The number of antelope in a certain herd increases every

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?

Let the rate of increase be r (so after 1 year the number of antelopes will be \(500*(1+r)\), after 2 years the number of antelopes will be \(500*(1+r)^2\), and so on).

The question asks, if \(500*(1+r)^k=2*500\), or if \((1+r)^k=2\), then what is the value of k?

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

The above implies that \(500*(1+r)^{10}>10*500\), or that \((1+r)^{10}>10\). Different values of r satisfy 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.

The above implies that \(500*(1+2r)^2=980\). Since there is only one unknown, we can find its (acceptable) value thus we can calculate k. Sufficient.

Answer: B.
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Here's the explanation.

To answer your first question, why x^10 > 10 is not sufficient.
You are asked, in the question, to determine the number of years it will take for the antelope herd to double in number. i.e. currently there are 500 Antelopes, when will it become 1000?

Considering the statement x^10 > 10, x can be any number greater than or equal to 2.
This means that we cannot practically determine the exact time required for the herd to double in number. So Statement 1 is insufficient.

Let's consider statement 2. This states that if the herd were to grow at twice it's current rate, there would be 980 antelopes in 2 years.

Let the current rate of growth be x
As per statement 2, if the growth rate is double the current rate, there would be 980 antelopes in 2 years.
So, the new growth rate would be 2x
So, in 1 Year, the number of antelopes would be 500(2x)
And in 2 Years, the number of antelopes would be 500(2x)^2, which is given to be 980
Solving this equation, we get
4x^2 = 980/500
i.e. x^2 = 49/100
Solving for x, we get x = 0.7
Using this, we can definitely determine the time required for the herd to double in number.

Hence, statement 2 is sufficient on its own to answer the question.

Hence the correct option is B.

Hope this helps!

Thanks for the wonderful explanation Bunuel. You make each problem sound so simple! +1 to you!

Good problem that!!

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").
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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

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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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

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.
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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.

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

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.
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