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the faucets are kept running

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the faucets are kept running

by sanju09 » Wed Feb 29, 2012 12:20 am
Faucet C fills a tank in 3 hours, whereas faucet D independently fills the same tank in 9 hours. What fraction of the tank will be filled, if both the faucets are kept running for 1½ hours, and the tank is originally one-fourth full?
A. 1/6
B. 1/3
C. 4/9
D. 5/9
E. 11/12
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by Anurag@Gurome » Wed Feb 29, 2012 12:46 am
sanju09 wrote:Faucet C fills a tank in 3 hours, whereas faucet D independently fills the same tank in 9 hours. What fraction of the tank will be filled, if both the faucets are kept running for 1½ hours, and the tank is originally one-fourth full?
A. 1/6
B. 1/3
C. 4/9
D. 5/9
E. 11/12
1 hour work of faucet C = 1/3
1 hour work of faucet D = 1/9
1 hour work of faucets C and D together = 1/3 + 1/9 = 4/9
Work done by faucets C and D in 3/2 hours = 4/9 * 3/2 = 2/3
Therefore, fraction of the tank filled = 1/4 + 2/3 = [spoiler]11/12[/spoiler]

The correct answer is E.
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by GMATGuruNY » Wed Feb 29, 2012 4:44 am
sanju09 wrote:Faucet C fills a tank in 3 hours, whereas faucet D independently fills the same tank in 9 hours. What fraction of the tank will be filled, if both the faucets are kept running for 1½ hours, and the tank is originally one-fourth full?
A. 1/6
B. 1/3
C. 4/9
D. 5/9
E. 11/12
Let the capacity of the tank be equal to the LCM of 3, 9, and 4: 36 units.
Amount currently in the tank = (1/4)36 = 9 units.
Rate for faucet C = w/t = 36/3 = 12 units per hour.
Rate for faucet D = w/t = 36/9 = 4 units per hour.
Combined rate for the two faucets = 12+4 = 16 units per hour.
In 3/2 hours, amount filled = r*t = (16)(3/2) = 24 units.
Fraction of the tank filled = (9+24)/36 = 33/36 = 11/12.

The correct answer is E.
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