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# Wes works at a science lab that conducts experiments on

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Wes works at a science lab that conducts experiments on [#permalink]

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30 Jun 2008, 13:16
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Wes works at a science lab that conducts experiments on bacteria. The population of the bacteria multiplies at a constant rate, and his job is to notate the population of a certain group of bacteria each hour. At 1 p.m. on a certain day, he noted that the population was 2,000 and then he left the lab. He returned in time to take a reading at 4 p.m., by which point the population had grown to 250,000. Now he has to fill in the missing data for 2 p.m. and 3 p.m. What was the population at 3 p.m.?
50,000
62,500
65,000
86,666
125,000

I am going through the above problem from a Manhattan CAT exam. Note the explanation says the following:If we decide to find a constant multiple by the hour, then we can say that the population was multiplied by a certain number three times from 1 p.m. to 4 p.m.: once from 1 to 2 p.m., again from 2 to 3 p.m., and finally from 3 to 4 p.m.

Let's call the constant multiple x.

2,000(x)(x)(x) = 250,000
2,000(x3) = 250,000
x3 = 250,000/2,000 = 125
x = 5

Therefore, the population gets five times bigger each hour.

My question is....The problem never states the frequency of increase. How can we say it increased every hr?
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30 Jun 2008, 13:23
jimmyjamesdonkey wrote:
The problem never states the frequency of increase. How can we say it increased every hr?

What's important is that the number of bacteria grows exponentially.

F(t) = F(0)*exp(at). The frequency doesn't matter, only coefficient a is going to change depending on the frequency.
For example, in hours are used for frequency exp(a*3hour) = 125 -> a = log(125) per 3 hours = log(5) per hour
if minutes are used for frequency exp(amin*3*60min) = 125 -> amin = log(125) per 3*60 min = log(5)/60 per min

you can easily verify that regardless of the frequency you will get the same result.
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Re: Can anyone explain this one for me? [#permalink]

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30 Jun 2008, 13:26
What if the population grows ever 65 mins instead of ever 60 mins?
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30 Jun 2008, 13:32
jimmyjamesdonkey wrote:
What if the population grows ever 65 mins instead of ever 60 mins?

growth is assumed continuous, not discrete. "The population of the bacteria multiplies at a constant rate"
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30 Jun 2008, 13:49
Couldn't a constant rate be...every 65 mins
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30 Jun 2008, 13:52
jimmyjamesdonkey wrote:
Couldn't a constant rate be...every 65 mins

No
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30 Jun 2008, 14:03
Sorry just do not understand...
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30 Jun 2008, 14:05
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jimmyjamesdonkey wrote:
Sorry just do not understand...

just picture a train going with a constant speed. you are asking, can we assume that the train jumps every 5 min or every 7 min. Does that help?
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30 Jun 2008, 14:09
Your saying the frequency doesn't matter because it is constantly growing, not just growing at a specific interval?
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30 Jun 2008, 14:48
The reason I'm asking is because of this problem...can you explain the difference between the two:

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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30 Jun 2008, 15:50
jimmyjamesdonkey wrote:
The reason I'm asking is because of this problem...can you explain the difference between the two:

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.

The number of bacteria is also growing exponentially here. To me those two problems are similar. F(t) = F(0)*exp(at). The question is about F(4)

1) F(t)/F(t-2) = 4 -> exp(2a) = 4 -> 2a = ln(4) = 2ln(2) -> a = ln(2)
F(t)-F(t-2) = 3750 -> F(t-2) = 3750/3, F(t) = 3750*4/3. //no info about F(0) -> insufficient info to calculate F(4)

2) F(3) = 2*F(0) -> exp(3a)=2 -> 3a = ln(2)-> a = ln(2)/3.
F(3) = 40,000 -> F(4) = F(3)*exp(a*1)=40,000*exp(3a*1/3)=40,000*(exp(3a))^1/3=40,000*2^1/3 -> sufficient -> B
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30 Jun 2008, 16:00
maratikus, answer is incorrect for the DS question..
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30 Jun 2008, 21:37
Hi guys, i think in the 1st problem the time is given from 1 to 4 and we are asked to find out the number at 3, hence we can use any constant rate that satisfies this ; for example take a constant rate of 30 mins in this case
2000(x)(x)(x)(x)(x)(x) = 250000
x=sqrt5 and the ans still turns out to be 50,000 you can even try and use a constant time interval of 10mins and the ans will still be the same !

Also is the ans for the 2nd one E ?
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01 Jul 2008, 05:21
jimmyjamesdonkey wrote:
maratikus, answer is incorrect for the DS question..

you are right. I misinterpreted the second statement. here intervals are important.
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01 Jul 2008, 08:02
is the answer A??? i dont see y this ? is very much complicated.. with a little trial and error it wd be fine.. its growing at a constant rate of 5 in 1PM it is 2000 at 2 pm its 10000 and 3 pm (*5) 50000 and 4 pm its 2,50,000...(Am i right???)
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01 Jul 2008, 13:33
moved second question by jimmyjamesdonkey to 7-t66531
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