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

     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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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0

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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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New post 29 Jul 2018, 07:27
14
4
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A
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D
E

Difficulty:

  95% (hard)

Question Stats:

34% (01:48) correct 66% (01:18) wrong based on 106 sessions

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Is x > y?


(1) |x| >|y|

(2) \(\frac{1}{4} x - \frac{1}{3} y >0\)

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Re: Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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New post Updated on: 29 Jul 2018, 08:35
4
kshitizbansal92 wrote:
is x > y?
1) |x| >|y|
2) \(\frac{1}{4}\) x - \(\frac{1}{3}\) y >0


Great question. Here's how I solved it.

is x > y?

Statement 1: |x| >|y| - Not Sufficient. Make x=2 and y=1, yes. Make x=-2 and y=-1, no.

Both of these options work for statement 1, but give different answers to the original question. So this statement has a "yes, no" answer. Therefore Not Sufficient.

Statement 2: \(\frac{1}{4}\)x - \(\frac{1}{3}\)y >0 - To make this one a little bit easier on the eyes, I multiplied both sides by the LCM just to eliminate fractions. That gave me 3x - 4y > 0.

Make x=2 and y=1 ----> 3(2) - 4(1) --> 6-4=2 --> 2>0. YES

Make x=-2 and y=-3 --> 3(-2) - 4(-3) --> -6+12=6 --> 6>0. NO

Both of these options for statement 2, but give different results for the original question. Therefore not sufficient.

Statement 1 and 2: 3x - 4y > 0 and |x| >|y|

x=2 and y=1. Works for both statements and answers the original question as YES.

x=-10 and y=-9. Works for both statements, answers the original question as NO.

Not Sufficient.

Answer E
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Originally posted by MikeScarn on 29 Jul 2018, 08:21.
Last edited by MikeScarn on 29 Jul 2018, 08:35, edited 4 times in total.
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Re: Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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New post 29 Jul 2018, 08:36
1
kshitizbansal92 wrote:
is x > y?
1) |x| >|y|
2) \(\frac{1}{4}\) x - \(\frac{1}{3}\) y >0


OA:E

Question stem : Is x>y

1) |x| >|y|
plugging the number (1,0) satisfies the condition |x|>|y|.
Is \(x>y?: yes\) as 1>0

plugging the number (-7,-6) satisfies the condition |x|>|y|.
Is \(x>y?: No\) as -7<-6

Statement 1 alone is not sufficient

2)\(\frac{1}{4}\) x - \(\frac{1}{3} y >0\)
plugging the number (1,0) satisfies the condition \(\frac{1}{4}\) x - \(\frac{1}{3} y >0\) as \(\frac{1}{4}>0\).
Is \(x>y?: yes\) as 1>0

plugging the number (-7,-6) satisfies the condition \(\frac{1}{4}\) x - \(\frac{1}{3} y >0\) as \(\frac{-7}{4}+\frac{6}{2}>0\).
Is \(x>y?: No\) as -7<-6
Statement 2 alone is not sufficient

Combining statement 1 and statement 2 will also be insufficient, as we have plugged the same values for yes answer and the same value for no answer in both statement 1 and 2.
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Re: Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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New post 29 Jul 2018, 08:55
is x > y?
1) |x| >|y|
2) 1/4 x - 1/3 y >0

1) |x| > |y| - This will work for positive values. But if the x and y are negative , the answer is No. Since there are two different answers this is Insufficient.

2) 1/4x - 1/3y > 0

If you take LCM , then it would 3x-4y > 0

If you substitute the numbers for x and y , you can see this option is Insufficient when x and y have negative values.

Together (1) and (2)
When you substitute negative values (-10, -9) you get No and for positive values (1,2) , you get Yes as an answer.

Ans: E
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Re: Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0 &nbs [#permalink] 29 Jul 2018, 08:55
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