Vincen wrote:Is b < 0 ?
(1) b³ < b
(2) b² > b
Target question: Is b < 0 ?
Statement 1: b³ < b
There are several values of b that satisfy statement 1. Here are two:
Case a: b = -2 (which means b³ = -8, and -8 < -2). In this case, the answer to the target question is
YES, b IS less than 0
Case b: b = 1/2 (which means b³ = 1/8, and 1/8 < 1/2). In this case, the answer to the target question is
NO, b is NOT less than 0
Since we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: b² > b
There are several values of b that satisfy statement 2. Here are two:
Case a: b = -1 (which means b² = 1, and 1 > -1). In this case, the answer to the target question is
YES, b IS less than 0
Case b: b = 2 (which means b² = 4, and 4 > 2). In this case, the answer to the target question is
NO, b is NOT less than 0
Since we cannot answer the
target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1 tells us that b³ < b
Statement 2 tells us that b² > b
Combine the inequalities to get: b³ < b < b²
This tells us that
b³ < b²
Subtract b² from both sides to get:
b³ - b² < 0
Factor to get:
b(b² - b) < 0
If
b(b² - b) < 0, then there are TWO POSSIBLE CASES:
Case a: b < 0 and (b² - b) > 0
OR
Case b: b > 0 and (b² - b) < 0
If we take statement 2 (b² > b) and subtract b from both sides, we get: b² - b > 0
This RULES OUT case b (above), which means
case a must be true.
If case a is true, then it must be the case that
b < 0
Since we can answer the
target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
