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1991 AJHSME Problems/Problem 19

Revision as of 04:08, 16 August 2015 by Cheap18 (talk | contribs) (→‎Solution)

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 $45$.

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See Also

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
All AJHSME/AMC 8 Problems and Solutions

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