Difference between revisions of "2022 AMC 8 Problems/Problem 12"
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==Solution 3== | ==Solution 3== | ||
Just try them! | 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== | ==Video Solution== | ||
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~savannahsolver | ~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
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.