A scientist is studying bacteria whose cell population

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A scientist is studying bacteria whose cell population [#permalink]

01 Jul 2008, 14:32

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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Re: multiplying bacteria Q2 [#permalink]

01 Jul 2008, 14:40

IMO answer is A.

given: population double every constant time interval.
1) population has quadrupled (double * double) in last two hours. => popuoation doubles every hr
lets say original population was x then now it is x+3750
and 3x=3750
therefore current population = 3750*4/3
hence the current population is know at the time interval when it double is known=1hr
therefore we can guess the population after 4hrs

2) if the population doubles 1 hr before the destruction then the constant time interval at which population increases is 3hr.
so the next split would be after 6hrs.... but the destruction was to happen aftr 4hr???
there is a contradiction over here. and hence we should ignore point (2

-pl share your views

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Re: multiplying bacteria Q2 [#permalink]

01 Jul 2008, 14:56

agree with A. I misinterpreted condition 1 again ...

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Re: multiplying bacteria Q2 [#permalink]

01 Jul 2008, 15:00

maratikus wrote:

agree with A. I misinterpreted condition 1 again ...

actually don't agree. what if interval is 45 min -> condition 1 still holds but results are going to be different

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Re: multiplying bacteria Q2 [#permalink]

01 Jul 2008, 15:06

the answer is C.

1) number of bacteria two hours ago was 1,250 and interval c: c > 2/3 and c <=1 -> insufficient
2) bacteria doubles at 3 hours, the number is equal to 40,000 -> insufficient

1&2: since bacteria doubles 2 hours before 0 and 3 hours after 0 -> 5*c = integer -> c = integer/5

since 40,000/1,250 = 32 = 2^5 -> bacteria doubled 5 times in 5 hours -> c = 5 -> sufficient

Re: multiplying bacteria Q2 [#permalink] 01 Jul 2008, 15:06

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A scientist is studying bacteria whose cell population

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A scientist is studying bacteria whose cell population

A scientist is studying bacteria whose cell population

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