1991 AJHSME Problems/Problem 19

Problem

The average (arithmetic mean) of $10$ different positive whole numbers is $10$ . The largest possible value of any of these numbers is

$\text{(A)}\ 10 \qquad \text{(B)}\ 50 \qquad \text{(C)}\ 55 \qquad \text{(D)}\ 90 \qquad \text{(E)}\ 91$

Solution

If the average of the numbers is $10$ , then their sum is $10\times 10=100$ .

To maximize the largest number of the ten, we minimize the other nine. Since they must be distinct, positive whole numbers, we let them be $1,2,3,4,5,6,7,8,9$ . Their sum is $45$ .

The sum of nine of the numbers is $45$ , and the sum of all ten is $100$ so the last number must be $100-45=55\rightarrow \boxed{\text{C}}$ .

1991 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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1991 AJHSME Problems/Problem 19

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