Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 23 Mar 2008
Posts: 214

If 1/2 of the air in a tank is removed with each stroke of a [#permalink]
Show Tags
Updated on: 13 May 2014, 06:32
2
This post received KUDOS
11
This post was BOOKMARKED
Question Stats:
63% (00:52) correct 37% (00:49) wrong based on 525 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by puma on 24 Jun 2008, 13:16.
Last edited by Bunuel on 13 May 2014, 06:32, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.



SVP
Joined: 30 Apr 2008
Posts: 1841
Location: Oklahoma City
Schools: Hard Knocks

Re: tank [#permalink]
Show Tags
24 Jun 2008, 13:18
3
This post was BOOKMARKED
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
_________________
 J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
GMAT Club Premium Membership  big benefits and savings



Intern
Joined: 30 Jul 2007
Posts: 35

Re: tank [#permalink]
Show Tags
24 Jun 2008, 13:58
1
This post received KUDOS
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



Intern
Joined: 04 May 2011
Posts: 10

If 1/2 of the air in a tank is removed with each stroke of a [#permalink]
Show Tags
Updated on: 24 Sep 2013, 02:43
1
This post received KUDOS
1
This post was BOOKMARKED
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
Originally posted by srivats212 on 09 Jun 2011, 11:26.
Last edited by Bunuel on 24 Sep 2013, 02:43, edited 1 time in total.
Renamed the topic and edited the question.



Current Student
Joined: 26 May 2005
Posts: 533

Re: Need Help  Fractions [#permalink]
Show Tags
09 Jun 2011, 11:49
4
This post received KUDOS
2
This post was BOOKMARKED
srivats212 wrote: 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 Hope this helps



Manager
Joined: 27 Apr 2008
Posts: 183

Re: Need Help  Fractions [#permalink]
Show Tags
09 Jun 2011, 11:52
1
This post received KUDOS
1
This post was BOOKMARKED
This is a decaytype problem...the tank is halved after each stroke. Therefore:
(0.5)^4 = 1/16 > 0.5 for halving, 4 for strokes
so 1/16 of air is left in the tank => 15/16 of the air was removed



Intern
Joined: 04 May 2011
Posts: 10

Re: Need Help  Fractions [#permalink]
Show Tags
09 Jun 2011, 12:10
1
This post received KUDOS
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



Manager
Joined: 08 Sep 2010
Posts: 138

Re: Need Help  Fractions [#permalink]
Show Tags
09 Jun 2011, 21:30
2
This post received KUDOS
1
This post was BOOKMARKED
A  15/16.. Left After 1st stroke = 1/2 Left After 2nd stroke = 1/2 * 1/2 = 1/4 Left After 3rd stroke = 1/2 * 1/4 = 1/8 Left After 4th stroke = 1/2 * 1/8 = 1/16 So removed = 1 1/16 = 15/16
_________________
My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own.
If you like my explanations award kudos.



Manager
Joined: 27 Apr 2008
Posts: 183

Re: Need Help  Fractions [#permalink]
Show Tags
10 Jun 2011, 09:40
As they say, the devil is in the details!



Manager
Joined: 12 Oct 2009
Posts: 174
Schools: Columbia, INSEAD, RSM, LBS

Re: Need Help  Fractions [#permalink]
Show Tags
10 Jun 2011, 11:42
By the 4th stroke the the air remaining is 1/16 , so what has been removed is 15/16



Manager
Joined: 05 Oct 2011
Posts: 165

Re: tank [#permalink]
Show Tags
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



Manager
Joined: 05 Oct 2011
Posts: 165

Re: Need Help  Fractions [#permalink]
Show Tags
02 Nov 2011, 10:46
1st stroke 1/2 2nd stroke 1/4 3rd stroke 1/8 4th stroke 1/16 total: (8+4+2+1)/16=15/16



Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 902
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

Re: Need Help  Fractions [#permalink]
Show Tags
24 Sep 2013, 02:31
Better use compound interest formula. Remaing amount = P ( 1  R/100)^n Let P is 100 ( 1  50/100)^4 Remaining amount = 100(1/16) if 1/16 of 100 is remaining means 11/16 = 15/16 is lost.
_________________
Piyush K
 Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press> Kudos My Articles: 1. WOULD: when to use?  2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".



Math Expert
Joined: 02 Sep 2009
Posts: 45380

Re: If 1/2 of the air in a tank is removed with each stroke of a [#permalink]
Show Tags
13 May 2014, 06:32



SVP
Status: The Best Or Nothing
Joined: 27 Dec 2012
Posts: 1837
Location: India
Concentration: General Management, Technology
WE: Information Technology (Computer Software)

Re: If 1/2 of the air in a tank is removed with each stroke of a [#permalink]
Show Tags
01 Aug 2014, 00:57
Let say total air = 1 Total air removed \(= \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4}\) \(= \frac{15}{16}\) Answer = A
_________________
Kindly press "+1 Kudos" to appreciate



Board of Directors
Joined: 17 Jul 2014
Posts: 2734
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

If 1/2 of the air in a tank is removed with each stroke of a [#permalink]
Show Tags
23 Jan 2017, 16:50
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 air has been removed after 4 strokes ?
A. 15/16 B. 7/8 C. 1/4 D. 1/8 E. 1/16 let's assign a nice number..16 (divisible by 2 and 4), for total Air.. after 1st stroke, we are left with 8 after 2nd stroke, we are left with 4 after 3rd stroke, we are left with 2 after 4th stroke, we are left with 1 we removed 15 parts out of 16. A: 15/16



Manager
Joined: 17 Aug 2012
Posts: 157
Location: India
Concentration: General Management, Strategy
GPA: 3.75
WE: Consulting (Energy and Utilities)

Re: If 1/2 of the air in a tank is removed with each stroke of a [#permalink]
Show Tags
25 Feb 2017, 10:40
Let initial air be X
after 1st stroke air left x/2 after 2nd stroke air left x/4 after 3rd stroke air left x/8 after 4th stroke air left x/16
air removed = original  air left =x  x/16 = 15x/16



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 11663
Location: United States (CA)
GRE 1: 340 Q170 V170

Re: If 1/2 of the air in a tank is removed with each stroke of a [#permalink]
Show Tags
19 Mar 2018, 11:11
Hi All, TESTing Values will work well on this question. A quick look at the answers tells us that the common denominator for all 5 options is 16; THAT is a huge clue that 16 would be a great number to TEST. We're told that 1/2 of the air in a tank is removed with each stroke. We're asked what fraction has been removed after 4 strokes…. Full = 16 1st = 168 = 8 left 2nd = 84 = 4 left 3rd = 42 = 2 left 4th = 21 = 1 left So 1/16 is LEFT; thus, 15/16 was REMOVED Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************




Re: If 1/2 of the air in a tank is removed with each stroke of a
[#permalink]
19 Mar 2018, 11:11






