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# Is |x+y|<|x|+|y|? 1) x+y<0 2) x<y

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7372
GMAT 1: 760 Q51 V42
GPA: 3.82
Is |x+y|<|x|+|y|? 1) x+y<0 2) x<y  [#permalink]

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30 Jun 2017, 01:01
00:00

Difficulty:

55% (hard)

Question Stats:

57% (01:42) correct 43% (01:17) wrong based on 63 sessions

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Is |x+y|<|x|+|y|?

1) x+y<0
2) x<y

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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7372 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is |x+y|<|x|+|y|? 1) x+y<0 2) x<y [#permalink] ### Show Tags 02 Jul 2017, 18:11 1 ==> If you modify the original condition and the question, for |x+y|<|x|+|y|?, you can square both sides, and if you expand from $$(|x+y|)^2<(|x|+|y|)^2?$$ Or, $$(x+y)^2<(|x|+|y|)^2?$$, you get $$x^2+y^2+2xy<|x|^2+|y|^2+2|xy|?$$, or $$x^2+y^2+2xy<x^2+y^2+2|xy|?, 2xy<2|xy|?, xy<|xy|?$$. However, in con 1) and con 2), xy<0 is not mentioned, so the answer is E. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Joined: 30 May 2013
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Re: Is |x+y|<|x|+|y|? 1) x+y<0 2) x [#permalink]

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30 Jun 2017, 03:25
IMO (E)

For a modulus equation, as we know if x&y both have same sign, then |x+y|= |x|+|y|. We can verify this by plugging nos.
Statement (1)x+y <0
2 cases are possible.
Case(i) Both x and y are negative hence |x+y| = |x|+|y|. i.e. |-2 + -1| = |-2| + |-1|
Hence question statement is false.
Case (ii) even-odd combination of x & y
|-2 + 1| < |-2| + |1|
Hence question statement is True.
Using both the cases, we can say that Statement (1) is not sufficient.

(2) x < y
This will be true for 2 types of cases
Case (1) when x& y both have same sign then |x+y| = |x| + |y|.
Hence questtion statement is false.
Case (2) when x is negative & y is positive. This is same as case (ii) of statement (1).
Hence question Statement is true.
Using both the cases, we say that statement (2) is not sufficient.

Combination of both the statements
Is not giving any additional information.

Sent from my GT-N7100 using GMAT Club Forum mobile app
Re: Is |x+y|<|x|+|y|? 1) x+y<0 2) x#permalink] 30 Jun 2017, 03:25
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