Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
First, figure out the value of the “strange symbol” for different small values of x.
#x# = smallest even integer greater than or equal to x2.
If x = 1, then we get 12 = 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 22 = 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 32 = 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 42 = 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 x2 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.
