What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,

Add 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,125
Grouping numbers together may quicken the addition
sum = 4959
4959/9 =551

Answer:
(B) 551

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.

The point of this question is not actually to use the formula average= sum of sample/ sample size - as Bunuel pointed out- because the rate of error, basically the chance of you miscalculating a number or two and having to recalculate all the numbers in the problem is too high. Because all these points are equidistant you can simply take the average of each

13 + 520 + 1120 and divide by 3.

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