Content check — Contemporary Math, Chapter 11: Voting and Apportionment

Your Turn 11.1 §11.1 source ↗

According to the Cook Political Report, a total of 158,394,605 votes were cast in the 2020 U.S. presidential election. Of those, 81,281,502 were cast for Joe Biden, 74,222,593 for Donald Trump, and 2,890,510 for other candidates. Although U.S. presidential elections are not determined by popular vote, did either Joe Biden or Donald Trump secure a majority of the votes?

Answer

A simple majority is a number equaling more than 50 percent of the total.

0.50(158,394,605) = 79,197,302.5

Yes, Joe Biden secured a majority of the votes because he had more than 79,197,302.5 votes.

Yes, Joe Biden won the majority.

Your Turn 11.2 §11.1 source ↗

Consider the results of the 2018 U.S. Senate primary for Maryland. Determine which candidate won the primary for the Republicans (R). Did the candidate win a majority or a plurality of Republican votes?

Answer

When no candidate has a majority of the votes, the one with the most votes is said to have a plurality. In the Maryland race, the candidate with the most votes (Tony Campbell) only secured 29.22 percent of the votes. To win by a majority of votes is winning by more than 50 percent. Tony Campbell only won by a plurality. Tony Cambell won a plurality of votes in the Republican primary.

Your Turn 11.3 §11.1 source ↗

The student government bylaws of a particular college require that a new president is elected annually by plurality voting. In the event of a tie, the bylaws require the candidate(s) with the fewest votes to be removed from the ballot and a runoff election to be held with the remaining candidates. This process is repeated until a single candidate receives a plurality and wins the election. The results of two voting rounds are given in the table below.

Candidate	Votes in First Round	Votes in Second Round
Ferguson	158	168
Garcia	103	104
Hearn	157	180
Isaac	123	123
Jackman	58	Eliminated
Kelly	72	74
Lim	158	180

Which two candidates tied in the second round? Does this mean there will be a third election?

Answer

Hearn and Lim tied. Yes, there must be a third election.

Your Turn 11.3 (b) §11.1 source ↗

The student government bylaws of a particular college require that a new president is elected annually by plurality voting. In the event of a tie, the bylaws require the candidate(s) with the fewest votes to be removed from the ballot and a runoff election to be held with the remaining candidates. This process is repeated until a single candidate receives a plurality and wins the election. The results of two voting rounds are given in the table below.

Candidate	Votes in First Round	Votes in Second Round
Ferguson	158	168
Garcia	103	104
Hearn	157	180
Isaac	123	123
Jackman	58	Eliminated
Kelly	72	74
Lim	158	180

If there was a winner, who won? If there should be a runoff, who must be removed from the ballot?

Answer

There was no winner because there was a tie. Kelly must be removed because Kelly had the least number of votes and there will be a third election. Kelly must be removed in the runoff.

Your Turn 11.4 §11.1 source ↗

There are six members on the board of a Parent Teacher Association (PTA) at a local elementary school: the president (P), the vice president (V), the recording secretary (R), the liaison to the administration (L), the treasurer (T), and the chief fundraiser (C). The board must decide which equipment to purchase for the classrooms with moneys from their annual fundraisers. The preferences of the board members are shown in the table below.

	P	V	R	L	T	C
Option A	3	4	1	3	4	1
Option B	1	3	2	1	3	2
Option C	2	2	3	2	1	4
Option D	4	1	4	4	2	3

The board uses plurality method and a runoff in the event of a tie, such that the option(s) with the least votes will be eliminated in each round. Which option will be chosen?

Answer

Step 1: Determine the number of votes necessary to have a majority. There are six members, so 50 percent of 6 is 0.5(6) = 3. A majority is at least one more than 50 percent, which is 4 or more votes.

Step 2: In the first round, A and B have 2 votes. C and D have 1 vote. No one has the majority. C and D tie for the least number of votes, so eliminate them both.

Step 3: Reorder the votes with the remaining options.

	P	V	R	L	T	V
Option A	2	2	1	2	2	1
Option B	1	1	2	1	1	2

Option B now has 4 votes, which wins the majority.

Option B.

Your Turn 11.5 §11.1 source ↗

A kindergarten class votes on their favorite colors using a ranked ballot. Use the results in the following table to answer the questions.

Number of Ballots	4	6	4	7
Red	2	6	2	5
Blue	1	2	1	4
Green	6	5	5	2
Yellow	5	4	6	3
Purple	4	3	4	1
Pink	3	1	3	6

How many students voted in total?

Answer

The numbers in the top row tell you the number of ballots.

4 + 6 + 4 + 7 = 21

4 + 6 + 4 + 7 = 21

Your Turn 11.5 (b) §11.1 source ↗

A kindergarten class votes on their favorite colors using a ranked ballot. Use the results in the following table to answer the questions.

Number of Ballots	4	6	4	7
Red	2	6	2	5
Blue	1	2	1	4
Green	6	5	5	2
Yellow	5	4	6	3
Purple	4	3	4	1
Pink	3	1	3	6

How many students voted for yellow as their favorite color?

Answer

Look at the Yellow row. There are no 1s in the Yellow row. A 1 would indicate that someone chose yellow as their favorite color, so the answer is “none.” None

Your Turn 11.5 (c) §11.1 source ↗

A kindergarten class votes on their favorite colors using a ranked ballot. Use the results in the following table to answer the questions.

Number of Ballots	4	6	4	7
Red	2	6	2	5
Blue	1	2	1	4
Green	6	5	5	2
Yellow	5	4	6	3
Purple	4	3	4	1
Pink	3	1	3	6

How many students in the class indicated that blue was their favorite color while green was their least favorite color?

Answer

There are six colors. The rank of 1 indicates a favorite color. The rank of 6 indicates a least favorite color. Look at the Blue row and the Green row. There is only one column where there is a 1 in the Blue row (indicating a favorite color) and a 6 in the Green row (indicating a least favorite color). The number at the top of that column is 4. That means that there are four ballots with those choices. 4

Your Turn 11.6 §11.1 source ↗

Suppose that 58 Star Wars fans were asked to vote for their favorite Star Wars character. They were given a ranked ballot, and the results are shown in the following table. Use ranked-choice voting to determine the winner.

Number of Ballots	7	6	10	8	4	5	6	7	2	3
Han Solo	3	5	1	3	2	4	2	2	1	4
Princess Leia	2	1	3	6	4	3	3	1	3	2
Luke Skywalker	6	6	6	1	5	5	5	4	6	1
Chewbacca	4	4	5	4	6	1	6	5	4	5
Yoda	1	2	4	2	3	2	1	3	2	3
R2-D2	5	3	2	5	1	6	4	6	5	6

Answer

Using ranked-choice voting, Yoda is determined to be the winner.

Step 1: Determine the number of votes needed to achieve a majority.

Add the number of ballots.

7 + 6 + 10 + 8 + 4 + 5 + 6 + 7 + 2 + 3 = 58

0.50(58) = 29

To win a majority requires 30 or more votes.

Step 2: Count the number of first-place votes for each candidate. If a candidate has a majority, then you are done. Otherwise, eliminate the candidate with the fewest votes and repeat this step.

Han Solo: 10 + 2 = 12

Princess Leia: 6 + 7 = 13

Luke Skywalker: 8 + 3 = 11

Chewbacca: 5

Yoda: 7 + 6 = 13

R2-D2 = 4

No one won. Eliminate R2-D2.

Second Round:

Number of Ballots	7	6	10	8	4	5	6	7	2	3
Han Solo	3	4	1	3	1	4	2	2	1	4
Princess Leia	2	1	2	5	3	3	3	1	3	2
Luke Skywalker	5	5	5	1	4	5	4	4	5	1
Chewbacca	4	3	4	4	5	1	5	5	4	5
Yoda	1	2	3	2	2	2	1	3	2	3

Han Solo: 10 + 4 + 2 = 16

Princess Leia: 6 + 7 = 13

Luke Skywalker: 8 + 3 = 11

Chewbacca: 5

Yoda: 7 + 6 = 13

Still, no one has the majority. Eliminate Chewbacca.

Third Round:

Number of Ballots	7	6	10	8	4	5	6	7	2	3
Han Solo	3	3	1	3	1	3	2	2	1	4
Princess Leia	2	1	2	4	3	2	3	1	3	2
Luke Skywalker	4	4	4	1	4	4	4	4	4	1
Yoda	1	2	3	2	2	1	1	3	2	3

Han Solo: 10 + 4 + 2 = 16

Princess Leia: 6 + 7 = 13

Luke Skywalker: 8 + 3 = 11

Yoda: 7 + 5 + 6 = 18

Still, no one has the majority. Eliminate Luke Skywalker.

Fourth Round:

Number of Ballots	7	6	10	8	4	5	6	7	2	3
Han Solo	3	3	1	2	1	3	2	2	1	3
Princess Leia	2	1	2	3	3	2	3	1	3	1
Yoda	1	2	3	1	2	1	1	3	2	2

Han Solo: 10 + 4 + 2 = 16

Princess Leia: 6 + 7 + 3 = 16

Yoda: 7 + 8 + 5 + 6 = 26

Still, no one has the majority. Eliminate both Han Solo and Princess Leia. Only Yoda remains and would have the majority.

Yoda wins.

Your Turn 11.7 §11.1 source ↗

Consider the color preferences of the kindergarten class once more. Using the Borda count method, determine the total number of points the color blue received.

Answer

Step 1: For each cell in the table, multiply the number of votes times (6 minus the rank assigned). The 6 comes from the number of colors in the table.

For the color blue, find the points for each column:

First column: 4 × (6 – 1) = 4 × 5 = 20

Second column: 6 × (6 – 2) = 6 × 4 = 24

Third column: 4 × (6 – 1) = 4 × 5 = 20

Fourth column: 7 × (6 – 4) = 7 × 2 = 14

Step 2: Add the points for each column.

20 + 24 + 20 + 14 = 78

Blue received 78 points.

Your Turn 11.8 §11.1 source ↗

Answer the following questions using the table below, which summarizes Imaginarian voter preferences.

Number of Ballots	12	19	27	29	31	21
Candidate A	1	1	2	2	3	3
Candidate B	2	3	1	3	1	2
Candidate C	3	2	3	1	2	1

Use the ranked-choice voting method to determine the winner of the election.

