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# PS Population (m01q09)

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PS Population (m01q09) [#permalink]

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07 Aug 2008, 04:26
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The population of Linterhast was 3,600 people in 1990 and 4,800 people in 1993. If the population growth rate per thousand is constant, then what will be the population in 1996?

(A) 6,000
(B) 6,400
(C) 7,200
(D) 8,000
(E) 9,600

[Reveal] Spoiler: OA
B

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Re: PS Population [#permalink]

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07 Aug 2008, 04:42
Growth Rate during 1990-93 = (4800-3600)*100/3600 = 100/3

1996 Population = 4800 + 4800*100/3*100 = 4800 + 1600 = 6400

Answer B.
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Re: PS Population [#permalink]

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07 Aug 2008, 05:00
abhijit,

how do you define population growth rate, can it be taken as increase in population over specific period of time...

for example, the distance growth rate is speed, which is distance covered divided by time...

going back to question

1990 --- 3600
1993 --- 4800
population growth rate (as per above definition ) = (4800-3600)/3 = 400 per year
Population in 1996 = population in 1993 + 3 * increase per year (constant as per question)
= 4800 + 3*400
= 6000

Option A.... what is a flaw in this logic... Or the question wordings needs to changed
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Re: PS Population [#permalink]

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07 Aug 2008, 05:16
durgesh79 wrote:
abhijit,

how do you define population growth rate, can it be taken as increase in population over specific period of time...

for example, the distance growth rate is speed, which is distance covered divided by time...

going back to question

1990 --- 3600
1993 --- 4800
population growth rate (as per above definition ) = (4800-3600)/3 = 400 per year
Population in 1996 = population in 1993 + 3 * increase per year (constant as per question)
= 4800 + 3*400
= 6000

Option A.... what is a flaw in this logic... Or the question wordings needs to changed

I can take a guess. What you are doing is taking population as a function of time only (dividing by 3). While in reality future population is a function of present population also.
Not only do the current ppl contribute to increase, but the newly added also start contributing.

A more appropriate analogy would be that of compound interest.
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Re: PS Population [#permalink]

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07 Aug 2008, 05:25
bhushangiri wrote:
durgesh79 wrote:
abhijit,

how do you define population growth rate, can it be taken as increase in population over specific period of time...

for example, the distance growth rate is speed, which is distance covered divided by time...

going back to question

1990 --- 3600
1993 --- 4800
population growth rate (as per above definition ) = (4800-3600)/3 = 400 per year
Population in 1996 = population in 1993 + 3 * increase per year (constant as per question)
= 4800 + 3*400
= 6000

Option A.... what is a flaw in this logic... Or the question wordings needs to changed

I can take a guess. What you are doing is taking population as a function of time only (dividing by 3). While in reality future population is a function of present population also.
Not only do the current ppl contribute to increase, but the newly added also start contributing.

A more appropriate analogy would be that of compound interest.

i agree with the logic and i understand the analogy between interest rate and population also... i dont think on GMAT questions we are expected to know how population growth rate is defined... i mean the same question if i just change the words (take out population and include something else) the answer will change...

try solving this one...

The amount of water in pools was 3,600 L at 12 noon. It was 4,800 L at 3 PM . What will water level at 6PM, if the water increaseing rate is constant?

6000
6400
7200
8000
9600
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Re: PS Population [#permalink]

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07 Aug 2008, 06:12
Bhusangiri has already explained the problem. Now to elaborate further. In original question population has a domino effect, population creates population, thereby compounding the net results, however, in case of later question there is most probably one inlet pipe and you cannot force more water through that pipe than its maximum capacity. Thereby while in first case you need to consider the percentage, in later you need to consider only the constant rate change over time.
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Re: PS Population [#permalink]

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07 Aug 2008, 06:23
abhijit_sen wrote:
Bhusangiri has already explained the problem. Now to elaborate further. In original question population has a domino effect, population creates population, thereby compounding the net results, however, in case of later question there is most probably one inlet pipe and you cannot force more water through that pipe than its maximum capacity. Thereby while in first case you need to consider the percentage, in later you need to consider only the constant rate change over time.

