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

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

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New post 10 Jun 2015, 04:15
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Re: What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,  [#permalink]

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New post 10 Jun 2015, 08:58
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Bunuel wrote:
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.


looking at the set of numbers ,we may save time to make use of the numbers..
13 is avg of first three numbers , 520 for next three and 1120 for last three..
so basically we have to find avg of these three numbers..
551 ans ..B.. we dont require to get into proper calculation .. total >1600.. ans has to be between 500 and 600..only B remains
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Re: What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,  [#permalink]

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New post 10 Jun 2015, 07:25
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Bunuel wrote:
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.


Average = Sum of {12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,125} / No. of terms

i.e. Average = Sum of {12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,125} / 9

i.e. Average = 4959 / 9 = 551

Answer: Option
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Re: What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,  [#permalink]

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New post 10 Jun 2015, 08:01
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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
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Re: What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,  [#permalink]

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New post 15 Jun 2015, 01:35
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1
Bunuel wrote:
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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Re: What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,  [#permalink]

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New post 18 Jun 2017, 03:14
Bunuel wrote:
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.


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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Re: What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,  [#permalink]

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Re: What is the average of 12, 13, 14, 510, 520, 530, 1,115, 1,120, and 1,   [#permalink] 26 Feb 2019, 22:53
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