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

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.

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.

@@ Line 10: / Line 10: @@
 ==Solution==
+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>
 ==See Also==
 {{AMC8 box|year=2009|num-b=20|num-a=22}}
+{{MAA Notice}}

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

Latest revision as of 10:37, 16 January 2022

Problem

Solution

See Also