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Re: The average (arithmetic mean) of four distinct positive integers is 10 [#permalink]
Let the distinct number be A,B,C, and D

Its given A > B > C > D
also A + B + C + D =40 and A + B = 16 means C + D = 24.

Since the question ask for the largest possible number we should choose the least value for both A and B. We cant choose 8 & 8 bcos numbers has been mentioned as distinct. Hence the least possible value for A an B is 9 & 7.
then we can have the least possible value for C as 10 if C is 10 the most possible value for D is 14
Hence the answer B.
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Re: The average (arithmetic mean) of four distinct positive integers is 10 [#permalink]
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Outstanding Question.
Here is what i did in this Question ->

Let the 4 distinct integers be ->
w1
w2
w3
w4

Now,Using -->

\(Mean = \frac{Sum}{#}\)


Sum(4) = 10*4=40

Also Average of w1 and w2 is 8
Hence the sum -> w1+w2=8*2=16

Now w1+w2+w3+w4=40
16+w3+w4=40
Hence w4=24-w3
To maximise w4,we must minimise w3

If the Question didn't mention the word "DISTINCT"then we would get => w4=24-8=16

But as the integers involved are different -> the case we need to condor is 7,9,10
Hence,w4=24-10=14

Hence B
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Re: The average (arithmetic mean) of four distinct positive integers is 10 [#permalink]
goodyear2013 wrote:
The average (arithmetic mean) 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 possible value of the largest integer?
A) 12
B) 14
C) 15
D) 16
E) 17
M13-16


Keyword = four distinct positive integers

a +b +c + d= 40

a+b =16, means a can be 7 and b can be 9

d = 24 - c------------------from (40-16)
24 -12, 12, cant be possible
24-14, 10, can be possible, series becomes 7,9,10,14

24-15,9 cant be possible, because of Keyword.

B
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Re: The average (arithmetic mean) of four distinct positive integers is 10 [#permalink]
let a<b<c<d
a+b+c+d=40...........(i)
a+b=16.............(ii)
therefore, c+d=24..........(iii)

If d=12, then c=12, CANNOT be possible as they are distinct integers
If d=14, c=10, this is possible where the 4 numbers are 7,9,10,14
If d=15, c=9, this is NOT possible because the best case for a,b would be 7,9
We don't need to check d=16 and 17 as they will make the cases worse than when d=15

Answer: B
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Re: The average (arithmetic mean) of four distinct positive integers is 10 [#permalink]
Expert Reply
goodyear2013 wrote:
The average (arithmetic mean) 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 possible value of the largest integer?
A) 12
B) 14
C) 15
D) 16
E) 17



M13-16


Solution:

Since the sum of the four distinct integers is 10 x 4 = 40 and the sum of the two smallest integers is 8 x 2 = 16, the sum of the largest two integers is 40 - 16 = 24. Since we want the maximum possible value of the largest integer, we can let the second largest integer be as small as possible. The second integer can’t be 9; otherwise, the sum of the two smallest integers would be at most 7 + 8 = 15 (recall that all the integers are distinct). However, if the second smallest integer is 10 (and the two smallest integers are 7 and 9), the largest integer will then be 24 - 10 = 14.

Answer: B

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Re: The average (arithmetic mean) of four distinct positive integers is 10 [#permalink]
Good question because it tests averages and maximization of numbers

Basically once you realize it is 16 + c + d = 40

it is time to make cases for a b c d = 40

you can try 1 15 16 8 (But d cannot be less than previous number in sequence)

So you arrive at 7 9 + c + d = 40

to maximize d, you minimize c.

7 + 9 + 10 + d = 40

d = 14 max.

Answer B.
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Re: The average (arithmetic mean) of four distinct positive integers is 10 [#permalink]
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