Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The population in town A declined at a constant rate from 10 [#permalink]
09 Nov 2010, 11:23

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

100% (04:56) correct
0% (00:00) wrong based on 3 sessions

The population in town A declined at a constant rate from 10,000 in the year 1990 to 9,040 in the year 1998. The population in town B increased at a constant rate from 4,000 in the year 1994 to 4,560 in the year 1998. If the rates of change of population in towns A and B remain the same, in approximately what year will the populations in the two towns be equal?

Re: Solution anyone please? [#permalink]
09 Nov 2010, 13:07

Rate of decrease in population, in city A = (10000-9040)/8 = 120 persons/year Rate of increase in population, in city B = (4560-4000)/8 = 140 persons/year

Lets assume after t years, starting in 1990, the population becomes equals; => 10000 - 120*t = 4000 + 140*t => 260t = 6000 => t ~= 23 years

So, the populations for city A and B would be equal in year, 1990+23 = 2013.

Re: Solution anyone please? [#permalink]
09 Nov 2010, 13:21

anshumishra wrote:

Rate of decrease in population, in city A = (10000-9040)/8 = 120 persons/year Rate of increase in population, in city B = (4560-4000)/8 = 140 persons/year

Lets assume after t years, starting in 1990, the population becomes equals; => 10000 - 120*t = 4000 + 140*t => 260t = 6000 => t ~= 23 years

So, the populations for city A and B would be equal in year, 1990+23 = 2013.

Shouldn't you start at year 1998 because in city B had 4000 population in year 1994.

So I would use the equation

9040 - 120t = 4560 + 140t and solve for 17 < t < 18.

Re: Solution anyone please? [#permalink]
09 Nov 2010, 13:28

You are right. I made a mistake !

chaoswithin wrote:

anshumishra wrote:

Rate of decrease in population, in city A = (10000-9040)/8 = 120 persons/year Rate of increase in population, in city B = (4560-4000)/8 = 140 persons/year

Lets assume after t years, starting in 1990, the population becomes equals; => 10000 - 120*t = 4000 + 140*t => 260t = 6000 => t ~= 23 years

So, the populations for city A and B would be equal in year, 1990+23 = 2013.

Shouldn't you start at year 1998 because in city B had 4000 population in year 1994.

So I would use the equation

9040 - 120t = 4560 + 140t and solve for 17 < t < 18.

Re: Solution anyone please? [#permalink]
09 Nov 2010, 14:05

I think 2015 is the right answer. But what's wrong with this? An exponential function can be used to model population growth that has a constant percentage change in population: \(f(t)=ab^t\) Where f(t)= population after t years a=initial value b=growth factor t=time in years For town A: \(9040=10000b^8\) \(b=(9040/10000)^{1/8}=.987\) For town B: \(4560=4000b^4\) \(b=(4560/4000)^{1/4}=1.033\) Equating the two functions with an initial value corresponding to the year 1998 we get: \(9040(0.987)^t=4560(1.033)^t\) \(t=log1.98/(log1.033-log0.987)=14.99\) Therefore, 15 years after 1998, or in 2013, the populations of towns A and B will be the same.

Re: Solution anyone please? [#permalink]
09 Nov 2010, 14:44

1

This post received KUDOS

trx123 wrote:

The population in town A declined at a constant rate from 10,000 in the year 1990 to 9,040 in the year 1998. The population in town B increased at a constant rate from 4,000 in the year 1994 to 4,560 in the year 1998. If the rates of change of population in towns A and B remain the same, in approximately what year will the populations in the two towns be equal?

Town A : In 8 years declined from 10,000 to 9,040 ... So annualized decline of 1.2% Town B : In 4 years increased from 4,000 to 4,560 ... So annualized increase of 3.5%

Let it take x years to equate the two populations Current difference is 4480 Town A decreases at approx 90 people a year (since the pop changes from 9040 to lower .. we can calc this as approx 1.2% of around 8000) Town B increases at approx 210 people a year (since the pop changes from 4560 to higher .. we can calc this as approx 3.5% of around 6500) SO approx number of years = 4480/(300) = Approx 15

So answer should be 1998+15 = 2013

If you calculate this prcisely, you can verify it is 2013 as well _________________

Re: Solution anyone please? [#permalink]
09 Nov 2010, 15:35

Now I want to know what the actual answer is.

trx123 wrote:

I think 2015 is the right answer. But what's wrong with this? An exponential function can be used to model population growth that has a constant percentage change in population: \(f(t)=ab^t\) Where f(t)= population after t years a=initial value b=growth factor t=time in years For town A: \(9040=10000b^8\) \(b=(9040/10000)^{1/8}=.987\) For town B: \(4560=4000b^4\) \(b=(4560/4000)^{1/4}=1.033\) Equating the two functions with an initial value corresponding to the year 1998 we get: \(9040(0.987)^t=4560(1.033)^t\) \(t=log1.98/(log1.033-log0.987)=14.99\) Therefore, 15 years after 1998, or in 2013, the populations of towns A and B will be the same.