Answer

Ranked Choice Method:

Step 1: Determine the number of votes needed to achieve a majority.

Add the number of ballots.

12 + 19 + 27 + 29 + 31 +21 = 139

0.50(139) = 69.5

To win a majority requires 70 or more votes.

Step 2: Count the number of first-place votes for each candidate. If a candidate has a majority, then you are done. Otherwise, eliminate the candidate with the fewest votes and repeat this step.

A: 12 + 19 = 31

B: 27 + 31 = 58

C: 29 + 21 = 50

No one has the majority. Eliminate Candidate A.

Second Round:

Number of Ballots	12	19	27	29	31	21
Candidate B	1	2	1	2	1	2
Candidate C	2	1	2	1	2	1

B: 12 + 27 + 31 = 70

C: 19 + 29 + 21 = 69

Candidate B wins the majority vote using the ranked-choice method.

Candidate B would be considered the winner using the ranked-choice voting method.

Your Turn 11.8 (b) §11.1 source ↗

Answer the following questions using the table below, which summarizes Imaginarian voter preferences.

Number of Ballots	12	19	27	29	31	21
Candidate A	1	1	2	2	3	3
Candidate B	2	3	1	3	1	2
Candidate C	3	2	3	1	2	1

Use the Borda count method to determine the winner of the election.

Answer

To save subtracting all the time, a first-place vote earns 2 points. A second-place vote wins 1 point. A last place vote wins no points.

Number of Ballots	12	19	27	29	31	21
Candidate A	1	1	2	2	3	3
Candidate B	2	3	1	3	1	2
Candidate C	3	2	3	1	2	1

Candidate A: 12(2) + 19(2) + 27(1) + 29(1) + 31(0) + 21(0) = 118

Candidate B: 12(1) + 19(0) + 27(2) + 29(0) + 31(2) + 21(1) = 149

Candidate C: 12(0) + 19(1) + 27(0) + 29(2) + 31(1) + 21(2) = 150

Candidate C has the most points, so Candidate C wins.

Candidate C would be considered the winner using the Borda count method.

Your Turn 11.8 (c) §11.1 source ↗

Answer the following questions using the table below, which summarizes Imaginarian voter preferences.

Number of Ballots	12	19	27	29	31	21
Candidate A	1	1	2	2	3	3
Candidate B	2	3	1	3	1	2
Candidate C	3	2	3	1	2	1

Compare the results of the two methods. Did the same candidate win? What observations can you make about the results?

Answer

Different candidates won using the two methods. Candidate B won by a difference of just 1 vote in the ranked-choice method. Candidate C won by just 1 point in the Borda count method. When the difference between the candidates is so small, the choice of voting method changes who wins. Different candidates won. It appears that the vote counts were so close that a small shift in either direction could change the results of either method.

Your Turn 11.9 §11.1 source ↗

According to Variety magazine, there were 8,469 eligible Oscar voters in 2020. To make our matrix easier to work with, we've rounded this number up to 8,700. We'd never do this in a real voting situation. Suppose that the voter preferences for the ballot for Best Director had been as shown in the table below.

Number of Ballots	2,400	2,100	1,900	1,200	1,100
Martin Scorsese, The Irishman	3	1	4	5	2
Todd Phillips, Joker	1	4	5	2	4
Sam Mendes, 1917	5	5	3	4	5
Quentin Tarantino, Once Upon a Time in Hollywood	4	2	1	3	3
Bong Joon-ho, Parasite	2	3	2	1	1

Refer to the table to answer each question.

Construct a pairwise comparison matrix for the Best Director ballots.

Answer

A table shows the best director pairwise comparison matrix. The table has 6 rows and 7 columns. The row headers are opponent/runner, Scorsese (s), Phillips (P), Mendes (M), Tarantino (T), Bong (B), and points. Under opponent/runner it is labeled as Scorsese wins, Phillips wins, Mendes wins, Tarantino wins, and Bong wins. Under Scorsese (s) a symbol hyphen is labeled in a box, a box colored in blue and labeled as P S 3600, and the entire box is stricken out, a box labeled as M S 3100 is colored in yellow, a box colored in orange and labeled as B S 6600 and the entire box is stricken out. Under Phillips (P), a box colored in blue and labeled as S P 5100, a symbol hyphen is labeled in a box, a box labeled as M P 1900 is colored in blue, a box colored in yellow and labeled as T P 5100, a box colored in grey and labeled as B P 6300. Under Mendes (M), a box colored in yellow and labeled as S M 5600, a box labeled as P M 6800 is colored in blue, a symbol hyphen is labeled in a box, a box colored in blue and labeled as T M 8700, a box colored in yellow and labeled as B M 8700. Under Tarantino (T), a box colored in grey and labeled as S T 5600, a box labeled as P T 3600 is colored in yellow, a box colored in blue and labeled as M T 0, a hyphen is labeled in a box, a box colored in blue and labeled as B T 4700. Under Bong (B), a box colored in orange and labeled as S B 2100 and the entire box is stricken out, a box labeled as P B 2400 is colored in grey and the entire box is stricken out, a box colored in yellow and labeled as M B 0 and the entire box is stricken out, a box colored in blue and labeled as T B 4000 and the entire box is stricken out. Under points, it is labeled as 3, 1, 0, 2, 4.

Step 1: Create a grid with each actor on the horizontal and vertical headers. Indicate a win with a cross and a tie with a slash.

For instance, let’s find SP: Add the number of ballots in which Scorsese ranks higher than Phillips:

2,100 + 1,900 + 1,100 = 5,100

To find SM, add the number of ballots in which Scorsese ranks higher than Mendes:

2,400 + 2,100 + 1,100 = 5,600

To find ST, add the number of ballots in which Scorsese ranks higher than Tarantino.

2,400 + 2,100 + 1,100 = 5,600

To find SB, add the number of ballots in which Scorsese ranks higher than Bong.

2,100

To find PS, add the number of ballots in which Phillips ranks higher than Scorsese.

2,400 + 1,200 = 3,600

To find PM, add the number of ballots in which Phillips ranks higher than Mendes.

2,400 + 2,100 + 1,200 + 1,100 = 6,800

To find PT, add the number of ballots in which Phillips ranks higher than Tarantino.

2,400 + 1,200 = 3,600

To find PB, add the number of ballots in which Phillips ranks higher than Bong.

2,400

To find MS, add the number of ballots in which Mendes ranks higher than Scorsese.

1,900 + 1,200 = 3,100

To find MP, add the number of ballots in which Mendes ranks higher than Phillips.

1,900

To find MT, add the number of ballots in which Mendes ranks higher than Tarantino.

It does not happen, so record a zero.

To find MB, add the number of ballots in which MB ranks higher than Bong.

Again, it does not happen, so record a zero.

Continue to fill in the rest of the table.

It should look like this.

Opponent Runner	Scorsese (S)	Phillips (P)	Mendes (M)	Tarantino (T)	Bong (B)
Scorsese wins		SP 5100	SM 5600	ST 5600	SB 2100
Phillips wins	PS 3600		PM 6800	PT 3600	PB 2400
Mendes wins	MS 3100	MP 1900		MT 0	MB 0
Tarantino wins	TS 3100	TP 5100	TM 8700		TB 4000
Bong wins	BS 6600	BP 6300	BM 8700	BT 4700

Step 2: Add the total number of ballots: 2,400 + 2,100 + 1,900 + 1,200 + 1,100 = 8,700

Half of that is 4,350.

Anyone who did win over 4,350 votes is a loser. Put an X on the losers. Put a slash on the ties, who have exactly 4,350 (there are not any).

Each win (X) is one point. Each tie (slash) is a half point. Each loss is zero points.

Opponent Runner	Scorsese (S)	Phillips (P)	Mendes (M)	Tarantino (T)	Bong (B)	Points
Scorsese wins		SP 5100	SM 5600	ST 5600	SB 2100 Cross out SB 2100	3
Phillips wins	PS 3600 Cross out PS 3600		PM 6800	PT 3600 Cross out PT 3600	PB 2400 Cross out PB 2400	1
Mendes wins	MS 3100 Cross out MS 3100	MP 1900 Cross out MP 1900		MT 0 Cross out MT 0	MB 0 Cross out MB 0	0
Tarantino wins	TS 3100 Cross out TS 3100	TP 5100	TM 8700		TB 4000 Cross out TB 4000	2
Bong wins	BS 6600	BP 6300	BM 8700	BT 4700		4

Step 3: The winner is the one with the most points. Bong Joon-ho has 4 points.

Your Turn 11.9 (b) §11.1 source ↗

According to Variety magazine, there were 8,469 eligible Oscar voters in 2020. To make our matrix easier to work with, we've rounded this number up to 8,700. We'd never do this in a real voting situation. Suppose that the voter preferences for the ballot for Best Director had been as shown in the table below.

Number of Ballots	2,400	2,100	1,900	1,200	1,100
Martin Scorsese, The Irishman	3	1	4	5	2
Todd Phillips, Joker	1	4	5	2	4
Sam Mendes, 1917	5	5	3	4	5
Quentin Tarantino, Once Upon a Time in Hollywood	4	2	1	3	3
Bong Joon-ho, Parasite	2	3	2	1	1

Refer to the table to answer each question.

Who is the “Best Director” according to the pairwise comparison method?

Answer

Bong Joon-ho won according to the pairwise comparison method. The winner is Bong Joon-ho with 4 points using the pairwise comparison method.

Your Turn 11.9 (c) §11.1 source ↗

According to Variety magazine, there were 8,469 eligible Oscar voters in 2020. To make our matrix easier to work with, we've rounded this number up to 8,700. We'd never do this in a real voting situation. Suppose that the voter preferences for the ballot for Best Director had been as shown in the table below.

Number of Ballots	2,400	2,100	1,900	1,200	1,100
Martin Scorsese, The Irishman	3	1	4	5	2
Todd Phillips, Joker	1	4	5	2	4
Sam Mendes, 1917	5	5	3	4	5
Quentin Tarantino, Once Upon a Time in Hollywood	4	2	1	3	3
Bong Joon-ho, Parasite	2	3	2	1	1

Refer to the table to answer each question.

Is the winner a Condorcet candidate?

Answer

Yes, the winner is a Condorcet candidate.

