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If x≠y, (x+y)/(x-y)=?

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4332

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27 Sep 2017, 01:38
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Difficulty:

55% (hard)

Question Stats:

67% (01:37) correct 33% (01:40) wrong based on 45 sessions

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[GMAT math practice question]

If x≠y, $$\frac{(x+y)}{(x-y)}$$=?

1) $$x^2-4xy=-4y^2$$
2) $$(x^2+1)(x-2y)=0$$
[Reveal] Spoiler: OA

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Joined: 25 Feb 2013
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Schools: Mannheim"19 (S)
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27 Sep 2017, 01:48
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MathRevolution wrote:
[GMAT math practice question]

If x≠y, $$\frac{(x+y)}{(x-y)}$$=?

1) $$x^2-4xy=-4y^2$$
2) $$(x^2+1)(x-2y)=0$$

$$\frac{(x+y)}{(x-y)}$$ to know the value we need a relationship between $$x$$ & $$y$$

Statement 1: implies $$x^2-4xy+4y^2=0$$ or $$(x-2y)^2=0$$

therefore $$x=2y$$. Substitute the value of $$x$$ in question stem equation we will get $$3$$. Hence sufficient

Statement 2:$$(x^2+1)(x-2y)=0$$

$$x^2+1$$ is a sum of two positive number and hence cannot be equal to $$0$$

so $$x-2y=0$$ or $$x=2y$$ (Same as statement 1). Hence sufficient

Option D

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4332

Kudos [?]: 3046 [0], given: 0

GPA: 3.82

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29 Sep 2017, 01:22
=>

Forget conventional ways of solving math questions. In DS, VA (Variable Approach) method is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Using VA (Variable Approach) method, since we have 2 variables and 0 equation, C could be the answer most likely.

Condition 1)
x^2-4xy=-4y^2
x^2-4xy+4y^2= 0
(x-2y)^2 = 0
x – 2y = 0
x = 2y

( x + y ) / ( x – y ) = ( 2y + y ) / ( 2y – y ) = 3y / y = 3

This is sufficient.

Condition 2)

(x^2+1)(x-2y)=0
x^2 + 1 = 0 or x – 2y = 0.
We can take x – 2y = 0, since x^2 + 1 ≠ 0.
Thus, we have x = 2y.

( x + y ) / ( x – y ) = ( 2y + y ) / ( 2y – y ) = 3y / y = 3

This is sufficient too.

Therefore, the answer is D unlike our expectation..

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore, C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

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Re: If x≠y, (x+y)/(x-y)=?   [#permalink] 29 Sep 2017, 01:22
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