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?
Tricky absolute question..... help needed
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The portion in red does not account for the sign of y.Mo2men wrote: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
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.
Last edited by GMATGuruNY on Thu Aug 31, 2017 5:11 pm, edited 1 time in total.
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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.
|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.
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A solution with an alternate line of reasoning for Statement 2:Is |x+y|>|x-y|?
a) |x|>|y|
b) |x-y|<|x|
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.
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We could also treat this as a DISTANCE/NUMBER LINE problem.Is |x+y|>|x-y|?
a) |x|>|y|
b) |x-y|<|x|
|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.
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GMATGuruNY wrote:We could also treat this as a DISTANCE/NUMBER LINE problem.Is |x+y|>|x-y|?
a) |x|>|y|
b) |x-y|<|x|
|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
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In the blue illustration, x and y are both to the left of 0, implying that x and y are both negative.Mo2men wrote:Dear Mitch,GMATGuruNY wrote:We could also treat this as a DISTANCE/NUMBER LINE problem.Is |x+y|>|x-y|?
a) |x|>|y|
b) |x-y|<|x|
|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.
Thanks for your wonderful approaches you presented.
Should not the illustration be as follows:
x...-y..............0.............y
.....-y.....x.......0.............y
Thanks
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.
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Thanks Mitch for your awesome explanation.GMATGuruNY wrote:In the blue illustration, x and y are both to the left of 0, implying that x and y are both negative.Mo2men wrote:Dear Mitch,GMATGuruNY wrote:We could also treat this as a DISTANCE/NUMBER LINE problem.Is |x+y|>|x-y|?
a) |x|>|y|
b) |x-y|<|x|
|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.
Thanks for your wonderful approaches you presented.
Should not the illustration be as follows:
x...-y..............0.............y
.....-y.....x.......0.............y
Thanks
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.