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Data for a certain biology experiment are given in the table [#permalink]
 29 Sep 2009, 20:29
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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 Please point out my flaw:
We know,
in 6 hrs. bacteria increased 14.4-10=4.4
So,since it is increasing by a constant,
in 3 hrs>> 4.4 x 3/6 = 2.2
Therefore,
1:00 P.M. 10.0 grams
4:00 P.M. 12.2 grams
7:00 P.M. 14.4 grams IMO:C _________________
countdown-beginshas-ended-85483-40.html#p649902
Last edited by Bunuel on 04 Dec 2012, 03:33, edited 1 time in total.
Renamed the topic and edited the question.
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
29 Sep 2009, 21:00
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tejal777 wrote: 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
Please point out my flaw:
We know,
in 6 hrs. bacteria increased 14.4-10=4.4
So,since it is increasing by a constant,
in 3 hrs>> 4.4 x 3/6 = 2.2
Therefore,
1:00 P.M. 10.0 grams
4:00 P.M. 12.2 grams
7:00 P.M. 14.4 grams IMO:C IMO A. Note that the bacteria did not increase by same amount, but by same FRACTION. So the answer has to be less than 12.2, i.e. either A or B. Assuming A is correct: 10+10*F = 12 F = 1/512+12*F = 14.4So answer is A. Please let me know if you want me to explain further. ___________________________ Consider KUDOS for good posts 
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
29 Sep 2009, 21:32
I am sorry its not clear..could you please give a detailed expalnation? got my mistake though..its says same fraction.. 
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
29 Sep 2009, 21:37
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Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm:
Fractional increase from 1 PM to 4 PM = X / 10.0
Fractional increase from 4 PM to 7 PM = 14.4 / X
\frac{X}{10.0} = \frac{14.4}{X}
X^2 = (14.4)(10.0)
X^2 = 144
X = 12
Therefore, the correct answer is A: 12.0.
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
29 Sep 2009, 21:46
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tejal777 wrote: I am sorry its not clear..could you please give a detailed expalnation? got my mistake though..its says same fraction.. Sure. As we know that the bacterias are increasing at a constant Fraction, we can write the following equation: 10 + 10*F = X, where F = Fraction by which bacterias increases, and X = amount of bacterias after 3 hrs. After 6 hrs, number of bacterias will be: X+X*F = 14.4Substituting value of X from above equation, X+X*(X-10)/10 = 14.4X^2=144X=12I hope it is clear now  _________________________ Consider KUDOS for good posts
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
20 Jul 2011, 08:14
I was using a rather stupid way..and I got B. Can anyone pls tell what got wrong with my calculation? thanks .. Step 1: 14.4 - 10.00 = 4.4 (the total increase of amount) Step 2: since there are 6 hours, I took 4.4/6 = approx. 0.73 (rounding up to 0.7) So at 2pm -----> 10.0 + 0.7 = 10.7 3pm -----> 10.7 + 0.7 = 11.4 4pm -----> 11.4 + 0.7 = 12.1 (Here you go answer B) < But it isn't not right answer on OA  WHY??  Can anyone pls enlighten? thanks...
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
20 Jul 2011, 09:47
miweekend wrote: I was using a rather stupid way..and I got B. Can anyone pls tell what got wrong with my calculation? thanks .. Step 1: 14.4 - 10.00 = 4.4 (the total increase of amount) Step 2: since there are 6 hours, I took 4.4/6 = approx. 0.73 (rounding up to 0.7) So at 2pm -----> 10.0 + 0.7 = 10.7 3pm -----> 10.7 + 0.7 = 11.4 4pm -----> 11.4 + 0.7 = 12.1 (Here you go answer B) < But it isn't not right answer on OA WHY?? Can anyone pls enlighten? thanks... The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours. Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm: Fractional increase from 1 PM to 4 PM = X / 10.0 Fractional increase from 4 PM to 7 PM = 14.4 / X X \ 10 = 14.4 \ X X^2 = 14.4 * 10 X^2 = 144 X = 12 Therefore, the correct answer is A: 12.0.
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
20 Jul 2011, 18:58
varunmaheshwari wrote: miweekend wrote: I was using a rather stupid way..and I got B. Can anyone pls tell what got wrong with my calculation? thanks .. Step 1: 14.4 - 10.00 = 4.4 (the total increase of amount) Step 2: since there are 6 hours, I took 4.4/6 = approx. 0.73 (rounding up to 0.7) So at 2pm -----> 10.0 + 0.7 = 10.7 3pm -----> 10.7 + 0.7 = 11.4 4pm -----> 11.4 + 0.7 = 12.1 (Here you go answer B) < But it isn't not right answer on OA WHY?? Can anyone pls enlighten? thanks... The question says that bacteria increased by same fraction, not by same amount in 2 intervals of 3 hours. Let X represent the amount of bacteria present at 4:00 PM. Since the fractional increase must remain constant from 1 to 4pm as it is from 4pm to 7pm: Fractional increase from 1 PM to 4 PM = X / 10.0 Fractional increase from 4 PM to 7 PM = 14.4 / X X \ 10 = 14.4 \ X X^2 = 14.4 * 10 X^2 = 144 X = 12 Therefore, the correct answer is A: 12.0. thank you very much....i don't want to sound v. silly but still very confused about the question asked. ...... 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.?i .... If 10.0 grams is the denominator so why would 14.4 be the numerator? since 10.0 is at the bottom, 14.4 should be remaining at the bottom too ya?
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
08 Nov 2011, 12:32
(x-10)/3=(14.4-x)/3
2x=24.4
x=12.2
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
03 Dec 2012, 15:59
Change/Initial value...
Let's call X the value we want to find
(X-10)/10=(14,4-X)/X
X^2-10x=144-10X
X^2=144
X=12
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Re: AMOUNT OF BACTERIA PRESENT
[#permalink]
03 Dec 2012, 15:59
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