Difference between revisions of "2015 AMC 12B Problems/Problem 4"
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Let <math>-</math> denote any of the 6 racers not named. Then the correct order following all the logic would look like: | Let <math>-</math> denote any of the 6 racers not named. Then the correct order following all the logic would look like: | ||
| − | <cmath>-, \text{Rand}, \text{Todd}, -, \text{Jack}, \text{Marta}, -, \text{Hikmet}, -, \text{David}, -, -</cmath> | + | <cmath>-, \text{Rand}, \text{Todd}, -, \text{Jack}, \text{Marta}, -, \text{Hikmet}, -, \text{David}, -, -</cmath> |
Clearly the 8th place runner is <math>\fbox{\textbf{(B)}\; \text{Hikmet}}</math>. | Clearly the 8th place runner is <math>\fbox{\textbf{(B)}\; \text{Hikmet}}</math>. | ||
Revision as of 13:51, 4 March 2015
Problem
David, Hikmet, Jack, Marta, Rand, and Todd were in a 12-person race with 6 other people. Rand finished 6 places ahead of Hikmet. Marta finished 1 place behind Jack. David finished 2 places behind Hikmet. Jack finished 2 places behind Todd. Todd finished 1 place behind Rand. Marta finished in 6th place. Who finished in 8th place?
Solution
Let 
denote any of the 6 racers not named. Then the correct order following all the logic would look like:
![\[-, \text{Rand}, \text{Todd}, -, \text{Jack}, \text{Marta}, -, \text{Hikmet}, -, \text{David}, -, -\]](http://latex.artofproblemsolving.com/3/8/6/386363337427005ea96973471ab2eab9e269df00.png)
Clearly the 8th place runner is 
.
See Also
| 2015 AMC 12B (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. 
