Last visit was: 25 Apr 2024, 16:47 It is currently 25 Apr 2024, 16:47

Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
avatar
Intern
Intern
Joined: 01 Jul 2010
Posts: 23
Own Kudos [?]: 223 [70]
Given Kudos: 15
Concentration: IB, Entreprenuership, Operations
Schools:LBS, Harvard, Booth, Stanford, ISB, NTU
GPA: 8.5
WE 1: S/W Engineer
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619024 [42]
Given Kudos: 81595
Send PM
User avatar
Manager
Manager
Joined: 09 Sep 2011
Status:Enjoying the MBA journey :)
Posts: 120
Own Kudos [?]: 229 [5]
Given Kudos: 16
Location: United States (DC)
Concentration: General Management, Entrepreneurship
GMAT 1: 710 Q49 V38
WE:Corporate Finance (Other)
Send PM
General Discussion
avatar
Intern
Intern
Joined: 01 Jul 2010
Posts: 23
Own Kudos [?]: 223 [2]
Given Kudos: 15
Concentration: IB, Entreprenuership, Operations
Schools:LBS, Harvard, Booth, Stanford, ISB, NTU
GPA: 8.5
WE 1: S/W Engineer
Send PM
Re: Herd of antelope [#permalink]
2
Kudos
Thanks for the wonderful explanation Bunuel. You make each problem sound so simple! +1 to you!
User avatar
Manager
Manager
Joined: 14 Jun 2010
Posts: 130
Own Kudos [?]: 68 [0]
Given Kudos: 7
Send PM
Re: Herd of antelope [#permalink]
Good problem that!!
User avatar
Senior Manager
Senior Manager
Joined: 09 May 2012
Affiliations: UWC
Posts: 314
Own Kudos [?]: 8799 [2]
Given Kudos: 100
Location: Canada
GMAT 1: 620 Q42 V33
GMAT 2: 680 Q44 V38
GPA: 3.43
WE:Engineering (Entertainment and Sports)
Send PM
The number of antelope in a certain herd increases every [#permalink]
2
Kudos
On population type questions, do we always assume that we are dealing with exponential growth? On the GMAT, how would we differentiate between exponential growth and linear growth?
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619024 [2]
Given Kudos: 81595
Send PM
Re: The number of antelope in a certain herd increases every [#permalink]
1
Kudos
1
Bookmarks
Expert Reply
macjas wrote:
On population type questions, do we always assume that we are dealing with exponential growth? On the GMAT, how would we differentiate between exponential growth and linear growth?


Growth at some rate means that we have exponential growth.
Growth at some constant amount means linear growth (for example if we were told that "the number of antelope in a certain herd increases every year by 1,000").
User avatar
Intern
Intern
Joined: 28 Dec 2013
Posts: 23
Own Kudos [?]: 71 [3]
Given Kudos: 18
Location: United States
GPA: 3
WE:Information Technology (Insurance)
Send PM
The number of antelope in a certain herd increases every [#permalink]
2
Kudos
1
Bookmarks
The Official Explanation in MGMAT CAT Test is :

To solve a population growth question, we can use a population chart to track the growth. The annual growth rate in this question is unknown, so we will represent it as x. For example, if the population doubles each year, x = 2; if it grows by 50% each year, x = 1.5. Each year the population is multiplied by this factor of x.



Time
Population
Now
500
in 1 year
500x
in 2 years
500x2
:
:
in n years
500xn

The question is asking us to find the minimum number of years it will take for the herd to double in number. In other words, we need to find the minimum value of n that would yield a population of 1000 or more.

We can represent this as an inequality:
500xn > 1000
xn > 2

In other words, we need to find what integer value of n would cause xn to be greater than 2. To solve this, we need to know the value of x. Therefore, we can rephrase this question as: “What is x, the annual growth factor of the herd?”

(1) INSUFFICIENT: This tells us that in ten years the following inequality will hold:
500x10 > 5000
x10 > 10

There are an infinite number of growth factors, x, that satisfy this inequality.
For example, x = 1.5 and x = 2 both satisfy this inequality.
If x = 2, the herd of antelope doubles after one year.
If x = 1.5, the herd of antelope will be more than double after two years 500(1.5)(1.5) = 500(2.25).

