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?