Difference between revisions of "2022 AMC 8 Problems/Problem 12"
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==Solution 3== | ==Solution 3== | ||
Just try them! | Just try them! | ||
| − | <math>If we spin a 5 on the first spinner, there are no solutions. | + | <math>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 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 a 7 on the first spinner, there are no solutions. | ||
If we spin an 8 on the first spinner, there is one solution (81). | 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 | + | Therefore, there are 2 solutions and </math>4 \cdot 4 = 16$ total possibilities, so <cmath>\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}</cmath> |
==Video Solution== | ==Video Solution== | ||
Revision as of 10:14, 17 February 2023
Contents
Problem
The arrows on the two spinners shown below are spun. Let the number 
equal 
times the number on Spinner 
, added to the number on Spinner 
. What is the probability that 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]](http://latex.artofproblemsolving.com/5/5/2/552b70fd0a4e9d6d464f53f730e82883afd5b278.png)
Solution 1
First, we calculate that there are a total of 
possibilities. Now, we list all of two-digit perfect squares. 
and 
are the only ones that can be made using the spinner. Consequently, there is a 
probability that the number formed by the two spinners is a perfect square.
~MathFun1000
Solution 2
There are 
total possibilities of . We know 
, which 
is a number from spinner , and 
is a number from spinner . Also, notice that there are no perfect squares in the 
s or 
s, so only 
values of N work, namely and . Hence, .
~MrThinker
Solution 3
Just try them!
4 \cdot 4 = 16$ total possibilities, so ![\[\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}\]](http://latex.artofproblemsolving.com/b/7/1/b715fab0034a163c9c900b4c37a0a68ceeac0d39.png)
Video Solution
https://youtu.be/Ij9pAy6tQSg?t=1008
~Interstigation
Video Solution
https://youtu.be/p29Fe2dLGs8?t=58
~STEMbreezy
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
