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Question

After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?
A) 8/27
B) 1/27
C) 2/3
D) 19/27
E) 27/28
OAD

Anurag@Gurome · Answer

shikh wrote:After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?
A) 8/27
B) 1/27
C) 2/3
D) 19/27
E) 27/28
OAD

Let the tank has x units
No. of units removed by 1st pump = (1/3) * x = x/3
Remaining units = x - x/3 = 2x/3
No. of units removed by 2nd pump = (1/3) * (2x/3) = 2x/9 units
Remaining units = 2x/3 - 2x/9 = 4x/9
No. of units removed by 3rd pump = (1/3) * (4x/9) = 4x/27 units
Remaining units = 4x/9 - 4x/27 = 8x/27
Since only 8x/27 units are left, total no. of units removed = x - 8x/27 = 19x/27
Therefore, required fraction = 19/27

The correct answer is D.

Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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GMATGuruNY · Answer

shikh wrote:After installing a powerful pump onto a large gasoline tank, John pumped gasoline out of the tank. Each time John ran the pump, it removed 1/3 of the gas remaining in the tank. Assuming the gas tank was full when John began, what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?
A) 8/27
B) 1/27
C) 2/3
D) 19/27
E) 27/28
OAD

Let the tank has x units
No. of units removed by 1st pump = (1/3) * x = x/3
Remaining units = x - x/3 = 2x/3
No. of units removed by 2nd pump = (1/3) * (2x/3) = 2x/9 units
Remaining units = 2x/3 - 2x/9 = 4x/9
No. of units removed by 3rd pump = (1/3) * (4x/9) = 4x/27 units
Remaining units = 4x/9 - 4x/27 = 8x/27
Since only 8x/27 units are left, total no. of units removed = x - 8x/27 = 19x/27
Therefore, required fraction = 19/27

The correct answer is D.

GmatMathPro · Answer

Each time he runs the pump, the amount that remains is 2/3 of the previous amount. Thus, if he runs it x times, the fraction that remains is (2/3)^x, and the fraction that was pumped out is 1-(2/3)^x. He runs it three times, so 1-(2/3)^3 = 1-8/27 = 19/27

Ans: D

Pete Ackley
GMAT Math Pro
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