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OG12 PS#89

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by nasir » Thu Sep 02, 2010 11:07 pm
Can someone help with the explanation

Time Amount
1.00PM 10.0 gm
4.00PM x gm
7.00PM 14.4gm

Data for a certain biology experiment are given in the table. if the amount of bacteria present increased by the same factor during each of the two 3-hrs periods shown, how many gms of bacteria were present at 4:00PM

Thanks
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by Rahul@gurome » Thu Sep 02, 2010 11:18 pm
nasir wrote:Can someone help with the explanation

Time Amount
1.00PM 10.0 gm
4.00PM x gm
7.00PM 14.4gm

Data for a certain biology experiment are given in the table. if the amount of bacteria present increased by the same factor during each of the two 3-hrs periods shown, how many gms of bacteria were present at 4:00PM

Thanks
That's PS#81.
Let the factor by which the amount of bacteria present increased every 3 hrs be f.
From 1 PM to 4 PM, that is 3 hrs, the bacteria increases by f. So, 10 * f = x is 1st equation.
Similarly from 4 PM to 7 PM, again at a difference of 3 hrs, bacteria increases by f. So, x * f = 14.4 is 2nd equation.
Solving the two equations, (10 * f) * f = 14.4 or 10 * f^2 = 14.4
f^2 = 1.44 or f = 1.2
So, x = 10 * 1.2 = 12 gm

The correct answer is [spoiler](A)[/spoiler].
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by this_time_i_will » Thu Sep 02, 2010 11:46 pm
we know that the increment is caluclated by: (new-old)/old.
so. (x-10)/10 = (14.4-x)/x
solving x = 12.
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