Your Turn 11.10 §11.1 source ↗

A pairwise comparison matrix is given. Determine the winner by the pairwise comparison method. If there is not a winner, explain why. If there is a winner, tell whether the winner is a Condorcet candidate.

A table shows the pairwise comparison between P, Q, R, S, T, and Z. The data given in the table are as follows. The table shows six rows and seven columns. The column headers are Runner and Opponent, P, Q, R, S, T, and Z. Column one shows P wins, Q wins, R wins, S wins, T wins, and Z wins. Column two shows Nil, Q P 5, R P 2, S P 0, T P 2, and Z P 1. Column three shows P Q 1, Nil, R Q 0, S Q 2, T Q 5, and Z Q 4. Column four shows P R 4, Q R 6, Nil, S R 1, T R 2, and Z R 5. Column five shows P S 6, Q S 4, R S 5, Nil, T S 0, and Z S 1. Column six shows P T 4, Q T 1, R T 4, S T 6, Nil, and Z T 2. Column seven shows P Z 5, Q Z 2, R Z 1, S Z 5, T Z 4, and Nil.

A Pairwise Comparison Matrix

Answer

Compare PQ 1 to QP 5. Because 1 is less than 5, PQ 1 is the loser. Cross out PQ 1.

Compare PR 4 to RP 2. Because 2 is less than 4, RP 2 is the loser. Cross out RP 2.

Compare PS 6 to SP 0. Because 0 is less than 6, SP 0 is the loser. Cross it out.

Continue comparing in this manner until you have completed the table. Because there are no 3s in the table, there will be no ties.

A win earns 1 point, a tie earns a half point, and a loss earns nothing.

P has 4 wins (1 point each) for 4 points.

Q has 3 wins (1 point each) for 3 points.

R has 2 wins (1 point each) for 2 points.

S has 2 wins (1 point each) for 2 points.

T has 2 wins (1 point each) for 2 points.

Z has 2 wins (1 point each) for 2 points.

Candidate P wins with 4 points. P is not a Condorcet winner because P lost to Q.

Candidate P wins. Candidate P is not a Condorcet candidate because P lost to Candidate Q.

Your Turn 11.11 §11.1 source ↗

Suppose there is an election with 26 candidates, A through Z, and that Candidate C is a Condorcet candidate.

How many points did Candidate C win?

Answer

A Condorcet candidate wins every pairwise matchup. Candidate C had to win 25 pairwise matchups with the other letters to be the Condorcet candidate. Thus, Candidate C has 25 points. Candidate C won 25 points.

Your Turn 11.11 (b) §11.1 source ↗

Suppose there is an election with 26 candidates, A through Z, and that Candidate C is a Condorcet candidate.

What is the greatest number of points that any one of the other candidates could win?

Answer

The greatest number of points another candidate can earn is 24 points. Another candidate could win 24 matchups with everyone but C. The greatest number of points another candidate could win is 24 points.

Your Turn 11.11 (c) §11.1 source ↗

Suppose there is an election with 26 candidates, A through Z, and that Candidate C is a Condorcet candidate.

Is it possible for Candidate C to lose or tie?

Answer

To meet the qualification to be a Condorcet candidate, you must win every pairing. That excludes the possibility of a tie or a loss. No, Candidate C cannot lose or tie.

Your Turn 11.12 §11.1 source ↗

The Chionilis family is trying to decide on a restaurant again, but now they don’t want to deal with multiple runoffs or even ranking. They will use the approval voting method shown in the following table. Each family member will approve their top two choices. Rainbow China won when multiple runoffs were used. Find the winner for tonight’s dinner.

Options	A	R	S	T	G	E	M	D
Rainbow China	Yes	No	No	Yes	No	No	Yes	Yes
Dough Boys Pizza	Yes	Yes	Yes	Yes	No	Yes	No	Yes
Taco City	No	No	Yes	No	Yes	Yes	No	No
Caribbean Flavor	No	Yes	No	No	Yes	No	Yes	No

Answer

Count the number of Yes votes earned by each option.

Rainbow China: 4

Dough Boys Pizza: 6

Taco City: 3

Caribbean Flavor: 3

The option with the most Yes votes wins.

Dough Boys Pizza wins with the most points.

Dough Boys Pizza is the winner for dinner.

Your Turn 11.13 §11.1 source ↗

What would be the outcome of the election if every member of the Chionilis family approved their top three choices from the table below?

Options	A	R	S	T	G
Rainbow China	1	3	3	1	3
Dough Boys Pizza	2	2	1	2	1
Taco City	3	4	2	4	2
Caribbean Flavor	4	1	4	3	4

Answer

Count the number of Yes votes earned by each option. Anything with a score of 1, 2, or 3 counts as a Yes.

Rainbow China: 5

Dough Boys Pizza: 5

Taco City: 3

Caribbean Flavor: 2

The option with the most Yes votes wins.

Rainbow China and Dough Boys Pizza have a tie this time.

There is a tie between Rainbow China and Dough Boys Pizza.

Your Turn 11.14 §11.1 source ↗

Extrapolate the results of an approval method election using Table 11.3, the assumptions from Example 11.14, and the additional assumptions that the supporters of Al Gore would all approve Ralph Nader and that half of the supporters of George Bush would approve Pat Buchanan while half would approve Harry Brown.

Table 11.3 in the book ↗

Candidate	Party	Votes	Percentage
(G) George W. Bush	Republican	2,912,790	48.85%
(A) Al Gore	Democrat	2,912,253	48.84%
(R) Ralph Nader	Green	97,488	1.63%
(P) Pat Buchanan	Reform	17,484	0.29%
(H) Harry Brown	Libertarian	16,415	0.28%
(O) 7 Other Candidates	Other	6,680	0.11%
	Total	5,963,110

Florida Results in the 2000 U.S. Presidential Election (source: https://www.fec.gov/resources/cms-content/documents/FederalElections2000_PresidentialGeneralElectionResultsbyState.pdf)

Answer

Candidate	Party	Votes	Percentage
(G) George W. Bush	Republican	2,912,790	48.85%
(A) Al Gore	Democrat	2,912,253	48.84%
(R) Ralph Nader	Green	97,488	1.63%
(P) Pat Buchanan	Reform	17,484	0.29%
(H) Harry Brown	Libertarian	16,415	0.28%
(O) 7 Other Candidates	Other	6,680	0.11%

100 percent of Pat Buchanan supporters would approve of George W. Bush.

100 percent of Ralph Nader supporters would approve Al Gore.

72 percent of Libertarians would approve of George W. Bush.

28 percent of Libertarians would approve Al Gore (as was roughly the known percentage at the time according to the Cato Institute).

50 percent of the supporters of other candidates would approve of George Bush, whereas 50 percent would approve of Al Gore.

100 percent of Al Gore supporters would approve of Ralph Nader.

50 percent of George Bush supporters would approve of Pat Buchanan.

50 percent of 2,912,790 is 1,456,395.

50 percent of George Bush supporters would approve of Harry Brown.

George W. Bush: 2,912,790 + 17,484 + 11,819 + 3340 = 2,945,433

Al Gore: 2,912,253 + 97,488 + 4,596 + 3,340 = 3,017,677

Ralph Nader: 97,488 + 2,912,253 = 3,009,741

Pat Buchanan: 17,484 + 1,456,395 = 1,473,879

Harry Brown: 11,819 + 4,596 + 1,456,395 = 1,472,810

Other Candidates: 6,680

Al Gore would win.

Al Gore is the winner.

Your Turn 11.15 §11.1 source ↗

The juniors at a high school in Central Florida are voting for a theme park to visit for an end of the year field trip. The options are Disney’s four theme parks: Animal Kingdom, Magic Kingdom, EPCOT Center, or Hollywood Studios.

Percentage of Voters	35%	25%	30%
Animal Kingdom	1	3	4
Magic Kingdom	2	1	3
Epcot Center	3	4	1
Hollywood Studios	4	2	2

Determine the outcome of the vote by each of the given voting methods. Plurality

Answer

Plurality considers only the top choice of each voter. The percentages for first-place votes show that the Animal Kingdom wins.

Animal Kingdom: 35%

Magic Kingdom: 25%

Epcot: 30%

Hollywood Studios: no first-place votes

Animal Kingdom

Your Turn 11.15 (b) §11.1 source ↗

The juniors at a high school in Central Florida are voting for a theme park to visit for an end of the year field trip. The options are Disney’s four theme parks: Animal Kingdom, Magic Kingdom, EPCOT Center, or Hollywood Studios.

Percentage of Voters	35%	25%	30%
Animal Kingdom	1	3	4
Magic Kingdom	2	1	3
Epcot Center	3	4	1
Hollywood Studios	4	2	2

Determine the outcome of the vote by each of the given voting methods. Ranked-choice method

Answer

A park needs more than 50 percent of the first-place votes to win in the ranked-choice method.

Animal Kingdom: 35%

Magic Kingdom: 25%

Epcot: 30%

Hollywood Studios: no first-place votes

No one wins the first round. Eliminate Hollywood Studios and redo the ranks.

Second Round:

Percentage of Voters	35%	25%	30%
Animal Kingdom	1	2	3
Magic Kingdom	2	1	2
Epcot Center	3	3	1

Animal Kingdom: 35%

Magic Kingdom: 25%

Epcot: 30%

No one wins. Eliminate Magic Kingdom and redo the ranks.

Third Round:

Percentage of Voters	35%	25%	30%
Animal Kingdom	1	1	2
Epcot Center	2	2	1

Animal Kingdom: 35% + 25% = 60%

Epcot: 30%

Animal Kingdom wins.

Animal Kingdom

Your Turn 11.15 (c) §11.1 source ↗

The juniors at a high school in Central Florida are voting for a theme park to visit for an end of the year field trip. The options are Disney’s four theme parks: Animal Kingdom, Magic Kingdom, EPCOT Center, or Hollywood Studios.

Percentage of Voters	35%	25%	30%
Animal Kingdom	1	3	4
Magic Kingdom	2	1	3
Epcot Center	3	4	1
Hollywood Studios	4	2	2

Determine the outcome of the vote by each of the given voting methods. Borda count

Answer

Percentage of Voters	35%	25%	30%
Animal Kingdom	1	3	4
Magic Kingdom	2	1	3
Epcot Center	3	4	1
Hollywood Studios	4	2	2

First place gets (4 – 1) = 3 points, second place gets 2 points, third place gets 1 point, and fourth place gets 0 points.

