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VP  Joined: 02 Jul 2012
Posts: 1098
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42 GPA: 3.8
WE: Engineering (Energy and Utilities)

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Question Stats: 100% (00:56) correct 0% (00:00) wrong based on 88 sessions

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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

Director  Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 986
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

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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

$$y(t)=y(0) * 4^[t/I]$$
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.
_________________
VP  Joined: 02 Jul 2012
Posts: 1098
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42 GPA: 3.8
WE: Engineering (Energy and Utilities)

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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

$$y(t)=y(0) * 4^[t/I]$$
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
Manager  Joined: 04 Oct 2011
Posts: 166
Location: India
GMAT 1: 440 Q33 V13 GPA: 3

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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

$$y(t)=y(0) * 4^[t/I]$$
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
Director  Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 986
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

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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.
_________________
Manager  Joined: 04 Oct 2011
Posts: 166
Location: India
GMAT 1: 440 Q33 V13 GPA: 3

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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 ??
Director  Status: Been a long time guys...
Joined: 03 Feb 2011
Posts: 986
Location: United States (NY)
Concentration: Finance, Marketing
GPA: 3.75

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With proper practice, this relation can be accurately used as the formula.
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Intern  Joined: 24 Jul 2013
Posts: 2

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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

$$y(t)=y(0) * 4^[t/I]$$
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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_________________ Re: A certain rabbit population quadruples every three years.   [#permalink] 08 Jun 2015, 15:28
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