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

(Video Solution)
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==Solution 2==
 
==Solution 2==
 
Follow the first few steps of Solution 1. We must have <math>M\square\square\square\square</math>, and also have <math>AS\text{(anything)}D \text{ and } K\dots D</math>. There are only <math>4</math> spaces available for <math>A, S, D, K</math>, so the only possible arrangement of them is <math>KASD</math>, so the arrangement is <math>MKASD</math>, so the person in the middle car is <math>\boxed{\textbf{(A) }\text{Aaron}}</math>.
 
Follow the first few steps of Solution 1. We must have <math>M\square\square\square\square</math>, and also have <math>AS\text{(anything)}D \text{ and } K\dots D</math>. There are only <math>4</math> spaces available for <math>A, S, D, K</math>, so the only possible arrangement of them is <math>KASD</math>, so the arrangement is <math>MKASD</math>, so the person in the middle car is <math>\boxed{\textbf{(A) }\text{Aaron}}</math>.
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==Video Solution by WhyMath==
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https://youtu.be/G04TnUc5iAA
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~savannahsolver
  
 
==Video Solution==
 
==Video Solution==

Revision as of 18:15, 26 February 2021

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 1

Write the order of the cars as $\square\square\square\square\square$, where the left end of the row represents the back of the train and the right end represents the front. 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 it results in there being no way for $D$ to sit in front of $A$. It follows that, with the only possible arrangement being $MKASD$, the person sitting in the middle car is $\boxed{\textbf{(A) }\text{Aaron}}$.

Solution 2

Follow the first few steps of Solution 1. We must have $M\square\square\square\square$, and also have $AS\text{(anything)}D \text{ and } K\dots D$. There are only $4$ spaces available for $A, S, D, K$, so the only possible arrangement of them is $KASD$, so the arrangement is $MKASD$, so the person in the middle car is $\boxed{\textbf{(A) }\text{Aaron}}$.

Video Solution by WhyMath

https://youtu.be/G04TnUc5iAA

~savannahsolver

Video Solution

https://youtu.be/61c1MR9tne8

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

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