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Difference between revisions of "2008 AMC 8 Problems/Problem 24"

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==Solution==
 
==Solution==
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The numbers can at mos multiply to be <math>60</math>. The squares less than <math>60</math> are <math>1,4,9,16,25,36,</math> and <math>49</math>. The possible pairs are <math>(1,1),(1,4),(2,2),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6),</math> and <math>(9,4)</math>. There are <math>11</math> choices and <math>60</math> possibilities giving a probability of <math>\boxed{\textbf{(C)}\ \frac{11}{60}}</math>.
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The numbers can at most multiply to be <math>60</math>. The squares less than <math>60</math> are <math>1,4,9,16,25,36,</math> and <math>49</math>. The possible pairs are <math>(1,1),(1,4),(2,2),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6),</math> and <math>(9,4)</math>. There are <math>11</math> choices and <math>60</math> possibilities giving a probability of <math>\boxed{\textbf{(C)}\ \frac{11}{60}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=23|num-a=25}}
 
{{AMC8 box|year=2008|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:34, 6 October 2024

Problem

Ten tiles numbered $1$ through $10$ 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?

$\textbf{(A)}\ \frac{1}{10}\qquad\textbf{(B)}\ \frac{1}{6}\qquad\textbf{(C)}\ \frac{11}{60}\qquad\textbf{(D)}\ \frac{1}{5}\qquad\textbf{(E)}\ \frac{7}{30}$


Video Solution

https://www.youtube.com/watch?v=ZqPFm9cU0MY ~David

Solution

The numbers can at most multiply to be $60$. The squares less than $60$ are $1,4,9,16,25,36,$ and $49$. The possible pairs are $(1,1),(1,4),(2,2),(4,1),(3,3),(9,1),(4,4),(8,2),(5,5),(6,6),$ and $(9,4)$. There are $11$ choices and $60$ possibilities giving a probability of $\boxed{\textbf{(C)}\ \frac{11}{60}}$.

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

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