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# M01-09

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 00:14
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35% (medium)

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67% (01:29) correct 33% (01:36) wrong based on 256 sessions

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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

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16 Sep 2014, 00:14
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Official Solution:

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

This is a set rate problem. If the population grew by 1,200 people in the past three years, then it grew by 33 percent:
$$\frac{4800-3600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%$$

Therefore in the next three years the population will grow at the same rate of 33% because the growth rate has been constant.

$$4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400$$

Another approach is to backsolve by comparing the ratio of each answer to 4,800. For example, the ratio of 7,200 to 4,800 is not the same as the ratio of 4,800 to 3,600.

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24 Nov 2014, 07:58
Ah this makes it so simple.

Bunuel, if the question asked what would be the population 4 years from 93 i.e. in 97/gap not being uniform, how do we arrive at the answer quickly - assuming the answer choices are widely spread?
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11 Oct 2015, 06:04
Bunuel wrote:
Official Solution:

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

This is a set rate problem. If the population grew by 1,200 people in the past three years, then it grew by 33 percent:
$$\frac{4800-3600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%$$

Therefore in the next three years the population will grow at the same rate of 33% because the growth rate has been constant.

$$4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400$$

Another approach is to backsolve by comparing the ratio of each answer to 4,800. For example, the ratio of 7,200 to 4,800 is not the same as the ratio of 4,800 to 3,600.

It is mentioned in the question that
Population growth rate per thousand is constant
but @Buneul you took % of increase in population as constant.
Is this correct?
I think u need to correct the question.
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Joined: 02 Sep 2009
Posts: 55271

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11 Oct 2015, 06:13
Mechmeera wrote:
Bunuel wrote:
Official Solution:

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

This is a set rate problem. If the population grew by 1,200 people in the past three years, then it grew by 33 percent:
$$\frac{4800-3600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%$$

Therefore in the next three years the population will grow at the same rate of 33% because the growth rate has been constant.

$$4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400$$

Another approach is to backsolve by comparing the ratio of each answer to 4,800. For example, the ratio of 7,200 to 4,800 is not the same as the ratio of 4,800 to 3,600.

It is mentioned in the question that
Population growth rate per thousand is constant
but @Buneul you took % of increase in population as constant.
Is this correct?
I think u need to correct the question.

The rate of increase in multiplying by some constant the same as the percentage increase.
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09 Sep 2016, 03:20
1
Ah this makes it so simple.

Bunuel, if the question asked what would be the population 4 years from 93 i.e. in 97/gap not being uniform, how do we arrive at the answer quickly - assuming the answer choices are widely spread?

Let x = population growth rate per thousand each year

1990: 3.6 (unit: thousand people)
1991: $$3.6 + 3.6x = 3.6(x+1)$$
1992: $$3.6(x+1) + 3.6x(x+1) = 3.6(x+1)(x+1) = 3.6(x+1)^2$$
Similarly:
1993: $$3.6(x+1)^3 = 4.8 => (x+1)^3 = 4/3 => x+1$$

Now we can calculate the population in any year. E.g. 1997: $$3.6(x+1)^7$$ (unit: thousand people)
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03 Dec 2016, 00:09
1
Pop in 1990 = 3600, Pop in 1993 = 4800 Time = 3 yrs. Let rate of growth be r% per year

This implies $$4800 = 3600(1+\frac{r}{100})^3$$

This gives $$(1+\frac{r}{100})^3 = \frac{4800}{3600}= \frac{4}{3}$$

Now Pop in 1996 i.e. 3 years after 1993 will be $$= 4800(1+\frac{r}{100})^3$$

$$=4800*\frac{4}{3} = 6400$$
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23 Jul 2017, 11:54
I solved this question using the concept of linear growth and got the answer as 6000 as follows:
y=mx+c where y is the final value,x is time,m is growth and c is the constant.
at x=0,y=3600 so c=3600
Now when y=4800,x=3 so 4800=m*3+3600 which implies m=400
Thus at x=6,i.e in 1996, y=400*6+3600=6000.

