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

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<math> \textbf{(A)}\ a \qquad\textbf{(B)}\ b \qquad\textbf{(C)}\ c \qquad\textbf{(D)}\ d \qquad\textbf{(E)}\ e </math>
 
<math> \textbf{(A)}\ a \qquad\textbf{(B)}\ b \qquad\textbf{(C)}\ c \qquad\textbf{(D)}\ d \qquad\textbf{(E)}\ e </math>
 
==Solution==
 
==Solution==
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We work thorough the equation step by step, simplifying as follows:
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<cmath>\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*}</cmath>
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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==See Also==
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{{AHSME box|year=1994|num-b=9|num-a=11}}
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{{MAA Notice}}

Latest revision as of 17:33, 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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