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Rates Question

by RBF3 » Sun Jul 17, 2011 3:30 pm
Pumps A, B, and C operate at their respective constant rates. Pumps A and B, operating simultaneously, can fill a certain tank in 6/5 hours; pumps A and C, operating simultaneously, can fill the tank in 3/2 hours; and pumps B and C, operating simultaneously, can fill a tank in 2 hours. How many hours does it take pumps A, B, and C operating simultaneously, to fill the tank?

1/3
1/2
2/3
5/6
1

The correct answer is 1, just not sure how to get there?
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by goalevan » Sun Jul 17, 2011 4:32 pm
Our answer will eventually be calculated in this form:

Time = work / rate

How many hours (time) does it take pumps A, B, and C to fill 1 tank (work)?

t hours = 1 tank / rate in tanks/hour of A, B, C working simultaneously

A: rate of A working alone, B: rate of B working alone, C: rate of C working alone:

A + B = 1 tank / (6/5) hours = 5/6 tanks/hr
B + C = 1 tank / (3/2) hours = 2/3 tanks/hr
A + C = 1 tank / (2) hours = 1/2 tanks/hr

If we add the three two-pump combined rates, we get A + B + B + C + A + C = 5/6 + 2/3 + 1/2
A + A + B + B + C + C = (5 + 4 + 3)/6
2(A + B + C) = 12/6 = 2
A + B + C = 1 tanks/hr. This is the combined rate for all three pumps.

Returning to the original question and the formula time = work / rate:

t = 1 tank / 1 tanks/hr = 1 hour.
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by RBF3 » Sun Jul 17, 2011 4:55 pm
Thanks again goalevan.
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by GMATGuruNY » Mon Jul 18, 2011 3:24 am
I posted a solution here:

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by amit2k9 » Mon Jul 18, 2011 9:56 pm
pumps A and B can fill the tank per hour = 5/6 = 1/A+ 1/B

similarly B and C and C and A can fill the tank/hour

1/B+1/C = 2/3
1/C+1/A = 1/2

hence, 2(1/A+1/B+1/C) = 5/6 + 1/2 + 2/3

thus (1/A + 1/B + 1/C) = 1.
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