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

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Manager
Joined: 28 Jun 2018
Posts: 138
Location: India
Concentration: Finance, Marketing
GPA: 4
Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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29 Jul 2018, 08:27
13
4
00:00

Difficulty:

95% (hard)

Question Stats:

35% (01:48) correct 65% (01:19) wrong based on 104 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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Manager
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Joined: 04 Sep 2017
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Re: Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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Updated on: 29 Jul 2018, 09:35
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.

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Originally posted by MikeScarn on 29 Jul 2018, 09:21.
Last edited by MikeScarn on 29 Jul 2018, 09:35, edited 4 times in total.
Senior Manager
Joined: 22 Feb 2018
Posts: 355
Re: Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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29 Jul 2018, 09:36
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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Posts: 322
Re: Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0  [#permalink]

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29 Jul 2018, 09:55
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, 09:55
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