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GMATbuster92
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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0 29 Jul 2018, 08:27
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

33% (01:57) correct 67% (01:40) wrong based on 266 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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E
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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0 Updated on: 29 Jul 2018, 09:35
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.
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kshitizbansal92
is x > y?
1) |x| >|y|
2) \(\frac{1}{4}\) x - \(\frac{1}{3}\) y >0
Show more

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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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0 29 Jul 2018, 09:36
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kshitizbansal92
is x > y?
1) |x| >|y|
2) \(\frac{1}{4}\) x - \(\frac{1}{3}\) y >0
Show more

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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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0 29 Jul 2018, 09:55
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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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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0 29 Feb 2020, 11:59
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GMATbuster92
Is x > y?


(1) |x| >|y|

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


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Visit https://www.mathrevolution.com/gmat/lesson for details.

Since we have 2 variables (x and y) and 0 equations, C is most likely the answer. So, we should consider conditions 1) & 2) together first. After comparing the number of variables and the number of equations, we can save time by considering conditions 1) & 2) together first.

Conditions 1) & 2)

If x = 2, y = 1, then the answer is "yes".
If x = -5, y = -4, then the answer is "no".

Since both conditions together do not yield a unique solution, they are not sufficient.

Therefore, E is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B, or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D, or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Is x > y? (1) |x| >|y| (2) 1/4 x + 1/3 y > 0 23 Mar 2024, 09:15
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