Difference between revisions of "2004 AMC 8 Problems/Problem 1"
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== Problem == | == Problem == | ||
| − | + | On a map, a <math>12</math>-centimeter length represents <math>72</math> kilometers. How many kilometers does a <math>17</math>-centimeter length represent? | |
<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224 </math> | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 102\qquad\textbf{(C)}\ 204\qquad\textbf{(D)}\ 864\qquad\textbf{(E)}\ 1224 </math> | ||
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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 == | ||
| + | Since we know <math>12</math> centimeters on the map is equal to <math>72</math> kilometers, <math>1</math> centimeter on the map is simply <math>72/12</math>, which is <math>6</math>. The problem is asking how many kilometers a <math>17</math>-centimeter length represents, so we multiply <math>6</math> by <math>17</math> which is simply 102, your answer. | ||
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| + | ==See Also== | ||
| + | {{AMC8 box|year=2004|before=First <br />Question|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 03:21, 24 July 2018
Contents
Problem
On a map, a 
-centimeter length represents 
kilometers. How many kilometers does a 
-centimeter length represent?
Solution
We set up the proportion 
. Thus 
Solution 2
Since we know centimeters on the map is equal to kilometers, 
centimeter on the map is simply 
, which is 
. The problem is asking how many kilometers a -centimeter length represents, so we multiply by which is simply 102, your answer.
See Also
| 2004 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
