Difference between revisions of "2016 AMC 10A Problems/Problem 13"

Revision as of 22:53, 3 February 2016

Five friends sat in a movie theater in a row containing $5$ seats, numbered $1$ to $5$ from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seats to the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?

$\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution

We see that the following configuration works: Bea - Ada - Ceci - Dee - Edie After moving, it becomes Ada - Ceci - Bea - Edie - Dee. Thus, Ada was in seat $\boxed{2}$ .

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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 AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 2: / Line 2: @@
 <math>\textbf{(A) }1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math>
+==Solution==
+We see that the following configuration works:
+Bea - Ada - Ceci - Dee - Edie
+After moving, it becomes
+Ada - Ceci - Bea - Edie - Dee.
+Thus, Ada was in seat <math>\boxed{2}</math>.
+==See Also==
+{{AMC10 box|year=2016|ab=A|num-b=14|num-a=16}}
+{{MAA Notice}}

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Difference between revisions of "2016 AMC 10A Problems/Problem 13"

Revision as of 22:53, 3 February 2016

Solution

See Also