After installing a powerful pump onto a large gasoline tank, John pump

Let Total Gasoline \(= 27\)

Gasoline removed in first cycle \(= \frac{1}{3} * 27 = 9\)

Gasoline left after first cycle = \(27 - 9 = 18\)

Gasoline removed in second cycle \(= \frac{1}{3} * 18 = 6\)

Gasoline left after second cycle = \(18 - 6 = 12\)

Gasoline removed in third cycle \(= \frac{1}{3} * 12 = 4\)

Gasoline left after third cycle = \(12 - 4 = 8\)

Total Gasoline removed in 3 cycles = 9 + 6 + 4 = 19

Fraction of Gasoline Removed to Total Gasoline \(= \frac{19}{27}\) . Answer (D) ...

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?

Assume that the total gas was 900

1st run removed \(\frac{1}{3}\) of 900 = 300 : remaining gas = 900 - 300 = 600

2nd run removed \(\frac{1}{3}\) of 600 = 200: remaining gas = 600 - 200 = 400

3rd run removed \(\frac{1}{3}\) of 400 = 400/3: remaining gas = 400 - \(\frac{400}{3}\) = \(\frac{800}{3}\)

total removed gas = 500 + \(\frac{400}{3}\) = \(\frac{1900}{3}\)

fraction of removed gas = \(\frac{1900}{(3 * 900)}\)= \(\frac{19}{27}\)

Total gas = 27
After first run, Removed gas = 1/3 of 27 = 9, Remaining gas = 18
After second run, Removed gas = 1/3 of 18 = 6, Remaining gas = 12
After third run, Removed gas = 1/3 of 12 = 4, Remaining gas = 8
Total removed gas = 9 + 6 +4 = 19
Fraction of removed gas = 19/27 (D)

Given: 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.

Asked: 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?

1st run:-
Gas removed = 1/3
Gar remaining = 2/3

2nd run: -
Gas removed = 2/3*1/3 = 2/9
Gas remaining = 2/3 - 2/9 = 4/9

3rd run: -
Gas removed = 4/9*1/3 = 4/27
Gas remaining = 4/9 - 4/27 = 8/27

Total gas removed = 1 - 8/27 = 19/27

IMO D