Your approach is different from mine in that I used constant numerical rate approach whereas you used a constant percent rate approach.

The two approach has to lead to different answers.

Your approach would show that Town A's population is decreasing at 1.3% per year.

Whereas my approach would show that Town A's population is decreasing at 120pp/year.

The thing to note about percent rate of change is that the number of population change would either accelerate or decelerate as time goes on.

Therefore, since Town A's population is decreasing, the number of population decreasing would gradually become smaller, whereas the number of population increase in Town B's population would gradually accelerate.

Because of this reason, the crossing point for the two towns would come earlier in the constant percent rate analysis than the crossing point for the two towns in the constant numerical rate analysis.

If I was given this on the GMAT I would just use the numerical rate approach and move on. If I tried to go the percent route, I might stress myself too much and jeopardize the rest of the QUANT section Anyone else have a take on this?

Re: Solution anyone please? [#permalink]
09 Nov 2010, 16:45

chaoswithin wrote:

anshumishra wrote:

Rate of decrease in population, in city A = (10000-9040)/8 = 120 persons/year Rate of increase in population, in city B = (4560-4000)/8 = 140 persons/year

Lets assume after t years, starting in 1990, the population becomes equals; => 10000 - 120*t = 4000 + 140*t => 260t = 6000 => t ~= 23 years

So, the populations for city A and B would be equal in year, 1990+23 = 2013.

Shouldn't you start at year 1998 because in city B had 4000 population in year 1994.

So I would use the equation

9040 - 120t = 4560 + 140t and solve for 17 < t < 18.

So year 2015 would be my answer.

I believe 2015 is the correct answer. The problem states that the population growth is constant. I used exponential growth, which as you sated is not constant but accelerates or decelerates with time. The constant rate formula is \(f(t)=9040-120t\) The exponential rate formula is: \(f(t)=9040(.987)^t\) Thank you for answering

Re: Solution anyone please? [#permalink]
10 Nov 2010, 07:44

The problem states that the population (the number of people) and not the percent of the population, increased or decreased at a constant rate. The word rate is defined as a value describing one quantity in terms of another. In this case one quantity is time, the other is number of people. Initially, I also assumed that the rate was the yearly percent change in population. The answer really depends how you interpret the question.

Re: Solution anyone please? [#permalink]
10 Nov 2010, 08:19

1

This post received KUDOS

Expert's post

The rate at which population increases is the percentage at which it increases. If I say, 'The population increases at a constant rate', I mean it increases in every time period by a fixed percentage e.g. 10%. In fact population increase is a typical example of compounding. But that really made the calculations here torturous, which is definitely not a characteristic of GMAT questions. If they want to say that they are referring to a constant increase/decrease in number of people, they need to say 'the population increases/decreases by a constant number of people every year' or something. _________________

Re: Solution anyone please? [#permalink]
10 Nov 2010, 11:20

VeritasPrepKarishma wrote:

The rate at which population increases is the percentage at which it increases. If I say, 'The population increases at a constant rate', I mean it increases in every time period by a fixed percentage e.g. 10%. In fact population increase is a typical example of compounding. But that really made the calculations here torturous, which is definitely not a characteristic of GMAT questions. If they want to say that they are referring to a constant increase/decrease in number of people, they need to say 'the population increases/decreases by a constant number of people every year' or something.

I agree with you and initially that was exactly my approach. But this problem was taken from a multiple choice Arizona teacher proficiency test. The answer key gave the correct answer as 2015. So I presume that they meant constant number of people. Anyway, thanks for your help.

Re: Solution anyone please? [#permalink]
11 Nov 2010, 12:12

Quote:

The rate at which population increases is the percentage at which it increases. If I say, 'The population increases at a constant rate', I mean it increases in every time period by a fixed percentage e.g. 10%. In fact population increase is a typical example of compounding. But that really made the calculations here torturous, which is definitely not a characteristic of GMAT questions. If they want to say that they are referring to a constant increase/decrease in number of people, they need to say 'the population increases/decreases by a constant number of people every year' or something.

Re: Solution anyone please? [#permalink]
26 Nov 2013, 05:56

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Hey, Last week I started a few new things in my life. That includes shifting from daily targets to weekly targets, 45 minutes of exercise including 15 minutes of yoga, making...

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...