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Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|

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Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post Updated on: 05 Feb 2019, 04:35
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Is |x + y| < |x| + |y|?


(1) |x| ≠ |y|

(2) |x – y| > |x + y|


Appreciate views on how to solve this. Thank you.
I did this by picking numbers choosing various numbers positive, negative, fractions, 0.
Took make almost four minutes. Any other way?

Originally posted by BabySmurf on 23 Jan 2014, 13:56.
Last edited by Bunuel on 05 Feb 2019, 04:35, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post 23 Jan 2014, 22:02
7
14
BabySmurf wrote:
Is |x + y| < |x| + |y|?

(1) | x | ≠ | y |
(2) | x – y | > | x + y |





Appreciate views on how to solve this. Thank you.
I did this by picking numbers choosing various numbers positive, negative, fractions, 0.
Took make almost four minutes. Any other way?


This question can be done very easily if you are familiar with the properties of absolute value.
For all real x and y, \(|x + y| \leq |x| + |y|\)

\(|x + y| = |x| + |y|\) when x and y have the same sign (e.g. x = 4, y = 6 OR x = -2, y = -3) or at least one of them is 0.
When x and y have opposite signs, \(|x + y| < |x| + |y|\)

So the question asks us whether x and y have opposite signs or not.

(1) | x | ≠ | y |
Doesn't tell us anything about the signs of x and y. Not sufficient.

(2) | x – y | > | x + y |
When will | x – y | be greater than | x + y |? Only when x and y have opposite signs e.g. | x – y | = |3 - (-2)| = 5 but | x + y | = |3 - 2| = 1
If x and y have the same sign or one of them is 0, | x + y | will be greater or equal to | x – y |.
Hence this statement tells us that x and y have opposite signs. So it is sufficient alone.

Answer (B)
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post 23 Jan 2014, 14:19
5
BabySmurf wrote:
Is |x + y| < |x| + |y|?

(1) | x | ≠ | y |
(2) | x – y | > | x + y |





Appreciate views on how to solve this. Thank you.
I did this by picking numbers choosing various numbers positive, negative, fractions, 0.
Took make almost four minutes. Any other way?


The answer is B.
for |x+y| <|x|+|y|
we have four cases
1.x +ve ,y+ve
in that case
|x+y| =|x|+|y|
2.x-ve ,y -ve
in that case
|x+y| =|x|+|y|
3.x+ve and y -ve
in that case
|x+y| <|x|+|y|
4.x -ve and Y +ve
in that case
|x+y| <|x|+|y|
and if x=0 or y =0 (i.e. any one equal to zero)
|x+y| =|x|+|y|

so if x and y are of opposite signs then only |x + y| < |x| + |y|
now 1 gives
|x|=|y|
with this we cant answer the question as we donot know signs of x and y
2.| x – y | > | x + y |
this implies that x and y are of opposite signs. So with this we can answer the question.
Key point here is you should understand immediately after looking at the inequality that its valid only if x and y are of opposite signs.
Give me kudos if this helps
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post 23 Jan 2014, 14:42
cherukuri1011 wrote:
BabySmurf wrote:
Is |x + y| < |x| + |y|?

(1) | x | ≠ | y |
(2) | x – y | > | x + y |





Appreciate views on how to solve this. Thank you.
I did this by picking numbers choosing various numbers positive, negative, fractions, 0.
Took make almost four minutes. Any other way?


The answer is B.
for |x+y| <|x|+|y|
we have four cases
1.x +ve ,y+ve
in that case
|x+y| =|x|+|y|
2.x-ve ,y -ve
[....]


I am unsure with "ve" is... Thanks.
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post 23 Jan 2014, 15:00
BabySmurf wrote:
cherukuri1011 wrote:
BabySmurf wrote:
Is |x + y| < |x| + |y|?

(1) | x | ≠ | y |
(2) | x – y | > | x + y |





Appreciate views on how to solve this. Thank you.
I did this by picking numbers choosing various numbers positive, negative, fractions, 0.
Took make almost four minutes. Any other way?


The answer is B.
for |x+y| <|x|+|y|
we have four cases
1.x +ve ,y+ve
in that case
|x+y| =|x|+|y|
2.x-ve ,y -ve
[....]


I am unsure with "ve" is... Thanks.

Hi BabySmurf,
I used a shorter expression X +ve means x is positive number and x -ve means if x is a negative number
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post 24 Jan 2014, 00:53
1
2
BabySmurf wrote:
Is |x + y| < |x| + |y|?

(1) | x | ≠ | y |
(2) | x – y | > | x + y |





Appreciate views on how to solve this. Thank you.
I did this by picking numbers choosing various numbers positive, negative, fractions, 0.
Took make almost four minutes. Any other way?


Similar questions to practice:
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is-x-y-x-y-137050.html
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Hope this helps.
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post 14 Apr 2016, 09:07
Bunuel wrote:
BabySmurf wrote:
Is |x + y| < |x| + |y|?

(1) | x | ≠ | y |
(2) | x – y | > | x + y |





Appreciate views on how to solve this. Thank you.
I did this by picking numbers choosing various numbers positive, negative, fractions, 0.
Took make almost four minutes. Any other way?


Similar questions to practice:
is-x-y-x-y-87926.html
is-x-y-x-y-1-x-y-2-x-y-x-101660.html
is-x-y-x-y-123108.html
is-x-y-x-y-160531.html
is-x-y-x-z-1-y-z-2-x-86132.html
is-a-b-a-b-105457.html
is-x-y-x-y-1-x-y-2-x-y-132654.html
is-x-y-146991.html
is-x-y-x-y-137050.html
is-x-z-y-x-z-y-154245.html

Hope this helps.


If I square both sides, I am going to get xy<|x||y|. That would give B as an answer.

I have found squaring methods to help in inequalities (though not in every case)...
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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New post 13 Sep 2016, 07:05
BabySmurf wrote:
Is |x + y| < |x| + |y|?

(1) | x | ≠ | y |
(2) | x – y | > | x + y |




i picked B.

1. x has a different absolute value than y.
1st case: x=2, b=3 => |x+y|=|x|+|y| no
2nd case: x=2, b=-3 -> |x+y|<|x|+|y|

2. it must be true that at least one of the numbers is negative, in which case |x + y| < |x| + |y| is always true
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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|  [#permalink]

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Re: Is |x + y| < |x| + |y|? (1) |x| ≠ |y| (2) |x – y| > |x + y|   [#permalink] 05 Feb 2019, 04:33
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