Can anyone explain this one for me?

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Can anyone explain this one for me?

by jimmyjamesdonkey » Mon Jun 30, 2008 4:28 pm
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(x^3) = 250,000
x^3 = 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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Can anyone explain this one for me?

by simpdimp » Mon Jun 30, 2008 5:30 pm
It says that the "The population of the bacteria multiplies at a constant rate". It doesn't matter if you calculate at 1/2 hr rate or 1 hr rate.

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by jimmyjamesdonkey » Mon Jun 30, 2008 5:54 pm
Then can you explain why this one is C.

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by jimmyjamesdonkey » Mon Jun 30, 2008 5:54 pm
jimmyjamesdonkey wrote:Then can you explain why this one is C.
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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by simpdimp » Mon Jun 30, 2008 7:01 pm
C is correct, it certainly took me some time to figure that out.

1. The first option says, "Since the population divided two hours ago"; this could be at the end of 2nd hr or 3rd or even 4th. So not sufficient

2. Second one essentially means at the end of 3rd hour there are 40,000 cells, but we don't know the starting number. So not sufficient

Combine 1 and 2, if there are 40,000 at the end of 3rd, the first statement points to end of 2nd hour, which means the experiment started with 1250 (4x = x+ 3750).

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by asigheartau » Tue Jul 01, 2008 2:55 pm
simpdimp wrote:C is correct, it certainly took me some time to figure that out.

1. The first option says, "Since the population divided two hours ago"; this could be at the end of 2nd hr or 3rd or even 4th. So not sufficient

2. Second one essentially means at the end of 3rd hour there are 40,000 cells, but we don't know the starting number. So not sufficient

Combine 1 and 2, if there are 40,000 at the end of 3rd, the first statement points to end of 2nd hour, which means the experiment started with 1250 (4x = x+ 3750).
We may know the starting number, but that is not what the question is asking. The question was asking for the total number of bacteria before the experiment was destroyed.

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This may help...

by evansbd » Tue Jul 08, 2008 9:59 am
Notice at the end of the original question " He has to fill in the missing data for 2pm and 3pm". That tells you right there that you need to find the rate by the hour. If it said find data for 2:30, and 4:00pm, then the interval would be 90 minutes but either way if the numbers (2000, 250,000) were the same you would get the same answer since they say the rate is constant.