Suppose that S is a finite set of positive integers. If the greatest integer in S is removed from S, then the average value (arithmetic mean) of the integers remaining is 32. If the least integer in S is also removed, then the average value of the integers remaining is 35. If the greatest integer is then returned to the set, the average value of the integers rises to 40. The greatest integer in the original set S is 72 greater than the least integer in S. What is the average value of all the integers in the set S?

The average value of the all the integers in the set S is 36.8.

Expert Answers

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S is a finite set of positive integers. Let there be N integers in the set and the arithmetic mean of all the integers be M. The sum of all the integers in the set S is M*N.

When the greatest integer in S is removed from S, the arithmetic mean of the integers remaining is 32. Let the greatest integer be G.

This gives M*N - G = 32*(N-1).

When the least integer in S is also removed, the average value of the integers remaining is 35. Let the least integer in the set be S.

This gives M*N - G - S = 35*(N-2).

If the greatest integer is then returned to the set, the average value of the integers rises to 40.

This gives M*N - S = 40*(N-1).

The greatest integer in the original set S is 72 greater than the least integer in S, or G - S = 72.

M*N - G = 32*(N-1) ...(1)

M*N - G - S = 35*(N-2) ...(2)

M*N - S = 40*(N-1) ...(3)

G - S = 72 ...(4)

(1) - (3)

M*N - G - (M*N - S) = 32*(N-1) -40*(N-1)

S - G = (N - 1)*(32 - 40)

-72 = (N - 1)*(-8)

N - 1 = 9

N = 10

Substituting N = 10 and G = S + 72 in M*N - G - S = 35*(N-2).

10*M - S -72 - S = 35*8

=> 10M - 72 - 2S = 280

=> 10M - 2S = 352

=> S = (10M - 352)/2

M*N - S = 40*(N-1)

=> 10M - S = 360

Substituting S = (10M - 352)/2.

10M - (10M - 352)/2 = 360

=> 20M - 10M + 352 = 720

=> 10M = 368

=> M = 36.8

The average value of all the integers in the set S is 36.8

Last Updated by eNotes Editorial on
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