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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 Updated on: 20 Jul 2021, 00:43
Originally posted by study on 17 Nov 2009, 04:52.
Last edited by Bunuel on 20 Jul 2021, 00:43, edited 1 time in total.
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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61% (01:48) correct 39% (02:02) wrong based on 6743 sessions

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

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


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B
Every tricky question should trigger a follow up quiz. Build that habit with GMAT Club Forum Quiz →.

Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
https://gmatclub.com/forum/data-sufficiency-ds-141/
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 17 Nov 2009, 06:28
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Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.

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

(1) \(y<x\), 3 possible cases for \(|x-y|>|x|-|y|\):

    A. \(0<y<x\):
    ---------------\(0\)---\(y\)---\(x\)---

    In this case \(|x-y|>|x|-|y|\) becomes \(x-y>x-y\), which gives \(0>0\). Which is wrong;

    B. \(y<0<x\):
    ---------\(y\)---\(0\)---------\(x\)---
    In this case \(|x-y|>|x|-|y|\) becomes \(x-y>x+y\), which gives \(y<0\). Which is right, as we consider the range \(y<0<x\);

    C. \(y<x<0\)
    ---\(y\)---\(x\)---\(0\)--------------
    In this case \(|x-y|>|x|-|y|\) becomes \(x-y>-x+y\), which gives \(x>y\). Which is right, as we consider the range \(y<x<0\).

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

    A. \(y<0<x\)
    ----\(y\)-----\(0\)-------\(x\)---
    In this case \(|x-y|>|x|-|y|\) becomes \(x-y>x+y\), which gives \(y<0\). Which is right, as we consider the range \(y<0<x\);

    B. \(x<0<y\)
    ----\(x\)-----\(0\)-------\(y\)---
    In this case \(|x-y|>|x|-|y|\) becomes \(-x+y>-x-y\), which gives \(y>0\). Which is right, as we consider the range \(x<0<y\).

In both cases inequality holds true. Sufficient.

Answer: B.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 14 Apr 2010, 07:38
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raghavs
is |x-y|>|x|-|y|

1>y<x
2>x*y<0
Show more

IMO answer is 'B'

given expression, LHS = |x-y|; RHS = |x| - |y|

1) if y<x
Case I: x<0 => y<0 --> LHS = RHS
Case II: x>0, y>0 but <x --> LHS = RHS
Case III: x>0, y<0 --> LHS > RHS
Hence 1) alone is not sufficient

2) if x*y <0 => either x or y <0 and other has to be >0
Case I: x<0, y>0 --> LHS > RHS
Cae II: x>0, y<0 --> LHS > RHS
No other case.
Hence, 2) alone is sufficient

OA pls
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 12 May 2010, 01:35
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Is |x - y| > |x| - |y|?


(1) \(y<x\)

Try two positive number \(x=3>y=1\) --> is \(|3-1|>|3|-|1|\)? --> is \(2>2\)? Answer NO.

Try ANY other case but both positive: \(x=-5>y=-7\) --> is \(|-5-(-7)|>|-5|-|-7|\)? --> is \(2>-2\)? Answer YES.

Two different answers. Not sufficient.


(2) \(xy<0\), means \(x\) and \(y\) have different signs.

Now we can spot here that when \(x\) and \(y\) have different signs \(x-y\) always "contribute" to each other so that its absolute value will increase: \(x=3\), \(y=-1\) --> \(|x-y|=|3+1|=|4|=4\) or \(x=-3\), \(y=1\) --> \(|x-y|=|-3-1|=|4|=4\).

But \(|x|-|y|\) is difference (thus not "contributing") of two positive values (as neither equals to zero). \(x=3\), \(y=-1\) --> \(|x|-|y|=|3|-|-1|=2\) or \(x=-3\), \(y=1\) --> \(|x|-|y|=|-3|-|1|=2\).

So \(xy<0\) means \(|x-y|>|x|-|y|\) is always true.

Sufficient.


Answer: B.

Hope it helps.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 26 Jun 2010, 19:48
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Bunuel
Basically the question asks whether the distance between the two points x and y on the line is greater than the difference between the individual distances of x and y from 0.

\(|x-y|>|x|-|y|\)?

(1) \(y<x\), 3 possible cases for \(|x-y|>|x|-|y|\):

A. ---------------\(0\)---\(y\)---\(x\)---, \(0<y<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x-y\) --> \(0>0\). Which is wrong;
B. ---------\(y\)---\(0\)---------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
C. ---\(y\)---\(x\)---\(0\)--------------, \(y<x<0\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\). Which is right, as we consider the range \(y<0<x\).

Two different answers. Not sufficient.

(2) \(xy<0\), means \(x\) and \(y\) have different signs, hence 2 cases for \(|x-y|>|x|-|y|\):

A. ----\(y\)-----\(0\)-------\(x\)---, \(y<0<x\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);
B. ----\(x\)-----\(0\)-------\(y\)---, \(x<0<y\) --> in this case \(|x-y|>|x|-|y|\) becomes: \(-x+y>-x-y\) --> \(y>0\). Which is right, as we consider the range \(x<0<y\).

