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Data for a certain biology experiment are given in the table [#permalink]
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29 Sep 2009, 19:29
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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 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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Last edited by Bunuel on 04 Dec 2012, 02:33, edited 1 time in total.
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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
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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 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 [#permalink]
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29 Sep 2009, 20:32
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]
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29 Sep 2009, 20:37
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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]
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29 Sep 2009, 20:46
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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: \(X+X*F = 14.4\) 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 [#permalink]
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20 Jul 2011, 07:14
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]
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20 Jul 2011, 08:47
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]
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20 Jul 2011, 17:58
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 [#permalink]
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08 Nov 2011, 11:32
(x10)/3=(14.4x)/3 2x=24.4 x=12.2



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Re: AMOUNT OF BACTERIA PRESENT [#permalink]
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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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Re: Data for a certain biology experiment are given in the table [#permalink]
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17 Nov 2013, 14:55
Set fraction to equal X
from 1pm to 4pm is X*10 from 4pm to 7pm is 14.4/X
since the factor is consistent between the two increases we can equate the two numbers. X*10 = 14.4/X
X^2 = 14.4/10 X^2 = 144/100 X = 12/10 = 6/5
So 10*(6/5) = 12 and 12*(6/5) = 14.4
Answer is A



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Re: Data for a certain biology experiment are given in the table [#permalink]
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17 Jan 2015, 03:03
x/10 = 14.4/X => x^2 = 144 =>X=12
Dont simply take averages....as i see many people did....silly mistake but this will cost dear



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Re: Data for a certain biology experiment are given in the table [#permalink]
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15 Sep 2015, 23:50
"inreased by the same fraction" means multiplying by the same number
So, x/10=14.4/x => x^2=144 => x=12
A



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Re: Data for a certain biology experiment are given in the table [#permalink]
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28 Nov 2015, 04:58
We are looking for \(10*x^3\) We have \(10*x^6= 14,4\) So \(x^6= 1,44\) ==> We can see a root of 12 here \(x^3= 1,2\) \(=>10*1,2 = 12\)
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Data for a certain biology experiment are given in the table [#permalink]
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28 Nov 2015, 09:24
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 [#permalink]
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17 Aug 2016, 03:04
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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 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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Re: Data for a certain biology experiment are given in the table [#permalink]
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14 Aug 2017, 14:17
we can easily see that 14.4 is almost the same as 144, which is 12 squared. so 10*1.12 = 12 12*1.12 = 14.4 A is the answer.



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Data for a certain biology experiment are given in the table [#permalink]
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15 Aug 2017, 13:19
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 [#permalink]
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16 Aug 2017, 07:14
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 This is a geometric progression question Let the rate be x first term is 10 second term is x third term =14.4 first method The nth term is given by ar^n1 so 14.4=10r^31 14.4=10r^2 1.44=r^2 r=1.2 Now we can find out the second term ar^21=ar 10*1.2=12 There is one more method x/10=14.4/x or x^=144 or X=12 The answer is A
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