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Is ab < 0?

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by Max@Math Revolution » Fri Jan 18, 2019 3:53 am

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[Math Revolution GMAT math practice question]

Is ab < 0?

1) |a+b| = - ( a + b )
2) |a+b| + 1 = |a| + |b|
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Source: — Data Sufficiency |
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by fskilnik@GMATH » Fri Jan 18, 2019 1:54 pm
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is ab < 0?

1) |a+b| = - ( a + b )
2) |a+b| + 1 = |a| + |b|
Excellent problem, Max. Congrats!
$$ab\mathop {\,\, < }\limits^? \,\,0$$

$$\left( 1 \right)\,\,\,\left| {a + b} \right| = - \left( {a + b} \right)\,\,\,\, \Leftrightarrow \,\,\,\,\,a + b \le 0$$
$$\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {a,b} \right) = \left( { - 2, 1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,\left| {a + b} \right| + 1 = \left| a \right| + \left| b \right|\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,ab < 0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle $$
$$\left( * \right)\,\,ab \ge 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| {a + b} \right| = \left| a \right| + \left| b \right|\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| {a + b} \right| + 1 \ne \left| a \right| + \left| b \right|\,\,\,\,,\,\,\,{\rm{impossible}}$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Sun Jan 20, 2019 5:25 pm
=>

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.

You should remember that the inequality |x+y| < |x| + |y| is equivalent to the inequality xy < 0.

Condition 2) tells us that |a+b| + 1 = |a| + |b|. Thus, |a + b| < |a| + |b| and ab < 0.
Thus, condition 2) is sufficient.

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

Therefore, the answer is B.
Answer: B
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