Difference between revisions of "2007 AMC 12A Problems/Problem 4"
(wik) |
Armalite46 (talk | contribs) m (→Solution) |
||
| (4 intermediate revisions by 4 users not shown) | |||
| Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
| − | + | <cmath>16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14</cmath> | |
| − | + | <cmath>\frac{14}2=7\Rightarrow\boxed{A}</cmath> | |
| + | |||
== See also == | == See also == | ||
{{AMC12 box|year=2007|ab=A|num-b=3|num-a=5}} | {{AMC12 box|year=2007|ab=A|num-b=3|num-a=5}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 18:36, 27 October 2013
Problem
Kate rode her bicycle for 30 minutes at a speed of 16 mph, then walked for 90 minutes at a speed of 4 mph. What was her overall average speed in miles per hour?
Solution
![\[16 \cdot \frac{30}{60}+4\cdot\frac{90}{60}=14\]](http://latex.artofproblemsolving.com/1/0/f/10f984f12d433181bf4e70d5aba17bf45718855a.png)
![\[\frac{14}2=7\Rightarrow\boxed{A}\]](http://latex.artofproblemsolving.com/e/c/8/ec8d365a6c0eb1087db3b2f36dbfb57ab28a2d4d.png)
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 3 |
Followed by Problem 5 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