Animal Kingdom: 35(3) + 25(1) + 30(0) = 130

Magic Kingdom: 35(2) + 25(3) + 30(1) = 175

Epcot Center: 35(1) + 25(0) + 30(3) = 125

Hollywood Studios: 35(0) + 25(3) + 30(3) = 165

Magic Kingdom wins.

Magic Kingdom

Check Your Understanding 11.1.1 §11.1 source ↗

Name three voting methods that use a ranked ballot.

Answer

Answers may vary. Example: Ranked-choice, Borda count, and pairwise comparison.

Check Your Understanding 11.1.2 §11.1 source ↗

Determine whether the following statement is true or false: The same ranked ballots may result in a different winner depending on which voting method is used.

Answer

True.

Check Your Understanding 11.1.3 §11.1 source ↗

Determine whether the following statement is true or false. A majority candidate is always a Condorcet candidate.

Answer

True

Check Your Understanding 11.1.4 §11.1 source ↗

The _______________ method is a system of voting using ranked ballots in which each candidate is awarded points corresponding to the number of candidates ranked lower on each ballot.

Answer

Borda count

Check Your Understanding 11.1.5 §11.1 source ↗

The ________________ method is a system of voting using ranked ballots (or multiple elections) in which each candidate receives a point for each candidate they would beat in a one-on-one election and half a point for each candidate they would tie.

Answer

Pairwise comparison

Check Your Understanding 11.1.6 §11.1 source ↗

The __________________ method is a runoff voting system in which only the candidate(s) with the very least votes are eliminated.

Answer

Hare

Check Your Understanding 11.1.7 §11.1 source ↗

Explain the differences between two-round voting and ranked-choice voting.

Answer

In two-round voting, only the top two candidates from Round 1 move on to Round 2, and there are only two rounds. In ranked-choice voting, all candidates except those in last place move on to the next round, and there can be many rounds of voting.

Your Turn 11.16 §11.2 source ↗

Use the information in the table to answer the following questions.

Voters	L	M	N	O	P
Option A	2	2	3	2	1
Option B	1	4	1	4	2
Option C	3	1	4	1	3
Option D	4	1	2	3	4

Which option is the winner when using the plurality voting method?

Answer

Option A: 1 vote

Option B: 2 votes

Option C: 2 votes

Option D: 1 vote

There is a tie between Options B and C.

Using the plurality voting method, there would be a tie between Option B and Option C.

Your Turn 11.16 (b) §11.2 source ↗

Use the information in the table to answer the following questions.

Voters	L	M	N	O	P
Option A	2	2	3	2	1
Option B	1	4	1	4	2
Option C	3	1	4	1	3
Option D	4	1	2	3	4

Which option is the winner when using the ranked-choice voting method?

Answer

There are five voters.

50 percent of 5 is 2.5.

The majority is 3 votes or more.

First Round:

Voters	L	M	N	O	P
Option A	2	2	3	2	1
Option B	1	4	1	4	2
Option C	3	1	4	1	3
Option D	4	1	2	3	4

Option A: 1 vote

Option B: 2 votes

Option C: 2 votes

Option D: 1 vote

No one wins. Eliminate A and D.

Second Round:

Voters	L	M	N	O	P
Option B	1	2	1	2	1
Option C	2	1	2	1	2

Option B: 3 votes

Option C: 2 votes

Option B wins.

Using the ranked-choice voting method, Option B would be the winner.

Your Turn 11.16 (c) §11.2 source ↗

Use the information in the table to answer the following questions.

Voters	L	M	N	O	P
Option A	2	2	3	2	1
Option B	1	4	1	4	2
Option C	3	1	4	1	3
Option D	4	1	2	3	4

Does the majority criterion apply? If so, for which voting method(s), if any, did the majority criterion fail?

Answer

The majority criterion states that the winner should be in favor with the majority of the voters. That is not the case here because there was not a majority winner in the first round. No, there was not a majority candidate in Round 1.

Your Turn 11.17 §11.2 source ↗

Use the information in the following table to find the winner using each of the voting methods in parts 1 and 2, then answer the question in part 3.

Voters	L	M	N	O	P
Option A	2	2	3	2	2
Option B	1	4	1	4	1
Option C	3	1	4	1	3
Option D	4	3	2	3	4

Pairwise comparison

Answer

AB 2 vs BA 3. BA wins, so cross out AB.

AC 3 vs CA 2. AC wins, so cross out CA.

AD 4 vs DA 1. AD wins, so cross out DA

BC 3 vs CB 2. BC wins, so cross out CB.

BD 3 vs DB 2. BD wins, so cross out DB.

CD 3 vs DC 2. CD wins, so cross out DC.

Opponent Runner	A	B	C	D	POINTS
A		AB 2 Cross out AB 2	AC 3 wins	AD 4 wins	2
B	BA 3 wins		BC 3 wins	BD 3 wins	3
C	CA 2 Cross out CA 2	CB 2 Cross out CB 2		CD 3 wins	1
D	DA 1 Cross out DA 1	DB 2 Cross out DB 2	DC 2 Cross out DC 2		0

B wins.

Using the pairwise comparison voting method, Option B wins.

Your Turn 11.17 (b) §11.2 source ↗

Use the information in the following table to find the winner using each of the voting methods in parts 1 and 2, then answer the question in part 3.

Voters	L	M	N	O	P
Option A	2	2	3	2	2
Option B	1	4	1	4	1
Option C	3	1	4	1	3
Option D	4	3	2	3	4

Borda count

Answer

First place earns (4 – 1) or 3 points.

Second place earns (4 – 2) or 2 points.

Third place earns (4 – 3) or 1 point.

Fourth place earns (4 – 4) or 0 points.

Voters	L	M	N	O	P
Option A	2	2	3	2	2
Option B	1	4	1	4	1
Option C	3	1	4	1	3
Option D	4	3	2	3	4

Option A: 2 + 2 + 1 + 2 + 2 = 9

Option B: 3 + 0 + 3 + 0 + 3 = 9

Option C: 1 + 3 + 0 + 3 + 1 = 8

Option D: 0 + 1 + 2 + 1 + 0 = 4

Option A and B tie.

Using the Borda count method, Option A and Option B tie.

Your Turn 11.17 (c) §11.2 source ↗

Use the information in the following table to find the winner using each of the voting methods in parts 1 and 2, then answer the question in part 3.

Voters	L	M	N	O	P
Option A	2	2	3	2	2
Option B	1	4	1	4	1
Option C	3	1	4	1	3
Option D	4	3	2	3	4

Does the majority criterion apply? If so, for which voting method(s) did the majority criterion fail?

Answer

There are five votes.

50 percent of five is 2.5.

A majority would need 3 first-place votes.

In a simple ballot, Option B would win the majority criterion because Option B had 3 first-place votes. The majority criterion would fail in the Borda count method because Options A and B tied.

Yes. The majority criterion fails for Borda count.

Your Turn 11.18 §11.2 source ↗

Determine whether the Condorcet criterion applies based on the summary of ranked ballots given in the table below.

Votes	3	2
Option A	1	3
Option B	2	1
Option C	3	2

Answer

An X represents crossing out a cell.

Opponent Runner	A	B	C	Points
A		AB 3 wins	AC 2 wins	2
B	BA 2 Cross out BA 2		BC 5 wins	1
C	CA 2 Cross out CA 2	CB 0 Cross out CB 0		0

Option A is a Condorcet candidate because Option A won every matchup.

Yes, because Option A is a Condorcet candidate.

Your Turn 11.19 §11.2 source ↗

Use the summary of ranked ballots below to find the winner and determine whether the Condorcet criterion is satisfied when each of the following voting methods are used. Recall that Option A is a Condorcet candidate.

Votes	3	2
Option A	1	3
Option B	2	1
Option C	3	2

Plurality

Answer

Option A: 3 votes

Option B: 2 votes

Option C: 0 votes

Option A wins by having the most votes. The Condorcet criterion is satisfied.

The pairwise comparison had already shown that Option A was a Condorcet candidate.

Opponent Runner	A	B	C	Points
A		AB 3 wins	AC 2 wins	2
B	BA 2 Cross out BA 2		BC 5 wins	1
C	CA 2 Cross out CA 2	CB 0 CB 0		0

Option A. Yes, the Condorcet criterion is satisfied.

Your Turn 11.19 (b) §11.2 source ↗

Use the summary of ranked ballots below to find the winner and determine whether the Condorcet criterion is satisfied when each of the following voting methods are used. Recall that Option A is a Condorcet candidate.

Votes	3	2
Option A	1	3
Option B	2	1
Option C	3	2

Ranked-choice voting

Answer

First Round:

Total votes: 5

50 percent of 5 is 2.5.

To win the majority, an option needs 3 votes or more.

Votes	3	2
Option A	1	3
Option B	2	1
Option C	3	2

Option A: 3 votes

Option B: 2 votes

Option C: 0 votes

Option A has enough votes to satisfy winning by a majority. The Condorcet criterion is satisfied.

Option A. Yes, the Condorcet criterion is satisfied.

Your Turn 11.19 (c) §11.2 source ↗

Use the summary of ranked ballots below to find the winner and determine whether the Condorcet criterion is satisfied when each of the following voting methods are used. Recall that Option A is a Condorcet candidate.

Votes	3	2
Option A	1	3
Option B	2	1
Option C	3	2

Borda count

Answer

In Borda count, the first place out of three earns (3 – 1) or 2 points.

Second place earns (3 – 2) or 1 point.

Third place earns (3 – 3) or 0 points.

Votes	3	2
Option A	1	3
Option B	2	1
Option C	3	2

Option A: 3(2) + 2(0) = 6

Option B: 3(1) + 2(2) = 7

Option C: 3(0) + 2(1) = 2

Option B wins. No, the Condorcet criterion is not satisfied. The Condorcet candidate did not win.

Option B. No, the Condorcet criterion is not satisfied.

Your Turn 11.20 §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Determine the winner of the election by Borda count.

Answer

Standard Poodle.

Your Turn 11.20 (b) §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Suppose that the 61 voters in the third column increased their ranking of the winner by 1. Determine the winner by Borda count with the new rankings.

