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

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==Problem==
 
==Problem==
  
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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>?
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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>?
  
 
<math> \textbf{(A)}\ \frac{64}{225}    \qquad
 
<math> \textbf{(A)}\ \frac{64}{225}    \qquad

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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