2020 AMC 8 Problems/Problem 6: Difference between revisions
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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 1== | |||
The following arrangement works, if Darren is <math>D</math>, Aaron is <math>A</math>, Karen is <math>K</math>, Maren is <math>M</math>, and Sharon is <math>S</math>: <math>DSAKM</math>. | |||
Thus the answer is <math>\textbf{(A) }\text{Aaron}</math> | |||
~Windigo | |||
==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 10:01, 18 November 2020
Problem 6
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 1
The following arrangement works, if Darren is 
, Aaron is 
, Karen is 
, Maren is 
, and Sharon is 
: 
.
Thus the answer is 
~Windigo
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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