Answer

Standard Poodle.

Your Turn 11.20 (c) §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Does this election violate the monotonicity criterion?

Answer

This election does not violate the monotonicity criterion.

Your Turn 11.20 (d) §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Do you think the result of part 3 is true for Borda count and the monotonicity criterion in general? Why or why not?

Answer

Increasing the ranking for a winner of a Borda count election on a ballot will increase that candidate’s Borda score while decreasing another candidate’s Borda score, but leaving the remaining candidates’ Borda score unchanged. So, a Borda count election will never violate the monotonicity criterion.

Your Turn 11.21 §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Determine the winner of the election by the ranked-choice method.

Answer

Labrador retriever wins the election.

Your Turn 11.21 (b) §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Suppose that the 24 voters in the last column of the table increased their ranking of the winner by 1. Determine the winner by the ranked-choice method with the new rankings.

Answer

Golden retriever wins the election.

Your Turn 11.21 (c) §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Does this election violate the monotonicity criterion?

Answer

This election violates the monotonicity criterion.

Your Turn 11.21 (d) §11.2 source ↗

Use the Favorite Large Dog Breed Ballot Summary to answer each question.

Do you think the result in question 3 is true for the ranked-choice method and the monotonicity criterion in general? Why or why not?

Answer

It is possible that the monotonicity criterion would be met in other ranked-choice election scenarios, but overall, the ranked-choice voting method is said to fail the monotonicity criterion even if it failed in only one scenario.

Your Turn 11.22 §11.2 source ↗

The local animal shelter is having another vote-by-donation charity event! This time, for a $10 donation, an individual can complete a ranked ballot indicating their favorite small dog breed, miniature, from the ones provided: miniature poodle, Yorkshire terrier, or Chihuahua. Use the summary of ballots below to answer each question.

Votes	53	42	24	11
(P) Miniature Poodle	1	3	3	2
(Y) Yorkshire Terrier	2	2	1	1
(C) Chihuahua	3	1	2	3

Determine the winner of the election by the ranked-choice method.

Answer

The total number of votes: 130.

50 percent of 130 is 65.

To win the majority, you need at least one more than 65, which is 66.

First Round:

Votes	53	42	24	11
(M) Miniature Poodle	1	3	3	2
(Y) Yorkshire Terrier	2	2	1	1
(C) Chihuahua	3	1	2	3

Miniature Poodle: 53

Yorkshire Terrier: 24 + 11 = 35

Chihuahua: 42

No one has 66, so eliminate the Yorkshire Terrier.

Second Round:

Votes	53	42	24	11
(M) Miniature Poodle	1	2	2	1
(C) Chihuahua	2	1	1	2

Miniature Poodle: 53 + 11 = 64

Chihuahua: 42 + 24 = 66

The Chihuahua wins.

Chihuahua.

Your Turn 11.22 (b) §11.2 source ↗

The local animal shelter is having another vote-by-donation charity event! This time, for a $10 donation, an individual can complete a ranked ballot indicating their favorite small dog breed, miniature, from the ones provided: miniature poodle, Yorkshire terrier, or Chihuahua. Use the summary of ballots below to answer each question.

Votes	53	42	24	11
(P) Miniature Poodle	1	3	3	2
(Y) Yorkshire Terrier	2	2	1	1
(C) Chihuahua	3	1	2	3

Determine the winner of the election if poodles are removed from the ballot.

Answer

The total number of votes: 130.

50 percent of 130 is 65.

To win the majority, you need at least one more than 65, which is 66.

First Round:

Votes	53	42	24	11
(Y) Yorkshire Terrier	1	2	1	1
(C) Chihuahua	2	1	2	2

Yorkshire Terrier: 53 + 24 + 11 = 88

Chihuahua: 42

The Yorkshire Terrier wins.

Yorkshire Terrier.

Your Turn 11.22 (c) §11.2 source ↗

The local animal shelter is having another vote-by-donation charity event! This time, for a $10 donation, an individual can complete a ranked ballot indicating their favorite small dog breed, miniature, from the ones provided: miniature poodle, Yorkshire terrier, or Chihuahua. Use the summary of ballots below to answer each question.

Votes	53	42	24	11
(P) Miniature Poodle	1	3	3	2
(Y) Yorkshire Terrier	2	2	1	1
(C) Chihuahua	3	1	2	3

Does this election violate IIA?

Answer

Yes, this violates IIA, or Independence of Irrelevant Alternatives Criterion. This means that the introduction or removal of a third candidate should not change or reverse the rankings of the original two candidates relative to one another. In particular, if a losing candidate is removed from a race or a new candidate is added, the winner should not change. Yes, this election violates IIA.

Your Turn 11.23 §11.2 source ↗

Use the initial ballot summary from Example 11.23 to answer the questions.

Determine the winner of the election by the Borda count method.

Answer

First place earns (4 – 1) or 3 points.

Second place earns (4 – 2) or 2 points.

Third place earns (4 – 3) or 1 point.

Fourth place earns (4 – 4) or 0 points.

Number of Ballots	9	11	7	6	3
(J) Jim Halpert (John Krasinski)	1	2	4	2	4
(P) Pam Beesly-Halpert (Jenna Fischer)	4	1	2	4	3
(D) Dwight Schrute (Rainn Wilson)	2	3	3	1	2
(M) Michael Scott (Steve Carell)	3	4	1	3	1

Jim Halpert: 9(3) + 11(2) + 7(0) + 6(2) + 3(0) = 61

Pam Beesley-Halpert: 9(0) + 11(3) + 7(2) + 6(0) + 3(1) = 50

Dwight Schrute: 9(2) + 11(1) + 7(1) + 6(3) + 3(2) = 60

Michael Scott: 9(1) + 11(0) + 7(3) + 6(1) + 3(3) = 45

Jim Halpert wins.

Jim wins.

Your Turn 11.23 (b) §11.2 source ↗

Use the initial ballot summary from Example 11.23 to answer the questions.

Determine the winner of the election by the Borda count method if Michael Scott is removed from the ballot.

Answer

Pam wins.

Your Turn 11.23 (c) §11.2 source ↗

Use the initial ballot summary from Example 11.23 to answer the questions.

Does this election violate IIA?

Answer

Yes, this is a violation of IIA.

Check Your Understanding 11.2.8 §11.2 source ↗

Which fairness criterion is violated by all four of the main ranked voting methods presented in this chapter?

Answer

(Independence of) Irrelevant Alternatives Criterion

Check Your Understanding 11.2.9 §11.2 source ↗

Which of the four main ranked voting methods presented in this chapter satisfies the Condorcet criterion?

Answer

Pairwise comparison

Check Your Understanding 11.2.10 §11.2 source ↗

Which of the four main ranked voting methods presented in this section violates the majority criterion?

Answer

Borda count method

Check Your Understanding 11.2.11 §11.2 source ↗

Which of the four main ranked voting methods presented in this section violates the monotonicity criterion?

Answer

Pairwise comparison

Check Your Understanding 11.2.12 §11.2 source ↗

According to Arrow’s Impossibility Theorem, which of the four main ranked voting methods presented in this chapter violate at least one of the fairness criteria?

Answer

All of them: Plurality, ranked-choice, pairwise comparison, and Borda count

Check Your Understanding 11.2.13 §11.2 source ↗

Determine whether the following statement is true or false and explain your reasoning: Any ranked election that violates the majority criterion also violates the Condorcet criterion.

Answer

True, because a majority candidate is always the Condorcet candidate.

Check Your Understanding 11.2.14 §11.2 source ↗

Determine whether the following statement is true or false and explain your reasoning: Any ranked election that violates the Condorcet criterion also violates the majority criterion.

Answer

False, because the ranked-choice method violates the Condorcet criterion, but it doesn’t violate the majority criterion.

Check Your Understanding 11.2.15 §11.2 source ↗

Does Arrow’s Impossibility Theorem apply to approval voting? Why or why not?

Answer

No. Arrow’s Impossibility Theorem does not apply to approval voting because it is not a ranked voting system

Check Your Understanding 11.2.16 §11.2 source ↗

Determine whether each statement is true or false. Explain your reasoning.

Arrow’s Impossibility Theorem guarantees that ranked voting systems always lead to unfair elections.

Answer

False. Arrow’s Impossibility Theorem says that no ranked voting system is perfect and that voter profiles may arise that will lead to a violation of one or more fairness criteria, but it does not guarantee that those voter profiles will occur or are even likely to occur.

Check Your Understanding 11.2.17 §11.2 source ↗

Determine whether each statement is true or false. Explain your reasoning.

Approval voting is in the class of voting systems called Cardinal Voting systems.

Answer

True. Candidates are rated as approved or not approved, and voters can give multiple candidates the same rating.

Your Turn 11.24 §11.3 source ↗

The SAT is to be administered at a high school. In preparation, pencils have been distributed to each of the classrooms based on the room capacity. Use the information in the following table to answer each question.


Room Number	B	C	D	E
Room Capacity (number of student desks)	24	18	32	22
Number of Pencils	36	27	48	33

Find the part to part ratio of desks to pencils for each room. Represent it as both a reduced fraction and a decimal rounded to the nearest hundredth as needed.

Answer

Divide the first quantity by the second quantity.

	B	C	D	E
Desks to Pencils	$\frac{24}{36} = \frac{2}{3} \approx 0.67$	$\frac{18}{27} = \frac{2}{3} \approx 0.67$	$\frac{32}{48} = \frac{2}{3} \approx 0.67$	$\frac{22}{33} = \frac{2}{3} \approx 0.67$

B

\frac{2}{3}

; 0.67, C

\frac{2}{3}

; 0.67, D

\frac{2}{3}

; 0.67, and E

\frac{2}{3}

; 0.67

Your Turn 11.24 (b) §11.3 source ↗

The SAT is to be administered at a high school. In preparation, pencils have been distributed to each of the classrooms based on the room capacity. Use the information in the following table to answer each question.


Room Number	B	C	D	E
Room Capacity (number of student desks)	24	18	32	22
Number of Pencils	36	27	48	33

Find the part to part ratio of pencils to desks for each room. Represent it as both a reduced fraction and a decimal rounded to the nearest hundredth as needed.

