Difference between revisions of "2010 AMC 8 Problems/Problem 2"

Latest revision as of 10:23, 18 November 2024

If $a @ b = \frac{a\times b}{a+b}$ for $a,b$ positive integers, then what is $5 @10$ ?

$\textbf{(A)}\ \frac{3}{10} \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{10}{3} \qquad\textbf{(E)}\ 50$

Substitute $a=5$ and $b=10$ into the expression for $a @ b$ to get:

$5 @ 10 = \frac{5\times 10}{5+10} = \frac{50}{15} = \frac{10}{3}$

Thus, answer choice $\boxed{\textbf{(D)}\ \frac{10}{3}}$ is correct.

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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

@@ Line 15: / Line 15: @@
 https://youtu.be/EEbksvfujhk
+==Video Solution by WhyMath==
+https://youtu.be/ZCpSVVCejpU
 ==See Also==
 {{AMC8 box|year=2010|num-b=1|num-a=3}}
 {{MAA Notice}}