Is x > y?
(1) |x| > |y|
(2) x ‚+ y > 0
OA C
Is x > y?
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Statement 1:jack0997 wrote:Is x > y?
(1) |x| > |y|
(2) x ‚+ y > 0
OA C
|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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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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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.
|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.