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# M19-07

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Math Expert
Joined: 02 Sep 2009
Posts: 51215

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16 Sep 2014, 00:05
1
5
00:00

Difficulty:

45% (medium)

Question Stats:

54% (00:55) correct 46% (00:56) wrong based on 210 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) $$\frac{x}{y} - \frac{y}{x} = 0$$

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Math Expert
Joined: 02 Sep 2009
Posts: 51215

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16 Sep 2014, 00:05
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
Posts: 9

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18 Sep 2016, 12:27
1
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?
Current Student
Joined: 26 Jan 2016
Posts: 103
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: 435
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33

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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
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

Intern
Joined: 02 Jan 2018
Posts: 5

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29 Sep 2018, 15:35
Hello Bunuel

I could be wrong here but
Option B can give you y = modulus (x), which IMO is expressing y in terms of x
Intern
Joined: 18 May 2018
Posts: 17
Location: India
Concentration: Marketing, Strategy
GMAT 1: 730 Q49 V40

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03 Oct 2018, 13:44
1
For statement 2, Y can be expressed as a square root of $$x^2$$. Technically that is sufficient within the parameters of the question, no?
Intern
Joined: 06 Aug 2018
Posts: 2

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10 Nov 2018, 07:48
1
agree with others, why is X^2 = Y^2 not sufficient? is it not expressing X in terms of Y? Pls respond Bunuel why this is not the case. Thank you in advance for your help
Math Expert
Joined: 02 Sep 2009
Posts: 51215

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10 Nov 2018, 07:51
notoriousmnz wrote:
agree with others, why is X^2 = Y^2 not sufficient? is it not expressing X in terms of Y? Pls respond Bunuel why this is not the case. Thank you in advance for your help

Thank you guys. This question is removed to tests' data base for revision.
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Re: M19-07 &nbs [#permalink] 10 Nov 2018, 07:51
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# M19-07

Moderators: chetan2u, Bunuel

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