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.?
50,000
62,500
65,000
86,666
125,000
I am going through the above problem from a Manhattan CAT exam. Note the explanation says the following:If we decide to find a constant multiple by the hour, then we can say that the population was multiplied by a certain number three times from 1 p.m. to 4 p.m.: once from 1 to 2 p.m., again from 2 to 3 p.m., and finally from 3 to 4 p.m.
Let's call the constant multiple x.
2,000(x)(x)(x) = 250,000
2,000(x3) = 250,000
x3 = 250,000/2,000 = 125
x = 5
Therefore, the population gets five times bigger each hour.
My question is....The problem never states the frequency of increase. How can we say it increased every hr?
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- dhoomketu
You are correct in saying that we shouldn't assume that it multiples each hour, however even without this assumption we get the same result.
Assume : r as multiplication factor for that constant period (who knows the bacteria may be double, triple etc.)
Let the constant period be t hrs (could be 1/2 hr , one hour etc.)
Then in 1 hour the bacteria population will be:
2000 x r^(1/t)
Similarly in 3 hrs:
2000 x r^(3/t) = 250000
r^(3/t) = 125
Since r should be an integer, the only way you could represent the equation is by writing:
r^(3/t) = 5^3
Hence r is 5 and consequently t is 1.
Assume : r as multiplication factor for that constant period (who knows the bacteria may be double, triple etc.)
Let the constant period be t hrs (could be 1/2 hr , one hour etc.)
Then in 1 hour the bacteria population will be:
2000 x r^(1/t)
Similarly in 3 hrs:
2000 x r^(3/t) = 250000
r^(3/t) = 125
Since r should be an integer, the only way you could represent the equation is by writing:
r^(3/t) = 5^3
Hence r is 5 and consequently t is 1.
- Jimmy
The reason I asked this question was because there was a Manhattan GMAT DS question that was wrong based on this principal...the question explanation stated we needed the frequency and the population size after a specific time. So I guess my question is...when does frequency matter, and when does it not matter? How can I identify when it does on a question?
- RonPurewal
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Jimmy Wrote:The reason I asked this question was because there was a Manhattan GMAT DS question that was wrong based on this principal...the question explanation stated we needed the frequency and the population size after a specific time. So I guess my question is...when does frequency matter, and when does it not matter? How can I identify when it does on a question?
one fact that may be surprising to you here:
a constant growth rate is a constant growth rate. it actually doesn't matter at all what time unit you use as your base interval!
this may seem counterintuitive at first, but think about it this way: let's say that some quantity doubles every day. then think about what happens every week: the quantity will double seven times, so that it is 128 times as large as it was one week before. therefore, the following statements are equivalent:
* growing by a factor of 2 every hour
* growing by a factor of 128 every week
and also
* growing by a factor of (24th root of 2) every hour
* growing by a factor of (86400th root of 2) every second
etc.
in other words, to sum it up concisely: if you have constant-rate growth, you can select whatever time unit is the most convenient.
4 posts Page 1 of 1