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# M19-07

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Math Expert
Joined: 02 Sep 2009
Posts: 43789

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16 Sep 2014, 00:05
Expert's post
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Difficulty:

55% (hard)

Question Stats:

51% (00:55) correct 49% (00:59) wrong based on 150 sessions

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What is $$y$$ in terms of $$x$$?

(1) $$\frac{1}{\frac{x}{y} + \frac{y}{x}} = \frac{1}{2}$$

(2) $$\frac{x}{y} - \frac{y}{x} = 0$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43789

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16 Sep 2014, 00:05
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Official Solution:

Let $$z$$ equal $$\frac{x}{y}$$.

Statement (1) by itself is sufficient. S1 gives that $$\frac{1}{z + \frac{1}{z}} = \frac{1}{2}$$ or $$z + \frac{1}{z} = 2$$ or $$z^2 + 1 = 2z$$ or $$(z - 1)^2 = 0$$ from where $$z = 1$$ and $$y = x$$.

Statement (2) by itself is insufficient. S2 gives that $$z - \frac{1}{z} = 0$$ or $$z = \frac{1}{z}$$ or $$z^2 = 1$$. This means that either $$y = x$$ or $$y = -x$$.

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Intern
Joined: 27 Mar 2014
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18 Sep 2016, 12:27
1
KUDOS
Hi Bunuel,

Can you provide some helpful tips/tricks of when we should know/recognize to use the "substitute for z" technique? It makes sense when you know to do this, but how do you recognize that this technique is a more efficient way of approaching the problem?
Manager
Joined: 26 Jan 2016
Posts: 115
Location: United States
GPA: 3.37

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04 Nov 2016, 14:06
1) Here you can make a quadratic

For statement one add the two fractions in the denominator together and you get 1/(x²+y²/xy)--> xy/x²+y²=1/2 cross multiple and get 2xy=x²+y² or 0=x²+y²-2XY

(x-y)²=0
x-y=0
x=y
suff

2. Here we get x/y=y/x or x²=y², if we square root both sides we dont know if x and y are each negative, each positive, or one is negative and one is positive insuff
Senior Manager
Joined: 08 Jun 2015
Posts: 369
Location: India
GMAT 1: 640 Q48 V29

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01 Aug 2017, 05:19
+1 for option A.
_________________

" The few , the fearless "

Intern
Joined: 09 Dec 2014
Posts: 37

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01 Aug 2017, 07:26
1
KUDOS
Bunuel wrote:
What is $$y$$ in terms of $$x$$?

(1) $$\frac{1}{\frac{x}{y} + \frac{y}{x}} = \frac{1}{2}$$

(2) $$\frac{x}{y} - \frac{y}{x} = 0$$

Statement 1: $$\frac{1}{\frac{x}{y} + \frac{y}{x}} = \frac{1}{2}$$

==> $$\frac{1}{((x^2 + y^2)/xy)} = \frac{1}{2}$$

$$\frac{xy}{(x^2 + y^2)} = \frac{1}{2}$$

2xy = $$x^2 + y^2$$

$$x^2 + y^2$$ - 2xy=0 ==> $$(x-y)^2 = 0$$ ==> x-y = 0 ==> x=y
Sufficient.

Statement 2: $$\frac{x}{y} - \frac{y}{x} = 0$$

$$\frac{(x^2 - y^2)}{xy} = 0$$

Multiply both sides by xy

$$x^2 - y^2 = 0$$

$$x^2 = y^2$$ ==> y= +x OR y= -x
Insufficient.

_________________

Thanks,
Ramya

Re: M19-07   [#permalink] 01 Aug 2017, 07:26
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# M19-07

Moderators: chetan2u, Bunuel

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