Can someone help me with this? i'm lost when i get to probabilities.

If a 3-digit integer is selcted at random from the integers 100 thru 199, inclusive, what is the probability that the first digit and the last digit of the integer are each equal to one more than the middle digit?


A) 2/225
B) 1/111
C) 1/110
D) 1/100
E) 1/50

The answer should be 1/100. The question indicates that the first and last digit are equal to 1 more than the middle digit indicating that the first and last digits are equal. Since the range is less than 199, the first digit cannot be greater than 1. This implies that with 0 as the middle digit and 1 as the first and last digits, 101 is the only integer possible out of the 100 integers.

Guest, thanks for the reply. At Manhattan GMAT, many of the teachers use the following slogan to describe the proper approach to each section of the GMAT: The Math is the Verbal, and the Verbal is the Math. What this indicates is that, often times, the key to solving "difficult" math problems correctly is simply to read the problem carefully, as the difficulty is with the wording of the problem rather than the computational complexity. Conversely, the key to successfully answering "difficult" verbal problems frequently comes down to maintaining a structured and methodical (i.e. "mathematical") approach to each problem type.

-dan