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

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Problem

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 1

Bash: 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 $\boxed{2}$ .

Solution 2

Process of elimination of possible configurations.

Let's say that Ada= $A$ , Bea= $B$ , Ceci= $C$ , Dee= $D$ , and Edie= $E$ . Ssss Since $B$ moved more to the right than $C$ did left, this implies that $B$ was in a LEFT end seat originally: $B,-,C\rightarrow -,C,B$

This is affirmed because $DE\rightarrow ED$ , which there is no new seats uncovered. So $A,B,C$ are restricted to the same $1,2,3$ seats. Thus, it must be $B,A,C\rightarrow A,C,B$ , and more specifically: $B,A,C,D,E\rightarrow A,C,B,E,D$

So $A$ , Ada, was originally in seat $\boxed{\textbf{(B)}\text{ 2}}$ .

Solution 3

The seats are numbered 1 through 5, so let each letter ( $A,B,C,D,E$ ) correspond to a number. Let a move to the left be subtraction and a move to the right be addition.

We know that $1+2+3+4+5=A+B+C+D+E=15$ . After everyone moves around, however, our equation looks like $(A+x)+B+2+C-1+D+E=15$ because $D$ and $E$ switched seats, $B$ moved two to the right, and $C$ moved 1 to the left.

For this equation to be true, $x$ has to be -1, meaning $A$ moves 1 left from her original seat. Since $A$ is now sitting in a corner seat, the only possible option for the original placement of $A$ is in seat number $\boxed{\textbf{(B)}\text{ 2}}$ .

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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

2016 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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 12 Problems and Solutions

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

@@ Line 20: / Line 20: @@
 Let's say that Ada=<math>A</math>, Bea=<math>B</math>, Ceci=<math>C</math>, Dee=<math>D</math>, and Edie=<math>E</math>.
+Ssss
 Since <math>B</math> moved more to the right than <math>C</math> did left, this implies that <math>B</math> was in a LEFT end seat originally:
 <cmath>B,-,C\rightarrow -,C,B</cmath>

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