If |2x|>|3y|, is x >y? 1) x>0 2) y>0

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by GMATGuruNY » Tue Jun 18, 2019 3:17 am

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Max@Math Revolution wrote:If |2x|>|3y|, is x >y?

1) x>0
2) y>0
Statement 1:
Test x=3.
Plugging x=3 into |2x|>|3y|, we get:
|2*3|>|3y|
6 > 3|y|
|y| < 2
-2 < y < 2
Since x=3 and -2 < y < 2, we get:
x > y.

Test x=1/2.
Plugging x=1/2 into |2x|>|3y|, we get:
|2 * 1/2|>|3y|
1 > 3|y|
|y| < 1/3
-1/3 < y < 1/3
Since x=1/2 and -1/3 < y < 1/3, we get:
x > y.

Whether x is close to 0 or farther to the right of 0, x > y.
Implication:
Regardless of the value of x, the answer to the question stem is YES.
SUFFICIENT.

Statement 2:
Test y=2.
Plugging y=2 into |2x|>|3y|, we get:
|2x|>|3*2|
2|x| > 6
|x| > 3
Case 1: x=4, with the result that x > y.
Case 2: x=-4, with the result that x < y.
Since the answer is YES in Case 1 but NO in Case 2, INSUFFICIENT.

The correct answer is A.
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by Max@Math Revolution » Wed Jun 19, 2019 11:52 pm

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=>

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. We should simplify the conditions, if necessary.

Condition 1)
Since x > 0, 3x > 2x ≥ |2x|>|3y| ≥ 3y. It follows that x > y.
Thus, condition 1) is sufficient.

Condition 2)
If x = 3 and y = 1, then x > y, and the answer is "yes".
If x = -3 and y = 1, then x < y, and the answer is "no".
Thus, condition 2) is not sufficient since it doesn't yield a unique answer.

Therefore, A is the answer.
Answer: A