From (1)

y/w+z>1/2

(w+z)/y < 2 (Taking inverse inequality changes)

(w+y+z)/y < 3 (Adding 1 and taking LCM)

y/(w+y+z) > 1/3 (Again taking inverse)

Now similarly from (2) we have;

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

Hence P(y) > P(z) C

The question can be rephrased as , is y>z ?

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

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

Statement 1 and 2

As the inequalities are pointing in opposite directions, we can subtract the two, notice that
there are two scenarios, in one case, y>z and in the other y<z
y-z>(w+z)/2-(w+y)/2. simplifying, we get y>z
z-y<(w+y)/2-(w+z)/2. simplifying, we get y>z Sufficient

mehrasa 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

could you please give me the solution mathematically when we consider both stats.. i need some clarification on inequality problem..thnx

I dont know if this question really has solution.. Since we are dealing with animals, here only +ve integers comes into the picture..

Statement 1 is insufficient as it can take any values and equation still be fine
Statement 2:- this only works if y=0, and z<w/2 .. Again this statement is insufficient

Going by algebra, we can get proove that y>z, but then z needs to be -ve (as per the statement 2)..

Let me know if i am getting it wrong..

mehrasa 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

could you please give me the solution mathematically when we consider both stats.. i need some clarification on inequality problem..thnx

Almost no math is needed when we combine the two statements.
Linking them together, we get:
z/(w+y) < 1/2 < y/(w+z)
z/(w+y) < y/(w+z).

The inequality above compares the following ratios:
zebras : other animals < yaks : other animals.

Clearly, for the relationship above to hold true, there must be fewer zebras than yaks.
SUFFICIENT.
The correct answer is C.

Looking for an algebraic proof will only make the problem take LONGER.
Remember: the GMAT is not a math test; it's a REASONING test.

This problem is similar to the following from GMATPrep:
https://www.beatthegmat.com/probability-t74040.html