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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?

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

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 [....]

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

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.

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?

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?

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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