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The average (arithmetic mean) of the 11 numbers in a list is

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The average (arithmetic mean) of the 11 numbers in a list is [#permalink] New post 06 Mar 2018, 15:39
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The average (arithmetic mean) of the 11 numbers in a list is 14. If the average of 9 of the numbers in the list is 9, what is the average of the other 2 numbers?

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[Reveal] Spoiler: OA
36.5


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Re: The average (arithmetic mean) of the 11 numbers in a list is [#permalink] New post 06 Mar 2018, 19:17
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There are a few ways to do this. Here's one:

Call the total of the 9 numbers N. Since we know the 9 numbers average 9, we can use the average equation [(Sum)/(# of items)=(Avg)]. So:

N/9 = 9

and thus we know that N = 81. If we call the total of the two numbers T, we can make an equation out of the first sentence in the question:

(N + T)/11 = 14

So N + T = 154

Since we know N is 81, then T = 154 - 81, so T is 73.

Since T is actually 2 numbers, we should divide 73 by 2 to get the average, which is 36.5.
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Re: The average (arithmetic mean) of the 11 numbers in a list is [#permalink] New post 04 May 2018, 09:31
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Carcass wrote:
The average (arithmetic mean) of the 11 numbers in a list is 14. If the average of 9 of the numbers in the list is 9, what is the average of the other 2 numbers?

enter your value


Let's assign some variables:
Let a, b, c, d, e, f, g, h, i, j, k be the ORIGINAL 11 numbers.


The average (arithmetic mean) of the 11 numbers in a list is 14.
So, we can write: (a+b+c+d+e+f+g+h+i+j+k)/11 = 14
Multiply both sides by 11 to get: So, a+b+c+d+e+f+g+h+i+j+k = 154

If the average of 9 of the numbers in the list is 9 . . .
So, we can write: (a+b+c+d+e+f+g+h+i)/9 = 9
Multiply both sides by 9 to get: So, a+b+c+d+e+f+g+h+i = 81

. . . what is the average of the other 2 numbers?
We now have the following system:
a+b+c+d+e+f+g+h+i+j+k = 154
a+b+c+d+e+f+g+h+i = 81

Subtract the bottom equation by the top equation to get: j + k = 73
So, the AVERAGE of j and k = (j + k)/2
= 73/2
= 36.5

Answer: 36.5

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Re: The average (arithmetic mean) of the 11 numbers in a list is   [#permalink] 04 May 2018, 09:31
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