Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x/y < 0?

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

$${x \over y}\,\,\mathop < \limits^? \,\,0\,\,\,\,\left( {y \ne 0} \right)\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,xy\,\,\mathop < \limits^? \,\,0\,\,\,\,,\,\,\,{\rm{with}}\,\,y \ne 0$$
$$\left( 1 \right)\,\,\,\,\left| {x + y} \right|\,\,\, < \,\,\,\left| x \right| + \left| y \right|\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,xy < 0\,\,\,\,\,\,\,\left( {{\rm{SUFF}}.} \right)$$
$$\left( 2 \right)\,\,\,\,x + y < 0\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {0, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 2,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.\,\,\,\,\,\,\left( {{\rm{INSUFF}}.} \right)$$


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

Regards,
Fabio.

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x/y < 0?

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

Question stem, rephrased:
Do x and y have different signs?

Statement 1: |x+y| < |x|+|y|
Since an absolute value cannot be negative, both sides here 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.
SUFFICIENT.

Statement 2: x+y < 0
If x=-1 and y=-1, then x and y have the same sign.
If x=-10 and y=1, then x and y have different signs.
INSUFFICIENT.

The correct answer is A.

=>

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 is equivalent to asking if xy < 0. This can be seen by multiplying both sides of the inequality by y^2.

Condition 1) is equivalent to xy < 0 as shown below:
|x+y|<|x|+|y|
=> |x+y|^2 < (|x|+|y|)^2
=> (x+y)^2 < |x|2+2|x||y|+|y|^2
=> x^2+2xy+y^2 < x^2+2|xy|+y^2
=> 2xy < 2|xy|
=> xy < |xy|
=> xy < 0
Thus, condition 1) is sufficient.

Condition 2)
If x = -2 and y = 1, then the answer is 'yes'.
If x = -1 and y = -1, then the answer is 'no'.
Since it does not give a unique answer, condition 2) is not sufficient.

Therefore, the correct answer is A.
Answer: A