Difference between revisions of "2020 AMC 8 Problems/Problem 6"
Sevenoptimus (talk | contribs) (Wrote an actual proof (to replace the "solutions" which simply claim that an arrangement works)) |
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| − | ==Problem | + | ==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> | ||
| − | ==Solution | + | ==Solution== |
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| − | + | 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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| − | Let the carts look like < | ||
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==Video Solution== | ==Video Solution== | ||
https://youtu.be/G04TnUc5iAA | https://youtu.be/G04TnUc5iAA | ||
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==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 07:26, 20 November 2020
Contents
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?
Solution
Let the carts look like 
, where the left represents the back and the right represents the front of the train. Call the people 
, 
, 
, 
, and 
respectively. The first condition gives 
, so we try 
, 
, and 
. In the first case, as sat in front of , we must have 
or 
, both of which do not comply with the last condition. In the second case, we obtain 
, which works; the third case is obviously impossible, since there is no way for to sit in front of . Thus, with the only possible arrangement being , the person sitting in the middle car is 
.
Video Solution
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. 
