Difference between revisions of "2022 AMC 8 Problems/Problem 12"
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*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</math> total possibilities, so <cmath>\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}</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> | ||
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| + | ~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== | ==Video Solution== | ||
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~savannahsolver | ~savannahsolver | ||
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| + | ==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 13:44, 23 November 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!
- 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 total possibilities, so ![\[\frac{2}{16} = \boxed{\textbf{(B) }\dfrac{1}{8}}\]](http://latex.artofproblemsolving.com/b/7/1/b715fab0034a163c9c900b4c37a0a68ceeac0d39.png)
~ligonmathkid2
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=NvktZSXqcd8LE9yU&t=1604
~Math-X
Video Solution (CREATIVE THINKING!!!)
~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
~savannahsolver
Video Solution
~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. 
