Difference between revisions of "2022 AMC 8 Problems/Problem 12"
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==Problem== | ==Problem== | ||
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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> | + | The arrows on the two spinners shown below are spun. Let the number <math>N</math> equal <math>10</math> times the number on Spinner <math>\text{A}</math>, added to the number on Spinner <math>\text{B}</math>. What is the probability that <math>N</math> 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> | ||
+ | <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 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 | ||
+ | |||
+ | ==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= | + | {{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?
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
~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.