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Strategies and techniques for approaching featured GMAT topics

# M01-09

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Math Expert
Joined: 02 Sep 2009
Posts: 51229

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15 Sep 2014, 23:14
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Difficulty:

35% (medium)

Question Stats:

65% (01:02) correct 35% (00:58) wrong based on 225 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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Posts: 51229

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15 Sep 2014, 23:14
1
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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, 06:58
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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JA
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Current Student
Joined: 04 Jul 2014
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25 Nov 2014, 20:04
Bunuel look forward to your views!

joseph0alexander wrote:
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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Cheers!!

JA
If you like my post, let me know. Give me a kudos!

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24 Jul 2015, 09:54
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.

bumping an old ques..

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?

also the given population growth rate is per thousand. doesn't this make any difference..?
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Joined: 18 Aug 2015
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15 Sep 2015, 18:29
1
I think this is a poor-quality question.
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Joined: 18 Sep 2014
Posts: 1117
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11 Oct 2015, 05:04
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.
Math Expert
Joined: 02 Sep 2009
Posts: 51229

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11 Oct 2015, 05:13
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, 02:20
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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16 Oct 2016, 06:57
I think this is a high-quality question and I agree with explanation.
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02 Dec 2016, 23:09
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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16 Apr 2017, 09:16
The question doesn't ask for an increase by percentage rate. The question asks for an increase based on the "growth rate per thousands." I don't know what this means. I thought it was asking for the rate of population increase, which is 400 people per year (1990->1993 is an increase of 1200 people). At 400 people per year, the answer would be 6000. I had a feeling this question was too easy...

"Growth rate per thousand"...so I guess it is asking how much growth in population there was for everyone one thousand people? So from 3600 to 4800, the growth rate from 3.6 thousand to 4.8 thousand would be 1.2 thousand. Maybe this question is not well worded
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23 Jul 2017, 10:54
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.

Math Expert
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23 Jul 2017, 22:11
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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23 Jul 2017, 23:00
Hi ,

I think this must be a 500 level question
as growth rate is constant. say it is k
4800=k*3600
k=4/3
Now to calculate , population in 1996, growth rate is same , no of years is same
hence population =4800*(4/3)=6400
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23 Jul 2017, 23:04
mbaaspirant80 wrote:
Hi ,

I think this must be a 500 level question
as growth rate is constant. say it is k
4800=k*3600
k=4/3
Now to calculate , population in 1996, growth rate is same , no of years is same
hence population =4800*(4/3)=6400

Good that you find the question easy. But if you check the stats in the original post you'll see that empirically this is not true.
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18 Aug 2017, 03:26
@Bunnel

How do i know from the language of the question ?? whether its a linear growth or exponential.
Kindly help
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Joined: 09 Nov 2017
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20 Nov 2017, 18:43
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, 21:59
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, 17:52
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 ?

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# M01-09

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