Answer

	B	C	D	E
Pencils to Desks	$\frac{36}{24} = \frac{3}{2} = 1.5$	$\frac{27}{18} = \frac{3}{2} = 1.5$	$\frac{48}{32} = \frac{3}{2} = 1.5$	$\frac{33}{22} = \frac{3}{2} = 1.5$

B

\frac{3}{2}

; 1.5, C

\frac{3}{2}

; 1.5, D

\frac{3}{2}

; 1.5, and E

\frac{3}{2}

; 1.5

Your Turn 11.24 (c) §11.3 source ↗

The SAT is to be administered at a high school. In preparation, pencils have been distributed to each of the classrooms based on the room capacity. Use the information in the following table to answer each question.


Room Number	B	C	D	E
Room Capacity (number of student desks)	24	18	32	22
Number of Pencils	36	27	48	33

Have the pencils been distributed proportionally? If so, what is the constant ratio of pencils to desks? Give the units.

Answer

Yes, the constant ratio is

\frac{3}{2}

pencils per desk.

Your Turn 11.25 §11.3 source ↗

Refer again to the table providing information on classroom capacities and pencil distribution.


Room Number	B	C	D	E
Room Capacity (number of student desks)	24	18	32	22
Number of Pencils	36	27	48	33

Determine the number of pencils that would be allocated to a classroom F with 28 desks by multiplying the number of desks by the ratio of pencils to desks.

Answer

$\frac{Pencils}{Desks} = \frac{3}{2} = \frac{p}{28}$

$\begin{matrix} 2 p = 84 \\ p = 42 \end{matrix}$

You need to allocate 42 pencils.

42 pencils would be allocated.

Your Turn 11.25 (b) §11.3 source ↗

Refer again to the table providing information on classroom capacities and pencil distribution.


Room Number	B	C	D	E
Room Capacity (number of student desks)	24	18	32	22
Number of Pencils	36	27	48	33

Determine the number of pencils that would be allocated to a classroom F with 28 desks by dividing the number of desks by the ratio of desks to pencils.

Answer

$\frac{Desks}{Pencils} = \frac{2}{3} = \frac{28}{p}$

$\begin{matrix} 2 p = 84 \\ p = 42 \end{matrix}$

You need to allocate 42 pencils.

42 pencils would be allocated.

Your Turn 11.25 (c) §11.3 source ↗

Refer again to the table providing information on classroom capacities and pencil distribution.


Room Number	B	C	D	E
Room Capacity (number of student desks)	24	18	32	22
Number of Pencils	36	27	48	33

Assuming pencils continued to be distributed in the same manner, if there were another classroom, G, that received 51 pencils, how many desks would we expect to find in the room?

Answer

$\frac{Desks}{Pencils} = \frac{2}{3} = \frac{d}{51}$

$\begin{matrix} 3 d = 102 \\ d = 34 \end{matrix}$

You need to allocate 34 desks.

34 desks

Your Turn 11.26 §11.3 source ↗

Table 11.9 contains a list of the five U.S. states ranked sixth through tenth in the number of representatives in the U.S. House of Representatives, along with the population of that state in 2021. Use the information in the table to answer the questions. What is the ratio of State Population to Representative Seats for each state to the nearest hundred thousand?

Table 11.5 in the book ↗

State	Representative Seats	State Population
(IL) Illinois	18	12,804,100
(OH) Ohio	16	11,714,600
(MI) Michigan	14	9,992,430
(GA) Georgia	14	10,830,000
(NC) North Carolina	13	10,701,000

The Sixth through Tenth Ranked U.S. States by Number of Representatives (sources: https://www.census.gov/popclock/ [state population], https://www.britannica.com/topic/United-States-House-of-Representatives-Seats-by-State-1787120 [representative seats])

Answer

Make a ratio of $\frac{State Population}{Representative Seats}$ for each state to the nearest hundred thousand.

State	$\frac{State Population}{Representative Seats}$
IL	$\frac{12, 804, 100}{18} \approx 700, 000$
OH	$\frac{11, 714, 600}{16} \approx 700, 000$
MI	$\frac{9, 992, 430}{14} \approx 700, 000$
GA	$\frac{10, 830, 000}{14} \approx 800, 000$
NC	$\frac{10, 701, 000}{13} \approx 800, 000$

IL 700,000; OH 700,000; MI 700,000; GA 800,000; NC 700,000

Your Turn 11.26 (b) §11.3 source ↗

Table 11.9 contains a list of the five U.S. states ranked sixth through tenth in the number of representatives in the U.S. House of Representatives, along with the population of that state in 2021. Use the information in the table to answer the questions. What is the ratio of Representative Seats to State Population for each state rounded to seven decimal places?

Table 11.5 in the book ↗

State	Representative Seats	State Population
(IL) Illinois	18	12,804,100
(OH) Ohio	16	11,714,600
(MI) Michigan	14	9,992,430
(GA) Georgia	14	10,830,000
(NC) North Carolina	13	10,701,000

The Sixth through Tenth Ranked U.S. States by Number of Representatives (sources: https://www.census.gov/popclock/ [state population], https://www.britannica.com/topic/United-States-House-of-Representatives-Seats-by-State-1787120 [representative seats])

Answer

State	$\frac{Representative Seats}{State Population}$
IL	$\frac{18}{12, 804, 100} \approx 0.0000014$
OH	$\frac{16}{11, 714, 600} \approx 0.0000014$
MI	$\frac{14}{9, 992, 430} \approx 0.0000014$
GA	$\frac{14}{10, 830, 000} \approx 0.0000013$
NC	$\frac{13}{10, 701, 000} \approx 0.0000012$

IL 0.0000014; OH 0.0000014; MI 0.0000014; GA 0.0000013; NC 0.0000014

Your Turn 11.26 (c) §11.3 source ↗

Table 11.9 contains a list of the five U.S. states ranked sixth through tenth in the number of representatives in the U.S. House of Representatives, along with the population of that state in 2021. Use the information in the table to answer the questions. What is the ratio of Representative Seats to State Population for each state rounded to six decimal places?

Table 11.5 in the book ↗

State	Representative Seats	State Population
(IL) Illinois	18	12,804,100
(OH) Ohio	16	11,714,600
(MI) Michigan	14	9,992,430
(GA) Georgia	14	10,830,000
(NC) North Carolina	13	10,701,000

The Sixth through Tenth Ranked U.S. States by Number of Representatives (sources: https://www.census.gov/popclock/ [state population], https://www.britannica.com/topic/United-States-House-of-Representatives-Seats-by-State-1787120 [representative seats])

Answer

State	$\frac{Representative Seats}{State Population}$
IL	$\frac{18}{12, 804, 100} \approx 0.000001$
OH	$\frac{16}{11, 714, 600} \approx 0.000001$
MI	$\frac{14}{9, 992, 430} \approx 0.000001$
GA	$\frac{14}{10, 830, 000} \approx 0.000001$
NC	$\frac{13}{10, 701, 000} \approx 0.000001$

IL 0.000001; OH 0.000001; MI 0.000001; GA 0.000001; NC 0.000001

Your Turn 11.26 (d) §11.3 source ↗

Table 11.9 contains a list of the five U.S. states ranked sixth through tenth in the number of representatives in the U.S. House of Representatives, along with the population of that state in 2021. Use the information in the table to answer the questions. Does there appear to be a constant ratio? Are the results similar to the top five states?

Table 11.5 in the book ↗

State	Representative Seats	State Population
(IL) Illinois	18	12,804,100
(OH) Ohio	16	11,714,600
(MI) Michigan	14	9,992,430
(GA) Georgia	14	10,830,000
(NC) North Carolina	13	10,701,000

The Sixth through Tenth Ranked U.S. States by Number of Representatives (sources: https://www.census.gov/popclock/ [state population], https://www.britannica.com/topic/United-States-House-of-Representatives-Seats-by-State-1787120 [representative seats])

Answer

The ratio of State Population to Representative Seats seems to be either 700,000 or 800,000 to 1. There does appear to be a constant ratio of about 0.000001 to 1 of Representative Seats to State Population when rounding to six decimal places. This is the same as the top five states.

Your Turn 11.27 §11.3 source ↗

By the end of the first U.S. Congress in 1791, there were 13 states, 65 representative seats, and approximately 3,929,214 citizens. Find the standard divisor rounded to the nearest tenth.

Answer

The standard divisor is the total population divided by the house size. Divide the population by the 65 representative seats.

$\frac{3, 929, 214}{65} \approx 60, 449.4$

60,449.4

Your Turn 11.28 §11.3 source ↗

The Hernandez family and the Higgins family went trick-or-treating together for Halloween last year. They returned with 313 pieces of candy, which they will now apportion to the families. The Hernandez family has three children and the Higgins family has four children.

Identify the states, seats, house size, state population, and total population in this apportionment scenario.

Answer

The states are the recipients of the apportioned resource (the Hernandez family and the Higgins family).

The seats are the units of the resource being apportioned (the pieces of candy).

The house size is the total number of seats to be apportioned (the 313 pieces of candy).

The state population is the measurement of the state’s size (the size of the two families, which is three in the Hernandez family and four in the Higgins family).

The total population is the sum of the state populations $(3 + 4 = 7)$ .

The states are the Hernandez family and the Higgins family. The seats are the pieces of candy. The house size is 313. The state populations are three in the Hernandez family and four in the Higgins family. The total population is 7.

Your Turn 11.28 (b) §11.3 source ↗

The Hernandez family and the Higgins family went trick-or-treating together for Halloween last year. They returned with 313 pieces of candy, which they will now apportion to the families. The Hernandez family has three children and the Higgins family has four children.

Describe the ratio the standard divisor represents in this scenario and calculate the standard divisor to four decimal places.

Answer

The standard divisor is the total population divided by the house size.

In this scenario, the total population is the number of children in the two families (7).

The house size is the number of pieces of candy (313).

$\frac{7}{313} \approx 0.0224$ children per piece of candy.

The standard divisor is the ratio of the number of children to the number of pieces of candy. 0.0224 children per piece of candy.

Your Turn 11.29 §11.3 source ↗

By the end of the first U.S. Congress in 1791, there were 13 states, 65 representative seats, and approximately 3,929,214 citizens. In that year, the state of Delaware had a population of approximately 59,000. people. Use this information and the standard divisor you found in Your Turn 11.27 to find Delaware's standard quota rounded to two decimal places.

