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If 1/2 of the air in a tank is removed with each stroke of a [#permalink]

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24 Jun 2008, 13:16

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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 ?

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

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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?

If 1/2 of the air in a tank is removed with each stroke of a [#permalink]

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09 Jun 2011, 11:26

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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

This question is actually from the OG 12th edition. Am not able to understand the explanation given in OG for this problem. Could someone please explain in detail as to how to solve this

Last edited by Bunuel on 24 Sep 2013, 02:43, edited 1 time in total.

Could someone please help me out with the below problem:

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

This question is actually from the OG 12th edition. Am not able to understand the explanation given in OG for this problem. Could someone please explain in detail as to how to solve this

The best thing to do here is plug in Assume that there is 16 units of air in the tank first stroke = 1/2*16 = 8 ( we are left ith 8 now) second stroke = 1/2*8 = 4( left ith 4 units now) third stoke = 1/2*4 = 2( left with 2 units) 4th stroke = 1/2* 2 = 1 (left with 1 unit)

total units removed = 8+4+2+1 = 15 total unis in the tank at the begining = 16 thus its A . 15/16

Oh, I get it..! Thanks..!!! I think the catch over here is.."the tank is halved after each stroke"....failed to understand the question properly. Thought that a single stoke halves the air in it...so after 2 strokes guessed there wont be any air in it

Let P is 100 ( 1 - 50/100)^4 Remaining amount = 100(1/16)

if 1/16 of 100 is remaining means 1-1/16 = 15/16 is lost. _________________

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Re: if 1/2 of the air in a tank is removed with each stroke of a [#permalink]

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13 May 2014, 05:31

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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 ?

Re: If 1/2 of the air in a tank is removed with each stroke of a [#permalink]

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26 Oct 2015, 10:19

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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