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# Data for a certain experiment are given in the table. If the

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Data for a certain experiment are given in the table. If the [#permalink]  23 Sep 2008, 20:00
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Data for a certain experiment are given in the table. If the amount of microorganism present increased by the same fraction during each of the two 3-hour periods shown, how many grams of microorganism were present at 4:00 PM?
Time Amount
1 PM 10 grams
4 PM x grams
7 PM 14.4 grams

12.0
12.1
12.2
12.3
12.4
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Re: Analyze this! [#permalink]  23 Sep 2008, 20:21
12

10/x = x/14.4

x^2=144
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Re: Analyze this! [#permalink]  24 Sep 2008, 07:56
Yeah the trap answer I picked is 12.2
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Last edited by spiridon on 24 Sep 2008, 09:53, edited 1 time in total.
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Re: Analyze this! [#permalink]  24 Sep 2008, 08:32
(x-10)/10 = (14.4-x)/x

x(x-10)=10(14.4-x)

x^2-10x=144-10x

x^2=144, x=12..

spiridon wrote:
Data for a certain experiment are given in the table. If the amount of microorganism present increased by the same fraction during each of the two 3-hour periods shown, how many grams of microorganism were present at 4:00 PM?
Time Amount
1 PM 10 grams
4 PM x grams
7 PM 14.4 grams

12.0
12.1
12.2
12.3
12.4
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Re: Analyze this! [#permalink]  24 Sep 2008, 09:44
Sorry I overlooked it

Thats the OA

everythings ok
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Last edited by spiridon on 24 Sep 2008, 09:49, edited 1 time in total.
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Re: Analyze this! [#permalink]  24 Sep 2008, 09:46
spiridon wrote:
however, OA is 1.2

They solved it using some crappy period formula

1.2 ? How did you determine the answer ?
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Re: Analyze this! [#permalink]  24 Sep 2008, 13:18
10*.2= 12
12*.2= 14.4

A is correct

The trap here is that the writers assume you will take the difference and divide by 2, this is not correct as the ratios are different

10 -> 12.2 does not equal 12.2 -> 14.4 in terms of ratios
Re: Analyze this!   [#permalink] 24 Sep 2008, 13:18
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