thanks for explaining in detail... i know that and i agree with your solution....
All i'm saying are we suposed to know this property of population growth or it should be mentioned in the question ... something like " the population growth rate per thousand people is constant"

The question is from GMAT club challenges. i forgot the test number and the question number... and the OA is B...

but i strongly believe that the language needs to be changed....
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Re: PS Population [#permalink]

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07 Aug 2008, 06:54
durgesh79 wrote:
All i'm saying are we suposed to know this property of population growth or it should be mentioned in the question ... something like " the population growth rate per thousand people is constant"
but i strongly believe that the language needs to be changed....

Durgesh yes we are supposed to know this property and surely you know it now
However, I do agree with you that wording of question can be changed to make it more clearer.
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Re: PS Population [#permalink]

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07 Aug 2008, 07:18
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good discussion guys.

You view the question in differt way to..

You can rephrase the original problem "% increase is constant over the periods"

% increase inpopulation from 1990-1993 = % increase in population from 1993-96

(4800-3600)/3600 = 33.3%

1996 population = 4800 *1.333 = 6400
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Re: PS Population [#permalink]

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07 Aug 2008, 17:54
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durgesh79 wrote:
The population of Linterhast was 3,600 people in 1990. It was 4,800 people in 1993. What will be the population in 1996, if the population growth rate has been constant over the years?

6000
6400
7200
8000
9600

This is a really good question.
Lets consider the growth of population being equal.
then lets call population in 1996 as x we get the eqn:

(4800-3600)/3600 = (x-4800)/4800
=> solving=> x=6400
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Re: PS Population [#permalink]

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08 Aug 2008, 23:41
durgesh79 wrote:
The population of Linterhast was 3,600 people in 1990. It was 4,800 people in 1993. What will be the population in 1996, if the population growth rate has been constant over the years?

6000
6400
7200
8000
9600

B) 4/3 of pop in 1993
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Re: PS Population (m01q09) [#permalink]

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02 Sep 2009, 09:22
Moderators, i dont think the wording of this question has changed yet. Today I fell into the same trap and picked A. If you agree that wording is misleading, please change the wording of the question....so there is no confusion.
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Re: PS Population (m01q09) [#permalink]

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15 Oct 2009, 01:46
Done. Here's the reworded question:

The population of Linterhast was 3,600 people in 1990 and 4,800 people in 1993. If the population growth rate per thousand is constant, then what will be the population in 1996?

(C) 2008 GMAT Club - m01#9

* 6,000
* 6,400
* 7,200
* 8,000
* 9,600

sdrandom1 wrote:
Moderators, i dont think the wording of this question has changed yet. Today I fell into the same trap and picked A. If you agree that wording is misleading, please change the wording of the question....so there is no confusion.

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Re: PS Population (m01q09) [#permalink]

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07 Nov 2009, 01:02
Sol1 : x^3 * 3600 = 4800
x^3 = 4800/3600

1996 = x ^6 * 3600 = 6400

Sol2: 4800 * x^3 = 4800 * 4800/3600 = 6400
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Re: PS Population (m01q09) [#permalink]

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07 Nov 2009, 18:49
durgesh79 wrote:
The population of Linterhast was 3,600 people in 1990. It was 4,800 people in 1993. What will be the population in 1996, if the population growth rate has been constant over the years?

6000
6400
7200
8000
9600

I think the wording is fiine. The rate of growth is always in %. If it were in numbers, it would be rate of growth in numbers.

Growth rate = (4800/3600)^(1/3) - 1 = (1.33)^(1/3) - 1 = 10% approx..
(however 1.33 is also fine for the even intervals of each 3 years)

Pop in 1996 = 4800 (1+0.10)^3 = 4800 (1.3333) = 6400 i.e. B.

dzyubam wrote:
The population of Linterhast was 3,600 people in 1990 and 4,800 people in 1993. If the population growth rate per thousand is constant, then what will be the population in 1996?

