Difference between revisions of "2009 AMC 8 Problems/Problem 21"

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.