Difference between revisions of "2009 AMC 8 Problems/Problem 21"
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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> | 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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==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}} | ||
Revision as of 22:43, 3 December 2015
Problem
Andy and Bethany have a rectangular array of numbers greater than zero with 
rows and 
columns. Andy adds the numbers in each row. The average of his sums is 
. Bethany adds the numbers in each column. The average of her sums is 
. Using only the answer choices given, What is the value of 
?
Solution
First, note that 
. Solving for , we get 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
