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# Each member of a pack of 55 wolves has either brown or blue

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Each member of a pack of 55 wolves has either brown or blue [#permalink]

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25 Mar 2006, 22:17
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Each member of a pack of 55 wolves has either brown or blue eyes and either a white or a grey coat. If there are more than 3 blue-eyed wolves with white coats, are there more blue-eyed wolves than brown-eyed wolves?

(1) Among the blue-eyed wolves, the ratio of grey coats to white coats is 4 to 3.
(2) Among the brown-eyed wolves, the ratio of white coats to grey coats is 2 to 1.

OPEN DISCUSSION OF THIS QUESTION IS HERE: each-member-of-a-pack-of-55-wolves-has-either-brown-or-blue-122299.html
[Reveal] Spoiler: OA
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25 Mar 2006, 22:55
My pick would be E.

1. Doesn't provide any info about Brown eyed wolves
2. Doesn't provide any info about Blu eyed wolves

Combining:
Since only the ratios are provided. There could be several pssobilities that satisfy 1 & 2.

BR:BLU = (6,49), (27,28), (48,7)

Hence E.
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25 Mar 2006, 23:10
Yeah, how I approached it is using table method.

if "x" is the total number of blue eyed wolves,

then (3x/7) > 3 ----------- (from the ratio given)
=> x > 7

Since, x > 7, the answer can be YES or NO.

Using Statement2, or even together, will not solve this problem!
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25 Mar 2006, 23:24
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Well, Vivek, the answer given is not (E). I guess that leaves you with only one other choice....

Try again, buddy.
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25 Mar 2006, 23:38
giddi77 wrote:
My pick would be E.

1. Doesn't provide any info about Brown eyed wolves
2. Doesn't provide any info about Blu eyed wolves

Combining:
Since only the ratios are provided. There could be several pssobilities that satisfy 1 & 2.

BR:BLU = (6,49), (27,28), (48,7)

Hence E.

Looks like MATT picked tis bad boy to trouble us. I am guessing this from MGMAT or Kaplan.

I would change it to C. since

BR:BLU = (6,49), (27,28), (48,7)

but (48,7) is not a possible solution as it would make the number of Blue Eyed White coats = 3 and hence is ruled out .

Now the only possibilities are

BR: BLU = (6,49) & (27,28)
In both cases BR < BLU, which answers the question.

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"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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25 Mar 2006, 23:42
Hehehehe,

I changed it to "C" too
Basically chose the pair of numbers that satisfy the requirement!

phew....MATT, this is brutal
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25 Mar 2006, 23:48
Yep, OA is (C)

OE: http://www.manhattangmat.com/ChallengeP ... cfm?ID=235

BTW Giddi: How did you determine it was a MGMAT problem? By difficulty? Or pattern recognition??
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25 Mar 2006, 23:55
GMATT73 wrote:
Yep, OA is (C)

OE: http://www.manhattangmat.com/ChallengeP ... cfm?ID=235

BTW Giddi: How did you determine it was a MGMAT problem? By difficulty? Or pattern recognition??

Just a guess based on the difficulty...
When you have more than 1 possible solution and still the answer is not E => It must be a MGMAT question!
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29 Mar 2006, 04:04
giddi

could you explain how you got the numbers (6,49) and (28,27)

thanks
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29 Mar 2006, 07:03
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myc2004 wrote:
giddi

could you explain how you got the numbers (6,49) and (28,27)

thanks

(1) Total of Blue eyed wolves = 7 parts. So the total number must be a multiple of 7. This could be 7,14,21,28...etc and then the remainder will be brown-eyed wolves. Insufficient as we can't compute the number of brown eyed wolves.

(2) Same problem as (1)

Using both,

Blue-eyed wolves have 7 parts
Brown-eye wolves has 3 parts.

So Blue-eyed wolves is a multiple of 7, such that remainder of wolves is a multiple of 3. Could be blue:brown = (7,48) (28,27) (49,6)

But then we're told number of blue-eyed white coat wolves is more than 3. So (7,48) is out. We're now left with blue:brown = (28,27) and (49,6).

Ans C
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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31 Jan 2014, 06:36
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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31 Jan 2014, 08:11
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Each member of a pack of 55 wolves has either brown or blue eyes and either a white or a grey coat. If there are more than 3 blue-eyed wolves with white coats, are there more blue-eyed wolves than brown-eyed wolves?

Look at the matrix below:

"There are more than 3 blue-eyed wolves with white coats" means that # of wolves which have blue eyes AND white coats is more than 3. The question asks whether there are more blue-eyed wolves (blue box) than brown-eyed wolves (brown box).

(1) Among the blue-eyed wolves, the ratio of grey coats to white coats is 4 to 3. Not sufficient on its own.
(2) Among the brown-eyed wolves, the ratio of white coats to grey coats is 2 to 1. Not sufficient on its own.

(1)+(2) When taken together we get the flowing matrix:

Wolves (1)+(2).png[/attachment]Notice that x and y must be integers (they represent some positive multiples for the ratios given in the statements).

So, we have that 3y+7x=55. After some trial and error we can find that this equation has only 3 positive integers solutions:
y=2 and x=7 --> 3y+7x=6+49=55;
y=9 and x=4 --> 3y+7x=27+28=55;
y=16 and x=1 --> 3y+7x=48+7=55;

Now, the third solution (x=1) is not valid, since in this case # of wolves which have blue eyes AND white coats becomes 3x=3, so not more than 3 as given in the stem. As for the first two cases, in both of them 7x is more than 3y (49>6 and 28>27), so we can answer definite YES, to the question whether there are more blue-eyed wolves (blue box) than brown-eyed wolves (brown box).

OPEN DISCUSSION OF THIS QUESTION IS HERE: each-member-of-a-pack-of-55-wolves-has-either-brown-or-blue-122299.html
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Re: Each member of a pack of 55 wolves has either brown or blue   [#permalink] 31 Jan 2014, 08:11
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# Each member of a pack of 55 wolves has either brown or blue

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