Bunuel wrote:
If x^2 = 131, which of the following is closest to a potential value of x?
A. -127
B. -11
C. 12
D. 13
E. 131
We need to determine the approximate value of x and are given that x^2 = 131.
Thus, x = √131 or x = -√131.
Since √121 < √131 < √144, we see that 11 < x < 12 (if x = √131) or -12 < x < -11 (if x = -√131). Therefore, we need to see whether x is closer to -11 or 12, since both of these numbers are in the given answer choices. We see that 131 is closer to 121 than it is to 144. So:
If x = √131, x is closer to 11 than it is to 12.
If x = -√131, x is closer to -11 than it is to -12.
In other words, regardless if x is √131 or -√131, it is closer to 11 or -11 than it is to 12 or -12.
Alternate Solution:
The two perfect squares between which 131 lies are 121 and 144. We see that 121 is closer to 131 than is 144, so the perfect square of interest is 121, which we can get as either 11^2 or (-11)^2. From the given answer choices, we see that our choice must be -11.
Answer: B
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