Difference between revisions of "2022 AMC 8 Problems/Problem 12"

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First, we realize that there are a total of <math>16</math> possibilities. Now, we list all of them that can be spun. This includes <math>64</math> and <math>81</math>. Then, our answer is <math>\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}</math>.
 
First, we realize that there are a total of <math>16</math> possibilities. Now, we list all of them that can be spun. This includes <math>64</math> and <math>81</math>. Then, our answer is <math>\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}</math>.
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~MathFun1000
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=10|num-a=12}}
 
{{AMC8 box|year=2022|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:17, 28 January 2022

Problem

The arrows on the two spinners shown below are spun. Let the number $N$ equal 10 times the number on Spinner A, added to the number on Spinner B. What is the probability that N is a perfect square number?

$\textbf{(A) } \dfrac{1}{16}\qquad\textbf{(B) } \dfrac{1}{8}\qquad\textbf{(C) } \dfrac{1}{4}\qquad\textbf{(D) } \dfrac{3}{8}\qquad\textbf{(E) } \dfrac{1}{2}$

Solution

First, we realize that there are a total of $16$ possibilities. Now, we list all of them that can be spun. This includes $64$ and $81$. Then, our answer is $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$.

~MathFun1000

See Also

2022 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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