SOLUTION: I think I picked the right area for this [?]. I am at a loss to set this one up in formulas. I get the rate/minute idea but get lost setting up an equation to get the answer.

Question 436290: I think I picked the right area for this [?].
I am at a loss to set this one up in formulas.
I get the rate/minute idea but get lost setting up an equation to get the answer.
Thank you.
A fire engine starts pumping water at 9:20 am at the rate of 800 gallons per minute. Another fire engine, pumping at the rate of 1,000 gallons per minute, starts at 9:30 am. At what time will the 2 engines have pumped the same number of gallons?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A fire engine starts pumping water at 9:20 am at the rate of 800 gallons per minute.
Another fire engine, pumping at the rate of 1,000 gallons per minute, starts at 9:30 am.
At what time will the 2 engines have pumped the same number of gallons?
:
Let m = no. of minutes first fire engine pumping time
then
(m-10) = pumping time of the 2nd fire engine, (started 10 min after the 1st engine)
:
2nd engine gal = 1st engine gal
1000(m-10) = 800m
1000m - 10000 = 800m
1000m - 800m = 10000
200m = 10000
m =
m = 50 minutes after 1st engine starts pumping
therefore
9:20 + :50 = 10:10, they will have pumped the same
:
:
Check this by finding the actual amt for each pump, they should be equal
50*800 = 40000
40*1000= 40000

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