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

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Is |x+y|<|x|+|y|?  [#permalink]

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New post 27 Sep 2018, 23:39
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Difficulty:

  15% (low)

Question Stats:

79% (00:55) correct 21% (01:03) wrong based on 91 sessions

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[Math Revolution GMAT math practice question]

Is \(|x+y|<|x|+|y|?\)

\(1) x<0\)
\(2) y>0\)

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Re: Is |x+y|<|x|+|y|?  [#permalink]

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New post 28 Sep 2018, 02:58
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

Is \(|x+y|<|x|+|y|?\)

\(1) x<0\)
\(2) y>0\)


Since an absolute value cannot be negative, both sides of the question stem are NONNEGATIVE, enabling us to safely square the inequality:
(|x+y|)²< (|x|+|y|)²
x² + y² + 2xy < x² + y² + 2|x||y|
xy < |xy|
The resulting inequality is valid only if x and y have DIFFERENT SIGNS.
Question stem, rephrased: Do x and y have different signs?

Clearly, neither statement alone is sufficient.
Statements combined: x < 0 < y
Thus, the answer to the rephrased question stem is YES.
SUFFICIENT.


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Re: Is |x+y|<|x|+|y|?  [#permalink]

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New post 28 Sep 2018, 03:58
Hi,

Already GMATguruNY has explained the perfect mathematical approach.

Let me try explaining the numbers approach.

It’s a straight forward modulus application question.

Question: |x+y| < |x|+|y| ?

Lets try x and y with different combination of numbers.

x and y both positive , x = 1 and y = 2 then we get, 3 < 3. Not possible.

x and y both negative, x = -1 and y = -2 then we get, 3 < 3. Not possible.

x positive and y negative, x = 1 and y = -2 then we get, 1 < 3. Works.

x negative and y positive, x = -1 and y = 2 then we get, 1 < 3. Works.

So x and y should have alternate signs, to get the YES answer.

Clearly individual statements are not sufficient.

Statement I is insufficient:

x < 0

We know x < 0, but we don’t know whether y > 0 or not.

So insufficient.

Statement I is insufficient:

y > 0

We know y > 0, but we don’t know whether x < 0 or not.

So insufficient.

Together we get, x < 0 and y > 0.

Alternate signs.

So sufficient.

Answer is C.
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Re: Is |x+y|<|x|+|y|?  [#permalink]

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New post 28 Sep 2018, 07:33
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Is \(|x+y|<|x|+|y|?\)

\(1) x<0\)
\(2) y>0\)

CONCEPTUAL (THEORETICAL) TOOL:

\(\left| {x + y} \right| \le \left| x \right| + \left| y \right|\)

is always valid and it is famous in Mathematics. It is the so-called first triangle inequality.

It is useful in the GMAT to know it and, also, to know that EQUALITY is valid if, and only if, we don´t have "opposing forces" (xy<0 does not occur):

\(\left| {x + y} \right| = \left| x \right| + \left| y \right|\,\,\,\,\,\, \Leftrightarrow \,\,\,\,xy \ge 0\)

(Try some values for x and y to feel comfortable about it. In less than 3min you will realize the result is pretty evident!)


SOLUTION: (Now absolutely trivial!)

\(\left| {x + y} \right|\,\,\mathop < \limits^? \,\,\,\left| x \right| + \left| y \right|\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\boxed{xy\,\,\mathop < \limits^? \,\,0}\)

\(\left( 1 \right)\,\,\,x < 0\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( 2 \right)\,\,\,y > 0\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,\,\left\{ \matrix{
x < 0 \hfill \cr
y > 0 \hfill \cr} \right.\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,xy < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Is |x+y|<|x|+|y|?  [#permalink]

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New post 30 Sep 2018, 21:54
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

The question \(|x+y|<|x|+|y|\) is equivalent to \(xy<0\) as shown below

\(|x+y|<|x|+|y|\)
\(=> |x+y|^2<(|x|+|y|)^2\)
\(=> |x+y|^2-(|x|+|y|)^2< 0\)
\(=> (x+y)^2-(|x|+|y|)^2< 0\)
\(=> x^2+2xy+y^2-(|x|^2 +2|x||y|+|y|^2) < 0\)
\(=> x^2+2xy+y^2-(x^2 +2|xy|+y^2) < 0\)
\(=> 2xy-2|xy| < 0\)
\(=> xy-|xy| < 0\)
\(=> xy < |xy|\)
\(=> xy < 0\)

This occurs if \(x < 0\) and \(y > 0\). Thus, both conditions together are sufficient.

Therefore, C is the answer.
Answer: C
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Re: Is |x+y|<|x|+|y|? &nbs [#permalink] 30 Sep 2018, 21:54
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