January 20, 2019  January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score. January 21, 2019 January 21, 2019 10:00 PM PST 11:00 PM PST Mark your calendars - All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday.
|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 06 Mar 2014
Posts: 239
Location: India
GMAT Date: 04-30-2015
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
 01 Oct 2014, 07:16
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.
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 52296
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
01 Oct 2014, 08:21
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 Growth2) The number of antelope in a certain herd increases every year by a constant factor: Linear GrowthBoth mean exponential growth: constant rate = constant factor.
_________________
New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Manager
Joined: 06 Mar 2014
Posts: 239
Location: India
GMAT Date: 04-30-2015
|
The number of antelope in a certain herd increases every
[#permalink]
Show Tags
03 Oct 2014, 04:02
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. Answer: B. 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. So in this Question i start with: 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. (B) is sufficient to answer. Is there anything wrong in this approach?
|
|
|
|
|
|
Intern
Joined: 28 Sep 2012
Posts: 11
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
06 Oct 2014, 09:15
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. Answer: B. 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??
|
|
|
|
|
|
Manager
Joined: 07 Apr 2015
Posts: 164
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
13 Jun 2015, 02:54
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!
|
|
|
|
|
|
Retired Moderator
Joined: 06 Jul 2014
Posts: 1231
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
13 Jun 2015, 03:46
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 noTh1ngYou 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
_________________
Simple way to always control time during the quant part.
How to solve main idea questions without full understanding of RC.
660 (Q48, V33) - unpleasant surprise
740 (Q50, V40, IR3) - anti-debrief 
|
|
|
|
|
|
Intern
Joined: 08 Jan 2018
Posts: 1
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
25 Mar 2018, 12:25
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 \(500*(1+r)^k=2*500\) --> 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. Question statement states that the number increases at a constant rate and not by a constant rate so when Statement 2 states that "If the herd were to grow in number at twice its current rate" doesn't it mean the new rate is 2*(1+ r) instead of (1+2r)??
|
|
|
|
|
|
DS Forum Moderator
Joined: 21 Aug 2013
Posts: 1432
Location: India
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
25 Mar 2018, 21:20
kush199191 wrote: 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 \(500*(1+r)^k=2*500\) --> 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. Question statement states that the number increases at a constant rate and not by a constant rate so when Statement 2 states that "If the herd were to grow in number at twice its current rate" doesn't it mean the new rate is 2*(1+ r) instead of (1+2r)?? Hello Earlier rate was presumed to be 'r' and not '1+r'. So when the new rate is twice the old rate, that means new rate is '2r' only, as has been done in the solution quoted by you.
|
|
|
|
|
|
Manager
Joined: 17 Jul 2017
Posts: 94
|
Re: The number of antelope in a certain herd increases every
[#permalink]
Show Tags
11 Nov 2018, 05:20
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 \(500*(1+r)^k=2*500\) --> 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. As per my understanding formula must be 500(1+r/100)^2 fr population after 2 years.how did u ignore 100 of denominator. Bunuel
|
|
|
|
|
|
|
Re: The number of antelope in a certain herd increases every &nbs
[#permalink]
11 Nov 2018, 05:20
|
|
|
|
|
Go to page
Previous
1 2
[ 29 posts ]
|
|
|
|
|
|
close
