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# The number of antelope in a certain herd increases every

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Current Student
Joined: 06 Mar 2014
Posts: 265
Location: India
GMAT Date: 04-30-2015
Re: The number of antelope in a certain herd increases every [#permalink]

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01 Oct 2014, 08:16
This is the Complete Question from MGMAT and i think it has a Flawed explanation or am i missing something here.

The number of antelope in a certain herd increases every year by a constant factor. If there are 500 antelope in the herd today, how many years will it take for the number of antelope to double?

(1) Ten years from now, there will be more than ten times the current number of antelope in the herd.

(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years.

Explanation: 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.

he 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

I am not pasting the second statement explanation as according to me the underlying difference between an Exponential and Linear Growth is not interpreted above.
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Joined: 02 Sep 2009
Posts: 44399
Re: The number of antelope in a certain herd increases every [#permalink]

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01 Oct 2014, 09:21
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.
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GMAT Date: 04-30-2015
The number of antelope in a certain herd increases every [#permalink]

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03 Oct 2014, 05:02
Bunuel wrote:
jananijayakumar wrote:
The number of antelope in a certain herd increases every year at a constant rate. If there are 500 antelope in the herd today, how many years will it take for the number of antelope to double?

(1) Ten years from now, there will be more than ten times the current number of antelope in the herd.

(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years.

Let the rate of increase be $$r$$ (so after 1 year # of antelope will be $$500*(1+r)$$, after 2 years # of antelope will be $$500*(1+r)^2$$ and so on). Question: if $$500*(1+r)^k=2*500$$ --> $$(1+r)^k=2$$, then $$k=?$$

(1) Ten years from now, there will be more than ten times the current number of antelope in the herd --> $$500*(1+r)^{10}>10*500$$ --> $$(1+r)^{10}>10$$ --> different values of $$r$$ satisfies this inequality hence we'll have different values of $$k$$. Not sufficient.

(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years --> $$500*(1+2r)^2=980$$ --> as there is only one unknown we can find its (acceptable) value thus we can calculate $$k$$. Sufficient.

My approach is exactly the same with the faint difference that i usually tend to get mixed up between t=0 and t=1 time intervals.

t=0, p (where p = 500)
t=1, pr ( where r is the rate of increase)
t=2, pr^2
.
.
.
t=10, pr^10

So as per (statement 2)
new rate (R) = 2r
at t=2, population would be 980.
Hence, at t=2, pR^2 = 980
where p = 500, R=2r and so from here we can calculate the rate of increase --> r = 7/10

So as per the question pr^x = 1000 (500*2) where "x" is the number of years when p doubles
x can be calculated here.

Is there anything wrong in this approach?
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Joined: 29 Sep 2012
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Re: The number of antelope in a certain herd increases every [#permalink]

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06 Oct 2014, 10:15
Bunuel wrote:
jananijayakumar wrote:
The number of antelope in a certain herd increases every year at a constant rate. If there are 500 antelope in the herd today, how many years will it take for the number of antelope to double?

(1) Ten years from now, there will be more than ten times the current number of antelope in the herd.

(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years.

Let the rate of increase be $$r$$ (so after 1 year # of antelope will be $$500*(1+r)$$, after 2 years # of antelope will be $$500*(1+r)^2$$ and so on). Question: if $$500*(1+r)^k=2*500$$ --> $$(1+r)^k=2$$, then $$k=?$$

(1) Ten years from now, there will be more than ten times the current number of antelope in the herd --> $$500*(1+r)^{10}>10*500$$ --> $$(1+r)^{10}>10$$ --> different values of $$r$$ satisfies this inequality hence we'll have different values of $$k$$. Not sufficient.

(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years --> $$500*(1+2r)^2=980$$ --> as there is only one unknown we can find its (acceptable) value thus we can calculate $$k$$. Sufficient.

Hi Bunuel ,

I understood that it increases in this fashion: 500, 500r,500r^2 and so on. However, i din't get how you converted this into 500(1+r)^2 for let's say 2 years. If we start at 500 then first year will be 500r ,second year will be 500r^2, which makes it 500+500r+500r^2 which is equal to 500(1+r+r^2). However when we simplify 500(1+r)^2 we get : 500(1+2r+r^2). Can you please explain where am i going wrong??
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13 Jun 2015, 03:54
I have a question concerning this question and more precise the second statement. I understand B is correct and that the second statement actually gives us enough information so solve for x, however I get a value for x that is not in line with the statement.

The problem says that "a certain herd increases every year by a constant factor". However I yield a value of x = 0,7 which contradicts the original statement as with a value of 0,7 the herd would obviously decrease every year.

I calculated the following:
500 * 2x^2 = 980
500 * 4x^2 = 980
4x^2 = 980 / 500
x^2 = 980 / 500 * (1/4)
x^2 = 245 / 500
x^2 = 49 / 100
x = 7 / 100 = 0,7

Where did i go wrong? Thx!
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Re: The number of antelope in a certain herd increases every [#permalink]

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13 Jun 2015, 04:46
noTh1ng wrote:
I have a question concerning this question and more precise the second statement. I understand B is correct and that the second statement actually gives us enough information so solve for x, however I get a value for x that is not in line with the statement.

The problem says that "a certain herd increases every year by a constant factor". However I yield a value of x = 0,7 which contradicts the original statement as with a value of 0,7 the herd would obviously decrease every year.

I calculated the following:
500 * 2x^2 = 980
500 * 4x^2 = 980
4x^2 = 980 / 500
x^2 = 980 / 500 * (1/4)
x^2 = 245 / 500
x^2 = 49 / 100
x = 7 / 100 = 0,7

Where did i go wrong? Thx!

Hello noTh1ng
You did all right.

But to calculate 70% increase (x=0,7) you should multiply 500 on the (1+x)
because 500*x gives you only increase (antilopes' cubs) without 500 old antilopes that you have initially
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