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25 Mar 2006, 22:17
Kudos
Bookmarks
C
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Difficulty:
95%
(hard)
Question Stats:
45% (03:01) correct
55%
(03:13)
wrong
based on 343
sessions
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Not Attempted Yet
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
(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
31 Jan 2014, 08:11
Kudos
Bookmarks
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).
Answer: C.
OPEN DISCUSSION OF THIS QUESTION IS HERE: each-member-of-a-pack-of-55-wolves-has-either-brown-or-blue-122299.html
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).
Answer: C.
OPEN DISCUSSION OF THIS QUESTION IS HERE: each-member-of-a-pack-of-55-wolves-has-either-brown-or-blue-122299.html
29 Mar 2006, 07:03
Kudos
Bookmarks
myc2004
(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
