The population in town A declined at a constant rate from 10

You are right. I made a mistake !

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.

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

Now I want to know what the actual answer is.



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?

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

The thing is that "rate" generally means percentage (think rate of interest got instance)
So I think correct answer is 2013

Posted from GMAT ToolKit

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.

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.

Nice explaination Karishma..

Population difference = 6000

Annual decrease in A's population \(= \frac{960}{8} = 120\)

Annual increase in B's population \(= \frac{560}{4} = 140\)

Total change \(= 120+140= 260\)

Required time \(= \frac{6000}{260}= 23\) (approx)

1990+23= year 2013

10000-120t=4000+140t
solving we get t=23(aprox)

Just gave you your Kudos #200 for this one, Good job
Cheers
J

I think the number would be 18>t>17.

The difference is not 6000 because the initial population of A (10000) and B (4000) in two different timelines. The gap should be considered from 1998 onwards.

If we follow the relativity formula.

A' rate is : 120 and B's rate is 140.

Here total rate is 120+140 = 260

Now the remaining gap of the population: 9040-4560 = 4480.

Number of years required: 4480/260= 17.23.

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