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A certain rabbit population quadruples every three years. [#permalink]

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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. [#permalink]

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11 Dec 2012, 23:47

Expert's post

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. _________________

Re: A certain rabbit population quadruples every three years. [#permalink]

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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 _________________

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Re: A certain rabbit population quadruples every three years. [#permalink]

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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)

i'm getting x= 81*2 _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: A certain rabbit population quadruples every three years. [#permalink]

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10 Jan 2013, 22:06

Expert's post

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.

Re: A certain rabbit population quadruples every three years. [#permalink]

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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 ?? _________________

GMAT - Practice, Patience, Persistence Kudos if u like

Re: A certain rabbit population quadruples every three years. [#permalink]

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28 Jan 2014, 14:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A certain rabbit population quadruples every three years. [#permalink]

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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!

Re: A certain rabbit population quadruples every three years. [#permalink]

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08 Jun 2015, 15:28

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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