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

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Math Revolution GMAT Instructor
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Is |x+y|<|x|+|y|? 1) x+y<0 2) x<y  [#permalink]

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New post 30 Jun 2017, 01:01
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

58% (01:47) correct 42% (01:22) wrong based on 61 sessions

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

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

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Re: Is |x+y|<|x|+|y|? 1) x+y<0 2) x [#permalink]

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New post 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.
Hence answer is (E)

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Math Revolution GMAT Instructor
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Re: Is |x+y|<|x|+|y|? 1) x+y<0 2) x<y  [#permalink]

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New post 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
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Re: Is |x+y|<|x|+|y|? 1) x+y<0 2) x<y   [#permalink] 02 Jul 2017, 18:11
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