SOLUTION: Hi again, I need some assistance with this mixture problem: It takes a pipe 10 minutes less than another one to fill a tank of water. Both pipes together can fill the tank in

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Question 45662: Hi again,
I need some assistance with this mixture problem:
It takes a pipe 10 minutes less than another one to fill a tank of water. Both pipes together can fill the tank in 12 minutes. How long will it take each one to fill the tank separately?
I am having trouble on how to start setting the equation up.
Thanks again,
Lou

Found 2 solutions by Fermat, stanbon:
Answer by Fermat(136) About Me  (Show Source):
You can put this solution on YOUR website!
Let the two pipes have fill rates of R1 and R2 m^3/min
Let the volume of the tank be V m^3
The time for pipe1 to fill the tank is,
T1 = V/R1 min
similiarly,
T2 = V/R2
The difference in times is 10 min, so we can write,
T1 - T2 = 10
(at the moment it doesn't matter whether you say T1-T2 or T2-T1, since we haven't said which of R1 and R2 is the greater. It will all work out in the end)
substituting for T1 and T2,
V/R1 - V/R2 = 10
V(1/R1 - 1/R2) = 10
====================
If pipe1 fills the tank at R1 m^3/min and pipe2 fills the tank at R2 m^3/min, then together they both fill the tank at a rate of (R1+R2) m^3/min.
T3 = V/(R1+R2)
12 = V/(R1 + R2)
================
You now have two equations in R1 and R2 from which you can solve for R1 and R2 separately, in terms of V.
You should end up with T1 = 30 mins, T2 = 20 mins.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
It takes a pipe 10 minutes less than another one to fill a tank of water. Both pipes together can fill the tank in 12 minutes. How long will it take each one to fill the tank separately?
Faster Pipe DATA:
Time to do the job = x minutes; Rate= (1/x) job/min
Slower Pipe Data:
Time to do the job = x-10 minutes: Rate = 1/(x-10) job/min
Filling Together Data:
Time to do the job = 12 minutes: Rate = 1/12 job/min
EQUATION:
rate + rate = rate together
1/x + 1/(x-10) = 1/12
[x-10+x]/[x(x-10} = 1/12
12(2x-10) = x^2-10x
x^2-34x+120=0
(x-30)(x-4)=0
x= 30 minutes or x= 4 minutes
Only the x=30 minute answer applies because the slower pipe time is then 20 min.
Slower pipe time is 30 minutes.
Faster pipe time is 20 minutes.
Cheers,
Stan H.