Bunuel wrote:
If #x# represents the smallest even integer greater than or equal to x^2, is #x# less than 12?
(1) |x – 3| ≤ 1
(2) |x – 1| ≤ 2
Kudos for a correct solution.
MANHATTAN GMAT OFFICIAL SOLUTION:First, figure out the value of the “strange symbol” for different small values of x.
#x# = smallest even integer greater than or equal to x^2.
If x = 1, then we get 1^2 = 1, and the smallest even integer greater than or equal to that is 2. So #1# = 2, which is less than 12.
If x = 2, then we get 2^2 = 4, and the smallest even integer greater than or equal to that is 4. So #2# = 4, which is less than 12.
If x = 3, then we get 3^2 = 9, and the smallest even integer greater than or equal to that is 10. So #3# = 10, which is less than 12.
If x = 4, then we get 4^2 = 16, and the smallest even integer greater than or equal to that is 16. So #4# = 16, which is NOT less than 12.
Larger values of x will also give us results greater than 12.
The value of x is not restricted to integers (or to positives, for that matter), but we should have a sense of what values work. If x is 1, 2, or 3, we get an answer of “Yes.” If x is 4 or bigger, we get “No.”
Statement 1: INSUFFICIENT. The condition |x – 3| ≤ 1 can be translated as follows: x is no more than 1 unit away from 3 on a number line. Thus, the largest value of x is 4, while the smallest is 2. (Plug back in and verify this.) Since we have possible values of 2, 3, and 4, we have some “Yes’s” and some “No’s.”
Statement 2: SUFFICIENT. The condition |x – 1| ≤ 2 can be translated to this: x is no more than 2 units away from 1. Thus, the largest value of x is 3, while the smallest is –1. All the values in this range have an x^2 less than or equal to 9, so the smallest even integer above those values is 10, which is definitely less than 12. In other words, for every value of x allowed by this statement, the answer to the question is “Yes.” That’s enough to answer the question definitively.
The correct answer is B.