Difference between revisions of "2008 AMC 8 Problems/Problem 24"
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<math>\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\ \frac{1}{5}\qquad\textbf{(E)}\ \frac{7}{30}</math> | <math>\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\ \frac{1}{5}\qquad\textbf{(E)}\ \frac{7}{30}</math> | ||
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| + | == Video Solution == | ||
| + | https://www.youtube.com/watch?v=ZqPFm9cU0MY ~David | ||
==Solution== | ==Solution== | ||
Latest revision as of 17:34, 6 October 2024
Contents
Problem
Ten tiles numbered 
through 
are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?
Video Solution
https://www.youtube.com/watch?v=ZqPFm9cU0MY ~David
Solution
The numbers can at most multiply to be 
. The squares less than are 
and 
. The possible pairs are 
and 
. There are 
choices and possibilities giving a probability of 
.
See Also
| 2008 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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. 
