GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Dec 2018, 12:04

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Happy Christmas 20% Sale! Math Revolution All-In-One Products!

December 20, 2018

December 20, 2018

10:00 PM PST

11:00 PM PST

This is the most inexpensive and attractive price in the market. Get the course now!
• ### Key Strategies to Master GMAT SC

December 22, 2018

December 22, 2018

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

# M01-09

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51280

### Show Tags

23 Nov 2017, 20:03
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.
_________________
Study Buddy Forum Moderator
Joined: 04 Sep 2016
Posts: 1270
Location: India
WE: Engineering (Other)

### Show Tags

03 Dec 2017, 15:28
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!
_________________

It's the journey that brings us happiness not the destination.

Intern
Joined: 16 Jul 2013
Posts: 33
Location: Hungary
Concentration: Entrepreneurship, Marketing
GMAT 1: 720 Q48 V40
GPA: 3.37

### Show Tags

19 Dec 2017, 23:06
For me this was easier and quicker to solve using the ratios given in the question stem:

4800/3600 = x/4800
x = 6400
Intern
Joined: 25 Nov 2017
Posts: 4

### Show Tags

19 Dec 2017, 23:16
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

Sent from my ZUK Z2132 using GMAT Club Forum mobile app
Intern
Joined: 02 Jun 2013
Posts: 21

### Show Tags

26 Mar 2018, 09:25
I think this is a high-quality question and I agree with explanation.
_________________

Nothing comes easy ! Neither do i want !!

Manager
Joined: 26 Feb 2018
Posts: 53
Location: India
GMAT 1: 560 Q41 V27
WE: Web Development (Computer Software)

### Show Tags

07 Jun 2018, 00:56
I think this is a high-quality question and I agree with explanation.
Intern
Joined: 17 Sep 2013
Posts: 2

### Show Tags

18 Jun 2018, 09:32
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.
_________________

regards,

honcho

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

### Show Tags

18 Jun 2018, 09:47
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.

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
Intern
Joined: 19 Nov 2012
Posts: 29

### Show Tags

23 Jun 2018, 03:43
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.
Math Expert
Joined: 02 Sep 2009
Posts: 51280

### Show Tags

23 Jun 2018, 05:20
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.
_________________
Intern
Joined: 19 Nov 2012
Posts: 29

### Show Tags

23 Jun 2018, 05:35
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.

Last thing, just to confirm there is no difference between "growth rate is constant" and "growth rate per 1000 is constant", correct?
Math Expert
Joined: 02 Sep 2009
Posts: 51280

### Show Tags

23 Jun 2018, 05:38
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.

Last thing, just to confirm there is no difference between "growth rate is constant" and "growth rate per 1000 is constant", correct?

____________________
Right.
_________________
Intern
Joined: 14 Feb 2015
Posts: 16

### Show Tags

03 Oct 2018, 01:49
Bunuel I still don't understand what it means by growth rate per 1000 is constant. I did understand how to do the question, but I didn't understand what the literal translation of this sentence is.
_________________

RJ

Intern
Joined: 25 Sep 2018
Posts: 1

### Show Tags

23 Oct 2018, 07:15
Hi, Please explain how do we calculate the same for say 1997 ?
Manager
Joined: 09 Jun 2014
Posts: 219
Location: India
Concentration: General Management, Operations
Schools: Tuck '19

### Show Tags

04 Nov 2018, 05:51
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.

Can you please explain what is the meaning of ..population growth per thousand vs rate of population growth

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...

Re: M01-09 &nbs [#permalink] 04 Nov 2018, 05:51

Go to page   Previous    1   2   [ 35 posts ]

Display posts from previous: Sort by

# M01-09

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.