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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?

Correct Answer:

**D**If 1/3 of a full tank of gas is removed, 2/3 of a full tank remains. Amount removed is 1/3 of the amount that remains, which changes depending on the number of times the pump has been run.

Pump Run # | Still Left [as fraction of full tank] | Removed [as fraction of full tank] |

0 | 1 | 0 Cumulative: 0 |

1 | 1 - (1/3) = 2/3 | 1(1/3) = 1/3 = 9/27 Cumulative: 0 + (9/27) = 9/27 |

2 | 1 - (9/27) - (6/27) = 12/27 = 4/9 | (1/3)(2/3) = 2/9 = 6/27 Cumulative: (9/27) + (6/27) = 15/27 |

3 | 1 - (9/27) - (6/27) - (4/27) = 8/27 | (1/3)(4/9) = 4/27 Cumulative: (15/27) + (4/27) = 19/27 |

- Each time the pump is used, 2/3 of the remaining gas stays and 1/3 of the remaining gas is removed.
- The first time running the pump removed 1/3 of the full tank, leaving 2/3 of a full tank of gas. So, after running the pump once, 1/3 of the total gas in the tank has been removed.
- The second time running the pump removed 1/3 of the 2/3 of a full tank that remained after running the pump for the first time. So, after running the pump twice, (1/3) + (1/3 of the 2/3) of the remaining tank had been removed. Thus far, 5/9 of the total gas in the tank has been removed.
- The third time running the pump removed 1/3 of the 4/9 of a full tank that remained after running the pump for the second time. So, after running the pump three times, 5/9 + 1/3 of 4/9 of the remaining tank had been removed. In total, 5/9 + 4/27 = 19/27 of the total gas in the tank at the beginning was removed after running the pump three times.