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If x and y are integers, is x > y?
1) x + y > 0
2) y^x < 0
Target question: Is x > y?
Given: x and y are integers
Statement 1: x + y > 0
There are several pairs of values that meet this condition. Here are two:
Case a: x = 2 and y = 1, in which case x > y
Case b: x = 1 and y = 2, in which case x < y
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: y^x < 0
There are several pairs of values that meet this condition. Here are two:
Case a: x = 1 and y = -1, in which case x > y
Case b: x = -3 and y = -1, in which case x < y
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined:
IMPORTANT: Statement 2 tells us that y is negative. We know this because (some positive value)^x can never evaluate to be a negative value.
So, y is negative, and statement 1 tell us that x + y > 0
In other words, x + (some negative value) > 0
From this we can conclude that x is positive.
If x is positive and y is negative, then it must be true that x > y
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer = C
Cheers,
Brent
