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# M60-23

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 8235
GMAT 1: 760 Q51 V42
GPA: 3.82

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11 Jun 2018, 06:56
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Question Stats:

75% (01:01) correct 25% (01:43) wrong based on 4 sessions

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If $$x$$ and $$y$$ are integers, is $$4x^2-y^2$$ an odd number?

1) $$x$$ is an odd number

2) $$y$$ is an odd number

_________________
MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8235 GMAT 1: 760 Q51 V42 GPA: 3.82 Re M60-23 [#permalink] ### Show Tags 11 Jun 2018, 06:56 Official Solution: 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, and then recheck the question. Modifying the question: In order for $$4x^2 - y^2$$ to be odd, $$y^2$$ must be odd since $$4x^2$$ is always even. This is equivalent to $$y$$ being odd. So, the question asks if y is odd. Thus, the answer is B. Condition 1) If y is an odd number, then both $$2x + y$$ and $$2x - y$$ are odd numbers and $$(2x+y)(2x-y)$$ is an odd number. If y is an even number, both $$2x + y$$ and $$2x - y$$ are even numbers and $$(2x+y)(2x-y)$$ is an even number. Since the question does not have a unique answer, condition 1) is not sufficient. Therefore, the answer is B. Answer: B _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
"Free Resources-30 day online access & Diagnostic Test"
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Joined: 20 Apr 2019
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18 May 2019, 06:33
Suppose if x = 0 ? Then just by stating y is odd will not fetch anything meaningful.
Hence x is odd is required, just to make sure x is not 0.
Re: M60-23   [#permalink] 18 May 2019, 06:33
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# M60-23

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