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by C » Wed Apr 14, 2010 9:30 pm
A zoo has w wildebeests, y yaks, z zebras, and no other animals. If an animal is chosen at random from the zoo, is the probability of choosing a yak greater than the probability of choosing a zebra?

(1) y/w+z > 1/2

(2) z/w+y < 1/2

OA is C

Source is MGMAT Word Translations.

Hope someone can explain. Thanks.
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by liferocks » Wed Apr 14, 2010 9:51 pm
A zoo has w wildebeests, y yaks, z zebras, and no other animals. If an animal is chosen at random from the zoo, is the probability of choosing a yak greater than the probability of choosing a zebra?

(1) y/w+z > 1/2

(2) z/w+y < 1/2

probability of choosing a yak =y/(W+y+z)
probability of choosing a zebra =z/(W+y+z)

from 1 we get y/(w+z)>1/2 or y/(w+y+z)>1/3-->not sufficient
from 2 we get z/(w+y)<1/2 or z/(w+y+z)<1/3-->not sufficient

together we can say z/(w+y+z)<y/(w+y+z) -->sufficient ans C

note: I have assumed that condition is like y/(w+z)>1/2 and z/(w+y)<1/2 if this is not correct than above logic does not hold good.
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by sanju09 » Thu Apr 15, 2010 12:22 am
liferocks wrote:A zoo has w wildebeests, y yaks, z zebras, and no other animals. If an animal is chosen at random from the zoo, is the probability of choosing a yak greater than the probability of choosing a zebra?

(1) y/w+z > 1/2

(2) z/w+y < 1/2

probability of choosing a yak =y/(W+y+z)
probability of choosing a zebra =z/(W+y+z)

from 1 we get y/(w+z)>1/2 or y/(w+y+z)>1/3-->not sufficient
from 2 we get z/(w+y)<1/2 or z/(w+y+z)<1/3-->not sufficient

together we can say z/(w+y+z)<y/(w+y+z) -->sufficient ans C

note: I have assumed that condition is like y/(w+z)>1/2 and z/(w+y)<1/2 if this is not correct than above logic does not hold good.
great explanation

For those who can be misled by the bold part in the above post must know that w, y, and z are positive integers and:

when y/ (w + z) > 1/2; (w + z)/y < 2,

or [(w + z)/y] + 1 < 2 + 1,

or (w + y + z)/y < 3,

or y/(w + y + z) > 1/3; and

when z/ (w + y) < 1/2; (w + y)/z > 2,

or [(w + y)/z] + 1 > 2 + 1,

or (w + y + z)/z > 3,

or z/(w + y + z) < 1/3.

Therefore comparison is possible.

[spoiler]C[/spoiler]
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by digvijay.rajput » Fri Aug 02, 2013 1:00 am
In the above example, can you please explain these steps....

when y/ (w + z) > 1/2; (w + z)/y < 2,

or [(w + z)/y] + 1 < 2 + 1,

or (w + y + z)/y < 3,

Thanks in advance
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by sanju09 » Fri Aug 02, 2013 4:16 am
digvijay.rajput wrote:In the above example, can you please explain these steps....

when y/ (w + z) > 1/2; (w + z)/y < 2,

or [(w + z)/y] + 1 < 2 + 1,

or (w + y + z)/y < 3,

Thanks in advance
Take suitable numbers for the variables when you are in fix.

E.g. if 3/4 > 1/2, then 4/3 < 2 RIGHT!

OR

If 5 > 1/2, then 1/5 < 2; the statement would consistently remain true. Do the remaining math likewise.
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by [email protected] » Fri Aug 02, 2013 3:48 pm
Hi C,

This DS question can be beaten by TESTing values and a bit of logic:

The question asks if the probability of picking a yak is greater than the probability of picking a zebra.

Since we're dealing with a certain number of animals (and positive integers only), this question can be rewritten as: "Is Y > Z"?

Fact 1 tells us that Y/(W+Z) > 1/2

We could have
Y = 4
W = 1
Z = 1
The answer to the question is YES

We could also have
Y = 4
W = 1
Z = 5
The answer to the question is NO
Inconsistent = INSUFFICIENT

Fact 2 tells us that Z/(W+Y) < 1/2

We could have
Y = 4
W =1
Z = 1
The answer to the question is YES

We could also have
Y = 4
W = 10
Z = 5
The answer to the question is NO
Inconsistent = INSUFFICIENT

Now, we have to COMBINE Facts (and we get a number property shortcut). Since the W is a "fixed" value in both inequalities, it's the Y and Z that have to change to fit the > or <. In this scenario, by definition, the Y would have to be greater than the Z.

Final Answer: C

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by Matt@VeritasPrep » Fri Aug 02, 2013 4:31 pm
Maybe an easier way:

(If I'm reading this right, the original statements are y/(w+z) > 1/2 ... not (y/w) + z > 1/2. If I'm mistaken, ignore this post.)

The question is really "Are there more yaks than zebras?" or "is y > z?"

S1::
y/(w+z) > 1/2
2y > w + z
y + y > w + z

So y > w or y > z, but we don't know which. (Both could be true, for that matter.)

S2::
1/2 > z/(w+y)
w + y > 2z
w + y > z + z

Much like the above: w > z or y > z, but we don't know which.

Combining the two statements, we have two options:

Either y > z ... or y > w and w > z, implying y > z. So y > z.
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