Given

Four hours from now, the population will double and then be destroyed immediately.

Statement 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) = (2^6)*1,250=80,000.

Statement 2

One hour before it is destroyed (or, in other words, three hours from now), the population will double to 40,000 cells. We need to determine the population from the sample at the time it is destroyed.

3 hours from now--population will double to 40,000

4 hours from now--population will double again then be destroyed

It's possible that the population doubles every hour, thus at 4 hours from now the population would be 80,000. But it is possible that the population doubles every half hour.

We could have:

3 hours from now--population will double to 40,000

3.5 hours from now--population will double to 80,000

4 hours from now--population will double to 160,000 and be destroyed.

So we can't obtain the exact number of bacteria before it is destroyed and Statement 2 is

Insufficient.
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