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

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

A) 8/27
B) 1/27
C) 2/3
D) 19/27
E) 27/28

Bunuel · Answer

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pushpitkc · Answer

Considering that the powerful pump which pumps out 1/3rd of the tank's gas at every attempt gives
1- 1/3 = 2/3 in the first iteration
2/3 - 1/3(2/3) = 4/9 in the second iteration
4/9 - 1/3(4/9) = 8/27 in the third iteration.

Hence the tank which has 8/27 of the initial gasoline left, it has 19/27(1-8/27) of the gasoline(Option D)

Method 2

Assume numbers
Initial : 2700
First iteration : 1800 left, 900 lost
Second iteration : 1200 left, 600 lost
Third iteration : 800 left, 400 lost

Total lost = 1900
Fraction of lost/total =19/27(Option D)

sashiim20 · Answer

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) ...

Leo8 · Answer

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

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}

Jawad001 · Answer

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)

Kinshook · Answer

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

bumpbot · Answer

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

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