(2) SUFFICIENT: This will allow us to find the growth factor of the herd. We can represent the growth factor from the statement as y. (NOTE y does not necessarily equal 2x because x is a growth factor. For example, if the herd actually grows at a rate of 10% each year, x = 1.1, but y = 1.2, i.e. 20%)

Time
Population
Now
500
in 1 year
500y
in 2 years
500y2

According to the statement, 500y2 = 980
y2 = 980/500
y2 = 49/25
y = 7/5 OR 1.4 (y can’t be negative because we know the herd is growing)
This means that the hypothetical double rate from the statement represents an annual growth rate of 40%.
The actual growth rate is therefore 20%, so x = 1.2.

The correct answer is B.


So X can take a value of 0.7 or 1.2 right? If I consider y=2x. Why am I not allowed to take y = 2x? Someone plz explain
User avatar
Manager
Manager
Joined: 06 Mar 2014
Posts: 166
Own Kudos [?]: 475 [1]
Given Kudos: 84
Location: India
GMAT Date: 04-30-2015
Send PM
Re: The number of antelope in a certain herd increases every [#permalink]
1
Kudos
Bunuel wrote:
macjas wrote:
On population type questions, do we always assume that we are dealing with exponential growth? On the GMAT, how would we differentiate between exponential growth and linear growth?


Growth at some rate means that we have exponential growth.
Growth at some constant amount means linear growth (for example if we were told that "the number of antelope in a certain herd increases every year by 1,000").



Hi Bunuel,

Below are the two questions one from MGMAT and other some GMAT Paper test and as per your guidance, this is my response, kindly correct if i am wrong here.

1) The number of antelope in a certain herd increases every year at a constant rate : Exponential Growth

2) The number of antelope in a certain herd increases every year by a constant factor: Linear Growth
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619024 [0]
Given Kudos: 81595
Send PM
Re: The number of antelope in a certain herd increases every [#permalink]
Expert Reply
earnit wrote:
Bunuel wrote:
macjas wrote:
On population type questions, do we always assume that we are dealing with exponential growth? On the GMAT, how would we differentiate between exponential growth and linear growth?


Growth at some rate means that we have exponential growth.
Growth at some constant amount means linear growth (for example if we were told that "the number of antelope in a certain herd increases every year by 1,000").



Hi Bunuel,

Below are the two questions one from MGMAT and other some GMAT Paper test and as per your guidance, this is my response, kindly correct if i am wrong here.

1) The number of antelope in a certain herd increases every year at a constant rate : Exponential Growth

2) The number of antelope in a certain herd increases every year by a constant factor: Linear Growth


Both mean exponential growth: constant rate = constant factor.
Director
Director
Joined: 20 Apr 2022
Posts: 629
Own Kudos [?]: 254 [0]
Given Kudos: 316
Location: India
GPA: 3.64
Send PM
Re: The number of antelope in a certain herd increases every [#permalink]
Bunuel avigutman KarishmaB pls clarify why cant double growth rate be 2x if we take initial growth rate as x rather than taking 1+r and double as 1+2r


janxavier wrote:
The Official Explanation in MGMAT CAT Test is :

To solve a population growth question, we can use a population chart to track the growth. The annual growth rate in this question is unknown, so we will represent it as x. For example, if the population doubles each year, x = 2; if it grows by 50% each year, x = 1.5. Each year the population is multiplied by this factor of x.



Time
Population
Now
500
in 1 year
500x
in 2 years
500x2
:
:
in n years
500xn

The question is asking us to find the minimum number of years it will take for the herd to double in number. In other words, we need to find the minimum value of n that would yield a population of 1000 or more.

We can represent this as an inequality:
500xn > 1000
xn > 2

In other words, we need to find what integer value of n would cause xn to be greater than 2. To solve this, we need to know the value of x. Therefore, we can rephrase this question as: “What is x, the annual growth factor of the herd?”

