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 whether the answer matches 2,000.
1. I started with (C) 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 A, 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 A
_________________
SC Butler has resumed! Get
two SC questions to practice, whose links you can find by date,
here.Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has -- Margaret Mead