In both cases inequality holds true. Sufficient.

Answer: B.
Show more

Bunuel,
for 1.B when .. y ..0 .. x, you said \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\).
and
1.c when ... y ... x ... 0, you said \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\).

Can you explain this a little bit more? How did you go about removing the absolute signs for this scenarios?

Thanks for your help.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 27 Jun 2010, 05:50
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Bunuel,
for 1.B when .. y ..0 .. x, you said \(|x-y|>|x|-|y|\) becomes: \(x-y>x+y\).
and
1.c when ... y ... x ... 0, you said \(|x-y|>|x|-|y|\) becomes: \(x-y>-x+y\) --> \(x>y\).

Can you explain this a little bit more? How did you go about removing the absolute signs for this scenarios?

Thanks for your help.
Show more


Consider absolute value of some expression - \(|some \ expression|\):
If the expression in absolute value sign (||) is negative or if \(some \ expression<0\) then \(|some \ expression|=-(some \ expression)\);
If the expression in absolute value sign (||) is positive or if \(some \ expression>0\) then \(|some \ expression|=some \ expression\).

(It's the same as for \(|x|\): if \(x<0\), then \(|x|=-x\) and if \(x>0\), then \(|x|=x\))

We have \(|x-y|>|x|-|y|\):

For B: ---------\(y\)---\(0\)---------\(x\)---, \(y<0<x\) (\(x>y\)) --> so as \(x-y>0\), then \(|x-y|=x-y\). Also as \(x>0\), then \(|x|=x\) and as \(y<0\), then \(|y|=-y\). So in this case \(|x-y|>|x|-|y|\) becomes: \(x-y>x-(-y)\) or \(x-y>x+y\) --> \(2y<0\) --> \(y<0\). Which is right, as we consider the range \(y<0<x\);

The same for C.

Hope it's clear.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 25 Nov 2011, 15:00
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Answer is (B).

Here's how I did it.

|x-y| has a range of possible values , min being |x|-|y| and max being |x|+|y|

Statement 1 : x>y . Scenarioes :- x=+ve , y=-ve , |x-y|= |x|+|y| ;
x=+ve , y=+ve , |x-y|=|x|-|y| ;
x=-ve , y=-ve , |x-y|= |-(x-y)|=|x|-|y|

So we cannot definitely say that |x-y| is greater than |x|-|y| because the min value for |x-y| is also |x|-|y|. So, statement 1 is not sufficient.

Statement 2 : xy<0 . Scenarios :- x=+ve , y=-ve , |x-y|= |x|+|y|;
x=-ve , y=+ve , |x-y| = |-(|x|+|y|)|=|x|+|y|.

Now we know |x| + |y| is definitely greater than |x|-|y|. So statement (2) is sufficient.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 30 Nov 2011, 01:16
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Funny, but I remember form university that |a-b|>||a|-|b||>|a|-|b|, therefore the above inequality is valid for all numbers a,b can somebody verify the inequality?

https://math.ucsd.edu/~wgarner/math4c/de ... nequal.htm
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askou
Funny, but I remember form university that |a-b|>||a|-|b||>|a|-|b|, therefore the above inequality is valid for all numbers a,b can somebody verify the inequality?

https://math.ucsd.edu/~wgarner/math4c/de ... nequal.htm
Show more

If you notice, you have missed the 'equal to' sign.
Generalizing,\(|a-b|\geq|a|-|b|\)

In some cases, the equality will hold.
e.g. a = 3, b = 2
You get 1 = 1

In others, the inequality will hold.
e.g. a = -3, b = 4
7 > -1

In this question, you have to figure out whether the inequality will hold.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 14 Jan 2012, 12:56
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mmphf
Is |x-y|>|x|-|y| ?

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

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 14 Jan 2012, 13:02
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This is a very simple question. !x-y!>|x|-|y| can only happen if both the numbers are of different signs.

If xy<0 then these numbers are of opposite signs. Hope this clears.


X=2 y=3 then |x-y|=|x|-|y|
if x=-2 and y = 3 then |x-y|>|x|-|y|
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someone79
This is a very simple question. !x-y!>|x|-|y| can only happen if both the numbers are of different signs.

If xy<0 then these numbers are of opposite signs. Hope this clears.


X=2 y=3 then |x-y|=|x|-|y|
if x=-2 and y = 3 then |x-y|>|x|-|y|
Show more

Red part is not correct \(|x-y|>{|x|-|y|}\) also holds true when \(x\) and \(y\) have the same sign and the magnitude of \(y\) is more than that of \(x\) (so for \(|y|>|x|\)). Example:
\(x=2\) and \(y=3\) --> \(|x-y|=1>-1={|x|-|y|}\);
\(x=-2\) and \(y=-3\) --> \(|x-y|=1>-1={|x|-|y|}\).

