Difference between revisions of "2010 AMC 12B Problems/Problem 7"
m (→See also) |
(→Solution) |
||
| Line 5: | Line 5: | ||
== Solution == | == Solution == | ||
| + | |||
| + | Let <math>x</math> be the time it is not raining, and <math>y</math> be the time it is raining, in hours. | ||
| + | |||
| + | We have the system: <math>30x+20y=16</math> and <math>x+y=2/3</math> | ||
| + | |||
| + | Solving gives <math>x=\frac{4}{15}</math> and <math>y=\frac{2}{5}</math> | ||
| + | |||
| + | We want <math>y</math> in minutes, <math>\frac{2}{5}*60=24 \Rightarrow C</math> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2010|num-b=6|num-a=8|ab=B}} | {{AMC12 box|year=2010|num-b=6|num-a=8|ab=B}} | ||
Revision as of 18:20, 28 January 2011
Problem 7
Shelby drives her scooter at a speed of 
miles per hour if it is not raining, and 
miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of 
miles in 
minutes. How many minutes did she drive in the rain?
Solution
Let 
be the time it is not raining, and 
be the time it is raining, in hours.
We have the system: 
and 
Solving gives 
and 
We want in minutes, 
See also
| 2010 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 | |
