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 $45$.

The sum of nine of the numbers is $45$, and the sum of all ten is $100$ so the last number must be $100-45=55\rightarrow \boxed{\text{C}}$. Nowaday, a lot of people are intersed in Cosplay. Many people are wearing <a href="http://www.zentaiclothing.com/zentai-suits-sale/mens-pvc-zentai-suits.html">Mens PVC Zentai Suits</a>, <a href="http://www.zentaiclothing.com/lycra-spandex-zentai-suits.html">Lycra Spandex Zentai Suits</a> and <a href="http://www.zentaiclothing.com/zentai-suits-sale/mens-full-body-zentai-suits.html">Mens Full Body Zentai Suits</a>. Some young people will take <a href="http://www.zentaiclothing.com/cospaly-costumes/catwoman-cosplay-costumes.html">Catwoman Cosplay Costumes</a> and <a href="http://www.zentaiclothing.com/superhero-cosplay-costumes.html">Superhero Cosplay Costumes</a>. Among these <a href="http://www.zentaiclothing.com/">Cheap Cosplay Costumes</a>, the most popular ones are <a href="http://www.zentaiclothing.com/superhero-cosplay-costumes/superman-zentai-suits.html">Superman Zentai Suits</a>, <a href="http://www.zentaiclothing.com/superhero-cosplay-costumes/spiderman-zentai-suits.html">Spiderman Zentai Suits</a> and <a href="http://www.zentaiclothing.com/disney-cosplay-costumes.html">Disney Cosplay Costumes</a>. You will find them in <a href="http://www.zentaiclothing.com/">Zentai Suits Online</a> store.

See Also

1991 AJHSME (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS