Difference between revisions of "2009 AMC 10A Problems/Problem 20"
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In all, Lauren will have to bike <math>20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}</math>km. Because her speed is <math>\frac{1}{4} \textbf{km/min}</math>, the time elapsed will be <math>\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}</math> | In all, Lauren will have to bike <math>20 - \frac{15}{4} = \frac{80}{4} - \frac{15}{4} = \frac{65}{4}</math>km. Because her speed is <math>\frac{1}{4} \textbf{km/min}</math>, the time elapsed will be <math>\frac{\frac{65}{4}}{\frac{1}{4}} = \boxed{\textbf{D) 65}}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/YJigRBg0LIQ | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Revision as of 20:21, 21 December 2020
Problem
Andrea and Lauren are kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of kilometer per minute. After minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea?
Solution 1
Let their speeds in kilometers per hour be and . We know that and that . (The second equation follows from the fact that .) This solves to and .
As the distance decreases at a rate of kilometer per minute, after minutes the distance between them will be kilometers.
From this point on, only Lauren will be riding her bike. As there are kilometers remaining and , she will need exactly an hour to get to Andrea. Therefore the total time in minutes is .
Solution 2
Because the speed of Andrea is 3 times as fast as Lauren and the distance between them is decreasing at a rate of 1 kilometer per minute, Andrea's speed is , and Lauren's . Therefore, after 5 minutes, Andrea will have biked km.
In all, Lauren will have to bike km. Because her speed is , the time elapsed will be
Video Solution
~savannahsolver
See Also
2009 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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.