If x and y are integers, is x>y?
1) x+y>0
2) y^x<0
1) x+y>0; clearly not sufficient
2) y^x<0; y may be any -ve integer and x may be any -ve or +ve odd integer. Not sufficient
What next???
Help please!
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jaspreetsra
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Statement 1:jaspreetsra wrote:If x and y are integers, is x>y?
1) x+y>0
2) y^x<0
1) x+y>0; clearly not sufficient
2) y^x<0; y may be any -ve integer and x may be any -ve or +ve odd integer. Not sufficient
What next???
Help please!
It's possible that x=2 and y=1, in which case x>y.
It's possible that x=1 and y=1, in which case x=y.
INSUFFICIENT.
Statement 2:
y^x < 0 implies that y<0 and that x is ODD.
It's possible that y=-1 and x=1, in which case x>y.
It's possible that y=-1 and x=-1, in which case x=y.
INSUFFICIENT.
Combined:
Linking together x+y > 0 and 0 > y, we get:
x+y > 0 > y
x+y > y
x > 0.
Linking together x > 0 and 0 > y, we get:
x > 0 > y
x > y.
SUFFICIENT.
The correct answer is C.
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Brent@GMATPrepNow
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Target question: Is x > y?If x and y are integers, is X > Y?
1) x + y > 0
2) y^x < 0
C
Statement 1: x + y > 0
There are several possible cases. Here are two:
case a: x = 2, y = 1, in which case x is greater than y.
case b: x = 1, y = 2, in which case x is not greater than y.
Since we can't answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: y^x < 0
There are several possible cases. Here are two:
case a: x = 1, y = -1, in which case x is greater than y.
case b: x = -4, y = -1, in which case x is not greater than y.
Since we can't answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 & 2:
From statement 2 (y^x < 0), we can conclude that y must be negative and x must be odd.
From this one statement, there are only two possibilities.
- Possibility #1: y is negative, and x is an odd negative number.
- Possibility #2: y is negative, and x is an odd positive number.
Statement 1 rules out Possibility #1 (since two negative numbers cannot have a positive sum).
This leaves us with Possibility #2, which means x is definitely greater than y.
Answer = C
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Brent
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