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Difference between revisions of "2020 AMC 8 Problems/Problem 6"
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==Problem 6==
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==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?
 
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?
  
 
<math>\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}</math>
 
<math>\textbf{(A) }\text{Aaron} \qquad \textbf{(B) }\text{Darren} \qquad \textbf{(C) }\text{Karen} \qquad \textbf{(D) }\text{Maren}\qquad \textbf{(E) }\text{Sharon}</math>
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==Solution 1==
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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>, 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>.
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==Solution 2==
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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 NiuniuMaths (Easy to understand!)==
 +
https://www.youtube.com/watch?v=8hgK6rESdek&t=9s
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 +
~NiuniuMaths
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 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/wnTfhfhS1VA
 +
 +
~Education, the Study of Everything
 +
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/UnVo6jZ3Wnk?si=lE54gozu1zUD7xkh&t=567
 +
 +
~Math-X
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 +
==Video Solution by North America Math Contest Go Go Go==
 +
 +
https://www.youtube.com/watch?v=RR6svhjdPEA
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==Video Solution by WhyMath==
 +
https://youtu.be/G04TnUc5iAA
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 +
~savannahsolver
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==Video Solution==
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https://youtu.be/61c1MR9tne8 ~ The Learning Royal
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==Video Solution by Interstigation==
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https://youtu.be/YnwkBZTv5Fw?t=186
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 +
~Interstigation
 +
 +
==Video Solution by STEMbreezy==
 +
https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS
 +
 +
~STEMbreezy
  
 
==See also==  
 
==See also==  
 
{{AMC8 box|year=2020|num-b=5|num-a=7}}
 
{{AMC8 box|year=2020|num-b=5|num-a=7}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:15, 24 January 2024

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 $MASKD$, 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 NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=8hgK6rESdek&t=9s

~NiuniuMaths

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/wnTfhfhS1VA

~Education, the Study of Everything

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/UnVo6jZ3Wnk?si=lE54gozu1zUD7xkh&t=567

~Math-X

Video Solution by North America Math Contest Go Go Go

https://www.youtube.com/watch?v=RR6svhjdPEA

Video Solution by WhyMath

https://youtu.be/G04TnUc5iAA

~savannahsolver

Video Solution

https://youtu.be/61c1MR9tne8 ~ The Learning Royal

Video Solution by Interstigation

https://youtu.be/YnwkBZTv5Fw?t=186

~Interstigation

Video Solution by STEMbreezy

https://youtu.be/U27z1hwMXKY?list=PLFcinOE4FNL0TkI-_yKVEYyA_QCS9mBNS

~STEMbreezy

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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