If 1/2 of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after 4 strokes?
A. 15/16
B. 7/8
C. 1/4
D. 1/8
E. 1/16
The OA is A. 15/16
I got the answer by assuming that "32" is the capacity of the tank, so after
Stroke #1 16 air is remaining
Stroke #2 8 air is remaining
Stroke #3 4 air is remaining
Stroke #4 2 air is remaining
Total Air Removed 30
30/32 = 15/16
However, if you assume that "16" is the capacity of the tank, then you arrive to the answer of 7/8 (choice B)
Stroke #1 8 air is remaining
Stroke #2 4 air is remaining
Stroke #3 2 air is remaining
Stroke #4 0 air is remaining
Total Air Removed 14
14/16 = 7/8
So is there any way to solve the above question so you always arrive to the OA of 15/16????
A. 15/16
B. 7/8
C. 1/4
D. 1/8
E. 1/16
The OA is A. 15/16
I got the answer by assuming that "32" is the capacity of the tank, so after
Stroke #1 16 air is remaining
Stroke #2 8 air is remaining
Stroke #3 4 air is remaining
Stroke #4 2 air is remaining
Total Air Removed 30
30/32 = 15/16
However, if you assume that "16" is the capacity of the tank, then you arrive to the answer of 7/8 (choice B)
Stroke #1 8 air is remaining
Stroke #2 4 air is remaining
Stroke #3 2 air is remaining
Stroke #4 0 air is remaining
Total Air Removed 14
14/16 = 7/8
So is there any way to solve the above question so you always arrive to the OA of 15/16????

















