2009 AMC 8 Problems/Problem 21

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Problem

Andy and Bethany have a rectangular array of numbers greater than zero with $40$ rows and $75$ columns. Andy adds the numbers in each row. The average of his $40$ sums is $A$. Bethany adds the numbers in each column. The average of her $75$ sums is $B$. Using only the answer choices given, What is the value of $\frac{A}{B}$?

$\textbf{(A)}\ \frac{64}{225}     \qquad \textbf{(B)}\   \frac{8}{15}    \qquad \textbf{(C)}\    1   \qquad \textbf{(D)}\   \frac{15}{8}    \qquad \textbf{(E)}\    \frac{225}{64}$

Solution

First, note that $40A=75B=\text{sum of the numbers in the array}$. Solving for $\frac{A}{B}$, we get $\frac{A}{B}=\frac{75}{40} \Rightarrow \boxed{\textbf{(D)}\   \frac{15}{8}}$

Solution

Any solution would have to work for any choice of numbers. In particular the special choice of all numbers being zero. In that case, however, $B=0$ and the fraction $\frac{A}{B}$ is undefined. The problem as stated therefore has no solution.

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AJHSME/AMC 8 Problems and Solutions

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