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

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


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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 14 Jan 2012, 12:56
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Is |x-y|>|x|-|y| ?

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

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

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 18 Mar 2012, 00:56
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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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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 24 Nov 2019, 04:24
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Is |x-y| > |x| - |y| ?
1) y < x
2) xy < 0
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|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:

|x| - |y| = RED - BLUE.
|x-y| = RED - BLUE.
Thus, |x-y| = |x| - |y|.

Case 2:

|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.

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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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 26 May 2021, 04:57
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Solution:

There are two properties with respect to absolute values

(1)|x+y|≤|x|+|y|, "=" sign holds for xy≥0 ( when x and y have the same sign)

(2)|x−y|≥|x|−|y| "=" sign holds for xy>0(when x and y have the same sign) and |x|>|y| (simultaneously)

St(1):-y < x
We do not know the signs of x and y (Insufficient)

St(2)- xy < 0

=> |x−y|>|x|−|y| and not equal to from property (2) mentioned above. (Sufficient)

(option b)

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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 05 Feb 2020, 05:54
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Is |x - y| > |x| - |y|?

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


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Solution:

Notice that x = (x - y) + y; thus, using the triangle inequality, we have:

|x| = |(x - y) + y| ≤ |x - y| + |y|

|x - y| ≥ |x| - |y|

This tells me that |x - y| can be greater than |x| - |y| or can be equal to |x| - |y|, but it can never be less than |x| - |y|. So, if I look for counter examples, I won’t waste time trying to look for numbers where |x - y| is less than |x| - |y|.

Statement One Alone:

If x > y, then x - y is positive; hence the given question reduces to “Is x - y > |x| - |y|?”. We immediately notice that if x and y are both positive, then |x| - |y| is equal to x - y. In this case, the answer to the question is no. We can, if we want, verify by letting x = 2 and y = 1.

If x and y are not both positive (which is only possible when x is positive but y is negative), then we have: “Is x - y > x - (-y)?” The inequality simplifies to “Is -y > y?”. Since y is negative, the answer to this question is yes. Again we can verify by taking values such as x = 2 and y = -1.

Statement one alone is not sufficient.

Statement Two Alone:

Notice that the given information in statement two tells us that neither x nor y is zero, x and y are not equal and they have opposite signs. All we have to do is to consider the cases when x is positive and x is negative to find the answer.

If x is positive, then y is negative. Thus, the expression x - y is positive, and the question becomes:

Is x - y > x - (-y)?

Is -y > y?

Since y is negative, the answer is yes.

If x is negative, then y is positive. Thus, the expression x - y is negative, and the question becomes:

Is -(x - y) > -x - y?

Is -x + y > -x - y?

Is y > -y?

Since y is positive, the answer is again yes. We see that |x - y| > |x| - |y| is satisfied whenever xy < 0.

Thus, statement two is sufficient to answer the question.

Answer: B
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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 23 Nov 2019, 19:05
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Is |x - y| > |x| - |y|?

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


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Target question: Is |x - y| > |x| - |y|?

Statement 1: y < x
Let's test some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 2 and y = 1. In this case, |x - y| = |2 - 1| = 1 and |x| - |y| = |2| - |1| = 1. So, the answer to the target question is NO, |x - y| is NOT greater than |x| - |y|
Case b: x = 2 and y = -1. In this case, |x - y| = |2 - (-1)| = 3 and |x| - |y| = |2| - |-1| = 1. So, the answer to the target question is YES, |x - y| IS greater than |x| - |y|
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: xy < 0
This tells us that one value (x or y) is positive, and the other value is negative. This sets up two possible cases:

Case a: x is positive and y is negative.
So, we're taking a positive value (x) and subtracting a negative value (y). Doing so yields a positive value that is bigger than x.
In other words, we have: 0 < x < |x - y|

Now let's examine |x| - |y|
Since x is positive, we know that |x| = x
Since y ≠ 0, we know that 0 < |y|
So, |x| - |y| = x - |y| = some number less than x
In other words, |x| - |y| < x

When we combine the inequalities we get: |x| - |y| < x < |x - y|
In this case, the answer to the target question is YES, |x - y| IS greater than |x| - |y|


Case b: x is negative and y is positive.
Here, we're taking a negative value (x) and subtracting a positive value (y). Doing so yields a negative value that is less than x.
In other words, we have: x - y < x < 0 < |x|
Important: since the MAGNITUDE of x - y is greater than the MAGNITUDE of x, we can write: |x| < |x - y|

Now let's examine |x| - |y|
Since x ≠ 0, we know that 0 < |x|
Since |x| is a positive number, we know that subtracting |y| (another positive value) will yield a number that is LESS THAN |x|
In other word, |x| - |y| < |x|

When we combine the inequalities we get: |x| - |y| < |x| < |x - y|
In this case, the answer to the target question is YES, |x - y| IS greater than |x| - |y|

In both possible cases, the answer to the target question is the same: YES, |x - y| IS greater than |x| - |y|
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: B

Cheers,
Brent
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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 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.
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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.
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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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askou
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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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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 30 Nov 2011, 03:57
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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, 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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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 14 Jan 2012, 13:18
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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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Is |x - y| > |x| - |y|? (1) y < x (2) xy < 0 31 Mar 2012, 15:00
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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
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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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