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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 05 Dec 2020, 17:34
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 19 Jul 2021, 18:46
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Can anybody suggest what should be the method of solving absolute value inequalities? I seem to be struggling with such kind of questions.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 20 Jul 2021, 00:46
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anwesharath
Can anybody suggest what should be the method of solving absolute value inequalities? I seem to be struggling with such kind of questions.
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10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 20 Jul 2021, 00:54
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anwesharath
Can anybody suggest what should be the method of solving absolute value inequalities? I seem to be struggling with such kind of questions.

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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This is extremely helpful! Thank you so much :)
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 06 Nov 2021, 23:54
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I think this approach is wrong. Reasons
1) What is the reason behind assuming |x| > |y| please ? This ssumption will only
remove a scenario in which |y| can be greater than |x|. How can that be accounted for in this method please ?
2) How would a student know to make this (incorrect) under the 2 minute time pressure that GMAT Quant imposes ?




MathRevolution
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) y < x
(2) xy < 0

If we modify the original condition,
if |x|>=|y|,
|x-y|>|x|-|y|>=0, so we can square both sides, which gives us (|x-y|)^2>(|x|-|y|)^2, and (x-y)^2>(|x|-|y|)^2
This becomes x^2+y^2-2xy>x^2+y^2-2|xy|, and if we simplify the inequality, -2xy>-2|xy|, and we ultimately want to know whether xy<0. This makes condition 2 sufficient.
if |x|<|y|, then |x|-|y|<0,
so |x-y|>|x|-|y| always work. Hence we do not need to deal with this, so the answer becomes (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 18 Nov 2021, 04:40
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|x-y|^2 > {|x|-|y|}^2
x2 + y2 - 2xy > x2 + y2 - 2 |x||y|
=> xy < |x||y|

Only case when xy < |x||y| is when x is -ve and y is +ve or x is +ve and y is -ve

Hence, xy < 0

Also, y < x cant be deduced from xy < |x||y|

Can anyone please tell me whether this approach is correct or not?
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 14 Dec 2021, 05:46
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Quote:
Is |x-y| > |x| - |y| ?
1) y < x
2) xy < 0

|x|= the distance between x and 0 = the RED segment on the number lines below.
|y| = the distance between y and 0 = the BLUE segment on the number lines below.
|x-y| = the distance BETWEEN X AND Y.

Statement 1: y < x
Case 1:
[edq435/absolute_number_line_1.jpg[/img][/url]
|x| - |y| = RED - BLUE.
|x-y| = RED - BLUE.
Thus, |x-y| = |x| - |y|.

Case 2:
[mber_line_2.jpg[/img][/url]
|x| - |y| = RED - BLUE.
|x-y| = RED + BLUE.
Thus, |x-y| > |x| - |y|.
INSUFFICIENT.

Statement 2: xy < 0
Since x and y have different signs, they are on OPPOSITE SIDES OF 0.e_3.jpg[/img][/url]
In each case:
|x| - |y| = RED - BLUE.
|x-y| = RED + BLUE.
Thus, |x-y| > |x| - |y|.
SUFFICIENT.

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B
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kudos for this perfect graphic solution. Exact way the asian teachers taught us at school.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 21 May 2022, 11:03
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GMATNinja, could you please advice what standard method is worthwhile to remember to solve this type of question? What should be the first step to solve this type of questions?
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 14 Jul 2025, 00:02
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Statement (1): y < x
Try different cases:
  1. x = 2, y = 1
    |x - y| = |2 - 1| = 1
    |x| - |y| = 2 - 1 = 1
    → 1 > 1? No
  2. x = 2, y = -1
    |x - y| = |2 - (-1)| = 3
    |x| - |y| = 2 - 1 = 1
    → 3 > 1? Yes
So, sometimes yes, sometimes no
→ Statement 1 is Not Sufficient

Statement (2): xy < 0
→ This means x and y have opposite signs
Try different cases:
  1. x = 2, y = -1
    |x - y| = |2 + 1| = 3
    |x| - |y| = 2 - 1 = 1
    → 3 > 1 Yes
  2. x = -3, y = 1
    |x - y| = |-3 - 1| = 4
    |x| - |y| = 3 - 1 = 2
    → 4 > 2 Yes
  3. x = -4, y = 4
    |x - y| = |-8| = 8
    |x| - |y| = 4 - 4 = 0
    → 8 > 0 Yes
Always true when x and y are of opposite signs
→ Statement 2 is Sufficient

Answer: (B) Statement (2) alone is sufficient





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

(1) y < x
(2) xy < 0

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

Probably the best way to solve this problem is plug-in method. Though there are two properties worth to remember:
1. Always true: \(|x+y|\leq{|x|+|y|}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign);

2. Always true: \(|x-y|\geq{|x|-|y|}\), note that "=" sign holds when \(x \leq y \leq 0\) or when \(0 \leq y \leq x\). So, essentially when \(xy \geq 0\) AND \(|x| \geq |y|\). (Our case)

So, the question basically asks whether we can exclude "=" scenario from the second property.

(1) y < x --> we cannot determine the signs of \(x\) and \(y\). Not sufficient.
(2) xy < 0 --> "=" scenario is excluded from the second property, thus \(|x-y|>|x|-|y|\). Sufficient.

Answer: B.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 27 Aug 2026, 04:22
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I squared the equation and tested it and arrived at the same answer.

Im sure my approach is wrong but not sure why
Bunuel


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

Probably the best way to solve this problem is plug-in method. Though there are two properties worth to remember:
1. Always true: \(|x+y|\leq{|x|+|y|}\), note that "=" sign holds for \(xy\geq{0}\) (or simply when \(x\) and \(y\) have the same sign);

2. Always true: \(|x-y|\geq{|x|-|y|}\), note that "=" sign holds when \(x \leq y \leq 0\) or when \(0 \leq y \leq x\). So, essentially when \(xy \geq 0\) AND \(|x| \geq |y|\). (Our case)

So, the question basically asks whether we can exclude "=" scenario from the second property.

(1) y < x --> we cannot determine the signs of \(x\) and \(y\). Not sufficient.
(2) xy < 0 --> "=" scenario is excluded from the second property, thus \(|x-y|>|x|-|y|\). Sufficient.

Answer: B.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 28 Aug 2026, 15:09
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Hi SanuBest,

You landed on B, but the squaring step is genuinely unsafe, and it is the same concern raised earlier in this thread.

The rule you skipped: you can square both sides of an inequality only when both sides are guaranteed non-negative. Squaring preserves an inequality only when nothing on either side is negative.

Look at the stem, |x - y| > |x| - |y|:

- The left side, |x - y|, is always ≥ 0. Fine.
- The right side, |x| - |y|, can be negative. Take x = 1, y = 2, which gives -1.

Once the right side can go negative, squaring can flip the truth of the statement.

A counterexample that exposes it:

- 3 > -5 is clearly true.
- Square both sides: 9 > 25 is clearly false.

Same illegal move, and it reverses the answer, so the squared version is not an equivalent of the original question.

So why did B still come out? Coincidence. Wherever the right side is negative, the original inequality is automatically true, a non-negative value beating a negative one, and those cases happen not to disturb the xy < 0 boundary your algebra produced. The conclusion survived; the reasoning behind it did not, which is exactly the discomfort you felt.

The safe route is the case-testing used throughout this thread: pick concrete numbers obeying each statement and check whether the yes/no answer can change.

Answer: B

SanuBest
I squared the equation and tested it and arrived at the same answer.

Im sure my approach is wrong but not sure why

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