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Q: Is z>1? Since we know z > -(x+y), it will suffice to know if -(x+y) > 1 that way we know z>1
In other words, treat -(x+y) as a constant K. If such constant K is >1 then z will be >1 because we already know that z > K
-(x+y) > 1 simplifies as x+y <1?
Simplifies as x+y <0 => x+y can be -0.1, -1, -6,..
Therefore x+y <1
SUFF => AD
Simplifies as x+y <-1 => x+y can be -1.1, -2, -6,..
This means x+y <1
SUFF => D
If x+y+z=0.5,z=0.6 -> Z>1 false
If x+y+z=9,z=10 -> Z>1 True
x+y<-1, Given x+y+z>0
x+y = -1.1, -2, -3.3...
To make x+y+z>0, z has to be >= |x+y| => z can be 1.1, 2, 3.3 ...
Hence z is always > 1
I just realised I made a mistake in the options I provided The question is correct but the options should have been
(1) Z>X+Y+1 (2) X+Y+1<0[/size]
Any help you can give me is appreciated.
I would still say the answer as B. B is sufficient as per the very elaborate explanations given before.
I find A insufficient because:-
Considering Z-1=X+Y and substituing it in X+Y+Z>0 ,we get Z-1+Z>0
2Z-1>0;Z>1/2. IF it was a number greater than 1, we could have said A is sufficient. But in this case when the sign is equality then we find Z>1/2 .So the real Z should be greater than 1/2. It still can be 3/4 or 1.2. So I would say A is insufficient.