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# Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7349
GMAT 1: 760 Q51 V42
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Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y  [#permalink]

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10 Apr 2018, 01:36
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Difficulty:

55% (hard)

Question Stats:

60% (01:34) correct 40% (01:53) wrong based on 72 sessions

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[GMAT math practice question]

Is $$x>0$$?

$$1) (x+y)^2 > (x-y)^2$$
$$2) x+y > x-y$$

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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 $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 14 Feb 2018 Posts: 390 Re: Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y [#permalink] ### Show Tags 10 Apr 2018, 04:37 1 According to statement 1, reduce the given equation to get 4xy > 0 Now, this could be possible only when x and y both are positive or x and y both are negative. So, we cannot say whether x is greater than zero. Insufficient. Statement 2 also gives 2y > 0 or y > 0. Insufficient again. Combining both, we know that y >0, there for 4xy > 0 to hold true x >0 has to be true. Hence, the answer is a clear Yes. Thus, C...both are together sufficient. Sent from my Lenovo K53a48 using GMAT Club Forum mobile app Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7349 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y [#permalink] ### Show Tags 12 Apr 2018, 01:55 => 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. Since we have 2 variables (x and y) and 0 equations, C is most likely to be the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first. Conditions 1) & 2) Condition 1) tells us that $$(x+y)^2 > (x-y)^2$$ $$=> x^2 + 2xy + y^2 > x^2 - 2xy + y^2$$ $$=> 2xy > -2xy$$ $$=> 4xy > 0$$ $$=> xy > 0$$ Condition 2) tells us that $$x + y > x –$$y $$=> y > -y$$ $$=> 2y > 0$$ $$=> y > 0.$$ Since $$xy > 0$$ and $$y > 0$$, we have $$x > 0$$. Thus, both conditions 1) & 2) together are sufficient. In general, there are many questions involving integers and statistics to which we need to apply CMT(Common Mistake Type) 4. Condition 1): If $$x = -2$$ and $$y = -1$$, then the answer is “yes”. If $$x = 1$$ and $$y = 2$$, then the answer is “no”. Thus, condition 1) is not sufficient. Condition 2): If $$x = -3$$ and $$y = -1$$, then the answer is “yes”. If $$x = -1$$ and $$y = -3$$, then the answer is “no”. Thus condition 2) is not sufficient. Therefore, C is the answer. Answer: C Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E. _________________ 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$149 for 3 month Online Course"
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Re: Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y  [#permalink]

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12 Apr 2018, 09:43
Top Contributor
MathRevolution wrote:
[GMAT math practice question]

Is $$x>0$$?

$$1) (x+y)^2 > (x-y)^2$$
$$2) x+y > x-y$$

Target question: Is x POSITIVE?

Statement 1: (x + y )² > (x - y)²
Take: (x + y )² > (x - y)²
Expand and simplify both sides to get: x² + 2xy + y ² > x² - 2xy + y ²
Subtract x² from both sides: 2xy + y ² > -2xy + y ²
Subtract y² from both sides: 2xy > -2xy
Add 2xy both sides: 4xy > 0
Divide both sides by 4 to get: xy > 0
If the product xy> 0, then there are two possibilities:
Case a: x is positive and y is positive. In this case, the answer to the target question is YES, x is positive
Case b: x is negative and y is negative. In this case, the answer to the target question is NO, x is not positive
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x + y > x - y
Take: x + y > x - y
Add y to both sides: x + 2y > x
Subtract x from both sides: 2y > 0
Divide both sides by 2 to get: y > 0
So, we know that y is positive, but we have no information about x
So, we cannot answer the target question with certainty.
Statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that there are two possible cases (case a: x and y are both positive OR case b: x and y are both negative)
Statement 2 tells us that y is positive
So, statement 2 ELIMINATES case b, which means x and y are both positive
The answer to the target question is YES, x is positive
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

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Re: Is x > 0 ? (1) (x + y)^2 > (x - y)^2 (2) x + y > x - y   [#permalink] 12 Apr 2018, 09:43
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