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If x and y are integers, is X > Y?

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Source: — Data Sufficiency |
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by vk_vinayak » Mon Sep 03, 2012 9:02 am
PGMAT wrote:If x and y are integers, is X > Y?

1) x + y > 0
2) y^x < 0

Can some one explain this with examples?

C
1. x=3, y=1 => x+y>0. Is X > Y ? YES
x=1, y=2 => x+y>0. Is X > Y ? NO

INSUFFICIENT

2. (-3)^-1 <0. Is X > Y ? YES
(-1)^-3 <0 Is X > Y ? NO

INSUFFICIENT

Combining 1 and 2:
x+y>0 => At least one of them has to be positive.

If we consider both as positive, y^x < 0 cant be possible. So, one has to positive and one has to be negative.

Consider: X -ve and Y to be +ve: y^x < 0 can't be possible.

So, Y is negative and X is positive. => X>Y.

(-2)^3 <0 and x+y>0 => Is X > Y ? YES

SUFFICIENT
- VK

I will (Learn. Recognize. Apply)
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by Brent@GMATPrepNow » Tue Sep 04, 2012 5:40 am
PGMAT wrote:If x and y are integers, is X > Y?
1) x + y > 0
2) y^x < 0
C
Target question: Is x > y?

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

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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