Actually the only case when \(|x-y|>{|x|-|y|}\) does not hold true is when \(xy>{0}\) (so when \(x\) and \(y\) have the same sign) and \(|x|>|y|\) (simultaneously). In this case \(|x-y|={|x|-|y|}\) (as shown in my previous post). Example:
\(x=3\) and \(y=2\) --> \(|x-y|=1={|x|-|y|}\);
\(x=-3\) and \(y=-2\) --> \(|x-y|=1={|x|-|y|}\).

Hope it's clear.
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study
Is |x - y| > |x| - |y|?

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


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That's just asking: is the distance between x and y, plus the distance from y to zero, greater than the distance from x to zero. If we have 0 < y < x or x < y < 0, these distances will be equal- the inequality will be false. If y is further from zero than x, or if y and x are on opposite sides of zero, the inequality will be true. This is easy to see if you draw a number line.

From 1), it is possible y and x are both to the right of zero, and the inequality is false, or that y and x are on the opposite sides of zero, and the inequality is true (among other possibilities): insufficient.

From 2), we know y and x are on opposite sides of zero (they have opposite signs): sufficient.
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I found talking through this one to be helpful.

Namely:

|x-y| represents the distance between x and y on the number line.

|x|-|y|, on the other hand, first takes the absolute value of both numbers - and thereby moving them both to the positive side of the number line - and THEN calculates the difference between x and y

Visually, it makes sense that if x and y are of different signs (for example, x=-5, y=5), then the difference between the two numbers on a number line is greater if measured before moving them both to the positive side of the number line.

At this point I logically deduced that it is impossible for |x-y| to be less than |x|-|y|. I also deduced at this point that if x and y have the same sign, it does not matter when the absolute value is taken because the difference between them will be the same either way.

After this thought process, the problem becomes much easier.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 25 Mar 2013, 03:08
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mmphf
Is |x-y|>|x|-|y| ?

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


How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 25 Mar 2013, 05:29
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mmphf
Is |x-y|>|x|-|y| ?

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


How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6
Show more

It is B because if you use the data of statement 2, you can say, "Yes, |x-y| is greater than |x|-|y|"

(2) xy < 0
This means that one of x and y is positive and the other is negative. You cannot take x = 1/2 and y = 1/3.
It is not about fractions/integers. It is about positive/negative numbers (most mod questions are about positive/negative numbers)
When xy < 0, |x-y|>|x|-|y| always holds.
Only when x and y both are positive or both are negative and |x|>|y|, then |x-y|=|x|-|y|
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mmphf
Is |x-y|>|x|-|y| ?

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


How it is B? Did they mention that X and Y are integers? No right, the answer should be E. If they provide details about X and Y as integers then it will be B otherwise it will be E.

can anyone help me about the scenario whether we consider fractions or not in this case?

Scenario:

x=1/2, y=1/3 ==> |1/2-1/3|=1/6 and |1/2|-|1/3|=1/6
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Fractions and integers have no role to play here. Check Bunuel's post above.

Whenever xy < 0, i.e. x is negative but y is positive OR x is positive but y is negative, |x-y| is greater than |x|-|y|.

e.g. x = -1/2, y = 1/3
|x-y| = |-1/2-1/3| = 5/6
|x|-|y| = 1/2 - 1/3 = 1/6

So |x - y| > |x|-|y|

Do you see the logic here? If one of x and y is positive and the other is negative, in |x - y|, absolute values of x and y get added and the sum is positive. While in |x|-|y|, the absolute values are subtracted.
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 02 Jan 2014, 23:36
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Is |x - y| > |x| - |y|?

(1) y < x
(2) xy < 0
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Statement I is insufficient

x....y.....(|x|-|y|)....|x-y|.....YES/NO
3....2.......1..............1........NO
3..-3........0..............6........YES

Statement II is sufficient
xy<0 means that either one of the numbers is negative and other is positive

|x-y| will always result to be a greater value as negative - positive or positive - negative will result in addition of the numbers and |x| - |y| will always result in subtraction.

Hence the answer is B
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 16 Nov 2014, 22:22
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Is |x-y|>|x|-|y| ?

(1) y < x
(2) xy < 0
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Responding to a pm:
You can solve this question easily if you understand some basic properties of absolute values. They are discussed in detail here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/02 ... -the-gmat/

One of the properties is

(II) For all real x and y, \(|x - y| \geq |x| - |y|\)
\(|x - y| = |x| - |y|\) when (1) x and y have the same sign and x has greater (or equal) absolute value than y (2) y is 0
\(|x - y| > |x| - |y|\) in all other cases

Question: Is |x-y|>|x|-|y|?

We need to establish whether the "equal to" sign can hold or not.

(1) y < x
Doesn't tell us whether they have the same sign or opposite. So we don't know whether the equal to sign will hold or greater than. Not sufficient.

(2) xy < 0
Tells us that one of x and y is positive and the other is negative (they do not have same sign). Also tells us that neither x nor y is 0. Hence, the "equal to" sign cannot hold. Sufficient to answer 'Yes'

Answer (B)
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 19 Oct 2015, 23:55
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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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