If x, y and z are positive integers, is x > z-y?
1) x-y-z>0
2) z^2=x^2+y^2
inequality
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- Vemuri
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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.
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.
Last edited by Vemuri on Tue Jun 16, 2009 2:44 am, edited 1 time in total.
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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-
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-
Last edited by bluementor on Tue Jun 16, 2009 6:09 am, edited 1 time in total.
- rahulg83
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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
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
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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)
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)
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- Domnu
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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.
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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.
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.
- rahulg83
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It is from gmatscore.comrah_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.
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IMO Drahulg83 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
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
The powers of two are bloody impolite!!
- hk
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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.
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.
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