Difference between revisions of "2009 AMC 10A Problems/Problem 20"
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| − | Let their speeds in kilometers per hour be <math>v_A</math> and <math>v_L</math>. We know that <math>v_A=3v_L</math> and that <math>v_A+v_L=60</math>. (The second equation follows from the fact that <math>1 | + | Let their speeds in kilometers per hour be <math>v_A</math> and <math>v_L</math>. We know that <math>v_A=3v_L</math> and that <math>v_A+v_L=60</math>. (The second equation follows from the fact that <math>1,\unit{\mathrm km/min} = 60,\unit{\mathrm km/h}</math>.) This solves to <math>v_A=45</math> and <math>v_L=15</math>. |
As the distance decreases at a rate of <math>1</math> kilometer per minute, after <math>5</math> minutes the distance between them will be <math>20-5=15</math> kilometers. | As the distance decreases at a rate of <math>1</math> kilometer per minute, after <math>5</math> minutes the distance between them will be <math>20-5=15</math> kilometers. | ||
Revision as of 20:56, 16 January 2016
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
Let their speeds in kilometers per hour be 
and 
. We know that 
and that 
. (The second equation follows from the fact that $1,\unit{\mathrm km/min} = 60,\unit{\mathrm km/h}$ (Error compiling LaTeX. Unknown error_msg).) 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 
.
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. 
