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Data for a certain biology experiment are given in the table
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Updated on: 04 Dec 2012, 03:33
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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 3hour 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 14.410=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
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Originally posted by tejal777 on 29 Sep 2009, 20:29.
Last edited by Bunuel on 04 Dec 2012, 03:33, edited 1 time in total.
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Re: AMOUNT OF BACTERIA PRESENT
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29 Sep 2009, 21:37
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
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29 Sep 2009, 21:00
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 3hour 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 14.410=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: \(10+10*F = 12\) \(F = 1/5\) \(12+12*F = 14.4\) So 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
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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
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29 Sep 2009, 21:46
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: \(X+X*F = 14.4\) Substituting value of X from above equation, \(X+X*(X10)/10 = 14.4\) \(X^2=144\) \(X=12\) I hope it is clear now _________________________ Consider KUDOS for good posts



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Re: AMOUNT OF BACTERIA PRESENT
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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
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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
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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 3hour 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
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03 Dec 2012, 15:59
Change/Initial value...
Let's call X the value we want to find
(X10)/10=(14,4X)/X X^210x=14410X X^2=144 X=12



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Data for a certain biology experiment are given in the table
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28 Nov 2015, 10:24
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 3hour 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
The most important part of this question is highlighted in red above, so I am taking the same into consideration for calculating the amount of bacteria. Since it is given  Quote: the amount of bacteria present increased by the same fraction \(\frac{x}{10}\) = \(\frac{14.4}{x}\) Or,\(x^2\) = \(144\) Or, \(x\) = \(12\) hence at 4:00 PM , the amount of bacteria present is 12 gm , Answer is (A)
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Re: Data for a certain biology experiment are given in the table
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17 Aug 2016, 04:04
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 3hour 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
The way I solved the question: Goal: amount of bacteria @ 4.00 PM which means 10 multiplied by something equals answer.
The difference between 7.00 PM and 1.00 PM is \(14.4  10 = 4.4\) We know that the original amount 10 is increased (or in this case  multiplied) by a fraction EACH 3 hours and there are 2 periods. \(10 * x * x = 14.4\) \(x^2 = \frac{14.4}{10} = x^2 =1.44\) \(\sqrt{x^2} = \sqrt{1.44}\) \(x = 1.2\) From this we solve for X which equals to 1.2. \(x * 10 = 1.2 * 10 = 12.0\) I then multiplied 10 by 1.2 thereby arriving at the answer choice (A)
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Data for a certain biology experiment are given in the table
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15 Aug 2017, 14:19
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 3hour 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 14.410=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 You can also use answer choices. Starting with C. 12.2 Original 10 times some factor = 12.2 10 * x = 12.2, x = 1.22 Per prompt, 12.2 also must increase by factor of 1.22. 12.2 * 1.22 = 4.884 ==> too large by quite a bit. So I picked smallest, Answer A A. 12.0 Original 10 * x = 12, x = 1.2 12 must be multiplied by same factor. (12 * 1.2) = 14.4. That works. Answer A
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Re: Data for a certain biology experiment are given in the table
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04 Apr 2018, 06:27
AKProdigy87 wrote: 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. Hi pushpitkc can you please explain why here x is numerator Fractional increase from 1 PM to 4 PM = X / 10.0 ? and here x is in denominator Fractional increase from 4 PM to 7 PM = 14.4 / X ? many thanks!



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Re: Data for a certain biology experiment are given in the table
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04 Apr 2018, 06:36
dave13 wrote: AKProdigy87 wrote: 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. Hi pushpitkc can you please explain why here x is numerator Fractional increase from 1 PM to 4 PM = X / 10.0 ? and here x is in denominator Fractional increase from 4 PM to 7 PM = 14.4 / X ? many thanks! Hey dave13The answer to your question lies in the question stem 1:00 P.M. 10.0 grams 4:00 P.M. x grams 7:00 P.M. 14.4 grams It is given that the bacteria increased by the same fraction during each of the two 3hour periods The increase in bacteria in the first three hour period = Bacteria(4 PM)/Bacterial(1 PM) = x/10 The increase in bacteria in the second three hour period = Bacteria(7 PM)/Bacterial(4 PM) = 14.4/x Hope this helps you!
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Re: Data for a certain biology experiment are given in the table
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06 Apr 2018, 08:45
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 3hour 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 We can let n = the multiplier for each 3hour period. So, we have: 1 p.m. = 10 grams 4 p.m. = 10n 7 p.m. = 10n^2 We can create the following equation: 10n^2 = 14.4 n^2 = 1.44 n = 1.2 Thus, we have 10n = 10(1.2) = 12 grams of bacteria at 4 p.m. Alternate Solution: Since the fractional increase is the same for both 3hour periods, we must have: x/10 = 14.4/x x^2 = 144 x = 12 or x = 12 Since the number of bacteria cannot be negative, the answer is 12. Answer: A
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Data for a certain biology experiment are given in the table
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27 Sep 2018, 23:10
Bunuel, VeritasKarishma, chetan2u your take on this question please? Thanks.



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Re: Data for a certain biology experiment are given in the table
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28 Sep 2018, 01:16
sadikabid27 wrote: Bunuel, VeritasKarishma, chetan2u your take on this question please? Thanks. The question says ""If the amount of bacteria present increased by the same fraction during each of the two 3hour periods shown" 10, x, 14.4 This means that x increases by a fraction and 14.4 increases by the same fraction (it means they are multiplied by the same number to get the next term) So they form a Geometric Progression. x = Geometric mean of (10, 14.4) \(= \sqrt{10*14.4} = 12\) Answer (A)
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