Question

The average of eleven different positive integers is 12. Let the largest of these integers be A. What can be the maximum value of A? What can be the minimum value of A?

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Answer 1

Answer:

can't be determined.we can't say the value of A.

Answer 2

The average of 11 different positive integers is 12, it can be concluded that the sum of the integers will be 12 x 11 = 132.

Largest of these integers is A.

Let the sum of the remaining 10 integers be B, so that B + A = 132 or A 132 - B

Now, for A to have the maximum value, B must have the lowest possible value.

Given that Integers are distinct and positive, B =1+2+3+4+5+6+7+8+9+10 = 55

Therefore, A= 132 - 55 = 77

Now, similarly for A to have the minimum value, B must have the largest possible value such that A is still the largest of all integers.

Thinking of the situation where average of the integers is the same as the median of the integers, I took five integers preceding 12, the integer 12 itself and five integers succeeding 12.

Of all the combinations, I found that 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 is the combination where the sum of B is largest possible. Therefore, A = 17.

Hence the maximum value of A will be 77.

And the minimum possible value of A is 17.