Difference between revisions of "2015 AMC 12B Problems/Problem 4"
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<math>\textbf{(A)}\; \text{David} \qquad\textbf{(B)}\; \text{Hikmet} \qquad\textbf{(C)}\; \text{Jack} \qquad\textbf{(D)}\; \text{Rand} \qquad\textbf{(E)}\; \text{Todd}</math> | <math>\textbf{(A)}\; \text{David} \qquad\textbf{(B)}\; \text{Hikmet} \qquad\textbf{(C)}\; \text{Jack} \qquad\textbf{(D)}\; \text{Rand} \qquad\textbf{(E)}\; \text{Todd}</math> | ||
| − | ==Solution== | + | ==Solution 1== |
Let <math>-</math> denote any of the 6 racers not named. Then the correct order following all the logic looks like: | Let <math>-</math> denote any of the 6 racers not named. Then the correct order following all the logic looks like: | ||
| Line 10: | Line 10: | ||
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>. | ||
| + | |||
| + | ==Solution 2== | ||
| + | We can list these out vertically to ensure clarity, starting with Marta and working from there. | ||
| + | |||
| + | |||
| + | <cmath>1 - </cmath> | ||
| + | <cmath>2 R </cmath> | ||
| + | <cmath>3 T </cmath> | ||
| + | <cmath>4 - </cmath> | ||
| + | <cmath>5 J </cmath> | ||
| + | <cmath>6 M </cmath> | ||
| + | <cmath>7 - </cmath> | ||
| + | <cmath>8 H </cmath> | ||
| + | |||
| + | Thus our answer is <math>\fbox{\textbf{(B)}\; \text{Hikmet}}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=5|num-b=3}} | {{AMC12 box|year=2015|ab=B|num-a=5|num-b=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:40, 20 January 2018
Contents
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 1
Let 
denote any of the 6 racers not named. Then the correct order following all the logic looks 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 
.
Solution 2
We can list these out vertically to ensure clarity, starting with Marta and working from there.
![\[1 -\]](http://latex.artofproblemsolving.com/1/0/5/105aa601b74d255fdec68cea63a561c4634d492d.png)
![\[2 R\]](http://latex.artofproblemsolving.com/4/6/6/466ae5fa03fc4fa18e3651976692954ddf28011e.png)
![\[3 T\]](http://latex.artofproblemsolving.com/5/4/c/54c7561099ad5198488c3c852b2566e327bf4673.png)
![\[4 -\]](http://latex.artofproblemsolving.com/3/4/d/34dde0e8318825de00c6c696b1c5e715454d4420.png)
![\[5 J\]](http://latex.artofproblemsolving.com/1/0/3/103642629cfae33c8189fbe9ce9ceda673dff2ec.png)
![\[6 M\]](http://latex.artofproblemsolving.com/0/4/0/04009310065aac423d578bbb44071c0ec188ca08.png)
![\[7 -\]](http://latex.artofproblemsolving.com/1/4/f/14fa5dbaae18d5eaff20f23b696f5eac2a6254be.png)
![\[8 H\]](http://latex.artofproblemsolving.com/7/a/d/7ad7e8da6775a21ffe050e6c07c16afbed4dc774.png)
Thus our answer 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. 
