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Senior Manager  Joined: 25 Oct 2008
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Location: Kolkata,India
Data for a certain biology experiment are given in the table  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 65% (01:34) correct 35% (01:34) wrong based on 395 sessions

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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

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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http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

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.
Renamed the topic and edited the question.
Manager  Joined: 11 Sep 2009
Posts: 115
Re: AMOUNT OF BACTERIA PRESENT  [#permalink]

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13
2
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.
##### General Discussion
Manager  Joined: 18 Aug 2009
Posts: 237
Re: AMOUNT OF BACTERIA PRESENT  [#permalink]

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3
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

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:
$$10+10*F = 12$$
$$F = 1/5$$
$$12+12*F = 14.4$$

Please let me know if you want me to explain further.
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Re: AMOUNT OF BACTERIA PRESENT  [#permalink]

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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.. _________________
http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902
Manager  Joined: 18 Aug 2009
Posts: 237
Re: AMOUNT OF BACTERIA PRESENT  [#permalink]

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2
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, $$X+X*(X-10)/10 = 14.4$$
$$X^2=144$$
$$X=12$$

I hope it is clear now _________________________
Consider KUDOS for good posts Intern  Joined: 29 Jun 2011
Posts: 20
Re: AMOUNT OF BACTERIA PRESENT  [#permalink]

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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...
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Re: AMOUNT OF BACTERIA PRESENT  [#permalink]

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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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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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Posts: 18
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GMAT 1: 650 Q45 V35 Re: AMOUNT OF BACTERIA PRESENT  [#permalink]

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2
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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Data for a certain biology experiment are given in the table  [#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 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

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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2
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

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  [#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 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 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

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.

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Re: Data for a certain biology experiment are given in the table  [#permalink]

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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

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  [#permalink]

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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

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

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!
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Re: Data for a certain biology experiment are given in the table  [#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 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 number of bacteria cannot be negative, the answer is 12.

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Data for a certain biology experiment are given in the table  [#permalink]

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Re: Data for a certain biology experiment are given in the table  [#permalink]

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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) $$= \sqrt{10*14.4} = 12$$

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