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23 Feb 2007, 04:29
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if x,y,z are integers
is x + y - z > x - y + z
1. yz < 0
2. y + z < 0
is x + y - z > x - y + z
1. yz < 0
2. y + z < 0
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23 Feb 2007, 04:40
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(E) for me 
x + y - z > x - y + z ?
<=> y - z > z - y ?
<=> 2*y > 2*z ?
<=> y > z ?
Stat1
y*z < 0
o If y=-1 and z=1, then z > y
o If y=1 and z=-1, then y > z
INSUFF.
Stat2
y + z < 0
o If y=-1 and z=0, then z > y
o If y=0 and z=-1, then y > z
INSUFF.
(1) and (2)
o If y=-2 and z=1, then z*y <0, y+z <0 and y < z
o If z=-2 and y=1, then z*y <0, y+z <0 and y > z
INSUFF.
x + y - z > x - y + z ?
<=> y - z > z - y ?
<=> 2*y > 2*z ?
<=> y > z ?
Stat1
y*z < 0
o If y=-1 and z=1, then z > y
o If y=1 and z=-1, then y > z
INSUFF.
Stat2
y + z < 0
o If y=-1 and z=0, then z > y
o If y=0 and z=-1, then y > z
INSUFF.
(1) and (2)
o If y=-2 and z=1, then z*y <0, y+z <0 and y < z
o If z=-2 and y=1, then z*y <0, y+z <0 and y > z
INSUFF.
23 Feb 2007, 10:37
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if x,y,z are integers
is x + y - z > x - y + z
1. yz < 0
2. y + z < 0
is y-z>0is y>z
from one
yz<0 source of negativity could be either x or z...insuff
from 2
y+z<0 both could be negative ( no info about magnitude) or one +ve and other -ve.insuff
both together(add)
yz+y+z < 0
y(z+1) <-z we know that y,z have apposite signs
if z is -ve then y is +ve and if z is +ve then y is -ve remains a valid relation after adding the two inequalities together
thus E IS MY ANSWER TOO
is x + y - z > x - y + z
1. yz < 0
2. y + z < 0
is y-z>0is y>z
from one
yz<0 source of negativity could be either x or z...insuff
from 2
y+z<0 both could be negative ( no info about magnitude) or one +ve and other -ve.insuff
both together(add)
yz+y+z < 0
y(z+1) <-z we know that y,z have apposite signs
if z is -ve then y is +ve and if z is +ve then y is -ve remains a valid relation after adding the two inequalities together
thus E IS MY ANSWER TOO
