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AMOUNT OF BACTERIA PRESENT

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AMOUNT OF BACTERIA PRESENT

by uptowngirl92 » Tue Sep 29, 2009 7:29 pm
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
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by hitmewithgmat » Tue Sep 29, 2009 8:06 pm
let's put "a" as the rate.

then x=10a for 4pm
14.4=ax for 7pm

when you combine the two equation, you will get
14.4=a(10a)
14.4=10a^2
1.2=a

then, x=10a
x=10(1.2)
x=12

A is the answer.
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by xcusemeplz2009 » Tue Sep 29, 2009 8:12 pm
the q has said it is inc by some constant fraction thus it is a GP problem, your approach would have been correct if it was said that the inc is constant value.
in the first case it is getting multiplied by some fraction,
in the later case it is getting added by a constant value.
from the q we are having info.
x/10=14.4/x
x^2=144
x=12

if u will cross check then u will get the fractional inc value as 1.2

but in your approach its different in both the time.
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Re: AMOUNT OF BACTERIA PRESENT

by Stuart@KaplanGMAT » Wed Sep 30, 2009 11:53 am
uptowngirl92 wrote: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
This is a very common mistake that people make in multiple percent increase questions (which are really just compound interest questions in disguise).

Consider the following:

A cyclist is traveling at 10kph. If the cyclist increases her speed by 20%, then increases her speed by another 20%, what's her final speed?

The answer is not, of course, 14 kph. We have to take 20% of 10 (=2) to get an interim speed of 12 and then take 20% of 12 (=2.4) to get a final speed of 14.4 kph.

The scenario I posted above is actually the exact same as in the question you posted; the only difference is that in my question we got asked for the final speed, in your question you got asked for the interim speed.

Similarly, it could have been asked as a compound interest question:

Bob invests $10 at his local bank. At the end of 2 years, the total value of the investment is $14.40. If Bob's investment earns a fixed rate of interest, compounded anually, what was the value of the investement at the end of the first year?

As always, when you do a question it's not about getting THAT question right - it's about learning something new that will help you on future questions; connecting this question to other, similar, questions is what will get you those extra points on test day.
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by Jeff@TargetTestPrep » Tue Jan 02, 2018 6:39 am
uptowngirl92 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
We can let n = the multiplier for each 3-hour 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 3-hour periods, we must have:

x/10 = 14.4/x

x^2 = 144

x = 12 or x = -12

Since the bacteria cannot increase by -12, the answer is 12.

Answer: A

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[email protected]

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