(1) INSUFFICIENT: This tells us that in ten years the following inequality will hold:
500x10 > 5000
x10 > 10

There are an infinite number of growth factors, x, that satisfy this inequality.
For example, x = 1.5 and x = 2 both satisfy this inequality.
If x = 2, the herd of antelope doubles after one year.
If x = 1.5, the herd of antelope will be more than double after two years 500(1.5)(1.5) = 500(2.25).

(2) SUFFICIENT: This will allow us to find the growth factor of the herd. We can represent the growth factor from the statement as y. (NOTE y does not necessarily equal 2x because x is a growth factor. For example, if the herd actually grows at a rate of 10% each year, x = 1.1, but y = 1.2, i.e. 20%)

Time
Population
Now
500
in 1 year
500y
in 2 years
500y2

According to the statement, 500y2 = 980
y2 = 980/500
y2 = 49/25
y = 7/5 OR 1.4 (y can’t be negative because we know the herd is growing)
This means that the hypothetical double rate from the statement represents an annual growth rate of 40%.
The actual growth rate is therefore 20%, so x = 1.2.

The correct answer is B.


So X can take a value of 0.7 or 1.2 right? If I consider y=2x. Why am I not allowed to take y = 2x? Someone plz explain
Tutor
Joined: 16 Oct 2010
Posts: 14823
Own Kudos [?]: 64923 [0]
Given Kudos: 426
Location: Pune, India
Send PM
Re: The number of antelope in a certain herd increases every [#permalink]
Expert Reply
Elite097 wrote:
Bunuel avigutman KarishmaB pls clarify why cant double growth rate be 2x if we take initial growth rate as x rather than taking 1+r and double as 1+2r


janxavier wrote:
The Official Explanation in MGMAT CAT Test is :

To solve a population growth question, we can use a population chart to track the growth. The annual growth rate in this question is unknown, so we will represent it as x. For example, if the population doubles each year, x = 2; if it grows by 50% each year, x = 1.5. Each year the population is multiplied by this factor of x.



Time
Population
Now
500
in 1 year
500x
in 2 years
500x2
:
:
in n years
500xn

The question is asking us to find the minimum number of years it will take for the herd to double in number. In other words, we need to find the minimum value of n that would yield a population of 1000 or more.

We can represent this as an inequality:
500xn > 1000
xn > 2

In other words, we need to find what integer value of n would cause xn to be greater than 2. To solve this, we need to know the value of x. Therefore, we can rephrase this question as: “What is x, the annual growth factor of the herd?”

(1) INSUFFICIENT: This tells us that in ten years the following inequality will hold:
500x10 > 5000
x10 > 10

There are an infinite number of growth factors, x, that satisfy this inequality.
For example, x = 1.5 and x = 2 both satisfy this inequality.
If x = 2, the herd of antelope doubles after one year.
If x = 1.5, the herd of antelope will be more than double after two years 500(1.5)(1.5) = 500(2.25).

(2) SUFFICIENT: This will allow us to find the growth factor of the herd. We can represent the growth factor from the statement as y. (NOTE y does not necessarily equal 2x because x is a growth factor. For example, if the herd actually grows at a rate of 10% each year, x = 1.1, but y = 1.2, i.e. 20%)

Time
Population
Now
500
in 1 year
500y
in 2 years
500y2

According to the statement, 500y2 = 980
y2 = 980/500
y2 = 49/25
y = 7/5 OR 1.4 (y can’t be negative because we know the herd is growing)
This means that the hypothetical double rate from the statement represents an annual growth rate of 40%.
The actual growth rate is therefore 20%, so x = 1.2.

The correct answer is B.


So X can take a value of 0.7 or 1.2 right? If I consider y=2x. Why am I not allowed to take y = 2x? Someone plz explain


The growth rate is r or x, whatever you want to call it.
(1 + r) is the multiplying factor.

