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

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==Solution 3==
 
==Solution 3==
 
Just try them!  
 
Just try them!  
−
If we spin a 5 on the first spinner, there are no solutions.
+
<cmath>If we spin a 5 on the first spinner, there are no solutions.</cmath>
−
If we spin a 6 on the first spinner, there is one solution (64).
+
<cmath>If we spin a 6 on the first spinner, there is one solution (64).</cmath>
−
If we spin a 7 on the first spinner, there are no solutions.
+
<cmath>If we spin a 7 on the first spinner, there are no solutions.</cmath>
−
If we spin an 8 on the first spinner, there is one solution (81).
+
<cmath>If we spin an 8 on the first spinner, there is one solution (81).</cmath>
−
Therefore, there are 2 solutions and <math>4 \cdot 4 = 16</math> total possibilities, so <cmath>\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}</cmath>
+
<math></math>Therefore, there are 2 solutions and <math>4 \cdot 4 = 16</math> total possibilities, so <cmath>\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}</cmath><math></math>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 10:07, 17 February 2023

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 calculate that there are a total of $4\cdot4=16$ possibilities. Now, we list all of two-digit perfect squares. $64$ and $81$ are the only ones that can be made using the spinner. Consequently, there is a $\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}$ probability that the number formed by the two spinners is a perfect square.

~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

Solution 3

Just try them! \[If we spin a 5 on the first spinner, there are no solutions.\] \[If we spin a 6 on the first spinner, there is one solution (64).\] \[If we spin a 7 on the first spinner, there are no solutions.\] \[If we spin an 8 on the first spinner, there is one solution (81).\] $$ (Error compiling LaTeX. Unknown error_msg)Therefore, there are 2 solutions and $4 \cdot 4 = 16$ total possibilities, so \[\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}\]$$ (Error compiling LaTeX. Unknown error_msg)

Video Solution

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

~Interstigation

Video Solution

https://youtu.be/p29Fe2dLGs8?t=58

~STEMbreezy

Video Solution

https://youtu.be/Wrbpq4D5F18

~savannahsolver

See Also

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