pumping together
In words, the rate is:
(1 tank)/(time to fill 1 tank)
pipe A:
1%2F1 1 tank/1 hr
pipe B:
1%2Fx 1 tank/x hrs
A and B together:
1%2F1+%2B+1%2Fx+=+1%2F.75
.75x+%2B+.75+=+x
.25x+=+.75
x+=+3
Using pipe B alone, it takes 3 hrs to fill the tank
check:
1+%2B+1%2F3+=+1%2F.75
4%2F3+=+4%2F3
OK

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Pipes A and B can fill a tank in 45 minutes when used together.
Alone, pipe A can fill the tank in one hour.
How long would it take to fill the tank if only pipe B is used?
:
Change 1 hr to 60 min
:
Let b = time for B to fill it alone
:
Let the completed job = 1 (a full tank)
:
45%2F60 + 45%2Fb = 1
Reduce the fraction
3%2F4 + 45%2Fb = 1
multiply by 4b, results:
3b + 4(45) = 4b
:
3b + 180 = 4b
:
180 = 4b - 3b
:
180 = b
:
B can fill the tank alone in 180 minutes
:
:
Check:
45/60 + 45/180 =
.75 + .25 = 1