* 6,000
* 6,400
* 7,200
* 8,000
* 9,600

I do not see any better wording. The previous was better and clear.

If you use the % growth rate, the solution is the same as above. If you solve using some x value per 1000, it would be more complex.

Pop in 1990 = 1000 (here we do not know the increament rate in number if we do not use the rate derrivved using 3600 and 4800)

Per thousand increament in 1991 = 3.6x
Pop in 1991= (3600+3.6x)

Per thousand increament in 1992 = (x/1000) (3600+3.6x) = 3.6x + (3.6x^2)/1000
Pop in 1992= (3600+3.6x) + 3.6x + (3.6x^2)/1000

Per thousand increament in 1993 = (x/1000) [(3600+3.6x) + 3.6x + (3.6x^2)/1000]
Pop in 1993 = wow................ so complicate.....

If we have to use % grwoth rate to calculate the growth/1000, then the rewording doesnot add any value.
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Re: PS Population (m01q09) [#permalink]

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21 Jan 2010, 12:53
Is the rate of growth the same as the 'Factor' by which the numbers increase ?

i.e from 1990 to 1993, pop. increased by a factor of 4,800/3600 = 1.33

Therefore, in 1996, pop. will be 1.33 times '93 pop. = 1.33 * 4800 = 6400.

Is there anything wrong with this approach ? (i.e in case this approach doesn't fit other rate questions).
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Re: PS Population (m01q09) [#permalink]

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22 Jan 2010, 01:21
Your solution is absolutely right. Rate problems involving population growth really differ from other rate questions.
kaptain wrote:
Is the rate of growth the same as the 'Factor' by which the numbers increase ?

i.e from 1990 to 1993, pop. increased by a factor of 4,800/3600 = 1.33

Therefore, in 1996, pop. will be 1.33 times '93 pop. = 1.33 * 4800 = 6400.

Is there anything wrong with this approach ? (i.e in case this approach doesn't fit other rate questions).

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Re: PS Population [#permalink]

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29 Jan 2010, 21:22
durgesh79 wrote:
abhijit_sen wrote:
Bhusangiri has already explained the problem. Now to elaborate further. In original question population has a domino effect, population creates population, thereby compounding the net results, however, in case of later question there is most probably one inlet pipe and you cannot force more water through that pipe than its maximum capacity. Thereby while in first case you need to consider the percentage, in later you need to consider only the constant rate change over time.

thanks for explaining in detail... i know that and i agree with your solution....
All i'm saying are we suposed to know this property of population growth or it should be mentioned in the question ... something like " the population growth rate per thousand people is constant"

The question is from GMAT club challenges. i forgot the test number and the question number... and the OA is B...

but i strongly believe that the language needs to be changed....

Agreed with you!
I understood the way it was supposed to be at first, but after reading the questions carefully - the constant growth - I ended up with choice A

Otherwise, the rate per thousand can be calculated by easy proportion:
90) 3600
93) 4800
Growth /thousand or Change/Original
4800+3600 ( original) = 6400
Now 3600/6400= 1\3 ( this is the proportion we need rto calculate the next growth) Thus
1/3= x( change) /6400 ( new Original) = x=1600
or 6400+1600= 4800
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Re: PS Population (m01q09) [#permalink]

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25 Nov 2010, 04:56
B.
Pop growth rate = 4800 - 1200 / 3600 = 1/3
1996 pop = 4800 + 4800*1/3 = 6400

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Re: PS Population (m01q09) [#permalink]

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09 Jan 2011, 09:51
This was my method:

The increase from 1990 to 1993 was 1200. Therefore when expressed as a percentage it comes to about 33.33% which is really 33.33/100.

If this were converted to per thousand, it would be 333 per thousand. From there on, [4800 + 4800(333.3/1000)] to find the increase.
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Re: PS Population (m01q09)   [#permalink] 09 Jan 2011, 09:51

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# PS Population (m01q09)

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