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

(Solution)
(Solution)
 
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<math>5 @ 10 = \frac{5\times 10}{5+10} = \frac{50}{15} = \frac{10}{3}</math>
 
<math>5 @ 10 = \frac{5\times 10}{5+10} = \frac{50}{15} = \frac{10}{3}</math>
  
Thus, answer choice <math>\boxed{\textbf{(D)}\ \frac{10}{3}}</math> is correct
+
Thus, answer choice <math>\boxed{\textbf{(D)}\ \frac{10}{3}}</math> is correct.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2010|num-b=1|num-a=3}}
 
{{AMC8 box|year=2010|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:51, 2 November 2020

Problem

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

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

Solution

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.

See Also

2010 AMC 8 (ProblemsAnswer KeyResources)
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

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