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If 2/3 of the air in a tank is removed with each stroke of a

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enigma123
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If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   01 Mar 2012, 16:58
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If 2/3 of the air in a tank is removed with each stroke of a vacuum pump, how many strokes does it take before less than 1% of the original amount of air in the tank remains?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

What's the best way to solve this question guys? By picking a number say or algebraically?
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D
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Bunuel
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   01 Mar 2012, 17:18
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enigma123 wrote:
If 2/3 of the air in a tank is removed with each stroke of a vacuum pump, how many strokes does it take before less than 1% of the original amount of air in the tank remains?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

What's the best way to solve this question guys? By picking a number say or algebraically?


2/3 of the air in a tank is removed with each stroke means that 1/3 of the air is left after one stroke, 1/3*1/3 after two strokes and so on. Basically after n strokes there will be (1/3)^n of the air left in the tank. The question asks about such n for which (1/3)^n<1/100 --> 3^n>100 --> n=5.

Answer: D.

Similar question to practice: if-3-4-of-the-mineral-deposits-in-a-reservoir-of-water-are-97300.html

Hope it helps.
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   29 May 2013, 23:09
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I love the solutions provided by Bunuel, but not always can come up with such logic, which is simple and straightforward, but i am trying to.
My approach is to find more or less good number and plug in. Lets take 90 as the total volume of an air, and the question asks after how many stokes there will be less than 0,9 air in a tank, if each stroke takes 2/3 of an air.
So after he 1 stroke the volume will be 30 (90-60=30), and so on
2 stroke - 30-20=10
3 stroke - 10-6,7=3,3
4 stroke - 3,3-2,2=1,1
5 stroke - 1,1 - minus something that gets us less than 0,9 for sure. so the answer is: after 5 strokes there will be less than 1% of air left in a tank. Option D.
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   03 Jan 2014, 08:08
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enigma123 wrote:
If 2/3 of the air in a tank is removed with each stroke of a vacuum pump, how many strokes does it take before less than 1% of the original amount of air in the tank remains?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

What's the best way to solve this question guys? By picking a number say or algebraically?


Well yeah algebraically it would be (1/3)^x <1/100

So x>=5

There you go, answer is D

Cheers!
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   29 Mar 2014, 05:50
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This question is one of the standard-type.

Whenever something increases or decreases by some percent on each cycle, your mind should start thinking about application of compound interest formula.

\(Final-amount = Original-amount [ 1 - \frac{Rate}{100}]^n --- for -decreasing- rate.\)
\(Final-amount = Original-amount [ 1 + \frac{Rate}{100}]^n --- for- increasing- rate.\)

For current question solve following equation:
\(1 > 100 [ 1 - \frac{\frac{200}{3}}{100}]^n\)

\(1 > 100 [ 1 - \frac{200}{300}]^n\)

for n = 5,6... this relation exist.

Thus answer is D.
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   20 Aug 2017, 04:34
2/3 of the air in a tank is removed with each stroke, that's 1/3 of the air is left after one stroke, 1/3*1/3 after two strokes and so on.
After x strokes there will be (1/3)^x of the air left in the tank. The question asks about such x for which (1/3)^x<1/100 --> 3^x>100 --> x=5
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   20 Aug 2017, 17:44
Hi Bunuel !

I wonder if the answer is 4 because the stem clearly state does it take BEFORE less than 1%.

Please shed some light! Many thanks :)
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   21 Aug 2017, 02:26
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leanhdung wrote:
Hi Bunuel !

I wonder if the answer is 4 because the stem clearly state does it take BEFORE less than 1%.

Please shed some light! Many thanks :)


I think it's clear that the question asks about the number of strokes that will make amount of the air less than 1%.
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   12 Nov 2020, 00:21
Bunuel wrote:
enigma123 wrote:
If 2/3 of the air in a tank is removed with each stroke of a vacuum pump, how many strokes does it take before less than 1% of the original amount of air in the tank remains?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

What's the best way to solve this question guys? By picking a number say or algebraically?


2/3 of the air in a tank is removed with each stroke means that 1/3 of the air is left after one stroke, 1/3*1/3 after two strokes and so on. Basically after n strokes there will be (1/3)^n of the air left in the tank. The question asks about such n for which (1/3)^n<1/100 --> 3^n>100 --> n=5.

Answer: D.

Similar question to practice: https://gmatclub.com/forum/if-3-4-of-the ... 97300.html

Hope it helps.



Bunuel so can we say that if (1/3)^n < 1/100 Then (2/3)^n must be greater that 99%
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   12 Nov 2020, 00:35
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Hoozan wrote:
Bunuel wrote:
enigma123 wrote:
If 2/3 of the air in a tank is removed with each stroke of a vacuum pump, how many strokes does it take before less than 1% of the original amount of air in the tank remains?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

What's the best way to solve this question guys? By picking a number say or algebraically?


2/3 of the air in a tank is removed with each stroke means that 1/3 of the air is left after one stroke, 1/3*1/3 after two strokes and so on. Basically after n strokes there will be (1/3)^n of the air left in the tank. The question asks about such n for which (1/3)^n<1/100 --> 3^n>100 --> n=5.

Answer: D.

Similar question to practice: https://gmatclub.com/forum/if-3-4-of-the ... 97300.html

Hope it helps.



Bunuel so can we say that if (1/3)^n < 1/100 Then (2/3)^n must be greater that 99%



No. It's 1 - (1/3)^n > 99/100

But notice that 1 - (1/3)^n ≠ (2/3)^n.

1 - (1/3)^n = (3^n - 1)/3^n
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   12 Nov 2020, 00:40
Bunuel if we wish to find the answer using the amount removed (2/3) how do we go about with the sum?
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Re: If 2/3 of the air in a tank is removed with each stroke of a [#permalink] New post   16 Sep 2022, 01:51
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