2009 AMC 8 Problems/Problem 21

Problem

Andy and Bethany have a rectangular array of numbers 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 $B$ . 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.

2009 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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.

Page

Toolbox

Search

2009 AMC 8 Problems/Problem 21

Contents

Problem

Solution

Solution

See Also