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by adilka » Tue Jan 06, 2009 7:49 pm
The average of four distinct positive integers is 10. If the average of the smaller two of these four integers is 8, which of the following represents the maximum value of the largest integer?

13
14
15
16
17
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Re: averages

by logitech » Tue Jan 06, 2009 8:59 pm
a+b+c+d = 40

a+b = 16

so c+d = 24

We will make D MAX by finding the min C

Since all numbers are distinct and the average of the first two small numbers are 8, we can not choose 8 and 8 , but we can choose 7 and 9

so C can be minimim 10, which gives us 14 as the answer for D.
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by adilka » Wed Jan 07, 2009 9:11 am
OA B
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by DavoodBeater » Wed Jan 07, 2009 12:06 pm
2*8 + X + Y = 4*10
X+Y=24
distinct number => the second smallest is at least 9.
=> for maximizing the largest, the third smallest must be 10.
=> 24-10=14
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by welcome » Thu Jan 08, 2009 7:01 am
DavoodBeater, as all no. and POSITIVE DISTINCE integers, second smallest will be 7 not and third smallest will be 8.

So the answer will be 24-8 = 16. 'D'.
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