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Difference between revisions of "2009 AMC 8 Problems/Problem 21"
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==Solution==
 
==Solution==
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Note that <math>40A=75B=\text{sum of the numbers in the array}</math>. This means that <math>\frac{A}{B}=\boxed{\text{(D) } \frac{15}{8}}</math> using basic algebraic manipulation.
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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{A}{B}=\frac{75}{40} \Rightarrow \boxed{\textbf{(D)}\  \frac{15}{8}}</math>
 
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==Solution==
 
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Any solution would have to work for any choice of numbers.  In particular the special choice
 
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of all numbers being zero.  In that case, however, <math>B=0</math> and the fraction <math>\frac{A}{B}</math> is undefined.  The problem as stated therefore has no solution.
 
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2009|num-b=20|num-a=22}}
 
{{AMC8 box|year=2009|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 10:37, 16 January 2022

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

Note that $40A=75B=\text{sum of the numbers in the array}$. This means that $\frac{A}{B}=\boxed{\text{(D) } \frac{15}{8}}$ using basic algebraic manipulation.

See Also

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

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