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M19-07

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M19-07 [#permalink]

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New post 16 Sep 2014, 01:05
1
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

54% (00:56) correct 46% (01:00) wrong based on 178 sessions

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Re M19-07 [#permalink]

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New post 16 Sep 2014, 01:05
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 \(y = -x\).


Answer: A
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Re: M19-07 [#permalink]

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New post 18 Sep 2016, 13:27
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?
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Re: M19-07 [#permalink]

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New post 04 Nov 2016, 15:06
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
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Re: M19-07 [#permalink]

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New post 01 Aug 2017, 06:19
+1 for option A.
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Re: M19-07 [#permalink]

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New post 01 Aug 2017, 08:26
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.

Answer: A
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Re: M19-07   [#permalink] 01 Aug 2017, 08:26
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