December 16, 2018 December 16, 2018 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score. December 16, 2018 December 16, 2018 03:00 PM EST 04:00 PM EST Strategies and techniques for approaching featured GMAT topics
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51229

Question Stats:
65% (01:02) correct 35% (00:58) wrong based on 225 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Math Expert
Joined: 02 Sep 2009
Posts: 51229

Re M0109
[#permalink]
Show Tags
15 Sep 2014, 23:14
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: \(\frac{48003600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%\) 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
_________________
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Current Student
Joined: 04 Jul 2014
Posts: 298
Location: India
GMAT 1: 640 Q47 V31 GMAT 2: 640 Q44 V34 GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)

Re: M0109
[#permalink]
Show Tags
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?
_________________
Cheers!!
JA If you like my post, let me know. Give me a kudos!



Current Student
Joined: 04 Jul 2014
Posts: 298
Location: India
GMAT 1: 640 Q47 V31 GMAT 2: 640 Q44 V34 GMAT 3: 710 Q49 V37
GPA: 3.58
WE: Analyst (Accounting)

Re: M0109
[#permalink]
Show Tags
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?
_________________
Cheers!!
JA If you like my post, let me know. Give me a kudos!



Current Student
Joined: 14 May 2014
Posts: 41
GPA: 3.11

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: \(\frac{48003600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%\) 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 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..?



Intern
Joined: 18 Aug 2015
Posts: 1

Re M0109
[#permalink]
Show Tags
15 Sep 2015, 18:29
I think this is a poorquality question.



Retired Moderator
Joined: 18 Sep 2014
Posts: 1117
Location: India

Re: M0109
[#permalink]
Show Tags
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: \(\frac{48003600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%\) 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 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

Re: M0109
[#permalink]
Show Tags
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: \(\frac{48003600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%\) 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 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.
_________________
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 24 Apr 2016
Posts: 9

AdmitJA 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? 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)
_________________
You are your best guru!



Intern
Joined: 02 Mar 2016
Posts: 7

Re M0109
[#permalink]
Show Tags
16 Oct 2016, 06:57
I think this is a highquality question and I agree with explanation.



PS Forum Moderator
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82

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 \(= 4800(1+\frac{r}{100})^3\)
\(=4800*\frac{4}{3} = 6400\)



Manager
Joined: 01 Nov 2016
Posts: 66
Concentration: Technology, Operations

Re: M0109
[#permalink]
Show Tags
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



Intern
Joined: 21 Nov 2014
Posts: 20
Location: India
GMAT 1: 710 Q47 V40 GMAT 2: 690 Q47 V38 GMAT 3: 750 Q49 V44
GPA: 3.1

Re: M0109
[#permalink]
Show Tags
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.
Can't we use linear growth for this question?Where am I going wrong exactly?Please help



Math Expert
Joined: 02 Sep 2009
Posts: 51229

Re: M0109
[#permalink]
Show Tags
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.
Can't we use linear growth for this question?Where am I going wrong exactly?Please help 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.
_________________
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 14 Apr 2017
Posts: 248
Location: India
sandeep : sharma
Concentration: International Business, Entrepreneurship
GPA: 3.9

Re: M0109
[#permalink]
Show Tags
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
_________________
Thanks 71 tough IR questions218dsquestionsforpractice awapptthebestresourceforgmat newcriticalreasoningsupermegathreadupdated



Math Expert
Joined: 02 Sep 2009
Posts: 51229

Re: M0109
[#permalink]
Show Tags
23 Jul 2017, 23:04



Intern
Joined: 21 Sep 2015
Posts: 2

Re: M0109
[#permalink]
Show Tags
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



Intern
Joined: 09 Nov 2017
Posts: 1

Re M0109
[#permalink]
Show Tags
20 Nov 2017, 18:43
I think this is a highquality 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 48003600=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 according to this method i got 6000 as an answer please help me understand why i m wring what i m basically missing out...please explain



Math Expert
Joined: 02 Sep 2009
Posts: 51229

Re: M0109
[#permalink]
Show Tags
20 Nov 2017, 21:59
mallika123 wrote: I think this is a highquality 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 48003600=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 according to this method i got 6000 as an answer please help me understand why i m wring what i m basically missing out...please explain We are told that the population growth rate per thousand is constant, not that the population growth number is constant per year.
_________________
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.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 09 Aug 2017
Posts: 23

Re: M0109
[#permalink]
Show Tags
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: \(\frac{48003600}{3600} = \frac{1200}{3600} = \frac{1}{3} \approx 33%\) 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 ? Sent from my Redmi 3S using GMAT Club Forum mobile app







Go to page
1 2
Next
[ 35 posts ]



