Looks simple, but tricky one

[sspati]

9. If X, Y and Z are positive integers, is X greater than Z – Y?



(1) X – Z – Y > 0.



(2) Z2 = X2 + Y2.

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Official Answer Later
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[Ashoka]

'B' for me.

[kautilya_bharat]

C:
Combinging these two gives
1) x > z+y
2) X^2 = (z+y)(z-y)

=> As X is always positive and moreover X^2 will always be positive.
Either Z+y and Z-y are both positive or both negative
but Z and Y are also positive integer which means Z+y is positive => Z-y is also positive => X < z-y

[Ashoka]

C:
Combinging these two gives
1) x > z+y
2) X^2 = (z+y)(z-y)

=> x < z-y

Where are you getting the bolded part from ??

[Amorous]

i would say either statements are sufficient to answer the question...

D

-Amor

[kautilya_bharat]

Z^2 = X^2 + Y^2 => X^2 = Z^2 - Y^2

Where are you getting the bolded part from ??

[Ashoka]

Z^2 = X^2 + Y^2 => X^2 = Z^2 - Y^2

Should't Z2 = X2 + Y2 be Z*2 = X*2 + Y*2 ?


sspati, what is intended in the original question ??

[archangel88]

I think it's D.

(1) x-z-y>0 => x>z+y
since x, y, and z are positive, if x>z+y => x>z-y; sufficient
(2) z^2=x^2+y^2
For right triangle of sides x, y, and hypotenuse z, this is the
relationship between sides, and for ANY triangle the length of
any one side is greater than the difference of the other two=>
x>z-y; sufficient

So D, for me.

[coolprasant]

I think it's D.

(1) x-z-y>0 => x>z+y
since x, y, and z are positive, if x>z+y => x>z-y; sufficient
(2) z^2=x^2+y^2
For right triangle of sides x, y, and hypotenuse z, this is the
relationship between sides, and for ANY triangle the length of
any one side is greater than the difference of the other two=>
x>z-y; sufficient

So D, for me.

I completely agree with archangel88. In fact I had come to the same conclusion much before I found archangel88's posting. The answer must be D. Please pass on the Official Answer.

[gmat168]

Agree with D.

And archangel's triangle analysis is truly inspired. I just tried numbers.