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Difference between revisions of "2019 AMC 8 Problems/Problem 12"

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==Solution 1==
 
==Solution 1==
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<math>B</math> is on the top, and <math>R</math> is on the side, and <math>G</math> is on the right side. That means that (image <math>2</math>) <math>W</math> is on the left side. From the third image, you know that <math>P</math> must be on the bottom since <math>G</math> is sideways. That leaves us with the back, so the back must be <math>A</math>. The front is opposite of the back, so the answer is <math>\boxed{\textbf{(A)}\ R}</math>.~heeeeeeeheeeee
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<math>B</math> is on the top, and <math>R</math> is on the side, and <math>G</math> is on the right side. That means that (image <math>2</math>) <math>W</math> is on the left side. From the third image, you know that <math>P</math> must be on the bottom since <math>G</math> is sideways. That leaves us with the back, so the back must be <math>A</math>. The front is opposite of the back, so the answer is <math>\boxed{\textbf{(A)}\ R}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 09:33, 25 February 2021

Problem

The faces of a cube are painted in six different colors: red $(R)$, white $(W)$, green $(G)$, brown $(B)$, aqua $(A)$, and purple $(P)$. Three views of the cube are shown below. What is the color of the face opposite the aqua face?

2019AMC8Prob12.png

Solution 1

$B$ is on the top, and $R$ is on the side, and $G$ is on the right side. That means that (image $2$) $W$ is on the left side. From the third image, you know that $P$ must be on the bottom since $G$ is sideways. That leaves us with the back, so the back must be $A$. The front is opposite of the back, so the answer is $\boxed{\textbf{(A)}\ R}$.

Solution 2

Looking closely we can see that all faces are connected with $R$ except for $A$. Thus the answer is $\boxed{\textbf{(A)}\ R}$.

It is A, just draw it out! ~phoenixfire

Solution 3

Associated video - https://www.youtube.com/watch?v=K5vaX_EzjEM

See also

2019 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

Only two of the cubes are required to solve the problem.

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