Tricky absolute question..... help needed

[Mo2men]

Is |x+y|>|x-y|?

(1) |x| > |y|
(2) |x-y| < |x|

Source:GC
OA: B

|x+y|>|x-y|...........Square both sides

x^2 + 2xy + y^2 > x^2 - 2xy + y^2

4xy > 0 .....the question could be repharsed

Is xy > 0 or Do both x & y have same sign?

1) |x| > |y|

Clearly Insufficient

2) |x-y| < |x| ..........Square both sides

x^2 - 2xy + y^2 < x^2

y^2 - 2 xy < 0 ...............Because we are certain that 'y' can't equal zero. I can divide by 'y'. If y =0 statement 2 would be |x| < |x| which is invalid. So you does not equal 0

y < 2x

here y and x can both have same sign or different sign.

Insufficient

Combine both statements 1 & 2

still insiffecent ..............Answer is E

Where did I go wrong in the above?

[GMATGuruNY]

Do both x & y have same sign?

2) |x-y| < |x| ..........Square both sides

x^2 - 2xy + y^2 < x^2

y^2 - 2 xy < 0 ...............Because we are certain that 'y' can't equal zero. I can divide by 'y'. If y =0 statement 2 would be |x| < |x| which is invalid. So you does not equal 0

y < 2x

The portion in red does not account for the sign of y.
You correctly squared the inequality to yield the following:
y² - 2xy < 0
y² < 2xy.

If y>0, then dividing both sides by y yields the following:
y < 2x.
Since y>0, we get:
0 < y < 2x, implying that both x and y are positive.

If y<0, then dividing both sides by y requires that we flip the inequality, yielding the following:
y > 2x.
Since y<0, we get:
2x < y < 0, implying that both x and y are negative.

In each case, x and y have the same sign.
Thus, the answer to the rephrased question stem is YES.

[Matt@VeritasPrep]

It might be easier to do this conceptually.

|x + y| > |x - y| can be stated as "x is further from -y than it is from y".

S2 can be stated as "x is further from 0 than it is from y".

If x is further from 0 than it is from y, x and y MUST be on the same side of zero, i.e. x and y must share the same sign. (If they had different signs, x to y would be (x to 0) + (0 to y), but with S2 that would make (x to 0) > (x to 0) + (0 to y), and (0 to y) can't be negative!)

With that in mind, the distance from x to -y = (the distance from x to 0) + (the distance from 0 to -y).

Since (x to 0) is already bigger than (x to y), we MUST have (x to 0) + (0 to -y) > (x to y), and S2 alone is sufficient.

[GMATGuruNY]

Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|

A solution with an alternate line of reasoning for Statement 2:

Is |x+y| > |x-y|?
When there is absolute value notation on each side, we can square the inequality.

(x+y)² > (x-y)²
x² + 2xy + y² > x² - 2xy + y²
4xy > 0
xy > 0.

Question rephrased: Do x and y have the same sign?

Statement 1: |x| > |y|
Here, x and y could have the same sign or different signs.
INSUFFICIENT.

Statement 2: |x-y| < |x|
Squaring both sides, we get:
(x-y)² < x²
x² - 2xy + y² < x²
y² < 2xy
xy > y²/2.
Since the square of a value cannot be negative, y²/2 cannot be negative.
Thus, xy>0, implying that x and y have the same sign.
SUFFICIENT.

The correct answer is B.

[GMATGuruNY]

Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|

We could also treat this as a DISTANCE/NUMBER LINE problem.
|x-y| = the distance between x and y.
|x+y| = the distance between x and -y.
|x| = the distance between x and 0.

Is |x-y| < |x+y|?
In words:
Is the distance between between x and y less than the distance between x and -y?
For x to be CLOSER TO Y than to -y, x and y must be TO THE SAME SIDE OF 0.
To illustrate:
x.....y..............0.............-y
.......y.....x.......0.............-y
As these examples show, when x and y are to the same side of 0, x is closer to y than to -y.

Question rephrased: Are x and y to the same side of 0?

Statement 1: |x| > |y|
x and y could be to the same side of 0 or on opposite sides of 0.
INSUFFICIENT.

