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Difference between revisions of "2005 AMC 8 Problems/Problem 1"

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==Problem 1==
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==Problem==
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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?  
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Connie multiplies a number by <math>2</math> and gets <math>60</math> as her answer. However, she should have divided the number by <math>2</math> to get the correct answer. What is the correct answer?
  
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<math>\mathrm{(A)}\ 12 \qquad\mathrm{(B)}\ 16 \qquad\mathrm{(C)}\ 20 \qquad\mathrm{(D)}\ 22 \qquad\mathrm{(E)}\ 26</math>
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<math> \textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240 </math>
  
 
==Solution==
 
==Solution==
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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{(B)}\ 16} </math>.
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If <math>x</math> is the number, then <math>2x=60</math> and <math>x=30</math>. Dividing the number by <math>2</math> yields <math>\dfrac{30}{2} = \boxed{\textbf{(B)}\ 15}</math>.
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A second way to do it is to divide the number by <math>4</math>, as you multiplied by <math>2</math> when you were supposed to divide by <math>2</math>. So, <math>\dfrac{60}{4} = \boxed{\textbf{(B)}\ 15}</math>.
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==See Also==
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{{AMC8 box|year=2005|before=First <br/> Question|num-a=2}}
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{{MAA Notice}}

Latest revision as of 09:10, 8 January 2024

Problem

Connie multiplies a number by $2$ and gets $60$ as her answer. However, she should have divided the number by $2$ to get the correct answer. What is the correct answer?

$\textbf{(A)}\ 7.5\qquad\textbf{(B)}\ 15\qquad\textbf{(C)}\ 30\qquad\textbf{(D)}\ 120\qquad\textbf{(E)}\ 240$

Solution

If $x$ is the number, then $2x=60$ and $x=30$. Dividing the number by $2$ yields $\dfrac{30}{2} = \boxed{\textbf{(B)}\ 15}$.

A second way to do it is to divide the number by $4$, as you multiplied by $2$ when you were supposed to divide by $2$. So, $\dfrac{60}{4} = \boxed{\textbf{(B)}\ 15}$.

See Also

2005 AMC 8 (Problems • Answer Key • Resources)
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

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