So if growth rate r = 10%, we multiply the original amount by (1 + r) i.e. (1 + 10/100) to get the new amount.
Tutor
Joined: 17 Jul 2019
Posts: 1304
Own Kudos [?]: 2287 [0]
Given Kudos: 66
Location: Canada
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Send PM
Re: The number of antelope in a certain herd increases every [#permalink]
Expert Reply
Elite097 wrote:
pls clarify why cant double growth rate be 2x if we take initial growth rate as x rather than taking 1+r and double as 1+2r
This is a very common confusion with percent change, Elite097. Did you know that a 100% increase is the same thing as multiplying by a factor of 2, and that a 200% increase is the same thing is multiplying by a factor of 3?
A percent change is multiplicative in nature (meaning it operates on the original number's distance from zero, as opposed to an additive operation, whose impact does not depend on the original number's distance from zero).
But, just because percent change is multiplicative doesn't mean that it's easy to rephrase it into multiplying by a factor.
I have a whole chapter on this in my book, if you want to dive into the reasoning behind the formulae. Otherwise, you can just memorize them.
Director
Director
Joined: 20 Apr 2022
Posts: 629
Own Kudos [?]: 254 [0]
Given Kudos: 316
Location: India
GPA: 3.64
Send PM
The number of antelope in a certain herd increases every [#permalink]
avigutman KarishmaB thanks i do know about factor change and additive change but pertaining to this question my doubt still holds as to which one i an implication for this question whether we are doubling the factor or the percentage , i.e, is it 2x where x is a factor or 1+2r where r is the % change? Ex if r=100% and x =2, then 1+2r= 3 and 2x=4 so they arent the same things

avigutman wrote:
Elite097 wrote:
pls clarify why cant double growth rate be 2x if we take initial growth rate as x rather than taking 1+r and double as 1+2r
This is a very common confusion with percent change, Elite097. Did you know that a 100% increase is the same thing as multiplying by a factor of 2, and that a 200% increase is the same thing is multiplying by a factor of 3?
A percent change is multiplicative in nature (meaning it operates on the original number's distance from zero, as opposed to an additive operation, whose impact does not depend on the original number's distance from zero).
But, just because percent change is multiplicative doesn't mean that it's easy to rephrase it into multiplying by a factor.
I have a whole chapter on this in my book, if you want to dive into the reasoning behind the formulae. Otherwise, you can just memorize them.
Tutor
Joined: 16 Oct 2010
Posts: 14823
Own Kudos [?]: 64923 [0]
Given Kudos: 426
Location: Pune, India
Send PM
Re: The number of antelope in a certain herd increases every [#permalink]
Expert Reply
Elite097 wrote:
avigutman KarishmaB thanks i do know about factor change and additive change but pertaining to this question my doubt still holds as to which one i an implication for this question whether we are doubling the factor or the percentage , i.e, is it 2x where x is a factor or 1+2r where r is the % change? Ex if r=100% and x =2, then 1+2r= 3 and 2x=4 so they arent the same things

avigutman wrote:
Elite097 wrote:
pls clarify why cant double growth rate be 2x if we take initial growth rate as x rather than taking 1+r and double as 1+2r
This is a very common confusion with percent change, Elite097. Did you know that a 100% increase is the same thing as multiplying by a factor of 2, and that a 200% increase is the same thing is multiplying by a factor of 3?
A percent change is multiplicative in nature (meaning it operates on the original number's distance from zero, as opposed to an additive operation, whose impact does not depend on the original number's distance from zero).
But, just because percent change is multiplicative doesn't mean that it's easy to rephrase it into multiplying by a factor.
I have a whole chapter on this in my book, if you want to dive into the reasoning behind the formulae. Otherwise, you can just memorize them.


Double growth rate is 2x if the initial growth rate is x - Nothing wrong with this.

But given the growth rate, how do you calculate the final value? Think compound interest.

If something increases by x% every year, you say
Final = Initial*(1 + x/100)^n

If something increases by 2x% every year, you will say
Final = Initial*(1 + 2x/100)^n

(1 + r) or (1 + x/100) is the multiplying factor for every year, not the growth rate.
When the growth rate is x%, you multiply the initial value by (1 + x/100) to get the value of next year.

The increase in value in this one year is simply
initial value * x/100.
GMAT Club Bot
Re: The number of antelope in a certain herd increases every [#permalink]
Moderator:
Math Expert
92915 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne