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If 2/3 of the air in a tank is removed with each stroke of a
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Author:  enigma123 [ 01 Mar 2012, 15:58 ]
Post subject:  If 2/3 of the air in a tank is removed with each stroke of a

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?

Author:  Bunuel [ 01 Mar 2012, 16:18 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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.

Author:  ziko [ 29 May 2013, 22:09 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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.

Author:  jlgdr [ 03 Jan 2014, 07:08 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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!
If it helps consider providing some Kudos ok?
J :)

Author:  PiyushK [ 29 Mar 2014, 04:50 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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.

Author:  rever08 [ 20 Aug 2017, 03:34 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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

Author:  leanhdung [ 20 Aug 2017, 16:44 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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 :)

Author:  Bunuel [ 21 Aug 2017, 01:26 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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%.

Author:  Hoozan [ 11 Nov 2020, 23:21 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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%

Author:  Bunuel [ 11 Nov 2020, 23:35 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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

Author:  Hoozan [ 11 Nov 2020, 23:40 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

Bunuel if we wish to find the answer using the amount removed (2/3) how do we go about with the sum?

Author:  bumpbot [ 16 Sep 2022, 00:51 ]
Post subject:  Re: If 2/3 of the air in a tank is removed with each stroke of a

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