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> | ||
| + | |||
| + | ==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}} | ||
Revision as of 20:36, 5 February 2015
Contents
Problem
Andy and Bethany have a rectangular array of numbers 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 
. What is the value of 
?
Solution
First, note that 
. Solving for , we get 
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, 
and the fraction is undefined. The problem as stated therefore has no solution.
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. 
