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

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Thus, Ada was in seat <math>\boxed{\textbf{(B) }2}</math>.
 
Thus, Ada was in seat <math>\boxed{\textbf{(B) }2}</math>.
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==Solution 2==
 
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Process of elimination of possible configurations.
 
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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>.
 
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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:
 
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<cmath>B,-,C\rightarrow -,C,B</cmath>
 
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This is affirmed because <math>DE\rightarrow ED</math>, which there is no new seats uncovered. So <math>A,B,C</math> are restricted to the same <math>1,2,3</math> seats.
 
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Thus, it must be <math>B,A,C\rightarrow A,C,B</math>, and more specifically:
 
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<cmath>B,A,C,D,E\rightarrow A,C,B,E,D</cmath>
 
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So <math>A</math>, Ada, was originally in seat <math>\boxed{\textbf{(B)}\text{ 2}}</math>.
 
  
 
==Solution 3==
 
==Solution 3==

Revision as of 00:43, 18 November 2019

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 $\boxed{\textbf{(B) }2}$.

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 $\boxed{\textbf{(B)}\text{ 2}}$.

See Also

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

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