Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2020 AMC 8 Problems/Problem 6"

(Wrote an actual proof (to replace the "solutions" which simply claim that an arrangement works))
m (Improved solution appearance)
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
−
 
+
Write the order of the carts as <math>\square\square\square\square\square</math>, where the left represents the back and the right represents the front of the train. Call the people <math>A</math>, <math>D</math>, <math>K</math>, <math>M</math>, and <math>S</math> respectively. The first condition gives <math>M\square\square\square\square</math>, so we try <math>MAS\square\square</math>, <math>M\square AS\square</math>, and <math>M\square\square AS</math>. In the first case, as <math>D</math> sat in front of <math>A</math>, we must have <math>MASDK</math> or <math>MASKD</math>, both of which do not comply with the last condition. In the second case, we obtain <math>MKASD</math>, which works, while the third case is obviously impossible, since there is no way for <math>D</math> to sit in front of <math>A</math>. Thus, with the only possible arrangement being <math>MKASD</math>, the person sitting in the middle car is <math>\boxed{\textbf{(A) }\text{Aaron}}</math>.
−
Let the carts look like <math>\_,\_,\_,\_,\_</math>, where the left represents the back and the right represents the front of the train. Call the people <math>A</math>, <math>D</math>, <math>K</math>, <math>M</math>, and <math>S</math> respectively. The first condition gives <math>M,\_,\_,\_,\_</math>, so we try <math>M,A,S,\_,\_</math>, <math>M,\_,A,S,\_</math>, and <math>M,\_,\_,A,S</math>. In the first case, as <math>D</math> sat in front of <math>A</math>, we must have <math>M,A,S,D,K</math> or <math>M,A,S,K,D</math>, both of which do not comply with the last condition. In the second case, we obtain <math>M,K,A,S,D</math>, which works; the third case is obviously impossible, since there is no way for <math>D</math> to sit in front of <math>A</math>. Thus, with the only possible arrangement being <math>M,K,A,S,D</math>, the person sitting in the middle car is <math>\boxed{\textbf{(A) }\text{Aaron}}</math>.
 
  
 
==Video Solution==
 
==Video Solution==

Revision as of 07:23, 20 November 2020

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

Write the order of the carts as $\square\square\square\square\square$, 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 $M\square\square\square\square$, so we try $MAS\square\square$, $M\square AS\square$, and $M\square\square AS$. In the first case, as $D$ sat in front of $A$, we must have $MASDK$ or $MASKD$, both of which do not comply with the last condition. In the second case, we obtain $MKASD$, which works, while 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 $MKASD$, the person sitting in the middle car is $\boxed{\textbf{(A) }\text{Aaron}}$.

Video Solution

https://youtu.be/G04TnUc5iAA

See also

2020 AMC 8 (Problems • Answer Key • Resources)
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_8_Problems/Problem_6&oldid=137949"