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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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23 Oct 2011, 10:23
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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.
[Reveal] Spoiler: OA

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

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17 Mar 2012, 04:28
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Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:

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).

Hope it's clear.

[Reveal] Spoiler:
Attachment:

Wolves (1)+(2).png [ 5.23 KiB | Viewed 8590 times ]

Attachment:

Wolves.png [ 4.33 KiB | Viewed 8617 times ]

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23 Oct 2011, 11:39
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+1 for C.

With given information, you can construct the following grid:

White Grey Total

Brown 2x x 3x

Blue 3y 4y 7y

Total 55

So, 3x+7y=55

Now the above combinations are satisfied only for x=2 and y=7 or x=9 and y=4.

In both cases 7y > 3x, i.e. Blue eyed wolves are greater than brown eyed wolves.

Hope that helps.
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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11 Jun 2013, 07:28
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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14 Sep 2015, 03:49
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

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.

Transforming the original condition and the question, we have the below 2by2 question which is a typical question in GMAT test.

There are 4 variables (a,b,c,d), 2 equations (a+b+c+d=55, b>3) and we need 2 more equations to match the number of variables and equations. Since there is 1 each in 1) and 2), there is high probability that C is the answer, and it actually turns out that C is the answer.
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GC DS LalaB Each member of a pack of (20150913).png [ 3.07 KiB | Viewed 2634 times ]

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

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17 Mar 2012, 03:43
LalaB wrote:
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.

Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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17 Mar 2012, 10:13
Bunuel wrote:
Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:
Attachment:
Wolves.png
"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:
Attachment:
Wolves (1)+(2).png
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).

Hope it's clear.

Thanks Bunnel, excellent explanation. +1
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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21 Aug 2013, 03:46
Bunuel wrote:
Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:
Attachment:
Wolves.png
"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:
Attachment:
Wolves (1)+(2).png
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).

Hope it's clear.

There could be another solution to the equation:
y=11 and x=3 --> 3y+7x=33+21=55; and in this case, 7x < 3y => A & B together are insufficient => E is the answer
Am I missing something here?
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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21 Aug 2013, 03:48
divineacclivity wrote:
Bunuel wrote:

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:
Attachment:
Wolves.png
"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:
Attachment:
Wolves (1)+(2).png
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).

Hope it's clear.

There could be another solution to the equation:
y=11 and x=3 --> 3y+7x=33+21=55; and in this case, 7x < 3y => A & B together are insufficient => E is the answer
Am I missing something here?

Arithmetic: 33+21=54 not 55.
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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12 Sep 2015, 09:50
Bunuel wrote:
Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:
Attachment:
Wolves.png
"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:
Attachment:
Wolves (1)+(2).png
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).

Hope it's clear.

Hello Bunel ,
why should i see x only as integer , why can't it be fraction with denominator as 7 eg:18/7 ?
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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12 Sep 2015, 14:15
divya517 wrote:
Bunuel wrote:
Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:
Attachment:
Wolves.png
"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:
Attachment:
Wolves (1)+(2).png
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).

Hope it's clear.

Hello Bunel ,
why should i see x only as integer , why can't it be fraction with denominator as 7 eg:18/7 ?

Because if x = fraction , lets say =18/7, then 3x = NUMBER OF WOLVES with white coats = 54/7 = fraction . How can number of wolves be fraction? It does not make any sense to say we have 3/4 wolves or 22/7 wolves etc. Thus, x can only take integer values.
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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23 Oct 2015, 01:25
Bunuel wrote:
AmoyV wrote:
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.

Merging topics.

Please refer to the discussion above.

If we do it by taking fractions, i.e. 3/7 X , 4/7 X , 2/3Y AND 1/3 Y, we do not get the same answer. Could you please advice?

3x/7>3 --> 3x>21--x>7
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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23 Oct 2015, 01:34
Bunuel wrote:
Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:
Attachment:
The attachment Wolves.png is no longer available
"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:
Attachment:
The attachment Wolves (1)+(2).png is no longer available
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).

Hope it's clear.

with fractions i get E as the answer
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20151023013242.jpg [ 114.11 KiB | Viewed 2449 times ]

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

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23 Oct 2015, 04:48
rahulkashyap wrote:
Bunuel wrote:
Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:

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).

Hope it's clear.

with fractions i get E as the answer

Can you please give TWO examples which satisfy both statements and the stem?
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Re: Each member of a pack of 55 wolves has either brown or blue [#permalink]

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

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23 Mar 2017, 16:47
Bunuel wrote:
Not able to understand the line " If there are more than 3 blue-eyed wolves with white coats"

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:

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).

Hope it's clear.

[Reveal] Spoiler:
Attachment:
Wolves (1)+(2).png

Attachment:
Wolves.png

Yes- if we combine both statements then we can inevitably solve for X- there is only one such value that can satisfy the ratios in this matrix- I plugged in numbers and serendipitously arrived at the answer- we know that, for example, the value of x must be somewhere between 4-20 and can thus plug in values.
Re: Each member of a pack of 55 wolves has either brown or blue   [#permalink] 23 Mar 2017, 16:47
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