Difference between revisions of "2004 AMC 8 Problems/Problem 1"
m (→Problem) |
(→Solution) |
||
Line 6: | Line 6: | ||
== 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> | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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 03:18, 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
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.