Data for a certain biology experiment are given in the table

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.

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 number of bacteria cannot be negative, the answer is 12.

Answer: A
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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.
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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, \(X+X*(X-10)/10 = 14.4\)
\(X^2=144\)
\(X=12\)

I hope it is clear now
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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?

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

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 -

\(\frac{x}{10}\) = \(\frac{14.4}{x}\)

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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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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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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Method I - Algebra - Multiplier

"The amount of bacteria present increased by the same fraction," \(x\), twice:
\((x*x) = x^2\) = the multiplier, twice

\(10 * x^2 = 14.4\)
\(x^2 = \frac{14.4}{10}\)
\(x = \sqrt{1.44}= 1.2 =\) the multiplier

Number of bacteria at 4 p.m.?

\(10 * 1.2 = 12\) bacteria present at 4 p.m.
(And \(12 * 1.2 = 14.4\) at 7 p.m. Correct.)

Answer A

Method II - Use answer choices

Start with (C) 12.2

Find one fractional increase:
\(\frac{12.2}{10} = 1.22\)

Result for the second fractional increase:
\(12 * 1.22 = 14.64\)

C's 4 p.m. value is too great. Eliminate D and E, which will be even greater.

Try A, the round number: 12

Fractional increase: \(\frac{12}{10} = 1.2\)
Multiply 4 p.m. value, 12, by 1.2:
\(12 * 1.2 = 14.4\). Correct.

Answer A
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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 dave13

The 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 3-hour 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!

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 number of bacteria cannot be negative, the answer is 12.

Answer: A
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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 3-hour 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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VeritasKarishma, Bunuel
I solved by adding same fraction Eg., 10 + a = x; x + a = 14.4; a is constant fraction.
Can you please advise why 'increases by' can't signify addition of same fraction?