A zoo has w wildebeests, y yaks, z zebras, and no other animals. If

one method is explained above..
another method could be..
as discussed on top we are basically asked" is y>z?"..
1) Statement 1 tells us \(\frac{y}{(w + z)} > \frac{1}{2}\)
or \(2y > w+z\).. add y to each side..
3y>w+x+y or y>total/3... insuff
2) statement 2 tells us \(\frac{z}{(w + y)}< \frac{1}{2}\)..
or \(2z < w+y\).. add z to each side..
3yz<w+x+y or \(z< \frac{total}{3}\)... insuff

combined we know that y> 1/3rd of total whereas z< 1/3rd of total..
so y>z.. suff
ans C

Question : Is y / (w+y+z) > z / (w+y+z)

Question : Is y > z ?

Statement 1: y/(w + z) > 1/2
i.e y is greater than (1/2) of w+z
Case-1: @ y = 4, w = 5 and z = 2 i.e. y > z
Case-2: @ y = 4, w = 2 and z = 5 i.e. y < z
NOT SUFFICIENT

Statement 2: z/(w + z) < 1/2
i.e z is greater than (1/2) of w+z
i.e. 2z < (w+z)
i.e. z < w
BUT z and y can't be compared, hence
NOT SUFFICIENT

Combining the two statements
z < w
and y/(w + z) > 1/2
i.e. (w + z) < 2y

i.e. Since, z < w therefore, (2z) < (w+z)

Combining the above two inequation
2z < (w + z) < 2y
i.e. z < y
SUFFICIENT

Answer: Option C

Statement 1 says "the ratio of yaks to other animals is greater than 1 to 2", or in other words, more than 1/3 of the animals are yaks. So the probability of picking a yak is greater than 1/3.

Statement 2 says "the ratio of zebras to other animals is less than 1 to 2", or in other words, less than 1/3 of the animals are zebras. So the probability of picking a zebra is less than 1/3.

So the two statements are sufficient together. Neither statement is sufficient alone, since we can get yes or no answers to the question by changing the fraction of animals which are wildebeests.

probability of selecting 1 yak = y/x+y+z
probability of selecting 1 zebra= z/x+y+z
is p(y)>p(z) ie y > z ?

1. y/w+z >1/2
2y > w+z insufficient

2. z/w+y <1/2
2z<w+y insufficient

combine both:
2y > w+z
2z < w+y

this gives
2y-z >w and 2z - y <w

hence 2y-z >2z-y
which gives y>z
so option C

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.

There are 3 variables and 0 equation. Thus the answer E is most likely.

After setting up, we have the equivalent question if \(y > z\).

Condition 1)
\(\frac{y}{w+z} > \frac{1}{2}\) is equivalent to \(2y > w + z\).
This is not sufficient since we don't know the value of \(w\).

Condition 2)
\(\frac{z}{w+y} > \frac{1}{2}\) is equivalent to \(2z > w + y\).
This is not sufficient since we don't know the value of \(w\).

Conditions 1) & 2)
The difference of two inequality \(2y > w + z\) and \(2z > w + y\) is \(2(y-z) > z - y\) or \(3(y-z) > 0\).
Thus \(y > z\).
This is sufficient.

Therefore the answer is C.


For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.

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