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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

by prab.sahi06 » Sun Jul 14, 2019 4:47 am
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.

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

by Kasey00 » Fri Feb 21, 2020 10:24 am
prab.sahi06 wrote:
Sun Jul 14, 2019 4:47 am
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.
Among the brown-eyed wolves, the ratio of white coats to grey coats is 2 to 1. It is a problem from the NCERT.
Last edited by Kasey00 on Tue Apr 21, 2020 8:08 am, edited 1 time in total.

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

by gmatknight » Thu Feb 27, 2020 11:26 pm
Seems like a standard question to solve using a table. It would be interesting to know if there was a trick to it, though.

Just a guy with a Q50.
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### Re: Each member of a pack of 55 wolves has either brown or blue

by Ignite » Sat Aug 01, 2020 10:07 pm
We can make a table like the one in the attached figure to help solve this problem.

-----------Brown---Blue
White------A--------B
Grey--------C------- D
Total--------X--------Y------55

Referring to the table.
We are given that B > 3 and X + Y = 55
Is Y > X ?

1. D:B = 4:3
Or, for a constant 'P, the values are
D = 4P and B = 3P , where P is an integer.
So the value of Y is in terms of 7P.
Thus possible values of Y can be- 7, 14,21,28,35,42,49

We cannot determine the value of X from this. And X can be lower or greater than Y depending on the value of Y here.

Not Sufficient

2. A:C = 2:1
For a constant 'Q', the values are
A = 2Q and C = Q, where Q is an integer.
So the value of X is in terms of 3Q.
Thus possible values of X can be- 3,6,9,12,15,18.......54

We cannot determine the value of Y from this. And Y can be lower or greater than X depending on the value of X here.

Not Sufficient

1 + 2
X is in terms of 3Q and Y is in terms of 7P. Also, X + Y = 55

So if we substitute values in 7P and subtract it from 55, the difference should be in terms of 3Q.

For example, if P = 4
Y = 28 and X = 27 <- valid

We will find that the only valid values are when P = 4 and P = 7. The other values are invalid because the difference is not divisible by 3.
When P = 7
Y = 49 and X = 6

Now in both cases, we find that Y > X

Hence Sufficient.