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

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

Revision as of 10:58, 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 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 $\boxed{\textbf{(A) }\text{Aaron}}$.

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

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