Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1994 AHSME Problems/Problem 10"

(→‎Solution)
m (→‎Solution)
 
Line 9: Line 9:
  
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 +
 +
==See Also==
 +
 +
{{AHSME box|year=1994|num-b=9|num-a=11}}
 +
{{MAA Notice}}

Latest revision as of 16:33, 9 January 2021

Problem

For distinct real numbers $x$ and $y$, let $M(x,y)$ be the larger of $x$ and $y$ and let $m(x,y)$ be the smaller of $x$ and $y$. If $a<b<c<d<e$, then \[M(M(a,m(b,c)),m(d,m(a,e)))=\] $\textbf{(A)}\ a \qquad\textbf{(B)}\ b \qquad\textbf{(C)}\ c \qquad\textbf{(D)}\ d \qquad\textbf{(E)}\ e$

Solution

We work thorough the equation step by step, simplifying as follows:

\begin{align*}M(M(a,m(b,c)),m(d,m(a,e)))&=M(M(a,b),m(d,a))\\&=M(b,a)\\&=\boxed{\textbf{(B) }b}\end{align*}

--Solution by TheMaskedMagician

See Also

1994 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1994_AHSME_Problems/Problem_10&oldid=141788"