Math 2000 - Solutions 13
Fall 2001
Unit 8A 2. This is exponential growth. After one year the population will grow by $0.02 \times 100,000 = 2000$, making it 102,000; after a second year it will grow by $0.02 \times 102,000 = 2040$, to 104,040; and after a third year the population will grow by a further $0.02 \times 104,040 = 2080.8$, which we round off to 2081, making it 106,121. 4. This is linear decay; in ten weeks gas will cost $\$1.60 - (10\times \$0.02) = \$1.40$ per gallon. 8. This is exponential growth. After one year the value of your house will grow by $0.10 \times \$100,000 = \$10,000$, to $110,000; after a second year the value will grow by $0.10 \times \$110,000 = \$1100$, to $121,000; and after a third year the value will rise a further $0.10 \times \$121,000 = \$12,100$, making it worth $133,100. Unit 8B 6. Because the balance increases by a factor of 2 in 10 years, and 30 years represents $\frac{30}{10}=3$ doubling times, the balance will increase by a factor of $2^3=8$ in 30 years. Because 50 years represents $\frac{50}{10}=5$ doubling times, the balance will increase by a factor of $2^{5}=32$ in 50 years. 8. Because prices increase by a factor of 2 in 4 weeks, and a year represents $\frac{52}{4}=13$ doubling times, prices will increase by a factor of $2^{13} =8192$ in a year. 12. Because the number of cells increases by a factor of 2 in 6 months, and 3 years represents $\frac{3 \times 12}{6}=6$ doubling times, the number of cells will increase by a factor of $2^{6}=64$ in 3 years. Thus, 1 cell grows to 64 cells in 3 years. Because 6 years represents 12 doubling times, the number of cells will increase by a factor of $2^{12}=4096$ in 6 years. Thus, 1 cell grows to 4096 cells in 6 years. Unit 11A 16a. A total of 29 votes were cast. 16b. A received no first place votes, B received 9 first place votes, C received 4 first place votes, D received 13 first place votes, and E received 3 first place votes. Therefore, D is the plurality winner (but not by a majority). 16c. >From part (b), we see that B and D enter the runoff and the votes of A, C, and E are redistributed. Now B receives $9 + 4 = 13$ votes and D receives the remainder, or 16, of the votes. Thus D is the winner of the top two runoff. 16d. In the sequential runoff we eliminate only the candidate with the fewest first place votes at each stage. From part (b), we see that A is eliminated first with no first place votes. Then E has next fewest first place votes. Redistributing E's votes, D receives E's 3 first place votes; so at this point D has 16 votes, B has 9 votes, and C has 4 votes. Now C is eliminated and B picks up 4 more votes. The final runoff is between B with 13 votes and D with 16 votes. The winner by sequential runoff is D. 16e. For the Borda Count, we score 5 points for a first place vote, 4 points for a second place vote, 3 point for a third place vote, 2 points for a fourth place vote, and 1 point for a fifth place vote. The point totals are as follows:
A: $(9 \times 4) + (7 \times 3) + (6 \times 1) + {}$
$ (4 \times 4) + (3 \times 4) = 91$. B: $(9 \times 5) + (7 \times 4) + (6 \times 4) + {}$
$ (4 \times 3) + (3 \times 2) = 115$. C: $(9 \times 3) + (7 \times 2) + (6 \times 2) + {}$
$ (4 \times 5) + (3 \times 1) = 76$. D: $(9 \times 2) + (7 \times 5) + (6 \times 5) + {}$
$ (4 \times 2) + (3 \times 3) = 100$. E: $(9 \times 1) + (7 \times 1) + (6 \times 3) + {}$
$ (4 \times 1) + (3 \times 5) = 53$.
Note that the point total is 435, as it must be. We see that B is the winner by the Borda count. 16f. Here are the results of the 10 pairwise races: B over A, 22 to 7; A over C, 19 to 10; A over D, 16 to 13; A over E, 20 to 9; B over C, 25 to 4; D over B, 16 to 13; B over E, 26 to 3; D over C, 16 to 13; C over E, 20 to 9; and D over E, 26 to 3. Thus A scores 3 points, B scores 3 points, C scores 1 point, D scores 3 points, and E scores no points. The pairwise comparison method gives a tie between A, B, and D. 16g. Candidate D wins by three methods and ties by another method; so D should be declared the winner of the election.