Difference between revisions of "2010 AMC 12B Problems/Problem 7"
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Let <math>x</math> be the time it is not raining, and <math>y</math> be the time it is raining, in hours. | 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 | + | We have the system: <math>30x+20y=16</math> and <math>x+y=\frac{2}{3}</math> |
Solving gives <math>x=\frac{4}{15}</math> and <math>y=\frac{2}{5}</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} | + | We want <math>y</math> in minutes, <math>\frac{2}{5} \cdot 60=24 \Rightarrow C</math> |
== Solution 2 == | == Solution 2 == | ||
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Since we are calculating the time in minutes, it is best to convert the speeds in minutes. Thus, the speed per minute when it is not raining is <math>\frac{1}{2}</math> miles per minute, and <math>\frac{1}{3}</math> miles per minute when it is raining. | Since we are calculating the time in minutes, it is best to convert the speeds in minutes. Thus, the speed per minute when it is not raining is <math>\frac{1}{2}</math> miles per minute, and <math>\frac{1}{3}</math> miles per minute when it is raining. | ||
| − | Thus, we have the equation, <math>\frac{1}{2} | + | Thus, we have the equation, <math>\frac{1}{2} \cdot (40-x) + \frac{1}{3} \cdot x = 16</math> |
Solving, gives <math>x</math> = <math>24</math> , so the amount of time she drove in the rain is <math>24</math> minutes. | Solving, gives <math>x</math> = <math>24</math> , so the amount of time she drove in the rain is <math>24</math> minutes. | ||
~coolmath2017 | ~coolmath2017 | ||
| + | |||
| + | ==Video Solution== | ||
| + | https://youtu.be/7GezNKKIEl4 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==Video Solution== | ==Video Solution== | ||
Latest revision as of 15:22, 20 August 2024
- The following problem is from both the 2010 AMC 12B #7 and 2010 AMC 10B #10, so both problems redirect to this page.
Problem
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 1
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, 
Solution 2
Let be the time she drove in the rain. Thus, the number of minutes she did not drive in the rain is 
.
Since we are calculating the time in minutes, it is best to convert the speeds in minutes. Thus, the speed per minute when it is not raining is 
miles per minute, and 
miles per minute when it is raining.
Thus, we have the equation, 
Solving, gives = 
, so the amount of time she drove in the rain is minutes.
~coolmath2017
Video Solution
~Education, the Study of Everything
Video Solution
https://youtu.be/I3yihAO87CE?t=429
~IceMatrix
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 | |
| 2010 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
