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If the average (arithmetic mean) of five positive numbers is 30, how

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If the average (arithmetic mean) of five positive numbers is 30, how  [#permalink]

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New post 28 Dec 2017, 00:28
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If the average (arithmetic mean) of five positive numbers is 30, how many of the numbers are greater than 30?

(1) None of the five numbers is equal to 10.
(2) The average (arithmetic mean) of the four smallest numbers is 4.

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Re: If the average (arithmetic mean) of five positive numbers is 30, how  [#permalink]

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New post 28 Dec 2017, 04:26
Bunuel wrote:
If the average (arithmetic mean) of five positive numbers is 30, how many of the numbers are greater than 30?

(1) None of the five numbers is equal to 10.
(2) The average (arithmetic mean) of the four smallest numbers is 4.


We'll try a few numbers to help us understand the logic.
This is an Alternative approach.

(1) We can choose all our numbers to be 30 or have all 30 except for one 29 and one 31
Insufficient.

(2) Let's try making the 4 smallest numbers exactly 4. Then the last number is definitely larger than 30.
We've seen that we can make 1 number greater than 30; next we'll try to make 2 numbers greater than 30.
Say our two largest numbers are 31 and 31. Then the 4 smallest numbers are 31, x, y, z.
These must have an average of 4 so (31+x+y+z)/4 = 4 --> 31+x+y+z = 16 --> x+y+z=16-31<0
But all our numbers are positive! This is impossible, and therefore (2) is sufficient to answer the question.

(B) is our answer.
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Re: If the average (arithmetic mean) of five positive numbers is 30, how &nbs [#permalink] 28 Dec 2017, 04:26
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