Answer

The standard divisor is the total population divided by the house size.

Standard Divisor: $\frac{3, 929, 214}{65} \approx 60, 449.4$

Standard Quota: $\frac{State Population}{Standard Divisor} seats = \frac{59, 000}{60, 449.4} seats = 0 .98 seats$

0.98

Your Turn 11.30 §11.3 source ↗

This year the Hernandez family and the Higgins family were joined by the Ho family for Halloween trick-or-treating. The Hernandez family has three children, the Higgins family has four children, and the Ho family has two children. This time, they collected 527 pieces of candy, which they are going to apportion based on the number of children. Find the standard quota for each family. Round all values to four decimal places. If traditional rounding methods are applied to determine the actual number whole number values of pieces of candy received by each family, do the values sum to 527?

Answer

The standard divisor is the total population divided by the house size.

In this scenario, the total population is the number of children in the two families $(3 + 4 + 2 = 9)$ .

The house size is the number of pieces of candy (527).

$Standard Divisor = \frac{Total Population}{House Size} = \frac{9 children}{527 candies} \approx 0.0170778$ children per piece of candy.

Family	$Standard Quota = \frac{State Population}{Standard Divisor}$
Hernandez	$\frac{3 children}{0.0170778 children per piece of candy} = 175 .67 candies$
Higgins	$\frac{4 children}{0.0170778 children per piece of candy} = 234 .22 candies$
Ho	$\frac{2 children}{0.0170778 children per piece of candy} = 117 .11 candies$

Add the three standard quotas: $175 .67 + 234 .22 + 117 .11 = 527$

This sum is 527. Even if you round to whole numbers, which any parent would when counting candy, the sum is 527.

$176 + 234 + 117 = 527$

Family	Family’s Standard Quota
Hernandez	175.6667 candies
Higgins	234.2222 candies
Ho	117.1111 candies
Total	527

The sum is very close to 527.

Check Your Understanding 11.3.18 §11.3 source ↗

In the following questions, assume there is a constant ratio between units of A and units of B. Two students are having a discussion. Determine who is correct: Student 1, Student 2, both, or neither.

Student 1 says that the number of units of A is the product of the number of units of B times the ratio of A to B.
Student 2 says that the number of units of A is the quotient of the number of units of B divided by the ratio of A to B.

Answer

Student 1:

A = B \cdot \frac{A}{B}

This is true because the right side simplifies to A.

Student 2:

$A = B \div \frac{A}{B}$

This is the same as

A = B \cdot \frac{B}{A}

This is not true because the left side does not equal the right side.

Student 1 is correct.

Check Your Understanding 11.3.19 §11.3 source ↗

In the following questions, assume there is a constant ratio between units of A and units of B. Two students are having a discussion. Determine who is correct: Student 1, Student 2, both, or neither.

Student 1 says that the number of units of A is the quotient of the number of units of B divided by the ratio of B to A.
Student 2 says that the number of units of B is the quotient of the number of units of A divided by the ratio of A to B.

Answer

Student 1:

$A = B \div \frac{B}{A}$

This is the same as

A = B \cdot \frac{A}{B}

This is true because the right side simplifies to A.

Student 2:

$B = A \div \frac{A}{B}$

This is the same as

B = A \cdot \frac{B}{A}

This is true because the right side simplifies to B.

Both students are correct.

Both are correct.

Check Your Understanding 11.3.20 §11.3 source ↗

In the following questions, assume there is a constant ratio between units of A and units of B. Two students are having a discussion. Determine who is correct: Student 1, Student 2, both, or neither.

Suppose there are 110 of item A and 55 of item B.
Student 1 says the ratio of A to B is

\frac{1}{2}

Student 2 says the ratio of B to A is

\frac{1}{0.5}

.

Answer

Student 1:

$\frac{A}{B} = \frac{110}{55} = 2$

Student 1 is wrong.

Student 2:

$\frac{B}{A} = \frac{55}{110} = \frac{1}{2}$

$\frac{1}{0.5} = 2$

Student 2 is wrong.

Both students are incorrect.

Neither are correct.

Check Your Understanding 11.3.21 §11.3 source ↗

In the following questions, assume there is a constant ratio between units of A and units of B. Two students are having a discussion. Determine who is correct: Student 1, Student 2, both, or neither.

Suppose the constant ratio of A to B is 0.75. Student 1 says the ratio of A to B is

\frac{4}{3}

Student 2 says the ratio of B to A is

\frac{3}{4}

.

Answer

Given: $\frac{A}{B} = 0.75$

Student 1:

$\frac{A}{B} = \frac{4}{3} \approx 1.3333$

Student 1 is wrong.

Student 2:

B to A should be $\frac{1}{0.75} \approx 1.33333$ .

Student 2 says B to A is $\frac{3}{4} = 0.75$ .

Student 2 is wrong.

Both students are wrong.

Neither are correct.

Check Your Understanding 11.3.22 §11.3 source ↗

In the following questions, assume there is a constant ratio between units of A and units of B. Two students are having a discussion. Determine who is correct: Student 1, Student 2, both, or neither.

Student 1 says that

Standard Divisor = \frac{House Size}{Total Population}

. Student 2 says that

Standard Divisor = \frac{Sum of the State Populations}{Total Number of Seats Available}

.

Answer

Student 1 is incorrect. $Standard Divisor = \frac{Total Population}{House Size}$

Student 2 is correct. $Standard Divisor = \frac{Total Population}{House Size}$ . The total population is the sum of the state populations. The house size is the total number of seats available.

Student 2 is correct.

Check Your Understanding 11.3.23 §11.3 source ↗

In the following questions, assume there is a constant ratio between units of A and units of B. Two students are having a discussion. Determine who is correct: Student 1, Student 2, both, or neither.

Student 1 says that

State's Standard Divisor = \frac{Total Population}{State's Standard Quota}

.
Student 2 says that

State's Standard Quota = \frac{State Population}{Standard Divisor}

.

Answer

Student 1 is incorrect. The House Size should be in the denominator. $Standard Divisor = \frac{Total Population}{House Size}$

Student 2 is correct. $Standard Quota = \frac{State Population}{Standard Divisor}$

Student 2 is correct.

Check Your Understanding 11.3.24 §11.3 source ↗

In the following questions, assume there is a constant ratio between units of A and units of B. Two students are having a discussion. Determine who is correct: Student 1, Student 2, both, or neither.

Student 1 says that

State's Standard Quota = State Population \div \frac{Total Population}{House Size}

.
Student 2 says that

State's Standard Quota = \frac{State Population}{Standard Divisor}

Answer

Because the Standard Quota = $\frac{State Population}{Standard Divisor} seats$ and the $Standard Divisor = \frac{Total Population}{House Size}$ ,

both students are correct. Student 1 substituted the Standard Divisor expression into the Standard Quota formula.

Both are correct.

Your Turn 11.31 §11.4 source ↗

The science department of a high school has received a grant for 34 laptops. They plan to apportion them among their six classrooms based on each classroom’s student capacity. Use the standard quotas in the table below to answer each question.

Room	Room Capacity	Room’s Standard Quota
A	30	$30 \div 4.88 \approx 6.15$ laptops
B	25	$25 \div 4.88 \approx 5.12$ laptops
C	28	$28 \div 4.88 \approx 5.74$ laptops
D	32	$32 \div 4.88 \approx 6.56$ laptops
E	24	$24 \div 4.88 \approx 4.92$ laptops
F	27	$27 \div 4.88 \approx 5.53$ laptops

Use traditional rounding to determine the number of laptops assigned to each classroom.

Answer

The total population of students: $30 + 25 + 28 + 32 + 24 + 27 = 166$ students. $Standard Divisor = \frac{Total Population}{House Size} = \frac{166 students}{34 laptops} \approx 4.88 students per laptop$

Room	$Standard Quota = \frac{State Population}{Standard Divisor}$	Rounded Laptops
A	$\frac{30 students}{4.88 students per laptop} = 6 .15 laptops$	6
B	$\frac{25 students}{4.88 students per laptop} = 5 .13 laptops$	5
C	$\frac{28 students}{4.88 students per laptop} = 5 .74 laptops$	6
D	$\frac{32 students}{4.88 students per laptop} = 6 .56 laptops$	7
E	$\frac{24 students}{4.88 students per laptop} = 4 .92 laptops$	5
F	$\frac{27 students}{4.88 students per laptop} = 5 .53 laptops$	6

6, 5, 6, 7, 5, 6

Your Turn 11.31 (b) §11.4 source ↗

The science department of a high school has received a grant for 34 laptops. They plan to apportion them among their six classrooms based on each classroom’s student capacity. Use the standard quotas in the table below to answer each question.

Room	Room Capacity	Room’s Standard Quota
A	30	$30 \div 4.88 \approx 6.15$ laptops
B	25	$25 \div 4.88 \approx 5.12$ laptops
C	28	$28 \div 4.88 \approx 5.74$ laptops
D	32	$32 \div 4.88 \approx 6.56$ laptops
E	24	$24 \div 4.88 \approx 4.92$ laptops
F	27	$27 \div 4.88 \approx 5.53$ laptops

Find the sum of the values from part 1.

Answer

The total population of students: $30 + 25 + 28 + 32 + 24 + 27 = 166$ students. $Standard Divisor = \frac{Total Population}{House Size} = \frac{166 students}{34 laptops} \approx 4.88 students per laptop$

Room	$Standard Quota = \frac{State Population}{Standard Divisor}$	Rounded Laptops
A	$\frac{30 students}{4.88 students per laptop} = 6 .15 laptops$	6
B	$\frac{25 students}{4.88 students per laptop} = 5 .13 laptops$	5
C	$\frac{28 students}{4.88 students per laptop} = 5 .74 laptops$	6
D	$\frac{32 students}{4.88 students per laptop} = 6 .56 laptops$	7
E	$\frac{24 students}{4.88 students per laptop} = 4 .92 laptops$	5
F	$\frac{27 students}{4.88 students per laptop} = 5 .53 laptops$	6

Sum of last column: $6 + 5 + 6 + 7 + 5 + 6 = 35$

35

Your Turn 11.31 (c) §11.4 source ↗

The science department of a high school has received a grant for 34 laptops. They plan to apportion them among their six classrooms based on each classroom’s student capacity. Use the standard quotas in the table below to answer each question.

