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

Question

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?

A. 2012
B. 2013
C. 2014
D. 2015
E. 2016

A. 2012
B.  2013 
C. 2014
D. 2015
E. 2016

anshumishra · Answer

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.

chaoswithin · Answer

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.

anshumishra · Answer

You are right. I made a mistake !

trx123 · Answer

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.

shrouded1 · Answer

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

chaoswithin · Answer

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?

trx123 · 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

shrouded1 · Answer

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

https://gmatclub.com/static/posted_from.png     Posted from    GMAT ToolKit

trx123 · Answer

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.

KarishmaB · Answer

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.

trx123 · Answer

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.

dimitri92 · Answer

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

mehtakaustubh · Answer

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

Parvezahmed2210 · Answer

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.

Posted from my mobile device

Amity007 · Answer

Town A
Population decreases from 10,000 (1990) to 9,040 (1998).
Over 8 years:
10000(1−r)8=904010000(1-r)^8 
Approx 1−r≈0.9875
So Town A declines by about  1.25% per year .
 
Town B
Population increases from 4,000 (1994) to 4,560 (1998).
Over 4 years:
4000(1+g)4=4560
Approx 1+g≈1.0333
So Town B grows by about  3.33% per year .
 
Set populations equal
Using 1998 as the reference year:
A(t)=9040(0.9875)^t
B(t)=4560(1.0333)^t
Set equal:
9040(0.9875)^t =4560(1.0333)^t
4560/9040=(0.9875/1.0333)^t
Taking logs:
 t=ln(1.9825)/ln(1.0464)≈0.684/0.0454≈15.1
So the populations become equal about  15 years after 1998 :
1998+15≈≈2013

BraveMine · Answer

How can you subtract 10000 from 4000 when these are two different years population ???? You cannot subtract Town B's  1994  population (4,000) from Town A's  1990  population (10,000). It completely ignores the fact that those two numbers happened 4 years apart. In 1990, Town B's population was actually 3,440.
You cannot subtract Town B's  1994  population (4,000) from Town A's  1990  population (10,000). It completely ignores the fact that those two numbers happened 4 years apart. In 1990, Town B's population was actually 3,440.

egmat · Answer

Hi BraveMine,

Your instinct is genuinely sharp, and you've actually put your finger on the single thing that splits this whole thread into the "2013" camp and the "2015" camp. So let me separate the two readings cleanly, because the answer depends entirely on which one the question intends.

First, you are completely right about one thing.  If "declined/increased at a constant rate" means a  constant number of people per year  (a straight line), then  4,000  is Town B's  1994  population, and you cannot drop it into an equation as if it were Town B's  1990  population. Projected linearly back to 1990, Town B would indeed be 4,000 - 140x4 =  3,440 , exactly as you said. Do the algebra honestly under that reading and the crossing comes out around  2015 , not 2013. The popular "10000 - 120t = 4000 + 140t" shortcut only lands on 2013 because it quietly pairs a 1990 number with a 1994 number - the very mismatch you flagged.

Here's the part that resolves it.  For populations, "a constant rate" almost always means a  constant percentage rate , not a fixed headcount - that's the standard reading (and the one KarishmaB pointed to earlier in this thread). Under the percentage reading, your year-offset worry disappears, and here's why: each town's growth factor is computed  only inside its own data window .

- Town A: over  its  8 years, factor = (9040/10000)^(1/8) =  0.9875  per year.
- Town B: over  its  4 years, factor = (4560/4000)^(1/4) =  1.0333  per year.

Notice that Town B's factor is built purely from B's own 1994->1998 data - the number  4,000  never gets compared to A's  10,000  at all. You then project  both  towns forward from a shared anchor (1998): 9040x(0.9875)^t = 4560x(1.0333)^t, which gives t =  15 , i.e.  2013 .

The one idea to lock in:  a  rate  (percent) lives inside each town's own timeline, so the starting years never have to match. A fixed  headcount-per-year  is an absolute difference, so the years must align - and that's the model where your 3,440 correction is exactly right.

Answer: B

BraveMine · Answer

Hey thanks for explaining yes even I feel rate is the correct way to calculate as constant number of people doesnot make any sense but at the same time I don't accept your solution fully as to solve this  9040x(0.9875)^t = 4560x(1.0333)^t is little complex to solve it in just 2 mins and I dont know gmat will not make you solve a complex logs to get to solution so do you have any better approach to this question

Adit_ · Answer

A bit tricky but easy when you know how to approximate.

So lets just assume it is constant rate year on year.
Fir one is 960 in 8 years.
Second is 560 in 4 years.

Now take 2012 for example which is 14 year from 1998.

Now in exponential growth, although slow the subsequent 4 year/8 year intervals will always supercede the constant growth.

So we can take a ballpark assumption of lets say 2 sets of 8 years and 4 sets of 4 years for each case.

2 sets of 8 years would be:
1920 decrease -> 9040 - 1920 = 7120
4 sets of 4 years would be:
2240 increase -> 4560 + 2240 = 6800.

Now if you see these are extremely close.
And you can again assume that per year at the very least 560/4 around 150 and 960/8 = 120 change happens.

The current difference is 320.
Thus 150+120 would make it even closer to what we are looking at.

Thus the year after 2012 would be the answer. =  2013. 
Would like to know better approaches, or faster approach as compared to this one.

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

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