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Bunuel
If xyz ≠ 0, is x(y + z) = 0?


(1) |y + z| = |y| + |z|

(2) |x + y| = |x| + |y|



Are You Up For the Challenge: 700 Level Questions

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Visit https://www.mathrevolution.com/gmat/lesson for details.

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

Since xyz ≠ 0, we have x ≠ 0 and the question ask if y + z = 0 or not.

Condition 1)
Remind that |y+z| = |y|+|z| is equivalent to yz ≥ 0.
Since xyz ≠ 0, we have yz ≠ 0 and yz > 0.
Thus x(y+z) can not be zero since both y and z are positive or negative.
Since condition 1) yields a unique solution "No", it is sufficient.

Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, both conditions are sufficient, when used together.

Condition 2)

If x = 1, y = 1, z = -1, then x(y+z) = 0 and the answer is "yes".
If x = 1, y = 1, z = 1, then x(y+z) = 2 one the answer is "no".
Since condition 2) does not yield a unique solution, it is not sufficient.

Therefore, A is the answer.
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Bunuel
If xyz ≠ 0, is x(y + z) = 0?

(1) |y + z| = |y| + |z|

(2) |x + y| = |x| + |y|

xyz ≠ 0

We need to answer the question:

Is x(y + z) = 0 ?

Since x ≠ 0, we can rephrase the question into:

Is y + z = 0 ?

Note: To evaluate the statements, we’ll use the following rule:
If |a + b| = |a| + |b|, then ab ≥ 0.

Statement One Alone:

=> |y + z| = |y| + |z|

Since xyz ≠ 0, yz > 0. y and z must be both positive or both negative, so their sum cannot be equal to zero. The answer to the rephrased question is a definite No.

Statement one is sufficient. Eliminate answer choices B, C, and E.

Statement Two Alone:

=> |x + y| = |x| + |y|

Since xyz ≠ 0, xy > 0. So, x and y are both positive or both negative.

If x = 1, y = 1, z =1, then the answer is No. Whereas, if x = -1, y = -1, z = 1, then the answer is Yes. Statement two is not sufficient.

Answer: A
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