Bunuel wrote:

Tough and Tricky questions: Word Problems.

Wes works at a science lab that conducts experiments on bacteria. The population of the bacteria multiplies at a constant rate, and his job is to notate the population of a certain group of bacteria each hour. At 1 p.m. on a certain day, he noted that the population was 2,000 and then he left the lab. He returned in time to take a reading at 4 p.m., by which point the population had grown to 250,000. Now he has to fill in the missing data for 2 p.m. and 3 p.m. What was the population at 3 p.m.?

A. 50,000

B. 62,500

C. 65,000

D. 86,666

E. 125,000

I worked from the answer choices because finding the constant rate looked quick for four of the answers. It was quick.

If we have

1 p.m. -- 2,000

2 p.m. -- ???

3 p.m. -- ???

4 p.m. -- 250,000

Then because the answer choices are for 3 p.m., divide 4 p.m. amount by an answer choice to get the multiplication factor.

Use that factor: divide the 3 p.m. amount by the factor to get the 2 p.m. amount, then once more (2 p.m. amt / factor = 1 p.m. amt) to see if the answer matches 2,000.

1. I started with (E) where 3 p.m. amount is 125,000

\(\frac{250,000}{125,000}\) to get multiplication factor =

2 3:00 --> 2:00 is \(\frac{125,000}{2}\) = 62,500 at 2:00.

Stop. Much too big. The next step, \(\frac{65,000}{2}\) is not going to equal 2,000 at 1 p.m.

2. Try C, where 3 p.m. amount = 50,000

\(\frac{250,000}{50,000}\) = multiplication factor

5Hour between 2:00 and 3:00 is then \(\frac{50,000}{5}\) = 10,000 at 2:00

2:00 -> 1:00 is \(\frac{10,000}{5}\) = 2,000 at 1 p.m. Correct

Answer C

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