Data Sufficiency

[Math Revolution GMAT math practice question]

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

1) x<0
2) y>0

Max@Math Revolution 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.

The correct answer is C.

Max@Math Revolution 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.

=>

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