Is x > y?

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Is x > y?

by jack0997 » Thu Jun 08, 2017 6:00 am
Is x > y?

(1) |x| > |y|
(2) x ‚+ y > 0

OA C

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by Jay@ManhattanReview » Thu Jun 08, 2017 6:41 am
jack0997 wrote:Is x > y?

(1) |x| > |y|
(2) x ‚+ y > 0

OA C
Statement 1:
|x| > |y|

ƒ=> x > y . . . (i): The answer is 'Yes.'
OR
x < -y . . . (ii): The answer is 'No.'
Thus, the answer cannot be uniquely determined. - Insufficient

Statement 2:
x ‚+ y > 0
ƒ=> x > -y . . . (iii)

Case 1: If y < 0: -y > 0 ƒ=> x > -y > 0
Thus, x is positive, hence is greater than y, which is negative: The answer is 'Yes.'

Case 2: If y > 0
We cannot determine whether x is also greater than the positive number y.
The answer may be 'Yes' or 'No.'

Thus, we cannot uniquely determine whether x > y. - Insufficient

Statement 1 & 2 together:

Conditions (ii) and (iii) contradict each other.
Thus, (ii) is not a possible scenario since we know that condition (iii) is true.

Thus, we have:
x > y . . . (i)
AND
x > -y . . . (iii)

The above two conditions are possible simultaneously, for example, if
x ƒ = 2 and y ƒ= 1
x ƒ = 2 and y ƒ = -1

Thus, x > y.

The answer to the question is 'Yes.' - Sufficient

The correct answer: C

Hope this helps!

Relevant book: Manhattan Review GMAT Data Sufficiency Guide

-Jay
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by [email protected] » Thu Jun 08, 2017 1:21 pm
Hi jack0997,

This DS question can be solved with a mix of TESTing VALUES and Number Properties.

We're asked if X is greater than Y. This is a YES/NO question.

1) |X| > |Y|

IF...
X=1, Y=0
Then the answer to the question is YES.

X=-1, Y=0
Then the answer to the question is NO.
Fact 1 is INSUFFICIENT

2) X+Y > 0

This Fact tells us that AT LEAST one of the two variables MUST be greater than 0....

IF...
X=1, Y=0
Then the answer to the question is YES.

X=0, Y=1
Then the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, we know that at least one of the two variables is greater than 0, so there are 2 options to consider:
1) BOTH X and Y are greater than 0. In this case, since |X| > |Y|, so X would have to be greater than Y.
2) One variable is positive and one is negative. In this case, since X+Y > 0, the positive value would have to be "more positive" than the other number is "negative" (for example +3 and -2). With the added restriction that |X| > |Y|, X would have to be the positive value and Y would have to be the negative value. Thus, X would always be greater than Y.

In both options, the answer to the question is YES.
Combined, SUFFICIENT

Final Answer: C

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by Matt@VeritasPrep » Thu Jun 08, 2017 5:19 pm
S1:

|x| > |y|

In words, this says "x is further from 0 than y is". This is true if x = 3 and y = 2, but it's also true if x = 3 and y = -2. Not sufficient!

S2:

x + y > 0

x > -y

OK, so x is greater than -y ... but we want to know if it's greater than y itself. Not sufficient!

Together, let's suppose that y > x > -y. (We know from S2 that x > -y, so we're trying the negative case, the one in which y is somehow greater than x.) If that's true, then both y and -y are further from 0 than x is. That contradicts S1, so it's impossible! That means that y > x doesn't work, and we must have x > y. After all that, our answer is C.