Difference between revisions of "2004 AMC 8 Problems/Problem 1"

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== Solution ==
 
== Solution ==
 
We set up the proportion  <math>\frac{12  \text{cm}}{72  \text{km}}=\frac{17 \text{cm}}{x \text{km}}</math>. Thus <math>x=102 \Rightarrow \boxed{\textbf{(B)}\ 102}</math>
 
We set up the proportion  <math>\frac{12  \text{cm}}{72  \text{km}}=\frac{17 \text{cm}}{x \text{km}}</math>. Thus <math>x=102 \Rightarrow \boxed{\textbf{(B)}\ 102}</math>
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== Solution 2 ==
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Since we know
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2004|before=First <br />Question|num-a=2}}
 
{{AMC8 box|year=2004|before=First <br />Question|num-a=2}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:18, 24 July 2018

Problem

On a map, a $12$-centimeter length represents $72$ kilometers. How many kilometers does a $17$-centimeter length represent?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224$

Solution

We set up the proportion $\frac{12  \text{cm}}{72  \text{km}}=\frac{17 \text{cm}}{x \text{km}}$. Thus $x=102 \Rightarrow \boxed{\textbf{(B)}\ 102}$

Solution 2

Since we know

See Also

2004 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
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All AJHSME/AMC 8 Problems and Solutions

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