Is x>y ?
A. x^2 > y^2
B. y>0
DS
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1. x^2 > y^2
=> x^2-y^2 > 0
=> (x+y) (x-y) > 0
This just tells us that both the terms ie. (x+y) and (x-y) in the above equation are either +ve or -ve
a) If both +ve than x>y
b) If both -ve than x<y
So insufficient as we dont know what x and y are , +ve or -ve integers
2. y>0
Insufficient
And when 1. and 2 are combined, we get
From a) we deduct x>y because (x-y) has to be >0
From b) we deduct x<y because (x-y) has to be <0
So still no conclusion is drawn.
So Answer is E
=> x^2-y^2 > 0
=> (x+y) (x-y) > 0
This just tells us that both the terms ie. (x+y) and (x-y) in the above equation are either +ve or -ve
a) If both +ve than x>y
b) If both -ve than x<y
So insufficient as we dont know what x and y are , +ve or -ve integers
2. y>0
Insufficient
And when 1. and 2 are combined, we get
From a) we deduct x>y because (x-y) has to be >0
From b) we deduct x<y because (x-y) has to be <0
So still no conclusion is drawn.
So Answer is E