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

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Joined: 18 Apr 2006
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03 May 2006, 18:59
I'll go for D.

stmt2-> Z^2=X^2 + Y^2

hence it forms an right angle with sides X, Y and Z where in Z is hypotenuse and X & Y are other 2 sides.

we know that for any triangle, sum of 2 sides is greater than the 3rd side.

hence X + Y > Z

=> X > Z - Y (Sufficient)

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04 May 2006, 06:55
gmat_crack wrote:
Dilshod wrote:
laxieqv wrote:
laxieqv wrote:
Dilshod wrote:
If X, Y and Z are positive integers, is X greater than Z â€“ Y?

(1) X â€“ Z â€“ Y > 0.

(2) Z2 = X2 + Y2

(1) --> X> Z+Y > Z-Y (because Z and Y are positive integers ---> Z+Y> Z-Y) ---> answer to the question: YES ---> suff

(2) Z^2 = X^2 + Y^2 --> X^2 = Z^2 - Y^2 = (Z-Y) ( Z+Y) > (Z-Y) (Z-Y)---> X^2 > (Z-Y)^2 ---> (X- (Z-Y)) * (X+ (Z-Y)) > 0
---> 2 cases:

X> Z-Y AND X > Y-Z

OR X< Z-Y AND X < Y-Z ( This can't happen because either Z-Y or Y-Z must be smaller than 0 ---> X <0 ---> against the assumption in the problem)
---> The only case is X> Z-Y ---> suff

Go for D.

Well, i had some flaw in the red part.

Correct it :
Z^2 = X^2+ Y^2 --> X^2= Z^2 - Y^2 --> X^2= (Z-Y) * (Z+Y)
---> X= [(Z-Y) * (Z+Y)]/ X = (Z-Y) * ( ( Z+Y) / X)
+If Z-Y < 0 ---> Z-Y < 0 < X --> Z-Y< X
+If Z-Y > 0
If the bold part < 1 ( that is Z+Y< X) --> X< Z-Y ---> Z+Y < X< Z-Y --> Z+Y < Z-Y ---> unreasonable --> this case can't happen
If the bold part >1 ( that is Z+Y > X) ---> X > Z-Y
--> finally, we still conclude that X> Z-Y

I'm still stuck to D.

Honestly, I didn't get your "bold part" approach. One thing for sure is that you shouldn't take gmat questions that serious. Remember that you have less than 2 minutes for each question.

So can anybody explain, why st.2 is insufficient?

Here is my two cents

From st 2

X= [(Z-Y) * (Z+Y)]/ X

---> X = (Z-Y) * (Z+Y)/X --------(A)
Now the value of X dependeds on value of (Z+Y)/X.
Here (Z+Y)/X can be greater than 1 or less than depends on the values of X, Y, Z.

if Z+ Y > X then A will be greater than Xand if Z+ Y < X then A will be less than X.

So from St 2 we can't say any thing.

Impressive....Agree with 'A'
Re: DS: integers   [#permalink] 04 May 2006, 06:55

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