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M01-09

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Re: M01-09  [#permalink]

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New post 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.


Answer: B


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

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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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Re: M01-09  [#permalink]

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New post 03 Dec 2017, 16:28
Spongebob02

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


We have simply converted decimal (0.33%) in to fraction (1/3) for ease of calculations while dealing with fractions.
Hope this helps!
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New post 20 Dec 2017, 00:06
For me this was easier and quicker to solve using the ratios given in the question stem:

4800/3600 = x/4800
x = 6400
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New post 20 Dec 2017, 00:16
Easy question if ever comes up on GMAT:

We can see every 3 population grew up by 1 unit.

Divide by 400
We get the ratio 9:12

Next equivalent ratio should be (12*4)/3=16

So, 16*400=6400

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New post 26 Mar 2018, 10:25
I think this is a high-quality question and I agree with explanation.
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New post 07 Jun 2018, 01:56
I think this is a high-quality question and I agree with explanation.
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New post 18 Jun 2018, 10:32
I think this is a poor-quality question and I don't agree with the explanation. Hi, well the rate per year comes to be (4800-3600)/3 = 400/year, therefore increasing at a const. rate of 400/year, the population in 1996 should be 400/year * 3 + 4800 = 6000.
Well, i agree that the OA must be 'B'. But please help me to convince myself to the OA.
What made you consider the % instead ?
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Re: M01-09  [#permalink]

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New post 18 Jun 2018, 10:47
honcho11 wrote:
I think this is a poor-quality question and I don't agree with the explanation. Hi, well the rate per year comes to be (4800-3600)/3 = 400/year, therefore increasing at a const. rate of 400/year, the population in 1996 should be 400/year * 3 + 4800 = 6000.
Well, i agree that the OA must be 'B'. But please help me to convince myself to the OA.
What made you consider the % instead ?


Hi honcho11

check your calculation in the highlighted part.

also you can check an alternate approach mentioned in the below link for clarity

https://gmatclub.com/forum/m01-183520.html#p1770983
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M01-09  [#permalink]

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New post 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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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: M01-09  [#permalink]

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New post 23 Jun 2018, 06:35
Bunuel wrote:
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.


Thanks Bunuel. Your explanations are always very helpful.
Last thing, just to confirm there is no difference between "growth rate is constant" and "growth rate per 1000 is constant", correct?
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Re: M01-09  [#permalink]

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New post 23 Jun 2018, 06:38
sanjay1810 wrote:
Bunuel wrote:
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.


Thanks Bunuel. Your explanations are always very helpful.
Last thing, just to confirm there is no difference between "growth rate is constant" and "growth rate per 1000 is constant", correct?

____________________
Right.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M01-09 &nbs [#permalink] 23 Jun 2018, 06:38

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