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2005 CEMC Gauss (Grade 7) Problems/Problem 14

Revision as of 02:23, 23 October 2014 by RTG (talk | contribs) (Created page with "== Problem == The numbers on opposite sides of a die total <math>7</math>. What is the sum of the numbers on the unseen faces of the two dice shown? <math>\text{(A)}\ 14 \qqua...")
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Problem

The numbers on opposite sides of a die total $7$. What is the sum of the numbers on the unseen faces of the two dice shown?

$\text{(A)}\ 14 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 21 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 30$

[asy] import three; unitsize(1cm); size(100); currentprojection=orthographic(1/2,-1,1/2); // three - currentprojection, orthographic draw((0,0,0)--(0,0,1)); draw((1,1,0)--(1,1,1)); draw((0,0,0)--(1,0,0)); draw((1,1,0)--(1,0,0)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); dot((0,.5,0)); dot((0,1,0)); dot((0,1.5,0)); dot((-.5,1.5,0)); dot((0,1.5,-.5)); dot((1,.2,.3)); dot((1,.2,.7)); dot((1,.8,.3)); dot((1,.8,.7)); dot((.1,1.3,.6)); draw((-1,0,1)--(-1,0,0)--(-2,0,0)--(-2,0,1)); draw((-1,0,0)--(-1,1,0)); draw((-1,1,1)--(-1,1,0)); draw((-1,0,1)--(-2,0,1)--(-2,1,1)--(-1,1,1)--cycle); dot((-1.8,0,0.2)); dot((-1.5,0,0.2)); dot((-1.2,0,0.2)); dot((-1.8,0,0.8)); dot((-1.5,0,0.8)); dot((-1.2,0,0.8)); dot((-1,0.2,0.2)); dot((-1,0.8,0.8)); dot((-1.2,0.2,1)); dot((-1.5,0.5,1)); dot((-1.8,0.8,1)); [/asy]

Solution 1

Since the sum of the numbers on opposite faces on a die is $7$, then $1$ and $6$ are on opposite faces, $2$ and $5$ are on opposite faces, and $3$ and $4$ are on opposite faces. On the first die, the numbers on the unseen faces opposite the $6$, $2$, and $3$ are $1$, $5$, and $4$ respectively. On the second die, the numbers on the unseen faces opposite the $1$, $4$, and $5$ are $6$, $2$, and $3$ respectively. The sum of the missing numbers is $1 + 5 + 4 + 6 + 2 + 3 = 21$. Therefore, the answer is $C$.

Solution 2

The sum of the numbers on a die is $1 + 2 + 3 + 4 + 5 + 6 = 21$ and so the sum of the numbers on two die is $2\times 21 = 42$. Since there is a sum of $21$ showing on the six visible faces, the sum of the numbers on the six unseen faces is $42 - 21 = 21$. The correct answer is $C$.

See Also

2005 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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CEMC Gauss (Grade 7)
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