Bunuel
What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,125?
(A) 419
(B) 551
(C) 601
(D) 620
(E) 721
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:The simple average formula (Average = Sum/Number of terms) applies to this problem.
However, the chance of computational error is high on a problem with this many terms of such a large size.
A nice shortcut is possible if we group the similar terms:
Group A: 12, 13, 14 (equidistant terms with an average of 13, the middle term)
Group B: 510, 520, 530 (equidistant terms with an average of 520, the middle term)
Group C: 1,115, 1,120, 1,125 (equidistant terms with an average of 1,120, the middle term)
Since each group of terms consisted of three values (and thus were equally weighted in the set of
Since each group of terms consisted of three values (and thus were equally weighted in the set of nine terms), the average of all nine original terms is simply the average of the respective averages of Groups A, B, and C:
13 + 520 + 1,120 = 1,653.
Average = 1,653/3 = 551
The correct answer is B.
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