Time Amount 1:00 P.M. 10.0 grams 4:00 P.M. x grams 7:00 P.M.

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Time Amount 1:00 P.M. 10.0 grams 4:00 P.M. x grams 7:00 P.M. [#permalink]

22 Aug 2006, 22:29

Time Amount
1:00 P.M. 10.0 grams
4:00 P.M. x grams
7:00 P.M. 14.4 grams

Data for a certain biology experiment are given above. If the amount of bacteria present increased by the same fraction during each of the two 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?

A. 12.0

B. 12.1

C. 12.2

D. 12.3

E. 12.4

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23 Aug 2006, 00:11

A

Let that fraction is y
10*y = x
x*y = 14.4

substituting 2 in 1
10 * 14.4/x = x
x^2 = 144
x = 12
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23 Aug 2006, 01:44

A

(x/10)= (14.4/x)

=> x=12

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23 Aug 2006, 04:32

(x-10)/10=(14.4-x)/x
=>x^2-10x=144-10x
=>x^2=144=>x=12
hence answer is A

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23 Aug 2006, 05:21

good trap question, agree with A.

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23 Aug 2006, 21:24

it is confusing.. increased implies that it might be addition rather than multiplication..

Last edited by gk3.14 on 23 Aug 2006, 21:32, edited 1 time in total.

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23 Aug 2006, 21:29

gk3.14 wrote:

it is confusing.. increased implies that it might be addition rather than muliplication..

(x-10)/10 = (14.4-x)/x
x^2 - 10x = 144 - 10x
x^2 = 144
x = 12

A.

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23 Aug 2006, 21:33

Ah i see now.. parvathaneni had done the same thing

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Re: AMOUNT OF BACTERIA PRESENT [#permalink]

23 Aug 2006, 21:55

gmatornot wrote:

I would use process of elimination
(A) % increase formula beween 4pm & 1pm = (12-10)/10 = 1/5
% increase formula beween 7pm & 4pm = (14.4-12)/12 = 2.4/12 = 1/5

Hence A

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23 Aug 2006, 22:22

AK's method is the simplest and the best.

Correct answer is A.

A:B:C

B/A = C/B

[#permalink] 23 Aug 2006, 22:22

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Time Amount 1:00 P.M. 10.0 grams 4:00 P.M. x grams 7:00 P.M.

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