If x and y are integers, and x + y < 0, is x - y > 0?
1) |x| + |y| > |x|
2) x^y = 1
OA C
Source: Manhattan Prep
If x and y are integers, and x + y < 0, is x — y > 0
This topic has expert replies
-
- Moderator
- Posts: 7187
- Joined: Thu Sep 07, 2017 4:43 pm
- Followed by:23 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
- Ian Stewart
- GMAT Instructor
- Posts: 2621
- Joined: Mon Jun 02, 2008 3:17 am
- Location: Montreal
- Thanked: 1090 times
- Followed by:355 members
- GMAT Score:780
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
We can rephrase the question "is x > y?"
Subtracting |x| on both sides, Statement 1 only tells us that |y| > 0. But that's always true unless y = 0, so Statement 1 only tells us y is nonzero.
Assuming x is nonzero (something the question needs to tell us, so that the left side of Statement 2 is always defined), Statement 2 can be true in three situations when x and y are integers:
- y = 0: in this case, because x+y < 0, we have x < 0. So x < y is true
- x = 1: in this case, because x + y < 0, we have 1 + y < 0, so y < -1. So x > y is true
- x = -1 and y is even. In this case, because x + y < 0, we have -1 + y < 0, so y < 1. Since y is an even integer, as long as y is nonzero, then y is at most -2, so x > y is true (except if y = 0, which we considered in the first case above).
So when Statement 2 is true, the only situation where x < y is true is when y = 0. Otherwise x > y. Since Statement 1 rules out the possibility that y=0, the two Statements combined ensure that x > y, and are sufficient together.
Subtracting |x| on both sides, Statement 1 only tells us that |y| > 0. But that's always true unless y = 0, so Statement 1 only tells us y is nonzero.
Assuming x is nonzero (something the question needs to tell us, so that the left side of Statement 2 is always defined), Statement 2 can be true in three situations when x and y are integers:
- y = 0: in this case, because x+y < 0, we have x < 0. So x < y is true
- x = 1: in this case, because x + y < 0, we have 1 + y < 0, so y < -1. So x > y is true
- x = -1 and y is even. In this case, because x + y < 0, we have -1 + y < 0, so y < 1. Since y is an even integer, as long as y is nonzero, then y is at most -2, so x > y is true (except if y = 0, which we considered in the first case above).
So when Statement 2 is true, the only situation where x < y is true is when y = 0. Otherwise x > y. Since Statement 1 rules out the possibility that y=0, the two Statements combined ensure that x > y, and are sufficient together.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com
ianstewartgmat.com
ianstewartgmat.com