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

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The population in town A declined at a constant rate from 10 [#permalink]

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09 Nov 2010, 12:23
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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?
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09 Nov 2010, 14:07
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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.
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09 Nov 2010, 14:21
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.

So year 2015 would be my answer.
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09 Nov 2010, 14:28
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.

So year 2015 would be my answer.
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09 Nov 2010, 15:05
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.
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09 Nov 2010, 15:44
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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
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09 Nov 2010, 16:35
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.

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?
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09 Nov 2010, 17:45
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:
$$f(t)=9040(.987)^t$$
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09 Nov 2010, 21:33
The thing is that "rate" generally means percentage (think rate of interest got instance)
So I think correct answer is 2013

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10 Nov 2010, 08:44
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.
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10 Nov 2010, 09:19
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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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 30 Aug 2010 Posts: 10 Followers: 0 Kudos [?]: 3 [0], given: 0 Re: Solution anyone please? [#permalink] Show Tags 10 Nov 2010, 12:20 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. Manager Joined: 01 Nov 2010 Posts: 179 Location: Zürich, Switzerland Followers: 2 Kudos [?]: 45 [0], given: 20 Re: Solution anyone please? [#permalink] Show Tags 11 Nov 2010, 13:12 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. Nice explaination Karishma.. Senior Manager Affiliations: SPG Joined: 15 Nov 2006 Posts: 327 Followers: 16 Kudos [?]: 750 [1] , given: 28 Re: Solution anyone please? [#permalink] Show Tags 19 Jan 2011, 07:38 1 This post received KUDOS 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 _________________ press kudos, if you like the explanation, appreciate the effort or encourage people to respond. Download the Ultimate SC Flashcards Manager Joined: 11 Jul 2009 Posts: 168 WE: Design (Computer Software) Followers: 1 Kudos [?]: 52 [0], given: 69 Re: Solution anyone please? [#permalink] Show Tags 19 Jan 2011, 09:50 10000-120t=4000+140t solving we get t=23(aprox) _________________ Kaustubh GMAT Club Legend Joined: 09 Sep 2013 Posts: 15474 Followers: 649 Kudos [?]: 209 [0], given: 0 Re: Solution anyone please? [#permalink] Show Tags 26 Nov 2013, 06:56 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. _________________ Current Student Joined: 06 Sep 2013 Posts: 2005 Concentration: Finance Followers: 68 Kudos [?]: 643 [0], given: 355 Re: Solution anyone please? [#permalink] Show Tags 26 Nov 2013, 06:57 dimitri92 wrote: 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 Just gave you your Kudos #200 for this one, Good job Cheers J GMAT Club Legend Joined: 09 Sep 2013 Posts: 15474 Followers: 649 Kudos [?]: 209 [0], given: 0 Re: The population in town A declined at a constant rate from 10 [#permalink] Show Tags 15 Jan 2017, 11:20 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. _________________ Re: The population in town A declined at a constant rate from 10 [#permalink] 15 Jan 2017, 11:20 Similar topics Replies Last post Similar Topics: In a certain rural town, the population grew from 13,000 in 1981 to 19 4 16 May 2017, 08:26 2 John invests$10,000 at the monthly constant compounded rate of annual 1 06 Jan 2017, 02:43
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