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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]

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New post Updated on: 13 May 2014, 06:32
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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
Spoiler: OA

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.
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Re: Need Help - Fractions  [#permalink]

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New post 09 Jun 2011, 11:49
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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
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Re: tank  [#permalink]

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New post 24 Jun 2008, 13:18
3
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]

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New post 24 Jun 2008, 13:58
1
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: Need Help - Fractions  [#permalink]

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New post 09 Jun 2011, 11:52
1
1
This is a decay-type 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
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Re: Need Help - Fractions  [#permalink]

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New post 09 Jun 2011, 12:10
1
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 :x
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Re: Need Help - Fractions  [#permalink]

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New post 09 Jun 2011, 21:30
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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
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Re: Need Help - Fractions  [#permalink]

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

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

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New post 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
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Re: Need Help - Fractions  [#permalink]

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New post 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 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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New post 13 May 2014, 06:32
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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


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Hope it helps.
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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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New post 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
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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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New post 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
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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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New post 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
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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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New post 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 = 16-8 = 8 left
2nd = 8-4 = 4 left
3rd = 4-2 = 2 left
4th = 2-1 = 1 left

So 1/16 is LEFT; thus, 15/16 was REMOVED

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