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

Question

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?

Bunuel · Accepted Answer

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.

ziko · Answer

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 .

jlgdr · Answer

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

PiyushK · Answer

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  ^n --- for -decreasing- rate. 
 Final-amount = Original-amount  ^n --- for- increasing- rate.

For current question solve following equation:
 1 > 100  ^n

1 > 100  ^n

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

Thus answer is D.

rever08 · Answer

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

leanhdung · Answer

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

Bunuel · Answer

I think it's clear that the question asks about the number of strokes that will make amount of the air less than 1%.

Hoozan · Answer

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

Bunuel · Answer

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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