Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

29 Sep 2009, 20:29

5

This post received KUDOS

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

45% (medium)

Question Stats:

59% (01:06) correct
41% (01:12) wrong based on 404 sessions

HideShow timer Statistics

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.?

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

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

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??

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.
_________________

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?

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

Show Tags

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 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) _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

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

Show Tags

17 Aug 2016, 04:04

1

This post received KUDOS

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)
_________________

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

Show Tags

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 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?

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

Show Tags

16 Aug 2017, 08:14

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 3-hour periods shown, how many grams of bacteria were present at 4:00 P.M.?

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^n-1 so 14.4=10r^3-1 14.4=10r^2 1.44=r^2 r=1.2 Now we can find out the second term ar^2-1=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 _________________

We are more often frightened than hurt; and we suffer more from imagination than from reality