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  • Typical Day of a UCLA MBA Student - Recording of Webinar with UCLA Adcom and Student

     December 14, 2018

     December 14, 2018

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    Carolyn and Brett - nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
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M19-07

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M19-07  [#permalink]

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New post 16 Sep 2014, 00:05
1
5
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

54% (00:55) correct 46% (00:56) wrong based on 210 sessions

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Re M19-07  [#permalink]

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New post 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 \(y = -x\).


Answer: A
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Re: M19-07  [#permalink]

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New post 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?
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Re: M19-07  [#permalink]

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New post 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
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Re: M19-07  [#permalink]

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New post 01 Aug 2017, 05:19
+1 for option A.
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Re: M19-07  [#permalink]

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New post 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.

Answer: A
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Re: M19-07  [#permalink]

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New post 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
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Re: M19-07  [#permalink]

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New post 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?
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Re: M19-07  [#permalink]

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New post 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
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Re: M19-07  [#permalink]

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New post 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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Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M19-07 &nbs [#permalink] 10 Nov 2018, 07:51
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