Difference between revisions of "2015 AMC 12B Problems/Problem 4"

(Problem)
(Solution)
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==Solution==
 
==Solution==
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Let <math>x</math> denote the 6 racers not named. Then the correct order following all the logic would look like:
  
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<math>x</math>, Rand, Todd, <math>x</math>, Jack, Marta, <math>x</math>, Hikmet, <math>x</math>, David, <math>x</math>, <math>x</math>.
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Clearly the 8th place runner 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 07:20, 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?

$\textbf{(A)}\; \text{David} \qquad\textbf{(B)}\; \text{Hikmet} \qquad\textbf{(C)}\; \text{Jack} \qquad\textbf{(D)}\; \text{Rand} \qquad\textbf{(E)}\; \text{Todd}$

Solution

Let $x$ denote the 6 racers not named. Then the correct order following all the logic would look like:

$x$, Rand, Todd, $x$, Jack, Marta, $x$, Hikmet, $x$, David, $x$, $x$.

Clearly the 8th place runner is $\fbox{\textbf{(B)}\; \text{Hikmet}}$.

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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

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