2009 AMC 8 Problems/Problem 21

Revision as of 22:43, 3 December 2015 by Williamliang451 (talk | contribs) (Solution)

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 $\frac{A}{B}=\frac{75}{40} \Rightarrow \boxed{\textbf{(D)}\   \frac{15}{8}}$

See Also

2009 AMC 8 (ProblemsAnswer KeyResources)
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. AMC logo.png