Bunuel wrote:
If x < y, is x² < y² ?
(1) y > 0
(2) x > 0
Target question: Is x² < y² ?STRATEGY: When I scan the target question and the two statements, I see that I can REPHRASE the target question as Is x² - y² < 0? We can also factor the left side to get:
REPHRASED target question: Is (x + y)(x - y) < 0 ? Given: x < y If we subtract y from both sides, we get:
x - y < 0In other words,
x - y is NEGATIVESo, to answer the target question we need only determine whether
x+y is positive or negative.
Statement 1: y > 0 There are several values of x and y that satisfy statement 1 (and the given information). Here are two:
Case a: x = 1 and y = 2. In this case, the answer to the REPHRASED target question is
YES, (x + y)(x - y) is less than 0 Case b: x = -3 and y = 2. In this case, the answer to the REPHRASED target question is
NO, (x + y)(x - y) is not less than 0 Since we can’t answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x > 0 So, x is positive.
Since we're told x < y, we know y is also positive, which means x+y is POSITIVE
Since we know x - y is NEGATIVE, we get: (x + y)(x - y) = (NEGATIVE)(POSITIVE) = NEGATIVE
In other words, the answer to the REPHRASED target question is
YES, (x + y)(x - y) is less than 0 Since we can answer the
target question with certainty, statement 2 is SUFFICIENT
Answer: B