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

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==Problem==
 
==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 <math> 40</math> sums is <math> A</math>. Bethany adds the numbers in each column. The average of her <math> 75</math> sums is <math> B</math>. What is the value of <math> \frac{A}{B}</math>?
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Andy and Bethany have a rectangular array of numbers greater than zero with <math> 40</math> rows and <math> 75</math> columns. Andy adds the numbers in each row. The average of his <math> 40</math> sums is <math> A</math>. Bethany adds the numbers in each column. The average of her <math> 75</math> sums is <math> B</math>. Using only the answer choices given, What is the value of <math> \frac{A}{B}</math>?
  
 
<math> \textbf{(A)}\ \frac{64}{225}    \qquad
 
<math> \textbf{(A)}\ \frac{64}{225}    \qquad
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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{A}{B}=\frac{75}{40} \Rightarrow \boxed{\textbf{(D)}\frac{15}{8}}</math>
 
 
 
==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, <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 14:46, 6 January 2018

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

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
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All AJHSME/AMC 8 Problems and Solutions

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