Problem Solving

Hello,

Can you please assist with this:

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.?


OA: 50,000

This is from MGMAT CAT.

gmattesttaker2 wrote: 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

We can plug in the answers, which represent the population at 3pm.
The correct answer choice is almost certain to be both a factor of 500,000 (the population at 4pm) and a multiple of 2000 (the population at 1pm).
Only the value in A -- 50,000 -- is both a factor of 500,000 and a multiple of 2000.

Answer choice A: 50,000
(population at 4pm)/(population at 3pm) = 250,000/50,000 = 5.
Implication: the population increases by a factor of 5 each hour.
If A is the correct answer choice, a factor of 5 must increase the population from 2000 at 1pm to 50,000 at 3pm:
Population at 1pm = 2000.
Population at 2pm = 2000*5 = 10,000.
Population at 3pm = 10,000*5 = 50,000.
Success!

The correct answer is A.

GMATGuruNY wrote:

gmattesttaker2 wrote: 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

We can plug in the answers, which represent the population at 3pm.
The correct answer choice is almost certain to be both a factor of 500,000 (the population at 4pm) and a multiple of 2000 (the population at 1pm).
Only the value in A -- 50,000 -- is both a factor of 500,000 and a multiple of 2000.

Answer choice A: 50,000
(population at 4pm)/(population at 3pm) = 250,000/50,000 = 5.
Implication: the population increases by a factor of 5 each hour.
If A is the correct answer choice, a factor of 5 must increase the population from 2000 at 1pm to 50,000 at 3pm:
Population at 1pm = 2000.
Population at 2pm = 2000*5 = 10,000.
Population at 3pm = 10,000*5 = 50,000.
Success!

The correct answer is A.

Hello Mitch,

Thank you very much for explanation. This is clear now.

Best Regards,
Sri

[email protected] wrote:Hi gmattesttaker2,

Mitch's approach is a great way to solve certain PS Quant questions, and it's an idea that you should embrace.

Here' the algebra approach though. Since the bacteria multiplies at a constant rate, we can call that multiple "x".

Based on the data given, we'd have:

1pm = 2,000 bacteria
2pm = 2,000x bacteria
3pm = 2,000x(x) bacteria
4pm = 2,000x(x)(x) = 250,000 bacteria

Now we can solve for x...

2,000(x^3) = 250,000
(x^3) = 125
x = 5

Multiplying from 1pm, we'd have
1pm = 2,000
2pm = 10,000
3pm = 50,000

GMAT assassins aren't born, they're made,
Rich

Hello Rich,

Thank you very much for the explanation. It is clear now.

Best Regards,
Sri

I just plug in the first answer and I found that it was the correct one.
I didn't check the other choices.
Is that ok during the exam??

Did we not assume that the interval for which the population multiplies by a factor of x is 1 hr. Why is that not a far fetched assumption to make? I have assumed the interval at which the population multiplies by x to be 90 min.
Using this approach,
Pop at 1:00 pm = 2000 (given)
Pop at 2:30 pm = 2000x
Pop at 4:00 pm =(2000x^2)

Now, equating, (2000x^2)=250000
x^2=125
x=sqrt(125)

Now this will mean that pop at 3 pm = Same as pop at 2:30 pm = 2000*sqrt(125) = 2000 * 11 = 22000 approx.

What is the explanation??