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Bacteria!!!!

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by logitech » Wed Nov 26, 2008 8:50 pm
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.

OA C and I am interested in to know the # cells at the end of 4 hrs.
LGTCH
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Source: — Data Sufficiency |
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by rohangupta83 » Thu Nov 27, 2008 3:37 am
i am not sure whether you really need the second statement.

here is my solution

statement I
2 hours ago let the population of bacteria = x
now the population = 4x (quadrupled)
or
now the population is x+3750 (population increased by 3750 in 4 hrs)

4x = x + 3750
3x = 3750
x=1250

also, population became 4 times in 2 hours

therefore, the population became double in 1 hr and then doubled again in the next hour - hence, quadrupled in 2 hours

2 hrs ago - 1250
1 hr ago - 2500
now - 5000
1 hr from now - 10,000
2 hrs from now - 20,000
3 hrs from now - 40,000
4 hrs from now - 80,000

statement I is sufficient.

statement II

now population = x
3 hours from now population = 40,000

we have no way to find the rate so Statement II is insufficient.

imo A
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by logitech » Thu Nov 27, 2008 10:32 am
rohangupta83 wrote:i am not sure whether you really need the second statement.

here is my solution

statement I
2 hours ago let the population of bacteria = x
now the population = 4x (quadrupled)
or
now the population is x+3750 (population increased by 3750 in 4 hrs)

4x = x + 3750
3x = 3750
x=1250

also, population became 4 times in 2 hours

therefore, the population became double in 1 hr and then doubled again in the next hour - hence, quadrupled in 2 hours

2 hrs ago - 1250
1 hr ago - 2500
now - 5000
1 hr from now - 10,000
2 hrs from now - 20,000
3 hrs from now - 40,000
4 hrs from now - 80,000

statement I is sufficient.

statement II

now population = x
3 hours from now population = 40,000

we have no way to find the rate so Statement II is insufficient.

imo A
"If the population quadrupled during the last two hours, it doubled twice during that interval, but this does not necessarily mean that the population doubled at 60 minute intervals. It may have, for example, doubled at 50 or 55 minute intervals. We cannot determine from statement (1) how frequently the population is doubling." :twisted:
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by rohangupta83 » Thu Nov 27, 2008 2:38 pm
logitech wrote:
"If the population quadrupled during the last two hours, it doubled twice during that interval, but this does not necessarily mean that the population doubled at 60 minute intervals. It may have, for example, doubled at 50 or 55 minute intervals. We cannot determine from statement (1) how frequently the population is doubling." :twisted:
damn! i fell into the trap!
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by cramya » Thu Nov 27, 2008 3:42 pm
Rohan dont sweat too much. IMO this is a 900+ question(since it required verbal and qiuant skills) :-)

Nice problem I agree but I would guess and move on. It better be an experimental one :evil:
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by iamcste » Thu Nov 27, 2008 4:14 pm
can feel bacteria around...


In such cases, I either choose C or E

so, I still have a 50-50 chance :)
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by logitech » Thu Nov 27, 2008 4:28 pm
iamcste wrote:can feel bacteria around...


In such cases, I either choose C or E

so, I still have a 50-50 chance :)
I think the choice is between A and C ;-)

Oh well, if you see a problem like this in your test. Smile and remember me! ;-)
LGTCH
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