Is x^2 > 15?
(1) x > -4
(2) x^3 < 0
[spoiler]OA=E[/spoiler]
Source: Veritas Prep
Is x^2 > 15?
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Target question: Is x² > 15?Gmat_mission wrote:Is x² > 15?
(1) x > -4
(2) x³ < 0
[spoiler]OA=E[/spoiler]
Source: Veritas Prep
Statement 1: x > -4
Let's TEST some values.
There are several values of x that satisfy statement 1. Here are two:
Case a: x = -1. In this case, x² = (-1)² = 1. So, the answer to the target question is NO, x² is NOT greater than 15
Case b: x = -3.99999. In this case, x² = (-3.99999)² ≈ 16. So, the answer to the target question is YES, x² IS greater than 15
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: x³ < 0
In other words, statement 2 is telling us that x is negative
Let's TEST some values.
ASIDE: When testing values for statement 2, see if you can re-use any of the cases you tested for statement 1.
Here it turns out we can re-use BOTH cases:
Case a: x = -1. In this case, x² = (-1)² = 1. So, the answer to the target question is NO, x² is NOT greater than 15
Case b: x = -3.99999. In this case, x² = (-3.99999)² ≈ 16. So, the answer to the target question is YES, x² IS greater than 15
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
In other words,
Case a: x = -1. In this case, x² = (-1)² = 1. So, the answer to the target question is NO, x² is NOT greater than 15
Case b: x = -3.99999. In this case, x² = (-3.99999)² ≈ 16. So, the answer to the target question is YES, x² IS greater than 15
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT
Answer: E
Cheers,
Brent