Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2022 AMC 8 Problems/Problem 12: Difference between revisions

← Older editNewer edit →
Mrthinker (talk | contribs)
editor
543 edits
Mcdsouza2025 (talk | contribs)
editor
27 edits
No edit summary
Line 43: Line 43:


~Interstigation
~Interstigation
==Video Solution==
https://youtu.be/p29Fe2dLGs8?t=58
~STEMbreezy


==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}}

Revision as of 20:18, 25 December 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

Video Solution

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

~STEMbreezy

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_8_Problems/Problem_12&oldid=184827"