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))
(Video Solution by STEMbreezy)
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<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>
  
==Solution==
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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>, 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>.
  
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>.
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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>.
 +
 
 +
==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==
 
==Video Solution==
https://youtu.be/G04TnUc5iAA
+
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
 +
 
 +
==Video Solution (CREATIVE THINKING!!!)==
 +
https://youtu.be/wnTfhfhS1VA
 +
 
 +
~Education, the Study of Everything
  
 
==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}}

Revision as of 02:42, 2 March 2023

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 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

Video Solution (CREATIVE THINKING!!!)

https://youtu.be/wnTfhfhS1VA

~Education, the Study of Everything

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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