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.
  
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>.
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==See Also==
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{{AMC8 box|year=2009|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 11:37, 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 (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

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