If x + y >0, is x > |y|?
(1) x > y
(2) y <0
OA is D
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x > |y|?
This topic has expert replies
grandh01 wrote:If x + y >0, is x > |y|?
If x+y>0 then either x, y or both will be positive. (at least one will positive and greater than zero)
(1) x > y
If x >y then x is positive. Y can be either negative or positive
If y>0 then x would have to be greater than absolute value of y for example x is 36 whereas y is 15.
If y<0 then it cannot be greater than the absolute value of x. For example if x is 36 then y cannot be less than -36 as it would not satisfy the x+y statement. Hence absolute value of y would be less than 36. Therefore x>abs y
(2) y <0
Since y <0 then x would be positive and absolute value of y cannot be less than x
Therefore both equations are sufficient on their own.
OA is D
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given: x + y > 0.
let us take the 1st option x > y.
1. if x > 0, y >= 0 then x + y > 0. As x > y so will be x > |y|.
if x > 0, y < 0. for x + y > 0. positive value should be more than negative value.
x + y > 0 can be written as X - |y| > 0 -> x > |y|
2. if x < 0, y will be less than zero (as x > y). so x + y will be negative. so you can eliminate this case.
so it can be answered by option 1 alone.
let us take the 2nd option y < 0.
x + y > 0. Only when x is positive.
so x + y > 0 can be written as x - |y| > 0. => x > |y|
so it can be answered by option 2 alone.
hence, it is D
let us take the 1st option x > y.
1. if x > 0, y >= 0 then x + y > 0. As x > y so will be x > |y|.
if x > 0, y < 0. for x + y > 0. positive value should be more than negative value.
x + y > 0 can be written as X - |y| > 0 -> x > |y|
2. if x < 0, y will be less than zero (as x > y). so x + y will be negative. so you can eliminate this case.
so it can be answered by option 1 alone.
let us take the 2nd option y < 0.
x + y > 0. Only when x is positive.
so x + y > 0 can be written as x - |y| > 0. => x > |y|
so it can be answered by option 2 alone.
hence, it is D
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