If #x# represents the smallest even integer greater than or equal to x

Question

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 .

pankajvashist · Answer

(1) |x – 3| ≤ 1
= -1 ≤(x-3) ≤1 => -2 ≤ x ≤4....not sufficient.

2) |x – 1| ≤ 2
= -2≤ (x-1)≤2 => -1≤x≤3 ....sufficient ...

so answer is B.

anudeep133 · Answer

Solution : 
Statement1 : |x – 3| ≤ 1 ==> 2 ≤ x ≤ 4
#x# is more than 12 for x = 4
#x# is less than 12 for x = 2.
Not suffiecient.

Statement2 : |x – 1| ≤ 2 ==> -1 ≤ x ≤ 3
#x# is less than 12 for x = 3
#x# is less than 12 for x = -1.
So, sufficient.

Option B.

shailendra79s · Answer

#x# ≥ x²
Statement 1:
|x-3| ≤1 → -1≤x-3≤1 → 2≤x≤4;
Since 4² ≥ 12, Insufficient
Statement 2:
|x-1| ≤ 2 → -2 ≤ x-1 ≤ 2 → -1 ≤ x ≤ 3
Since 3² ≤ 12, Sufficient
Answer: B

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem.
Remember equal number of variables and independent equations ensures a solution.

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
 
In the original condition we have 1 variable (x) and in order to match the number of variables and equations we need 1 more equation. Since there is 1 each in 1) and 2), there is high probability that D is the answer.

In case of 1) -2≤x≤4, x=-2,-1,0,1,2,3,4 and x^2=0,1,4,9,16 #0#=0 gives us the answer yes while for #4#=16 the answer is no. Therefore the condition is not sufficient. (not unique)
In case of 2), -1≤x≤3, x=-1,0,1,2,3 and x^2=0,1,4,9. Here, #x#<12 and the answer is yes therefore the condition is sufficient. Therefore answer is B

Normally for cases where we need 1 more equation, such as original conditions with 1 variable, or 2 variables and 1 equation, or 3 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore D has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) separately. Here, there is 59 % chance that D is the answer, while A or B has 38% chance. There is 3% chance that C or E is the answer for the case. Since D is most likely to be the answer according to DS definition, we solve the question assuming D would be our answer hence using 1) and 2) separately. Obviously there may be cases where the answer is A, B, C or E.

Bunuel · Answer

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.

AmritaSarkar89 · Answer

it feels good when you start thinking as logically as Bunuel

MSIMBA · Answer

Option 1 : 
|x – 3| ≤ 1 ==> 2 ≤ x ≤ 4
#x# is more than 12 for x = 4
#x# is less than 12 for x = 2.
Not suffiecient.

Option 2 : 
|x – 1| ≤ 2 ==> -1 ≤ x ≤ 3
#x# is less than 12 for x = 3
#x# is less than 12 for x = -1.
So, sufficient.

Option B.

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If #x# represents the smallest even integer greater than or equal to x

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