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If a set of 6 positive integers has an average (arithmetic mean) of 53

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If a set of 6 positive integers has an average (arithmetic mean) of 53 [#permalink]

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New post 07 Jan 2018, 22:07
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If a set of 6 positive integers has an average (arithmetic mean) of 53, what percent of the integers is greater than 53?

(1) At least three of the integers are equal to 53.
(2) None of the integers are less than 53.
[Reveal] Spoiler: OA

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Re: If a set of 6 positive integers has an average (arithmetic mean) of 53 [#permalink]

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New post 10 Jan 2018, 22:29
Bunuel wrote:
If a set of 6 positive integers has an average (arithmetic mean) of 53, what percent of the integers is greater than 53?

(1) At least three of the integers are equal to 53.
(2) None of the integers are less than 53.


Lets say the integers are a,b,c,d,e,f. Given that a+b+c+d+e+f = 53*6 = 318.

(1) Say a=b=c=53. Now its possible that all other integers are also = 53 (since it says at least three are equal to 53, so there could be more). In that case 0% of integers will be > 53. But its possible that none of d,e,f are equal to 53. Eg, we could have d=e=52, and f=55. In this case, since only 1 integer is > 53, 16.66% of integers are greater than 53. So No unique answer can be determined. Insufficient.

(2) If none of the integers are less than 53, then to keep the average at 53 (or sum at 318) - none of the integers can be greater than 53 also. If we increase even one number from 53, we will have to decrease at least one other number from 53 so as to keep the sum constant as 318. So that means none of the integers is greater than 53. Thus answer is 0%. Sufficient.

Hence B answer
Re: If a set of 6 positive integers has an average (arithmetic mean) of 53   [#permalink] 10 Jan 2018, 22:29
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