2022 AMC 8 Problems/Problem 12: Difference between revisions

Revision as of 09:24, 7 October 2022

Problem

The arrows on the two spinners shown below are spun. Let the number $N$ equal $10$ times the number on Spinner $\text{A}$ , added to the number on Spinner $\text{B}$ . What is the probability that $N$ is a perfect square number? $[asy] //diagram by pog give me 1 billion dollars for this size(6cm); usepackage("mathptmx"); filldraw(arc((0,0), r=4, angle1=0, angle2=90)--(0,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((0,0), r=4, angle1=90, angle2=180)--(0,0)--cycle,lightgray); filldraw(arc((0,0), r=4, angle1=180, angle2=270)--(0,0)--cycle,mediumgray); filldraw(arc((0,0), r=4, angle1=270, angle2=360)--(0,0)--cycle,lightgray*0.5+mediumgray*0.5); label("$5$", (-1.5,1.7)); label("$6$", (1.5,1.7)); label("$7$", (1.5,-1.7)); label("$8$", (-1.5,-1.7)); label("Spinner A", (0, -5.5)); filldraw(arc((12,0), r=4, angle1=0, angle2=90)--(12,0)--cycle,mediumgray*0.5+gray*0.5); filldraw(arc((12,0), r=4, angle1=90, angle2=180)--(12,0)--cycle,lightgray); filldraw(arc((12,0), r=4, angle1=180, angle2=270)--(12,0)--cycle,mediumgray); filldraw(arc((12,0), r=4, angle1=270, angle2=360)--(12,0)--cycle,lightgray*0.5+mediumgray*0.5); label("$1$", (10.5,1.7)); label("$2$", (13.5,1.7)); label("$3$", (13.5,-1.7)); label("$4$", (10.5,-1.7)); label("Spinner B", (12, -5.5)); [/asy]$ $\textbf{(A)} ~\dfrac{1}{16}\qquad\textbf{(B)} ~\dfrac{1}{8}\qquad\textbf{(C)} ~\dfrac{1}{4}\qquad\textbf{(D)} ~\dfrac{3}{8}\qquad\textbf{(E)} ~\dfrac{1}{2}$

Solution 1

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

Solution 2

There are $4 \cdot 4 = 16$ total possibilities of $N$ . We know $N=10A+B$ , which $A$ is a number from spinner $A$ , and $B$ is a number from spinner $B$ . Also, notice that there are no perfect squares in the $50$ s or $70$ s, so only $4-2=2$ values of N work, namely $64$ and $81$ . Hence, $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$ .

~MrThinker

Video Solution

https://youtu.be/Ij9pAy6tQSg?t=1008

~Interstigation

@@ Line 27: / Line 27: @@
 <math>\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}</math>
-==Solution==
+==Solution 1==
 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>.
 ~MathFun1000
+==Solution 2==
+There are <math>4 \cdot 4 = 16</math> total possibilities of <math>N</math>. We know <math>N=10A+B</math>, which <math>A</math> is a number from spinner <math>A</math>, and <math>B</math> is a number from spinner <math>B</math>. Also, notice that there are no perfect squares in the <math>50</math>s or <math>70</math>s, so only  <math>4-2=2</math> values of N work, namely <math>64</math> and <math>81</math>. Hence, <math>\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}</math>.
+~MrThinker
 ==Video Solution==

2022 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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