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

Among X, Y and Z which one is greatest or which one is smallest is unknown.

(1) X – Z + Y > 0
Let X = 1, Y = 2 and Z = 3. Then X – Z + Y = 2 > 0
Thus, X = Z - Y. Hence X > Z - Y NO.

Let X = 3, Y = 2 and Z = 1. Then X – Z + Y = 4 > 0
Thus, X > Z - Y. Hence X > Z - Y YES.

INSUFFICIENT.

(2) Z^2 = X^2 + Y^2
This implies that Z is the greatest of all the three positive integers in fact they represent sides of a right angle triangle. So,

X^2 = Z^2 - Y^2
X^2 = (Z - Y) * (Z + Y)
In ether case here X would be such that Z - Y < X < Z + Y. Eg. in the set of 3,4,5 Z = 5 and X takes any value among 3 or 4.

Hence X > Z - Y Always.

SUFFICIENT.

Answer (B).
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X, Y and Z are positive integers, is X greater than Z—Y?

Statement1: X—Z+Y> 0
—> X> Z—Y (Always Yes)
Sufficient

Statement2: \(Z^{2}= X^{2} + Y^{2}\)
As chetan2u told, this equality is about right angled triangle (z is a hypotenuse and two other sides of a triangle)
—> according to features of a triangle,
X+Y> Z or
X+Z> Y or
Z+Y> X —>
——————-
X+Y> Z
—> X> Z—Y
(Always Yes)
Sufficient

The answer is D

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Known: X,Y,Z positive integers

Q. X > Z–Y ? or X-Z+Y > 0 ?

(1) X–Z+Y > 0
This statement directly answers the question.
SUFF

(2) Z^2 = X^2 + Y^2
If X=3, Y=4, Z=5, then X> Z–Y
If X=5, Y=12, Z=13, then X> Z–Y
SUFF

Final answer is (D)

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lnm87
If X, Y and Z are positive integers, is X greater than Z – Y?

Among X, Y and Z which one is greatest or which one is smallest is unknown.

(1) X – Z + Y > 0
Let X = 1, Y = 2 and Z = 3. Then X – Z + Y = 2 > 0
Thus, X = Z - Y. Hence X > Z - Y NO.

Let X = 3, Y = 2 and Z = 1. Then X – Z + Y = 4 > 0
Thus, X > Z - Y. Hence X > Z - Y YES.

INSUFFICIENT.

(2) Z^2 = X^2 + Y^2
This implies that Z is the greatest of all the three positive integers in fact they represent sides of a right angle triangle. So,

X^2 = Z^2 - Y^2
X^2 = (Z - Y) * (Z + Y)
In ether case here X would be such that Z - Y < X < Z + Y. Eg. in the set of 3,4,5 Z = 5 and X takes any value among 3 or 4.

Hence X > Z - Y Always.

SUFFICIENT.

Answer (B).

Again a silly error. :roll:
Either all of them are equal X = Y = Z = 1 then X > Z - Y
or such that X – Z + Y > 0. Thus, X > Z - Y always

For foolproof method it is best to say that statement 1 states that one side of a triangle is less than sum of other two sides.
X + Y > Z.

Hence Statement 1 is SUFFICIENT.
Answer (D).
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chetan2u
Bunuel
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z – Y > 0

(2) Z^2 = X^2 + Y^2

(1) X – Z – Y > 0
\(X – Z – Y > 0......X>Z+Y\)
As All are positive integers, X is greater than SUM of Z and Y, so Surely gtreater than their difference too...X>Z-Y
Suff

(2) Z^2 = X^2 + Y^2
This says that Z is a hypotenuse with X and Y as other two sides.. May not help here
\(X^2=Z^2-Y^2=(Z+Y)(Z-Y)=X*X\)
Surely for this to be true, one of (Z+Y) and (Z-Y) should be GREATER than X and other LESS than Y..
But Sum of Z and Y has to be greater than difference of Z and Y. So, so X<Z+Y, but X>Z-Y.
Suff

D, although statement I says X is the greatest while II says Z is the greatest, but statement II is well written to deduce the answer.


Beautiful solution for statement II
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Bunuel
If X, Y and Z are positive integers, is X greater than Z – Y?

(1) X – Z + Y > 0

(2) Z^2 = X^2 + Y^2

(x,y,z) = positive integers

\(x>z-y…x-z+y>0\)

(1) X – Z + Y > 0 sufic

(2) Z^2 = X^2 + Y^2 sufic

\(z^2=x^2+y^2…x^2=z^2-y^2…(x=positive.int)…x^2>0…z^2-y^2>0…z>y\)
\(x>z-y…x^2>(z-y)^2…(z^2-y^2)>(z^2+y^2-2zy)…-2y^2>-2zy…y<z\)

Ans (D)
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