2020 AMC 8 Problems/Problem 6

Revision as of 07:26, 20 November 2020 by Sevenoptimus (talk | contribs) (Wrote an actual proof (to replace the "solutions" which simply claim that an arrangement works))

Problem

Aaron, Darren, Karen, Maren, and Sharon rode on a small train that has five cars that seat one person each. Maren sat in the last car. Aaron sat directly behind Sharon. Darren sat in one of the cars in front of Aaron. At least one person sat between Karen and Darren. Who sat in the middle car?

$\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}$

Solution

Let the carts look like $\_,\_,\_,\_,\_$, where the left represents the back and the right represents the front of the train. Call the people $A$, $D$, $K$, $M$, and $S$ respectively. The first condition gives $M,\_,\_,\_,\_$, so we try $M,A,S,\_,\_$, $M,\_,A,S,\_$, and $M,\_,\_,A,S$. In the first case, as $D$ sat in front of $A$, we must have $M,A,S,D,K$ or $M,A,S,K,D$, both of which do not comply with the last condition. In the second case, we obtain $M,K,A,S,D$, which works; the third case is obviously impossible, since there is no way for $D$ to sit in front of $A$. Thus, with the only possible arrangement being $M,K,A,S,D$, the person sitting in the middle car is $\boxed{\textbf{(A) }\text{Aaron}}$.

Video Solution

https://youtu.be/G04TnUc5iAA

See also

2020 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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 AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png