Difference between revisions of "2003 AMC 8 Problems/Problem 1"

 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
==Problem==
+
==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>
 
<math>\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26</math>
 +
  
 
==Solution==
 
==Solution==
Line 9: Line 10:
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2003|before=First <br />Question|num-a=2}}
 
{{AMC8 box|year=2003|before=First <br />Question|num-a=2}}
 +
{{MAA Notice}}

Latest revision as of 13:00, 29 November 2020

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?

$\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26$


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)
Preceded by
First
Question
Followed by
Problem 2
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. AMC logo.png