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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] New post 25 Mar 2006, 21:17
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A
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C
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E

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46% (03:07) correct 54% (02:23) wrong based on 49 sessions
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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 [#permalink] New post 25 Mar 2006, 21:55
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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 [#permalink] New post 25 Mar 2006, 22:10
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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 [#permalink] New post 25 Mar 2006, 22:24
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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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 [#permalink] New post 25 Mar 2006, 22:38
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.

Is C the answer?

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 [#permalink] New post 25 Mar 2006, 22:42
Hehehehe,

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

phew....MATT, this is brutal ;)
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 [#permalink] New post 25 Mar 2006, 22:48
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?? :shock:
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 [#permalink] New post 25 Mar 2006, 22:55
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?? :shock:


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! :P

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 [#permalink] New post 29 Mar 2006, 03:04
giddi

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

thanks
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 [#permalink] New post 29 Mar 2006, 06:03
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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).


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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink] New post 31 Jan 2014, 05:36
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink] New post 31 Jan 2014, 07:11
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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:
Image
"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:
Image
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).

Answer: C.

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, 07:11
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