Solutions Manual for Excursions in Modern Mathematics, 10th edition by Peter Tannenbaum, Chapter 1-17 | All Chapters

SOLUTIONS MANUAL
Excursions in Modern Mathematics, 10th Edition
By Peter Tannenbaum
SC
O
R
EG
U
ID
ES

, Table of Content
PART I: SOCIAL CHOICE

1. The Mathematics of Elections: The Paradoxes of Democracy

2. The Mathematics of Power: Weighted Voting

3. The Mathematics of Sharing: Fair-Division Games

4. The Mathematics of Apportionment: Making the Rounds
SC
PART II: MANAGEMENT SCIENCE

5. The Mathematics of Getting Around: Euler Paths and Circuits

6. The Mathematics of Touring: Traveling Salesman Problems
O
7. The Mathematics of Networks: The Cost of Being Connected

8. The Mathematics of Scheduling: Chasing the Critical Path
R
PART III: GROWTH

9. Population Growth Models: There Is Strength in Numbers
EG
10. Financial Mathematics: Money Matters

PART IV: SHAPE AND FORM

11. The Mathematics of Symmetry: Beyond Reflection
U
12. Fractal Geometry: The Kinky Nature of Nature

13. Fibonacci Numbers and the Golden Ratio: Tales of Rabbits and Gnomons
ID
PART V: STATISTICS

14.Censuses, Surveys, Polls, and Studies: The Joys of Collecting Data

15. Graphs, Charts, and Numbers: The Data Show and Tell
ES
16. Probabilities, Odds, and Expectations: Measuring Uncertainty and Risk

17. The Mathematics of Normality: The Call of the Bell

, Chapter 1
WALKING
1.1. Ballots and Preference Schedules
SC
1. Number of voters 5 3 5 3 2 3
1st choice A A C D D B
2nd choice B D E C C E
3rd choice C B D B B A
4th choice D C A E A C
5th choice E E B A E D
This schedule was constructed by noting, for example, that there were five ballots listing candidate C as the
O
first preference, candidate E as the second preference, candidate D as the third preference, candidate A as the
fourth preference, and candidate B as the last preference.

2. Number of voters 4 5 6 2
R
1st choice A B C A
2nd choice D C A C
3rd choice B D D D
EG
th
4 choice C A B B
3. (a) 5 + 5 + 3 + 3 + 3 + 2 = 21

(b) 11. There are 21 votes all together. A majority is more than half of the votes, or at least 11.

(c) Chavez. Argand has 3 last-place votes, Brandt has 5 last-place votes, Chavez has no last-place votes,
Dietz has 3 last-place votes, and Epstein has 5 + 3 + 2 = 10 last-place votes.

4. (a) 202 + 160 + 153 + 145 + 125 + 110 + 108 + 102 + 55 = 1160
U
(b) 581; There are 1160 votes all together. A majority is more than half of the votes, or at least 581.

(c) Alicia. She has no last-place votes. Note that Brandy has 125 + 110 + 55 = 290 last-place votes, Cleo
ID
has 202 + 145 + 102 = 449 last-place votes, and Dionne has 160 + 153 + 108 = 421 last-place votes.

5. Number of voters 37 36 24 13 5
1st choice B A B E C
2nd choice E B A B E
ES
3rd choice A D D C A
4th choice C C E A D
5th choice D E C D B
Here Brownstein was listed first by 37 voters. Those same 37 voters listed Easton as their second choice,
Alvarez as their third choice, Clarkson as their fourth choice, and Dax as their last choice.

6. Number of voters 14 10 8 7 4
1st choice B B A D E
2nd choice A D B C B
3rd choice E A E B A
4th choice C E D E C
5th choice D C C A D

, 2 Chapter 1: The Mathematics of Elections

7. Number of voters 14 10 8 7 4
A 2 3 1 5 3
B 1 1 2 3 2
C 5 5 5 2 4
D 4 2 4 1 5
E 3 4 3 4 1
SC
Here 14 voters had the same preference ballot listing B as their first choice, A as their second choice, E as
their third choice, D as their fourth choice, and C as their fifth and last choice.

8. Number of voters 37 36 24 13 5
A 1 2 5 2 4
B 3 1 2 4 1
C 2 4 3 1 5
O
D 5 3 1 5 2
E 4 5 4 3 3
R
9. Number of voters 255 480 765
1st choice L C M
2nd choice M M L
3rd choice C L C
EG
(0.17)(1500) = 255; (0.32)(500) = 480; The remaining voters (51% of 1500 or 1500-255-480=765) prefer M
the most, C the least, so that L is their second choice.

10. Number of voters 450 900 225 675
1st choice A B C C
2nd choice C C B A
3rd choice B A A B
U
100% - 20% - 40% = 40% of the voters number 225 + 675 = 900. So, if N represents the total number of
voters, then (0.40)N = 900 . This means there are N = 2250 total voters. 20% of 2250 is 450 (these voters
have preference ballots A, C, B). 40% of 2250 is 900 (these voters have preference ballots B, C, A).
ID
1.2. Plurality Method
11. (a) C. A has 15 first-place votes. B has 11 + 8 + 1 = 20 first-place votes. C has 27 first-place votes. D has 9
first-place votes. C has the most first-place votes with 27 and wins the election.

(b) C, B, A, D. Candidates are ranked according to the number of first-place votes they received (27, 20, 15,
ES
and 9 for C, B, A, and D respectively).

12. (a) D. A has 21 first-place votes. B has 18 first-place votes. C has 10 + 1 = 11 first-place votes. D has 29
first-place votes. D has the most first-place votes with 29 and wins the election.

(b) D, A, B, C.

13. (a) C. A has 5 first-place votes. B has 4 + 2 = 6 first-place votes. C has 6 + 2 + 2 + 2 = 12 first-place votes.
D has no first-place votes. C has the most first-place votes with 12 and wins the election.

(b) C, B, A, D. Candidates are ranked according to the number of first-place votes they received (12, 6, 5,
and 0 for C, B, A, and D respectively).

14. (a) B. A has 6 + 3 = 9 first-place votes. B has 6 + 5 + 3 = 14 first-place votes. C has no first-place votes. D
has 4 first-place votes. B has the most first-place votes with 14 and wins the election.