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The number of antelope in a certain herd increases every [#permalink]
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26 Aug 2010, 02:40
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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.
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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 \(500*(1+r)^k=2*500\) > if \((1+r)^k=2\), 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. Answer: B.
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Re: Herd of antelope [#permalink]
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26 Aug 2010, 20:54
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Thanks for the wonderful explanation Bunuel. You make each problem sound so simple! +1 to you!



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27 Aug 2010, 09:29
Good problem that!!



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Population Growth  Antelope [#permalink]
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15 Nov 2011, 15:37
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. Ok guys  this is how I am trying to solve this. But I got stuck. So can someone please help to unstuck me? Considering question stem Let's assume that rate of growth be x So, Time Growth Today 500 In 1 year 500x In 2 years 500x^2 . . . In n years 500x^n Therefore, 500x^n>5000 => x^n > 10 (1) So now the question becomes what is the value of x and n? Now considering statement 1 500x^10>5000 x^10>10 > Why this is insufficient?Now considering statement 2 Let say growth be y Time Population Now 500 In 1 year 500y In 2 years 500y^2 Therefore 500y^2=980 Y^2 = 980/500 y = 7/4 as it can't be negative. Again I am stuck after this. Can someone please guide me through?
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Re: Population Growth  Antelope [#permalink]
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16 Nov 2011, 05:30
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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!
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Re: Population Growth  Antelope [#permalink]
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16 Nov 2011, 08:02
we need to find the rate. let it be K Statement. 1: we only know on year ten the number will be more than 5000. nothing else. no way to find the rate. Insufficient.
Statement 2: 500+ 500*2K=980 Sufficient to find the rate k.



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Re: Population Growth  Antelope [#permalink]
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16 Nov 2011, 16:19
Hello BDSunDevil  We have to find the time and not the rate.
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17 Nov 2011, 03:29
Yeah. But, only if you get the rate, you get the figure for 3rd year and so on. which, will lead you to the desired year. I hope i am doing it right.



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Re: Population Growth  Antelope [#permalink]
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17 Nov 2011, 16:12
Alright, this is how I did it. (1) is of absolutely no significance because the statement "more than ten times" is incredibly vague and can mean anything from 10.1 times to 1,000,000,000 times more.
So, if (1) is incidental then we can cancel out options A, C and D. All we need to do now is figure if (2) can give you an answer.
If x is the rate and the herd will grow at twice it's current rate, then we can find next year's population by the equation 500 + 2x(500) and the population the year after would be 500 + 2x(500) 2x(2x{500}) = 980. Now we can solve for the rate of growth x. From there we just count how many years it takes to increase the herd at that constant rate. I'd do the math, but since we don't have to solve the problem, just know that/how it can be solved, there's not much point.
Answer B.



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Re: The number of antelope in a certain herd increases every [#permalink]
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07 Feb 2012, 21:45
Thanks was a little confused on this one... Good clarification.
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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?



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03 Dec 2012, 18:49
Can someone pls. clarify how to decide if the question is indicating an arithmetic progression or geometric progression. I thought constant rate refers to a growth by a certain constant number...and constant ratio/ factor means it is more a multiplier..
IN this question for example...I interpreted the statement 2 as..
Let x be initial growth rate ...so first year 500 + x second year = 500+ 2x
Now if rate is doubled....first year = 500+2x second year = 500+4x
Where am I interpreting incorrectly?
why is second year = 500+x^2
Pls. provide clarification...thanks in advance



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23 Jun 2014, 06:44
audiogal101 wrote: Can someone pls. clarify how to decide if the question is indicating an arithmetic progression or geometric progression. I thought constant rate refers to a growth by a certain constant number...and constant ratio/ factor means it is more a multiplier..
IN this question for example...I interpreted the statement 2 as..
Let x be initial growth rate ...so first year 500 + x second year = 500+ 2x
Now if rate is doubled....first year = 500+2x second year = 500+4x
Where am I interpreting incorrectly?
why is second year = 500+x^2
Pls. provide clarification...thanks in advance Can someone please explain why this is not the case of linear growth when constant rate is written in the question. Sorry if not asking in correct forum.



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Re: The number of antelope in a certain herd increases every [#permalink]
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23 Jun 2014, 07:29
shukasa wrote: audiogal101 wrote: Can someone pls. clarify how to decide if the question is indicating an arithmetic progression or geometric progression. I thought constant rate refers to a growth by a certain constant number...and constant ratio/ factor means it is more a multiplier..
IN this question for example...I interpreted the statement 2 as..
Let x be initial growth rate ...so first year 500 + x second year = 500+ 2x
Now if rate is doubled....first year = 500+2x second year = 500+4x
Where am I interpreting incorrectly?
why is second year = 500+x^2
Pls. provide clarification...thanks in advance Can someone please explain why this is not the case of linear growth when constant rate is written in the question. Sorry if not asking in correct forum. Have you checked this post: thenumberofantelopeinacertainherdincreasesevery99810.html#p1113618 ?
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23 Jun 2014, 08:55
oh my mistake, I interpreted it wrong..



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09 Jul 2014, 00:56
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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
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Re: The number of antelope in a certain herd increases every [#permalink]
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01 Oct 2014, 08:04
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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 Growth2) The number of antelope in a certain herd increases every year by a constant factor: Linear Growth




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