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AMOUNT OF BACTERIA PRESENT
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 in the table 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
Can someone pls explain why A is the answer?
Thanks
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AMOUNT OF BACTERIA PRESENT 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 in the table 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
Can someone pls explain why A is the answer? Thanks
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The question stem is that the bacteria increase is the same fraction during each of the 3-hour.
Let's take 12
So the increase in first period is 12-10/10=2/10 or 1/5 or 20%
For second period the fraction should be same that is 1/5 of 12 become 2.4
Its a simple ratio problem - fraction increase from 10 to x is the same as fraction increase from x to 14.4 or like richardj mentions... geometric mean.
10/x = x/14.4 => x^2 = 10*14.4=144 => x = 12
Kind of redundant but just to point.. treat this (and I mean this specific problem) like a ratio problem.
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.