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enigma123
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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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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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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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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leanhdung
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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Bunuel
enigma123
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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Hoozan
Bunuel
enigma123
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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