It is currently 23 Nov 2017, 10:11

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A scientist is studying bacteria whose cell population

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Director
Joined: 16 May 2007
Posts: 547

Kudos [?]: 76 [0], given: 0

A scientist is studying bacteria whose cell population [#permalink]

### Show Tags

15 Jul 2007, 07:14
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 100% (00:00) wrong based on 2 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed?

(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.

(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.

Kudos [?]: 76 [0], given: 0

Manager
Joined: 27 May 2007
Posts: 128

Kudos [?]: 14 [0], given: 0

### Show Tags

15 Jul 2007, 08:05
I would answer A, (1) alone is sufficient, but (2) alone is not sufficient. We know the population divided 2 hours ago, has quadrupled since, and has since increased by 3750, and that is doubles at constant intervals. So if it quadruples in 2 hours it is doubling every hour. Let x stand for the quantity before doubling 2 hours ago, x + 3750 is the current quantity. Since the quantity has quadrupled, 4x is also the current quantity, so the equation is x + 3750=4x. Solved, x=1250. The current quantity is therefore 5000, and it will double each hour for the next 4 hours, ending at 80000.

(2) is not sufficient because we don't know the rate - it might be doubling every 10 minutes!

BTW, I'm having trouble remembering that I don't have to solve every DS problem. It should have been enough to know that (1) gave a rate and an amount, and (2) gave only an amount but no rate.

Kudos [?]: 14 [0], given: 0

Senior Manager
Joined: 27 Jul 2006
Posts: 292

Kudos [?]: 18 [0], given: 0

### Show Tags

16 Jul 2007, 09:56
A is correct here, but D is tempting since if you read "will double" it gives the impression that it will double from its current state, which would mean that the bacteria double every 3 hours, and an hour from that moment there will still be 40,000 in the test tube, and 40,000 innocent bacteria will then be decimated by the unfeeling laboratory doctors.

Kudos [?]: 18 [0], given: 0

Senior Manager
Joined: 03 Jun 2007
Posts: 376

Kudos [?]: 16 [0], given: 0

### Show Tags

16 Jul 2007, 10:53
I agree with A here. I initially went with D with the possibility of bacteria doubling every 1 hr without considering that the bacteria might double every 3 hrs too

Kudos [?]: 16 [0], given: 0

Senior Manager
Joined: 28 Jun 2007
Posts: 315

Kudos [?]: 53 [0], given: 0

Re: DS : Bacteria Population [#permalink]

### Show Tags

16 Jul 2007, 11:33
trahul4 wrote:
A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed?

(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.

(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.

I think the answer is C.

(1) gives three pieces of information:
- The bacteria doubled 2 hours ago.
- The increase is 3750. So the total bacteria after the quadruplation is 3,750 * 4/3 = 5000.
- The bacteria quadrupled in the past 2 hours. Note that it does not say when exactly they quadrupled. So the bacteria could be doubling every 60 mins or every 45 mins for all we know. Hence, insufficient.

(2) Tells us that in (4 - 1) = 3 hours the total bacteria will be 40,000. But we dont know what the number of bacteria is currently. So we dont know the rate at which the bacteria is doubling. Hence, insufficient.

If you put (1) and (2) together, you have to double 3 times in 3 hours to go from 5,000 to 40,000. This means the doubling is occuring every 60 mins. So 80,000 bacteria will be destroyed.

Is my thinking too convoluted? What is the OA?

Kudos [?]: 53 [0], given: 0

Re: DS : Bacteria Population   [#permalink] 16 Jul 2007, 11:33
Display posts from previous: Sort by

# A scientist is studying bacteria whose cell population

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.