GmatKiss 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
Please help me is solving this Q
1st Approach
Let the capacity of the tank = w
1st run
Quantity remaining = x = (2/3)*w, Quantity pumped out = (1/3)*w
2nd run
Quantity remaining = y = (2/3)*x, Quantity pumped out = (1/3)*x
3rd run
Quantity remaining = z = (2/3)*y, Quantity pumped out = (1/3)*y
Question asked : what fraction of the total gas in the tank at the beginning was removed if John ran the pump three times?
=> (((1/3)*w)+((1/3)*x)+((1/3)*y)))/w = ?
from the above (1/3)*x = (1/3)*(2/3)*w = (2/9)*w = (6/27)*w
from the above (1/3)*y)= (1/3)*(2/3)*(2/3)*w = (4/27)*w
=> (((1/3)*w)+((1/3)*x)+((1/3)*y))/w = (((9/27)*w)+((6/27)*w)+((4/27)*w)))/w =[spoiler]19/27[/spoiler]
2nd Approach
Let the quantity be 27,
1st run
Quantity remaining = (2/3)*27 =18, Quantity pumped out = (1/3)*27 =9
2nd run
Quantity remaining = (2/3)*18 = 12, Quantity pumped out = (1/3)*18 =6
3rd run
Quantity remaining = (2/3)*12 = 8, Quantity pumped out = (1/3)*12 =4
Total quantity pumped out = 9+6+4 = 19
Quantity or capacity in/of the tank = 27
So [spoiler]19/27[/spoiler]!
