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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| + | ==Video Solution== | ||
| + | https://youtu.be/dHY8gjoYFXU?t=664 | ||
| + | |||
| + | ~IceMatrix | ||
Revision as of 04:39, 19 May 2020
Contents
Problem
Five friends sat in a movie theater in a row containing 
seats, numbered 
to 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?
Solution 1
Assume that Edie and Dee were originally in seats 3 and 4. If this were so, there is no possible position for which Bea can move 2 seats to the right. The same applies for seats 2 and 3. This means that either Edie or Dee was originally in an edge seat. If Edie and Dee were in seats 1 and 2, then Bea must have been in seat 3, which would mean that seat 5 would now be occupied and the positioning would not work. So, Edie and Dee are in seats 4 and 5. This means that Bea was originally in seat 1. Ceci must have been in seat 3 to keep seat 1 open, which leaves seat 2.
Thus, Ada was in seat 
.
Video Solution
https://youtu.be/dHY8gjoYFXU?t=664
~IceMatrix
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
