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If x+z=y, is x>y?

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If x+z=y, is x>y? [#permalink]

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New post 14 Apr 2017, 07:34
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If x+z=y, is x>y?

1) y>0
2) z<0
[Reveal] Spoiler: OA

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Re: If x+z=y, is x>y? [#permalink]

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New post 14 Apr 2017, 10:36
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MathRevolution wrote:
If x+z=y, is x>y?

1) y>0
2) z<0


is x > y? i.e. is x - y > 0

x+z = y
x-y = -z

so question becomes is z < 0 ? because if x-y < 0 then x - y will be +ve.

Stmt - 2 exactly tells you z < 0.
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Re: If x+z=y, is x>y? [#permalink]

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New post 14 Apr 2017, 10:47
HKD1710 wrote:
MathRevolution wrote:
If x+z=y, is x>y?

1) y>0
2) z<0


is x > y? i.e. is x - y > 0

x+z = y
x-y = -z

so question becomes is z < 0 ? because if x-y < 0 then x - y will be +ve.

Stmt - 2 exactly tells you z < 0.


Nicely explained here! I tried putting in values for X, Y and Z. Took much more time to solve that way

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Re: If x+z=y, is x>y? [#permalink]

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New post 16 Apr 2017, 18:28
==> If you modify the original condition and the question, you get x>y?, x-y>0? Or –z>0? Or z<0?. Thus, from con 2), it is always yes and sufficient.

The answer is B.
Answer: B
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Re: If x+z=y, is x>y? [#permalink]

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New post 03 Oct 2017, 22:50
MathRevolution wrote:
If x+z=y, is x>y?

1) y>0
2) z<0


x+z = y

So, if z>0 , x+z add upto y (for all values of y) and hence x<y

if z<0, x+z = x -|z| = y (for all values of y) and hence x>y

Statement 1 : y >0. it doesn't matter if y<0 or y >0
NOT SUFFICIENT

Statement 2: z<0 . Yes if this is the case then x>y as explained above.
SUFFICIENT

Answer B
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Re: If x+z=y, is x>y?   [#permalink] 03 Oct 2017, 22:50
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