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A certain rabbit population quadruples every three years.
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11 Dec 2012, 23:06
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Going to have a go at Problem Solving.. My own question, so no official answer. Please provide your feedback on how you found the question to be and anyway that I can make it clearer..
A certain rabbit population quadruples every three years. The population today is 81. Exactly six years from today, the entire population will be fed to a pack of wolves. If each wolf eats exactly 12 rabbits, how many wolves can be fed using this rabbit population?
Re: A certain rabbit population quadruples every three years.
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11 Dec 2012, 23:47
MacFauz wrote:
Going to have a go at Problem Solving.. My own question, so no official answer. Please provide your feedback on how you found the question to be and anyway that I can make it clearer..
A certain rabbit population quadruples every three years. The population today is 81. Exactly six years from today, the entire population will be fed to a pack of wolves. If each wolf eats exactly 12 rabbits, how many wolves can be fed using this rabbit population?
\(y(t)=y(0) * 4^[t/I]\) where: y(t)=population after given number of years. y(0)=initial population t=time I=amount of the time for the quantity to double. Putting the respective values, we get population after 6 years as 1296. Divide this by the exact capacity of wolves, and we get 108 as the answer.
Logical method: the population is quadrupling two times. So find the population as soon as it quadruples for the first time. Then multiply the result again by 4 to get the population after 6 years. And then the same.
+1C.
Btw Macfauz, you may apply to GMAC. I am quite certain that within a year or two, you may be writing questions for future GMAT takers. Good Luck.
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Re: A certain rabbit population quadruples every three years.
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11 Dec 2012, 23:58
Marcab wrote:
MacFauz wrote:
Going to have a go at Problem Solving.. My own question, so no official answer. Please provide your feedback on how you found the question to be and anyway that I can make it clearer..
A certain rabbit population quadruples every three years. The population today is 81. Exactly six years from today, the entire population will be fed to a pack of wolves. If each wolf eats exactly 12 rabbits, how many wolves can be fed using this rabbit population?
\(y(t)=y(0) * 4^[t/I]\) where: y(t)=population after given number of years. y(0)=initial population t=time I=amount of the time for the quantity to double. Putting the respective values, we get population after 6 years as 1296. Divide this by the exact capacity of wolves, and we get 108 as the answer.
Logical method: the population is quadrupling two times. So find the population as soon as it quadruples for the first time. Then multiply the result again by 4 to get the population after 6 years. And then the same.
+1C.
Btw Macfauz, you may apply to GMAC. I am quite certain that within a year or two, you may be writing questions for future GMAT takers. Good Luck.
Haha.. Thanks Marcab.. Hopefully will be able to come up with more questions here before I can do that..
Btw.. For the question.. We can save some time on the multiplication by keeping the final population in the form
\(3^4 * 2^4\). Dividing this by 12 we get : \(\frac{3^4 * 2^4}{2^2 * 3}\) = \(3^3 * 2^2\) = 108
Concentration: Entrepreneurship, International Business
GMAT 1: 440 Q33 V13
GPA: 3
Re: A certain rabbit population quadruples every three years.
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10 Jan 2013, 20:08
Marcab wrote:
MacFauz wrote:
Going to have a go at Problem Solving.. My own question, so no official answer. Please provide your feedback on how you found the question to be and anyway that I can make it clearer..
A certain rabbit population quadruples every three years. The population today is 81. Exactly six years from today, the entire population will be fed to a pack of wolves. If each wolf eats exactly 12 rabbits, how many wolves can be fed using this rabbit population?
\(y(t)=y(0) * 4^[t/I]\) where: y(t)=population after given number of years. y(0)=initial population t=time I=amount of the time for the quantity to double. Putting the respective values, we get population after 6 years as 1296. Divide this by the exact capacity of wolves, and we get 108 as the answer.
Logical method: the population is quadrupling two times. So find the population as soon as it quadruples for the first time. Then multiply the result again by 4 to get the population after 6 years. And then the same.
+1C.
Btw Macfauz, you may apply to GMAC. I am quite certain that within a year or two, you may be writing questions for future GMAT takers. Good Luck.
Hi Marcab,
I solved this using logical method. But i tried to figure out ur algebraic method ... i couldn't
y(t)=population after given number of years. (x) y(0)=initial population (81) t=time (6) I=amount of the time for the quantity to double. (3)
Re: A certain rabbit population quadruples every three years.
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10 Jan 2013, 22:06
Hii Shan. If you are pretty comfortable with the logical method, then don't get confuse with this one. Anyways, here is the clarification of my method: \(y(t)=y(0)*4^{t/I}\)
where y(t)= population after 6 years. y(0)=current population=81 4- multiplying factor.( Since here its given that population is quadrupling, hence 4) t-time duration given=6 I-time interval during which the population quadruples=3
The relation becomes: \(y(t)=81*4^{6/3}\) \(y(t)=81*4^2\) \(y(t)=81*16\) or \(1296\).
On dividing this by # of rabbits, you will get the # of wolves.
Concentration: Entrepreneurship, International Business
GMAT 1: 440 Q33 V13
GPA: 3
Re: A certain rabbit population quadruples every three years.
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11 Jan 2013, 00:30
Marcab wrote:
Hii Shan. If you are pretty comfortable with the logical method, then don't get confuse with this one. Anyways, here is the clarification of my method: \(y(t)=y(0)*4^{t/I}\)
where y(t)= population after 6 years. y(0)=current population=81 4- multiplying factor.( Since here its given that population is quadrupling, hence 4) t-time duration given=6 I-time interval during which the population quadruples=3
The relation becomes: \(y(t)=81*4^{6/3}\) \(y(t)=81*4^2\) \(y(t)=81*16\) or \(1296\).
On dividing this by # of rabbits, you will get the # of wolves.
Hope that helps.
Ya thanks dude.. I got this now..
But i just wanna confirm is this standard formula for these population sums? or it depends on problem ??
Re: A certain rabbit population quadruples every three years.
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29 Jan 2014, 14:26
MacFauz wrote:
Marcab wrote:
MacFauz wrote:
Going to have a go at Problem Solving.. My own question, so no official answer. Please provide your feedback on how you found the question to be and anyway that I can make it clearer..
A certain rabbit population quadruples every three years. The population today is 81. Exactly six years from today, the entire population will be fed to a pack of wolves. If each wolf eats exactly 12 rabbits, how many wolves can be fed using this rabbit population?
\(y(t)=y(0) * 4^[t/I]\) where: y(t)=population after given number of years. y(0)=initial population t=time I=amount of the time for the quantity to double. Putting the respective values, we get population after 6 years as 1296. Divide this by the exact capacity of wolves, and we get 108 as the answer.
Logical method: the population is quadrupling two times. So find the population as soon as it quadruples for the first time. Then multiply the result again by 4 to get the population after 6 years. And then the same.
+1C.
Btw Macfauz, you may apply to GMAC. I am quite certain that within a year or two, you may be writing questions for future GMAT takers. Good Luck.
Haha.. Thanks Marcab.. Hopefully will be able to come up with more questions here before I can do that..
Btw.. For the question.. We can save some time on the multiplication by keeping the final population in the form
\(3^4 * 2^4\). Dividing this by 12 we get : \(\frac{3^4 * 2^4}{2^2 * 3}\) = \(3^3 * 2^2\) = 108
Good idea! I saw that all the answers had differing unit digits, so i just kept track of unit digit to arrive at the answer. The answer basically boils down 81* 4^6/12 i.e 27*4^5, and we know that unit digit of power of 4 alternates between 4 and 6 with unit digit of odd power of 4 being 4 , so the answer should end with 8!
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