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

(Created page with "==Problem== Andy and Bethany have a rectangular array of numbers with <math> 40</math> rows and <math> 75</math> columns. Andy adds the numbers in each row. The average of his <...")
 
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\textbf{(D)}\  \frac{15}{8}    \qquad
 
\textbf{(D)}\  \frac{15}{8}    \qquad
 
\textbf{(E)}\    \frac{225}{64}</math>
 
\textbf{(E)}\    \frac{225}{64}</math>
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==Solution==
 +
 +
First, note that <math>40A=75B=\text{sum of the numbers in the array}</math>. Solving for <math> \frac{A}{B}</math>, we get <math> \frac{75}{40} =\frac{15}{8} =\frac{A}{B}</math>. <math> \boxed{\text{D}}</math>.

Revision as of 08:31, 27 April 2012

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 $\frac{75}{40} =\frac{15}{8} =\frac{A}{B}$. $\boxed{\text{D}}$.

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