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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?

A. 54
B. 81
C. 108
D. 216
E. 324

[Reveal] Spoiler:

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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?

A. 54
B. 81
C. 108
D. 216
E. 324

[Reveal] Spoiler:

$$y(t)=y(0) * 4^[t/I]$$
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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Kudos [?]: 1632 [0], given: 62

Moderator
Joined: 02 Jul 2012
Posts: 1219

Kudos [?]: 1621 [0], given: 116

Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)

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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?

A. 54
B. 81
C. 108
D. 216
E. 324

[Reveal] Spoiler:

$$y(t)=y(0) * 4^[t/I]$$
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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GMAT Reading Comprehension: 7 Most Common Passage Types

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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?

A. 54
B. 81
C. 108
D. 216
E. 324

[Reveal] Spoiler:

$$y(t)=y(0) * 4^[t/I]$$
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
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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.

Hope that helps.
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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 ??
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11 Jan 2013, 04:42
With proper practice, this relation can be accurately used as the formula.
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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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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?

A. 54
B. 81
C. 108
D. 216
E. 324

[Reveal] Spoiler:

$$y(t)=y(0) * 4^[t/I]$$
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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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).

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Re: A certain rabbit population quadruples every three years.   [#permalink] 08 Jun 2015, 15:28
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