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A scientist is studying bacteria whose cell population

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A scientist is studying bacteria whose cell population [#permalink] New post 15 Jul 2007, 06:14
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A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed?

(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.

(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.
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 [#permalink] New post 15 Jul 2007, 07:05
I would answer A, (1) alone is sufficient, but (2) alone is not sufficient. We know the population divided 2 hours ago, has quadrupled since, and has since increased by 3750, and that is doubles at constant intervals. So if it quadruples in 2 hours it is doubling every hour. Let x stand for the quantity before doubling 2 hours ago, x + 3750 is the current quantity. Since the quantity has quadrupled, 4x is also the current quantity, so the equation is x + 3750=4x. Solved, x=1250. The current quantity is therefore 5000, and it will double each hour for the next 4 hours, ending at 80000.

(2) is not sufficient because we don't know the rate - it might be doubling every 10 minutes!


BTW, I'm having trouble remembering that I don't have to solve every DS problem. It should have been enough to know that (1) gave a rate and an amount, and (2) gave only an amount but no rate.
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 [#permalink] New post 16 Jul 2007, 08:56
A is correct here, but D is tempting since if you read "will double" it gives the impression that it will double from its current state, which would mean that the bacteria double every 3 hours, and an hour from that moment there will still be 40,000 in the test tube, and 40,000 innocent bacteria will then be decimated by the unfeeling laboratory doctors.
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 [#permalink] New post 16 Jul 2007, 09:53
I agree with A here. I initially went with D with the possibility of bacteria doubling every 1 hr without considering that the bacteria might double every 3 hrs too
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Re: DS : Bacteria Population [#permalink] New post 16 Jul 2007, 10:33
trahul4 wrote:
A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed?

(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.

(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.


I think the answer is C.

(1) gives three pieces of information:
- The bacteria doubled 2 hours ago.
- The increase is 3750. So the total bacteria after the quadruplation is 3,750 * 4/3 = 5000.
- The bacteria quadrupled in the past 2 hours. Note that it does not say when exactly they quadrupled. So the bacteria could be doubling every 60 mins or every 45 mins for all we know. Hence, insufficient.

(2) Tells us that in (4 - 1) = 3 hours the total bacteria will be 40,000. But we dont know what the number of bacteria is currently. So we dont know the rate at which the bacteria is doubling. Hence, insufficient.

If you put (1) and (2) together, you have to double 3 times in 3 hours to go from 5,000 to 40,000. This means the doubling is occuring every 60 mins. So 80,000 bacteria will be destroyed.

Is my thinking too convoluted? What is the OA?
Re: DS : Bacteria Population   [#permalink] 16 Jul 2007, 10:33
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