SOLUTION: Please help me. A company buys a new pump that can pump 10.0 gal./min faster than its old one. It can fill the company's 500.0 gallon tank 15.0 minutes faster than the old pump.

Question 1057921: Please help me.
A company buys a new pump that can pump 10.0 gal./min faster than its old one. It can fill the company's 500.0 gallon tank 15.0 minutes faster than the old pump. What was the speed of the old pump?
Round to three significant digits.
Answer by LinnW(1048) (Show Source): You can put this solution on YOUR website!
Set s = speed in gals per minute of the old pump
(s+10) = speed of the new pump
Set t = time to fill the tank with the old pump
s*t = 500, that is at the speed of the old pump times old time = 500 gallons pumped.
(s+10)(t-15) = 500
that is the new pump speed times a time of 15 minutes less fills the 500 gallons
Since s*t = 500, if we divide each side by s we get t = (500/s)
Substitute (500/s) for t in (s+10)(t-15) = 500

add -350 to each side

multiply each side by s

add -150s to each side

divide each side by -5

Using a quadratic solver we have two roots
approximately 13.93 and -23.93
Since the pump speed must be positive, the original
pump speed is 13.93 gallons per minute.
Notice that since s*t = 500, (13.93)(t) = 500
t = 500/13.93
t = 35.893
Let's verify that (s+10)(t-15) = 500 works
(13.93+10)(35.893-15) ?= 500
(23.93)(20.893) = 499.96949
So we must be on track
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