IMO D.

Stmt1: Can be re-written as x>y+z. Since x,y,z are all positive and if x is greater than the addition of y & z. Then x is surely greater than the subtraction of y & z. Sufficient.

Stmt2: The question can be reframed to as z>y-x. From this statement z^2=x^2+y^2. Lets consider some numbers that satisfy this equation z=5, x=3, y=4 & then try them to see if the question can be answered. 5>4-3, which is true. The same will be the case for any positive numbers you select. So, this statement also is sufficient.

Statement 1:

x - y - z > 0
x > y + z

If x is greater than the sum of y and z, then surely x will be greater than the difference between y and z (since x, y and z are all positive integers). Sufficient

Statement 2:

Notice that this is essentially the pythagorean theorem, and from this we can treat x, y and z as the sides of a right angled triangle, with z being the hypotenuse. In a triangle, we know that the sum of the two shorter sides must be larger than the longest side, in this case the hypotenuse z.

so,

x + y > z
x > z - y

Sufficient.

IMO D. Are you sure of the OA?

-BM-

very sorry guys...it should read as x>z-y and not x>y-z
It was a typo..plz see the corrected question now

OA is A

But my answer still D

ststement 2 infers that x, y and z are sides of right angled triangle and we know that sum of any two sides is always greater than third.
so x+y>z for all x,y,z
=> x>z-y

Guys any clarity on this??/ what should be the answer

I agree. I think the answer should be D as well.

I disagree.
For statement 2,

If x= 3, y=4, z=5
Then x>z-y

If x =6, y=8, z=10
Then x=z-y

NOT SUFFICIENT

I think, to my wild imagination, statement 2 is trying confuse us to include pythagoras theory of z^2 = x^2+y^2

And from triangle properties, side of triangle x, y, and z
Always x+y>z
So x>z-y

My answer is (A)

I miscalculated

if x = 6, y = 8, z =10
x>z-y

Pythagoras's Theorem would definitely be okay here. From mathematics, we can say that if x, y, z > 0 and x^2 + y^2 = z^2, then x, y, z can ALWAYS be the sides of a right triangle with x, y, z.

I got answer as A but i did not use the pythogorus theorem

from second statement

z-y=x^2/(y+z)

now let us compute

x-(z-y)=(x-(x^2/(y+z)))
=x*(y+z-x)/(y+z)

is positive or negative depending on value of y+z-x

now obviously this does not lead us anywhere so answer can be A

but if you use pythogorus than obviously answer is D. can some instructor throw some light on it. Is it a official guide question. Dont think so? The question is not well thought off.

rah_pandey wrote:I got answer as A but i did not use the pythogorus theorem

from second statement

z-y=x^2/(y+z)

now let us compute

x-(z-y)=(x-(x^2/(y+z)))
=x*(y+z-x)/(y+z)

is positive or negative depending on value of y+z-x

now obviously this does not lead us anywhere so answer can be A

but if you use pythogorus than obviously answer is D. can some instructor throw some light on it. Is it a official guide question. Dont think so? The question is not well thought off.

It is from gmatscore.com

rahulg83 wrote:If x, y and z are positive integers, is x > z-y?

1) x-y-z>0
2) z^2=x^2+y^2

IMO D
1) sufficient.
x-y-z>0
or,x>z+y
so obviously>z-y

2) z^2=x^2+y^2
pythagoras th:
so x+y>z
x>z-y

I got the answer as D..

But i dont think this is a good question!! The rule with the GMAT question is that St.1 and St.2 should not contradict!!!

In this case both the statements are contradicting each other!!

Statement 1 says x-y-z>0, this means that x has to be the largest of the three.
Statement 2 says z^2 = x^2 +y^2 this says that z is the largest.

This type of contradiction is never seen in the real GMAT, GMATPrep or the Official guides.