Ans shd be B - Coz

1) |x-3| >=y

IF y =2 then x can be 5 or 6
Not sufficient

2) |x-3| <= -y

we are already given y>=0
IF we take y >= 1 then the condition given in 2) doesnt hold true
e.g. -> y = 1
|x-3| <= -1 -----> cant be coz of modulus

So we need to take y = 0
and if we take y=0 then there is a ceratin value of x

hence B

Good question indeed...!!

Hey, that was pretty simple! And I sweated my brains out breaking the modulus sign and coming up with all sorts of inequalities. Thanks man!

Need some suggestions here. Should I rather substitute random values into modulus problems to figure out the solution or break it up into inequalities? Or is it essentially a case by case scenario?

puneetdua wrote:Ans shd be B - Coz

1) |x-3| >=y

IF y =2 then x can be 5 or 6
Not sufficient

2) |x-3| <= -y

we are already given y>=0
IF we take y >= 1 then the condition given in 2) doesnt hold true
e.g. -> y = 1
|x-3| <= -1 -----> cant be coz of modulus

So we need to take y = 0
and if we take y=0 then there is a ceratin value of x

hence B

Good question indeed...!!

how do you know on this inequality to plug #s instead of trying to get fancy and manipulate the equalities? i.e. test positive/negative case?

puneetdua wrote:Ans shd be B - Coz

1) |x-3| >=y

IF y =2 then x can be 5 or 6
Not sufficient

2) |x-3| <= -y

we are already given y>=0
IF we take y >= 1 then the condition given in 2) doesnt hold true
e.g. -> y = 1
|x-3| <= -1 -----> cant be coz of modulus

So we need to take y = 0
and if we take y=0 then there is a ceratin value of x

hence B

Good question indeed...!!

how do you know on this inequality to plug #s instead of trying to get fancy and manipulate the equalities? i.e. test positive/negative case?