|x-y| > |x-z| ?
A) |y| > |z|
y=2,z=1 and x>2,
x-2 > x-1 false
y=-2,z=-1, and x>2,
x+2 > x+1 true.
Insufficient
B) x<0
insufficient
A&B) x<0 and |y| > |z|
a=-x and a>0
|a+y| > |a+Z| ??
if y,z>0, true
if y,z<0, and a>|y|,
like y=-2,z=-1, a=3
1>2 false
insufficient
IMO E
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Cans!!
sparkle6 wrote:Is |x-y| > |x-z| ?
1) |y| > |z|
2) x < 0
|x-y| > |x-z|.. i think the question gets reduced to is |y| > |z|?.. which is what statement 1 is.. IMO a
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chetansharma
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IMO the answer is E
Case A) |y| > |z|
assuming y= 10 and z=1 => |x-10|> |x-1| --(a)
Here depending on the x value the result changes.
If x=7 then the condition (a) will fail. Also if value of x changes to 2, condition (a) will hold good.
So Insufficient
Case B) x<0
Without the knowledge of values abt y,z nothing can be concluded.
so Insufficient
combining A & B,
assuming x=-1 and y=10, z=2
the given equation will be equal to |-1-10|>|-1-2| which is true
Now if x=-100, y=-10 and z=9 then the eqn will be |-100+10|>|-100-9| which is false.
So even using both the conditions nothing can be concluded. so the answer shd be E
What is the OA?
Regards,
Chetan
Case A) |y| > |z|
assuming y= 10 and z=1 => |x-10|> |x-1| --(a)
Here depending on the x value the result changes.
If x=7 then the condition (a) will fail. Also if value of x changes to 2, condition (a) will hold good.
So Insufficient
Case B) x<0
Without the knowledge of values abt y,z nothing can be concluded.
so Insufficient
combining A & B,
assuming x=-1 and y=10, z=2
the given equation will be equal to |-1-10|>|-1-2| which is true
Now if x=-100, y=-10 and z=9 then the eqn will be |-100+10|>|-100-9| which is false.
So even using both the conditions nothing can be concluded. so the answer shd be E
What is the OA?
Regards,
Chetan
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