If x+y+z>0, is z>1?
1)z>x+y+1
2)x+y+1<0
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cans
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x+y+z>0. To find z>1?
a)z>x+y+1
Also x+y+z>0
thus x+y>-z
or z>-z+1 => z>1/2 Insufficient (z can be 3/4 also and 2 also)
b)x+y+1<0
x+y<-1
-1>x+y>-z
-1>-z
1<z sufficient
IMO B
a)z>x+y+1
Also x+y+z>0
thus x+y>-z
or z>-z+1 => z>1/2 Insufficient (z can be 3/4 also and 2 also)
b)x+y+1<0
x+y<-1
-1>x+y>-z
-1>-z
1<z sufficient
IMO B
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Cans!!
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Cans!!
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Frankenstein
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Hi,
From(1): if x=1,y=2,z=5, z>x+y+1. z>1
if x=0.1,y=-0.2,z=1, z>x+y+1. z=1
Not sufficient
From(2):x+y+z>0 => x+y>-z. x+y+1<0 =>1-z<0 =>z>1
Sufficient
Hence B
From(1): if x=1,y=2,z=5, z>x+y+1. z>1
if x=0.1,y=-0.2,z=1, z>x+y+1. z=1
Not sufficient
From(2):x+y+z>0 => x+y>-z. x+y+1<0 =>1-z<0 =>z>1
Sufficient
Hence B
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