Statement 2: |x-y| < |x|
In words:
The distance between x and y is less than the distance between x and 0.
This will not be true if x and y are on opposite sides of 0.
To illustrate:
x.........0...........y
y.........0..........x
In each case here, the distance between x and y is GREATER than the distance between x and 0.
Thus, to satisfy statement 2, x and y must be TO THE SAME SIDE OF 0.
SUFFICIENT.

The correct answer is B.

[Mo2men]

Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|

We could also treat this as a DISTANCE/NUMBER LINE problem.
|x-y| = the distance between x and y.
|x+y| = the distance between x and -y.
|x| = the distance between x and 0.

Is |x-y| < |x+y|?
In words:
Is the distance between between x and y less than the distance between x and -y?
For x to be CLOSER TO Y than to -y, x and y must be TO THE SAME SIDE OF 0.
To illustrate:
x.....y..............0.............-y
.......y.....x.......0.............-y
As these examples show, when x and y are to the same side of 0, x is closer to y than to -y.

Dear Mitch,

Thanks for your wonderful approaches you presented.

Should not the illustration be as follows:

x...-y..............0.............y
.....-y.....x.......0.............y

Thanks

[GMATGuruNY]

Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|

We could also treat this as a DISTANCE/NUMBER LINE problem.
|x-y| = the distance between x and y.
|x+y| = the distance between x and -y.
|x| = the distance between x and 0.

Is |x-y| < |x+y|?
In words:
Is the distance between between x and y less than the distance between x and -y?
For x to be CLOSER TO Y than to -y, x and y must be TO THE SAME SIDE OF 0.
To illustrate:
x.....y..............0.............-y
.......y.....x.......0.............-y
As these examples show, when x and y are to the same side of 0, x is closer to y than to -y.

Dear Mitch,

Thanks for your wonderful approaches you presented.

Should not the illustration be as follows:

x...-y..............0.............y
.....-y.....x.......0.............y

Thanks

In the blue illustration, x and y are both to the left of 0, implying that x and y are both negative.
Since y is negative, -y > 0, with the result that -y is to the right of 0.
The blue illustration positions x closer to y than to -y.
Since the distance between x and -y is greater than the distance between x and y, the following inequality is implied:
|x+y | > |x-y|.
Thus, the blue illustration accurately reflects the question stem.

In the red illustration, x and y are to different sides of 0, with the result that x is closer to -y and than to y.
Since the distance between x and -y is less than the distance between x and y, the following inequality is implied:
|x+y | < |x-y|.
Thus, the red illustration does not accurately reflect the question stem.

[Mo2men]

Is |x+y|>|x-y|?

a) |x|>|y|

b) |x-y|<|x|

We could also treat this as a DISTANCE/NUMBER LINE problem.
|x-y| = the distance between x and y.
|x+y| = the distance between x and -y.
|x| = the distance between x and 0.

Is |x-y| < |x+y|?
In words:
Is the distance between between x and y less than the distance between x and -y?
For x to be CLOSER TO Y than to -y, x and y must be TO THE SAME SIDE OF 0.
To illustrate:
x.....y..............0.............-y
.......y.....x.......0.............-y
As these examples show, when x and y are to the same side of 0, x is closer to y than to -y.

Dear Mitch,

Thanks for your wonderful approaches you presented.

Should not the illustration be as follows:

x...-y..............0.............y
.....-y.....x.......0.............y

Thanks

In the blue illustration, x and y are both to the left of 0, implying that x and y are both negative.
Since y is negative, -y > 0, with the result that -y is to the right of 0.
The blue illustration positions x closer to y than to -y.
Since the distance between x and -y is greater than the distance between x and y, the following inequality is implied:
|x+y | > |x-y|.
Thus, the blue illustration accurately reflects the question stem.

In the red illustration, x and y are to different sides of 0, with the result that x is closer to -y and than to y.
Since the distance between x and -y is less than the distance between x and y, the following inequality is implied:
|x+y | < |x-y|.
Thus, the red illustration does not accurately reflect the question stem.

Thanks Mitch for your awesome explanation.