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At the end of every hour a culture of bacteria becomes some [#permalink]
15 Sep 2005, 08:54

00:00

A

B

C

D

E

Difficulty:

(N/A)

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At the end of every hour a culture of bacteria becomes some number of times larger than it was the previous hour. If the number of bacteria was originally greater than 1 and if the rate of growth also increases every hour, what was the original number of bacteria?

1) 1/3 of the original culture would have resulted in a total of 385 bacteria after 3 hours.

2) The original number of bacteria was less than 4.

Last edited by acid_burn on 15 Sep 2005, 13:09, edited 1 time in total.

Ok, I want to say E. I have feeling this could be wrong, but here is my reasoning. The stem does not give very much information (as some DS problems do). It it not clear whether "some number larger" is the same number each hour. Maybe we can assume this. If so, it doesn't say that that the number of bacteria is an integer. All we know is that the original number is great than one and is increasing every hour

1. Insufficient. X=original amount, Y=amount increased each hour, then we have: XY^3=385. 2 variables that we can't figure out.

2. Insufficient. X<4. Combined with the stem, 1<X<4. Like I said, I'm assuming that the origainal amount is not an integer, so it could be any number between 1 and 4. We still don't know Y.

Together, X can be an infinite amount of numbers, so there is no way to figure out Y. E is my answer.

I have a gut feeling I am missing something important, and that I am wrong. _________________

Ok, I want to say E. I have feeling this could be wrong, but here is my reasoning. The stem does not give very much information (as some DS problems do). It it not clear whether "some number larger" is the same number each hour. Maybe we can assume this. If so, it doesn't say that that the number of bacteria is an integer. All we know is that the original number is great than one and is increasing every hour

1. Insufficient. X=original amount, Y=amount increased each hour, then we have: XY^3=385. 2 variables that we can't figure out.

2. Insufficient. X<4. Combined with the stem, 1<X<4. Like I said, I'm assuming that the origainal amount is not an integer, so it could be any number between 1 and 4. We still don't know Y.

Together, X can be an infinite amount of numbers, so there is no way to figure out Y. E is my answer.

I have a gut feeling I am missing something important, and that I am wrong.

Geometric funda wont work here since it is mentioned that the "rate of growth" also increases every hour..Answer anyway should be 'E' only.

[quote="acid_burn"]At the end of every hour a culture of bacteria becomes some number of times larger than it was the previous hour. If the number of bacteria was originally greater than 1 and if the rate of growth also increases every hour, what was the original number of bacteria?

1) of the original culture would have resulted in a total of 385 bacteria after 3 hours.

2) The original number of bacteria was less than 4.

I pick E and here is my reasoning:

First consider 1). It tells us after 3 hours, the original number of bacteria grows to 385. And from the question the original number of bacteria is greater than 1 and the rate of growth increases every hour. We can have either the original number of bateria as a small number, (ie. 2 for example) and have the rapid rate of growth to reach 385 after 3 hours or the original number of bacteria as a larger number, (ie. 90 for example) and have the slower rate of growth. So 1) is not sufficient and we eliminate A and D.

Consider 2). It tells us the originial number of bacteria was less than 4. And from the question we know the original number of bacteria is greater than 1. This leaves two possibilites, 2 and 3. Same rational as above, we cannot determine which number is the right one. We eliminate B.

Combining 1) and 2), we still cannot solve the problem we face above, and we eliminate C and the answer is E.

I like your reasoning guys, the official answer uses similar logic but is more lenghty. they use an equation (1/3x)abc = 385 where x is the number of bacteria and a&b&c are rate of growth.

thus xabc = 1155

1155 has four prime factors : 3,5,7,11, which is unsufficient to tell the original number of bacteria since it can be 3,5,7,11 or 5,3,7,11 or 7,3,5,11 . Each pair gives a number for the bacteria and an increasing growth rate.

From statement two, we know that the original number of bacteria is less than 4. When both are combined only one pair complies - 3,5,7,11 where x=3

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