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23 Jul 2017, 23:11
tanv4u wrote:
I solved this question using the concept of linear growth and got the answer as 6000 as follows:
y=mx+c where y is the final value,x is time,m is growth and c is the constant.
at x=0,y=3600 so c=3600
Now when y=4800,x=3 so 4800=m*3+3600 which implies m=400
Thus at x=6,i.e in 1996, y=400*6+3600=6000.

It's exponential growth. So, it should be 4,800 = 3,600*n^3, which gives n^3 =4/3 --> 4,800*4/3 = 6,400.
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20 Nov 2017, 19:43
I think this is a high-quality question. my issue with the explanation is that i agree to the way it is explained but i m not fully convinced that this clears my doubt about why this question is solved in this particular manner..
i solved like this and want to know why my approach is wrong
4800-3600=1200/3=400
1990=3600
1991=3600+400=4000
1992=4000+400=4400
1993=4400+400=4800
(now when the rate is same for each year so i did this)
1994=4800+400=5200
1995=5200+400=5600
1996=5600+400=6000
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20 Nov 2017, 22:59
mallika123 wrote:
I think this is a high-quality question. my issue with the explanation is that i agree to the way it is explained but i m not fully convinced that this clears my doubt about why this question is solved in this particular manner..
i solved like this and want to know why my approach is wrong
4800-3600=1200/3=400
1990=3600
1991=3600+400=4000
1992=4000+400=4400
1993=4400+400=4800
(now when the rate is same for each year so i did this)
1994=4800+400=5200
1995=5200+400=5600
1996=5600+400=6000

We are told that the population growth rate per thousand is constant, not that the population growth number is constant per year.
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23 Nov 2017, 18:52
Bunuel wrote:
Official Solution:

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

This is a set rate problem. If the population grew by 1,200 people in the past three years, then it grew by 33 percent:
$$\frac{4800-3600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%$$

Therefore in the next three years the population will grow at the same rate of 33% because the growth rate has been constant.

$$4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400$$

Another approach is to backsolve by comparing the ratio of each answer to 4,800. For example, the ratio of 7,200 to 4,800 is not the same as the ratio of 4,800 to 3,600.

$$4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400$$ I did not understand this equation ?

Sent from my Redmi 3S using GMAT Club Forum mobile app
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Joined: 02 Sep 2009
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23 Nov 2017, 21:03
Spongebob02 wrote:
Bunuel wrote:
Official Solution:

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

This is a set rate problem. If the population grew by 1,200 people in the past three years, then it grew by 33 percent:
$$\frac{4800-3600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%$$

Therefore in the next three years the population will grow at the same rate of 33% because the growth rate has been constant.

$$4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400$$

Another approach is to backsolve by comparing the ratio of each answer to 4,800. For example, the ratio of 7,200 to 4,800 is not the same as the ratio of 4,800 to 3,600.

$$4,800 + \frac{1}{3}*4,800 = 4,800 + 1,600 = 6,400$$ I did not understand this equation ?

Sent from my Redmi 3S using GMAT Club Forum mobile app

In three years from 1990 to 1993 the population grew by third. Since the growth rate is constant, then in the next three years, from 1993 to 1996, the population will also grow by third.

Hope it's clear.
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23 Jun 2018, 04:43
Hi Bunuel,

Is there any difference between "growth rate is constant" and "growth rate per 1000 is constant" ? If yes, can you please provide any simple example on how it's works out differently?
I guess they both mean the same only but I'm not completely sure if that's true or if I've made any mistake below.

growth rate is (4800-3600 )/ 3600 X 100 = 33.33%
growth rate per year = 33.33 % / 3 = 11.11 %
AND
growth rate per thousand per year is ((4.8 - 3.6) / 3.6 X 100 ) /3 = 33.33% / 3 = 11.11%

Also, here the time period before and after is 3. Let's say if for the same data if we'd to compute the population in 1997, how would you solve it?
I tried and I get 3600 ( 10/9)^7 which is clearly wrong because 3600( 10/9)^3 is not equal to 4800. What's the mistake here?
Seeing the explanation of the unitary method should help understand the original question and it's solution in a much more better way.

This is good question and particularly confusing if someone is reading it for the first time.

