Art of Problem Solving

2003 AMC 8 Problems/Problem 1: Difference between revisions

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==Problem 1==
==Problem 1==
Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?  
Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?
 
<math>\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26</math>


==Solution==
==Solution==
On a cube, there are <math> 12 </math> edges, <math> 8 </math> corners, and <math> 6 </math> faces. Adding them up gets <math> 12+8+6= \boxed{\mathrm{(E)}\ 26} </math>.
On a cube, there are <math> 12 </math> edges, <math> 8 </math> corners, and <math> 6 </math> faces. Adding them up gets <math> 12+8+6= \boxed{\mathrm{(E)}\ 26} </math>.
==See Also==
{{AMC8 box|year=2003|before=First <br />Question|num-a=2}}
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Latest revision as of 15:42, 19 August 2024

Problem 1

Jamie counted the number of edges of a cube, Jimmy counted the numbers of corners, and Judy counted the number of faces. They then added the three numbers. What was the resulting sum?

Solution

On a cube, there are $12$ edges, $8$ corners, and $6$ faces. Adding them up gets $12+8+6= \boxed{\mathrm{(E)}\ 26}$.

See Also

2003 AMC 8 (ProblemsAnswer KeyResources)
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First
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Followed by
Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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