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Source: Beat The GMAT — Data Sufficiency |
I did it this way -
xy+z=z
xy = 0
i.e x=0 or y = 0
so a. x!0 then y = 0, so |x-y| will be greater than 0
frm b. y=0 so x!0, so |x-y| again > 0.
so D but ans is A .. ??
xy+z=z
xy = 0
i.e x=0 or y = 0
so a. x!0 then y = 0, so |x-y| will be greater than 0
frm b. y=0 so x!0, so |x-y| again > 0.
so D but ans is A .. ??
or is it this way !!
from b . y=0 so even x could be x=0 . in tht case x-y =0 is not greater than 0.
hence the OA is A.
Hussh!! self solve ..
from b . y=0 so even x could be x=0 . in tht case x-y =0 is not greater than 0.
hence the OA is A.
Hussh!! self solve ..
I didn't get the answer until you put the OA up... but it's a good question.
From the statement xy + z = z, we can deduce that xy = 0
This gives us 2 conditions:
CASE A: x or y is 0
CASE B: x and y both are 0
The problem is knowing whether we have CASE A or B.
With Case A |x-y| > 0 for sure. With Case B |x-y| is 0 which is not greater than 0.
Statement I helps us identify that only one of the numbers is 0... So |x-y| is definitely greater than 0. So I is sufficient
Statement II states that y is 0... but that doesn't help as we don't know if x is 0 or x != 0 .. so II is insufficient. Answer is A
From the statement xy + z = z, we can deduce that xy = 0
This gives us 2 conditions:
CASE A: x or y is 0
CASE B: x and y both are 0
The problem is knowing whether we have CASE A or B.
With Case A |x-y| > 0 for sure. With Case B |x-y| is 0 which is not greater than 0.
Statement I helps us identify that only one of the numbers is 0... So |x-y| is definitely greater than 0. So I is sufficient
Statement II states that y is 0... but that doesn't help as we don't know if x is 0 or x != 0 .. so II is insufficient. Answer is A
