Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 13 - Voting and Apportionment - Chapter Summary, Review, and Test - Review Exercises: 22

Answer

Candidate A wins the election using the pairwise comparison method.

Work Step by Step

With the pairwise comparison method, each candidate is compared with every other candidate. For each pair of candidates, if one candidate is ranked higher than the other candidate on a majority of ballots, then the higher-ranked candidate receives 1 point. If the two candidates tie, then they each receive 0.5 points. After all of the comparisons have been made, the candidate who receives the most points is declared the winner. We can compare Candidate A and Candidate C. We can find the number of ballots with Candidate A ranked higher than Candidate C and the number of ballots with Candidate C ranked higher than Candidate A. Candidate A: 180 + 100 = 280 Candidate C: 40 + 30 = 70 Since Candidate A is ranked higher than Candidate C on more ballots, Candidate A receives 1 point. We can compare Candidate A and Candidate B. We can find the number of ballots with Candidate A ranked higher than Candidate B and the number of ballots with Candidate B ranked higher than Candidate A. Candidate A: 180 Candidate B: 100 + 40 + 30 = 170 Since Candidate A is ranked higher than Candidate B on more ballots, Candidate A receives 1 point. We can compare Candidate B and Candidate C. We can find the number of ballots with Candidate B ranked higher than Candidate C, and the number of ballots with Candidate C ranked higher than Candidate B. Candidate B: 180 + 100 + 40 = 320 Candidate C: 30 Since Candidate B is ranked higher than Candidate C on more ballots, Candidate B receives 1 point. We can compare Candidate A and Candidate D. We can find the number of ballots with Candidate A ranked higher than Candidate D and the number of ballots with Candidate D ranked higher than Candidate A. Candidate A: 180 + 30 = 210 Candidate D: 100 + 40 = 140 Since Candidate A is ranked higher than Candidate D on more ballots, Candidate A receives 1 point. We can compare Candidate D and Candidate C. We can find the number of ballots with Candidate D ranked higher than Candidate C and the number of ballots with Candidate C ranked higher than Candidate D. Candidate D: 100 + 40 = 140 Candidate C: 180 + 30 = 210 Since Candidate C is ranked higher than Candidate D on more ballots, Candidate C receives 1 point. We can compare Candidate B and Candidate D. We can find the number of ballots with Candidate B ranked higher than Candidate D and the number of ballots with Candidate D ranked higher than Candidate B. Candidate B: 180 + 100 + 30 = 310 Candidate D: 40 Since Candidate B is ranked higher than Candidate D on more ballots, Candidate B receives 1 point. After all of the pairwise comparisons have been made, we can add up the total number of points for each candidate. Candidate A: 1 + 1 + 1 = 3 points Candidate B: 1 + 1 = 2 points Candidate C: 1 point Candidate D: 0 points Since Candidate A received the most points, Candidate A is declared the winner. Candidate A wins the election using the pairwise comparison method.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.