Thanks.
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Joined: 02 Sep 2009
Posts: 55271

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23 Jun 2018, 06:20
sanjay1810 wrote:
Hi Bunuel,

Is there any difference between "growth rate is constant" and "growth rate per 1000 is constant" ? If yes, can you please provide any simple example on how it's works out differently?
I guess they both mean the same only but I'm not completely sure if that's true or if I've made any mistake below.

growth rate is (4800-3600 )/ 3600 X 100 = 33.33%
growth rate per year = 33.33 % / 3 = 11.11 %
AND
growth rate per thousand per year is ((4.8 - 3.6) / 3.6 X 100 ) /3 = 33.33% / 3 = 11.11%

Also, here the time period before and after is 3. Let's say if for the same data if we'd to compute the population in 1997, how would you solve it?
I tried and I get 3600 ( 10/9)^7 which is clearly wrong because 3600( 10/9)^3 is not equal to 4800. What's the mistake here?
Seeing the explanation of the unitary method should help understand the original question and it's solution in a much more better way.

This is good question and particularly confusing if someone is reading it for the first time.

Thanks.

1. If the growth rate is constant, then it's constant for any period of time.

2. The growth rate per 3 years is 1 + 1/3 = 4/3, so per year it's x^3 = 4/3 --> $$x = \sqrt[3]{\frac{4}{3}}$$

3. In 1993, the population would be $$3,600*(\sqrt[3]{\frac{4}{3}})^3 = 4,800$$

4. In 1997, the population would be $$3,600*(\sqrt[3]{\frac{4}{3}})^7 \approx 7,044$$

Hope it helps.
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28 Feb 2019, 02:56
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 --> correct
1990->1993->1996
=> 3.6k -> 4.8k -> 4.8 k *growth_rate = 4.8k * (4.8k/3.6k) = 4.8k * (4/3) = 1.6k * 4 = 6.4k
C. 7,200
D. 8,000
E. 9,600
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14 Apr 2019, 00:44
Bunuel can you please elaborate a little more on your reply below? Its still not clear to me.
What does population growth per 1000 mean? and How to identify this problem as exponential growth instead of linear growth??

Bunuel wrote:
mallika123 wrote:
I think this is a high-quality question. my issue with the explanation is that i agree to the way it is explained but i m not fully convinced that this clears my doubt about why this question is solved in this particular manner..
i solved like this and want to know why my approach is wrong
4800-3600=1200/3=400
1990=3600
1991=3600+400=4000
1992=4000+400=4400
1993=4400+400=4800
(now when the rate is same for each year so i did this)
1994=4800+400=5200
1995=5200+400=5600
1996=5600+400=6000

We are told that the population growth rate per thousand is constant, not that the population growth number is constant per year.
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Joined: 02 Sep 2009
Posts: 55271

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14 Apr 2019, 01:10
1
shruthiarvindh wrote:
Bunuel can you please elaborate a little more on your reply below? Its still not clear to me.
What does population growth per 1000 mean? and How to identify this problem as exponential growth instead of linear growth??

Bunuel wrote:
mallika123 wrote:
I think this is a high-quality question. my issue with the explanation is that i agree to the way it is explained but i m not fully convinced that this clears my doubt about why this question is solved in this particular manner..
i solved like this and want to know why my approach is wrong
4800-3600=1200/3=400
1990=3600
1991=3600+400=4000
1992=4000+400=4400
1993=4400+400=4800
(now when the rate is same for each year so i did this)
1994=4800+400=5200
1995=5200+400=5600
1996=5600+400=6000

We are told that the population growth rate per thousand is constant, not that the population growth number is constant per year.

Take an example: say we are told that a population is growing 100 people per 1000 per year (10%). This means that every 1000 people "produce" 100 more. For example, if the population is 3,600 people in 1990, then in 1991, the growth will be 3*100 + 600/1000*100 = 360, which is 10% of 3,600.

Another case would be if we are told that a population is growing 100 people per year. So, if the population is 3,600 people in 1990, then in 1991 it will be 3,600 + 100, in 1992 it will be 3,600 + 100 + 100, so basically each year adds another 100 people.

Hope it's clear.
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20 May 2019, 01:44
I think this is a poor-quality question and I agree with explanation.
Re M01-09   [#permalink] 20 May 2019, 01:44
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# M01-09

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