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# S95-06

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Math Expert
Joined: 02 Sep 2009
Posts: 49300

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16 Sep 2014, 01:49
00:00

Difficulty:

55% (hard)

Question Stats:

50% (01:04) correct 50% (01:26) wrong based on 52 sessions

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If the average (arithmetic mean) of six different numbers is 25, how many of the numbers are greater than 25?

(1) None of the six numbers is greater than 50.

(2) Three of the six numbers are 7, 8, and 9, respectively.

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Joined: 02 Sep 2009
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16 Sep 2014, 01:49
1
Official Solution:

We must determine how many numbers in a set of six distinct numbers are greater than 25. Since the average of the six numbers is 25, these six numbers must sum to $$6 \times 25 = 150$$.

Statement 1 tells us that none of the six numbers is greater than 50. Many combinations of numbers satisfy this statement. For example, the six numbers could be 19, 20, 21, 22, 23, and 45. In this case, only 1 number is greater than 25. On the other hand, if the six numbers are 1, 2, 3, 47, 48, and 49, then 3 of the numbers are greater than 25. Since we cannot find a unique value, statement 1 alone is NOT sufficient to answer the question. Eliminate answer choices A and D. The correct answer choice is B, C, or E.

Statement 2 tells us that three of the six numbers are 7, 8, and 9. This means that the other three numbers must sum to $$150 - (7 + 8 + 9) = 150 - 24 = 126$$. However, since there is no limit to what any of the numbers can be, we can have different combinations. For example, the three numbers could be 1, 2, and 123. In this case, only 1 number is greater than 25. But if the three numbers are 41, 42, and 43, then all 3 are greater than 25. Statement 2 alone is also NOT sufficient. Eliminate answer choice B. The correct answer choice is either C or E.

Both statements together tell us that three of the numbers are 7, 8, and 9, that the other three numbers must sum to 126, and that none of the numbers is greater than 50. If all three remaining numbers are 25 or less (remember, the numbers must be different), then these remaining numbers can sum to, at maximum, $$25 + 24 + 23 = 72$$. If two numbers are 25 or less, and one is 50 (the upper bound for numbers), the sum is at most $$25 + 24 + 50 = 99$$. If one number is 25 or less, and the other two numbers are 50 and 49, then the maximum sum is $$25 + 50 + 49 = 124$$. In order to reach 126, all three numbers MUST be greater than 25. Both statements together are sufficient.

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Joined: 01 Nov 2016
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13 Apr 2017, 19:56
Very hard question. I understand the official answer, but I don't know how it is possible to do this within two minutes unless you have incredible number sense
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Joined: 02 Sep 2009
Posts: 49300

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14 Apr 2017, 00:36
joondez wrote:
Very hard question. I understand the official answer, but I don't know how it is possible to do this within two minutes unless you have incredible number sense

You can check other solutions here: https://gmatclub.com/forum/if-the-avera ... 88461.html
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Re: S95-06 &nbs [#permalink] 14 Apr 2017, 00:36
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# S95-06

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