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if 1/2 of the air in a tank is removed with each stroke of a

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if 1/2 of the air in a tank is removed with each stroke of a [#permalink] New post 24 Jun 2008, 13:16
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 the air has been removed after 4 strokes?

A) 15/16
B) 7/8
C) 1/4
D) 1/8
E) 1/16
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Re: tank [#permalink] New post 24 Jun 2008, 13:18
A

First you have 1

so
first stroke = 1/2
second stroke = 1/4
third stroke = 1/8
fourth stroke = 1/16

8/16 + 4/16 + 2/16 + 1/16 = 15/16

puma wrote:
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 the air has been removed after 4 strokes?

A) 15/16
B) 7/8
C) 1/4
D) 1/8
E) 1/16

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Re: tank [#permalink] New post 24 Jun 2008, 13:58
puma wrote:
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 the air has been removed after 4 strokes?

A) 15/16
B) 7/8
C) 1/4
D) 1/8
E) 1/16


I get A, too.

Let X = air in the tank.

Plug in 240 = X.

First stroke = 1/2 * 240 = 120
Second stroke = 1/2 * 120 = 60
Third stroke = 1/2 * 60 = 30
Fourth stroke = 1/2 * 30 = 15

So 15 remains, which is the amount of air left.

240 - 15 = 225 (amount of air removed). 225/240 = 15/16
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Re: tank [#permalink] New post 02 Nov 2011, 10:44
1st stroke 1/2
2nd stroke 1/4
3rd stroke 1/8
4th stroke 1/16
total: (8+4+2+1)/16=15/16
Re: tank   [#permalink] 02 Nov 2011, 10:44
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