Data Sufficiency

Hi,

I am facing problems with that particular question.

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) The population just divided and, 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.

[spoiler]
We need two additional pieces of information to solve this problem, which can be rephrased as "How frequently does the population double, and what is the population size at any given time immediately after it has doubled?"

(1) SUFFICIENT: If the population quadrupled during the last two hours, it doubled twice during that interval, meaning that the population doubled at 60 minute intervals. Since it has increased by 3,750 bacteria, we have:

Population (Now) - Population (2 hours ago) = 3750
Population (Now) = 4·Population (2 hours ago)

Substituting, we get 4·Population (2 hours ago) - Population (2 hours ago) = 3750
Population (2 hours ago) = 1250.
The population will double 6 times from that point to 4 hours from now

Population (4 hours from now) = 26·1,250=80,000.


(2) INSUFFICIENT: This statement does not give any information about how frequently the population is doubling.

The correct answer is A.
[/spoiler]

I don't quite understand their explanation. Even if we know the quantity "two hours ago", we still don't know what point of time "two hours ago" equals, right?

Thanks.

Both are sufficient. So answer will be D.

From question stem you know the scientist destroys 4 hours from now AND the population just divided. So now is hour 0.

1st statement: Tells when destroyed, 2 hours before it divided. So as it 2 hour for division to double. If x population at hour 0, 2x at 2 hours and 4x at 4 hour. Difference from hour 4 to hour 0 is 3x=3750. So you can find 4x.

2nd statement: The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.
As we know it just divided before destroying, so period to multiply is 1. 1 hour before it was 40,000, at hour 4 it will be 80,000.


Both statements are sufficient.

Hi,

thanks, but OA is A... (see black spoiler)