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.
Statement 1: The population just divided and, since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.
When the population increases by a factor of 4, the result is 3750 more cells:
4p = p + 3750
p = 1250.
Thus:
2 hours ago, p=1250.
Now, p = 1250+3750 = 5000.
Since the population quadruples every 2 hours, over the next 4 hours it will quadruple twice:
p = 4*4*5000 = 80,000.
SUFFICIENT.
Statement 2: The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.
No information about HOW OFTEN the population doubles.
If the population doubles EVERY HOUR, then in the final hour it will double from 40,000 to 80,000.
If the population doubles EVERY HALF-HOUR, then in the final hour it will double from 40,000 to 80,000 and then from 80,000 to 160,000.
INSUFFICIENT.
The correct answer is
A.
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