guerrero 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?
12
14
15
16
17
OAB
Sum = number * average.
Since the average of the 4 integers is 10, their sum = 4*10 = 40.
In ascending order, let the 4 integers = A+B+C+D.
Since A+B+C+D = 40, we get:
D = 40 - (A+B+C).
To MAXIMIZE the value of D, we must MINIMIZE the value of A+B+C.
To minimize the value of C, we must minimize the value of B.
Since the average of A and B is 8, their sum = 2*8 = 16.
Thus, the least option B is 9:
A+B = 7+9.
Thus, the least option for C is 10.
Thus:
Greatest possible value for D = 40 - (A+B+C) = 40 - (7+9+10) = 14.
The correct answer is
B.
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