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

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<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>
 
<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>.
+
First, we calculate that there are a total of <math>4\cdot4=16</math> possibilities. Now, we list all of two-digit perfect squares. <math>64</math> and <math>81</math> are the only ones that can be made using the spinner. Consequently, there is a <math>\frac{2}{16}=\boxed{\textbf{(B) }\dfrac{1}{8}}</math> probability that the number formed by the two spinners is a perfect square.
  
 
~MathFun1000
 
~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
 +
 +
==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).
 +
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>
 +
 +
~ligonmathkid2
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/oUEa7AjMF2A?si=NvktZSXqcd8LE9yU&t=1604
 +
 +
~Math-X
 +
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/Flq6oB23_WM
 +
 +
~Education, the Study of Everything
 +
 +
==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
 +
 +
==Video Solution==
 +
https://youtu.be/U6eqFrD7BXA
 +
 +
~harungurcan
  
 
==See Also==  
 
==See Also==  
 
{{AMC8 box|year=2022|num-b=11|num-a=13}}
 
{{AMC8 box|year=2022|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:44, 23 November 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).

Therefore, there are 2 solutions and $4 \cdot 4 = 16$ total possibilities, so \[\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}\]

~ligonmathkid2

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/oUEa7AjMF2A?si=NvktZSXqcd8LE9yU&t=1604

~Math-X

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/Flq6oB23_WM

~Education, the Study of Everything

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

Video Solution

https://youtu.be/U6eqFrD7BXA

~harungurcan

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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