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by bblast » Thu Jun 09, 2011 5:45 am
An adjustable intake valve fills the first 12 gallons of a 24-gallon tank at a rate of 28 gallons per minute. What average rate, in gallons per minute, must the valve allow while filling the remaining 12 gallons in order to have an average rate of 40 gallons per minute for filling the whole tank?
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72

D

cant this be solved by weighted averages ?
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by cans » Thu Jun 09, 2011 5:51 am
An adjustable intake valve fills the first 12 gallons of a 24-gallon tank at a rate of 28 gallons per minute. What average rate, in gallons per minute, must the valve allow while filling the remaining 12 gallons in order to have an average rate of 40 gallons per minute for filling the whole tank?
12 gallons at rate of 28 g/m
thus time taken for first 12G= 12/28 min = 3/7 min
avg rate=40g/min
time for 24 gallons = 24/40 = 6/10 min
thus time left for next 12 gallons = 3/5 - 3/7 = 6/35
no. of gallons = 12
rate = 12/(6/35) = 70 gallons per minute
IMO D
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by GMATGuruNY » Thu Jun 09, 2011 6:28 am
bblast wrote:An adjustable intake valve fills the first 12 gallons of a 24-gallon tank at a rate of 28 gallons per minute. What average rate, in gallons per minute, must the valve allow while filling the remaining 12 gallons in order to have an average rate of 40 gallons per minute for filling the whole tank?
.
52
60
65
70
72

D

cant this be solved by weighted averages ?
The size of the tank is immaterial.
What matters is that the rate to fill the first half of the tank is 28 gallons per minute, the rate to fill the whole tank is 40 gallons per minute.

Let tank = 560 gallons.
Time to fill the whole tank at 40 gallons per minute = w/r = 560/40 = 14 minutes.
Time to fill the first half of the tank at 28 gallons per minute = w/r = 280/28 = 10 minutes.
Time to fill the remaining half of the tank = total time - time to fill the first half = 14-10 = 4 minutes.
Rate to fill the second half = w/t = 280/4 = 70 gallons per minute.

The correct answer is D.
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by bblast » Thu Jun 09, 2011 6:47 am
Awesome approach on this Mitch.
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