Room	Room Capacity	Room’s Standard Quota
A	30	$30 \div 4.88 \approx 6.15$ laptops
B	25	$25 \div 4.88 \approx 5.12$ laptops
C	28	$28 \div 4.88 \approx 5.74$ laptops
D	32	$32 \div 4.88 \approx 6.56$ laptops
E	24	$24 \div 4.88 \approx 4.92$ laptops
F	27	$27 \div 4.88 \approx 5.53$ laptops

Does the sum found in part 2 equal the number of laptops available?

Answer

The total population of students: $30 + 25 + 28 + 32 + 24 + 27 = 166$ students. $Standard Divisor = \frac{Total Population}{House Size} = \frac{166 students}{34 laptops} \approx 4.88 students per laptop$

Room	$Standard Quota = \frac{State Population}{Standard Divisor}$	Rounded Laptops
A	$\frac{30 students}{4.88 students per laptop} = 6 .15 laptops$	6
B	$\frac{25 students}{4.88 students per laptop} = 5 .13 laptops$	5
C	$\frac{28 students}{4.88 students per laptop} = 5 .74 laptops$	6
D	$\frac{32 students}{4.88 students per laptop} = 6 .56 laptops$	7
E	$\frac{24 students}{4.88 students per laptop} = 4 .92 laptops$	5
F	$\frac{27 students}{4.88 students per laptop} = 5 .53 laptops$	6

Sum of last column: $6 + 5 + 6 + 7 + 5 + 6 = 35$

No. The rounded values added up to 35. There are only 34 available laptops. This is a problem.

No.

Your Turn 11.32 §11.4 source ↗

The reading coach at an elementary school has 13 gift cards to distribute to their three students as a reward for time spent reading. When they calculated the standard quota for each student based on the number of minutes they student had read, the results were: 4.49 gift cards, 4.03 gift cards, and 4.48 gift cards. How does this demonstrate the apportionment problem?

Answer

You cannot give out fractional gift cards. Round the gift card amounts using traditional rounding and add the results.

$4 + 4 + 4 = 12$

There are 13 cards. This leaves out one gift card.

It is not possible to give a fractional part of a gift card. Also, traditional rounding to the nearest integer results in 4 gift cards for each student, which leaves one extra gift card.

Your Turn 11.33 §11.4 source ↗

The apportionment of 70 new emergency blue lights in three parking lots is based on acreage. The standard quota for each lot is listed in the table below. Use this information to answer each question.

Lot	Acres	Lot’s Standard Quota
A	15	$15 \div 0.4857 \approx 30.88$ emergency blue lights
B	9	$9 \div 0.4857 \approx 18.53$ emergency blue lights
C	10	$10 \div 0.4857 = 20.59$ emergency blue lights

Give the lower quota for each parking lot.

Answer

The total population $= 15 + 9 + 10 = 34$ acres

$Standard Divisor = \frac{Total Population}{House Size} = \frac{34 acres}{70 lights} \approx 0.4857 acres per light$

Hamilton’s method is to always round down, which results in the lower quota.

Lot	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
A	$\frac{15 acres}{0.4857 acres per light} = 30 .88 lights$	30
B	$\frac{9 acres}{0.4857 acres per light} = 18 .53 lights$	18
C	$\frac{10 acres}{0.4857 acres per light} = 20 .59 lights$	20

30, 18, 20

Your Turn 11.33 (b) §11.4 source ↗

The apportionment of 70 new emergency blue lights in three parking lots is based on acreage. The standard quota for each lot is listed in the table below. Use this information to answer each question.

Lot	Acres	Lot’s Standard Quota
A	15	$15 \div 0.4857 \approx 30.88$ emergency blue lights
B	9	$9 \div 0.4857 \approx 18.53$ emergency blue lights
C	10	$10 \div 0.4857 = 20.59$ emergency blue lights

Find the sum of the lower quotas.

Answer

68 Add the lower quotas:

30 + 18 + 20 = 68

.

Your Turn 11.33 (c) §11.4 source ↗

The apportionment of 70 new emergency blue lights in three parking lots is based on acreage. The standard quota for each lot is listed in the table below. Use this information to answer each question.

Lot	Acres	Lot’s Standard Quota
A	15	$15 \div 0.4857 \approx 30.88$ emergency blue lights
B	9	$9 \div 0.4857 \approx 18.53$ emergency blue lights
C	10	$10 \div 0.4857 = 20.59$ emergency blue lights

By how much does this sum fall short of the actual number of emergency lights?

Answer

2

Add the lower quotas: $30 + 18 + 20 = 68$ .

There are 70 lights available.

This is a difference of 2.

Your Turn 11.34 §11.4 source ↗

In the country of Imaginaria, there will be four states: Fictionville, Pretendstead, Illusionham, and Mythbury. Suppose there will be 35 seats in the legislature of Imaginaria. Use Hamilton’s method of apportionment to determine the number of seats in each state based on the populations in the following table.

	Fictionville	Pretendstead	Illusionham	Mythbury	Total
Population	71,000	117,000	211,000	1,194,000	1,593,000

Answer

Use Hamilton’s method.

Step 1: Find the standard divisor.

$Standard Divisor = \frac{Total Population}{House Size} = \frac{1,593,000}{35 seats} \approx 45, 514.28571 people per seat$

Step 2: Find each state’s standard quota.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$	1
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$	2
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$	4
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$	26

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $1 + 2 + 4 + 26 = 33$ .

There are two more seats to allot. The largest fraction parts are 0.64 for Illusionham and 0.57 for Pretendstead. Give them the extra seats.

Fictionville: 1

Pretendstead: 3

Illusionham: 5

Mythbury: 26

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$1 + 3 + 5 + 26 = 35$

The final apportionment:

Fictionville: 1

Pretendstead: 3

Illusionham: 5

Mythbury: 26

The final Hamilton apportionment is Fictionville as follows: 1, Pretendstead 3, Illusionham 5, and Mythbury 26.

Your Turn 11.35 §11.4 source ↗

Apportionment Method V has been used to allocate 125 seats to ten states as shown in the table below. Determine whether the apportionment satisfies the quota rule and justify your answer.

	State A	State B	State C	State D	State E	State F	State G
Standard Quota	41.26	10.70	16.00	13.11	17.00	5.77	2.64
Lower Quota	41	10	16	13	17	5	2
Upper Quota	42	11	17	14	18	6	3
Method V	42	11	17	13	18	4	2

Answer

If a scenario exists in which a particular apportionment allocates a value greater than the upper quota or less than the lower quota, then that apportionment violates the quota rule and the apportionment method that was used violates the quota rule.

State F’s lower quota is 5 and upper quota is 6.

State F violates the quota rule because it receives 4 seats.

4 is lower than the lower quota of 5.

The other states are compliant with the quota rule.

Method V violates the quota rule because State F receives 4 seats instead of 5 or 6.

Your Turn 11.36 §11.4 source ↗

Suppose the population of a state is 12 and the standard divisor is 0.225.

Find the state’s standard quota.

Answer

Standard Quota = \frac{State Population}{Standard Divisor} = \frac{12}{0 .225} = 53 .33 seats

53.33

Your Turn 11.36 (b) §11.4 source ↗

Suppose the population of a state is 12 and the standard divisor is 0.225.

Decrease the standard divisor by 0.200 units and use the modified divisor to determine the modified quota for the state.

Answer

Decrease 0.225 by 0.2:

0.225 – 0.2 = 0.025, the new standard divisor

$Standard Quota = \frac{State Population}{Standard Divisor} = \frac{12}{0 .025} = 480 seats$

480

Your Turn 11.36 (c) §11.4 source ↗

Suppose the population of a state is 12 and the standard divisor is 0.225.

Increase the modified divisor from part 2 by 0.100 units and use the new modified divisor to determine the modified quota for the state.

Answer

Increase 0.025 by 0.10.

0.025 + 0.10 = 0.125, the new standard divisor

$Standard Quota = \frac{State Population}{Standard Divisor} = \frac{12}{0 .125} = 96 seats$

96

Your Turn 11.36 (d) §11.4 source ↗

Suppose the population of a state is 12 and the standard divisor is 0.225.

Choose any value of divisor between the value of the modified divisor from part 2 and the value of the modified divisor from part 3 and use it to determine the modified quota for the state.

Answer

Answers may vary. With a modified divisor of 0.100, the modified quota is 120.

Your Turn 11.36 (e) §11.4 source ↗

Suppose the population of a state is 12 and the standard divisor is 0.225.

Which modified quota was the smallest, the modified quota from part 2, from part 3, or from part 4? Explain why.

Answer

The modified quota from part 3 was the smallest, because the divisor was the largest of the three. Dividing the same number by a larger value gives a smaller result.

Your Turn 11.37 §11.4 source ↗

Let’s return to the Imaginarian states of Fictionville, Pretendstead, Illusionham, and Mythbury. Suppose that there are going to be 35 seats in the legislature. This time use Jefferson’s method of apportionment to determine the number of seats in each state based on the populations in the table below. How many seats would each state receive?

	Fictionville	Pretendstead	Illusionham	Mythbury	Total
Population	71,000	117,000	211,000	1,194,000	1,593,000

Answer

Step 1: Find the standard divisor.

$Standard Divisor = \frac{Total Population}{House Size} = \frac{1,593,000}{35 seats} \approx 45, 514.28571 people per seat$

Step 2: Find each state’s standard quota.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down. Find the sum, too.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$	1
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$	2
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$	4
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$	26

Add up the lower quotas: $1 + 2 + 4 + 26 = 33$ .

Step 4: If the sum of the lower quotas equals the number of seats, you are done. If the sum of the lower quotas is less than the number of seats, reduce the standard divisor. If the sum is greater, increase the standard divisor. Return to Step 2 using the modified divisor.

Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, lower quota, and the sum of the lower quotas based on the modified divisor.

Second Round:

Find each state’s standard quota using 43,000 people per seat.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{43, 000 people per seat} = 1 .65 seats$
Pretendstead	$\frac{117,000 people}{43, 000 people per seat} = 2 .72 seats$
Illusionham	$\frac{211,000 people}{43, 000 people per seat} = 4 .91 seats$
Mythbury	$\frac{1,194,000 people}{43, 000 people per seat} = 27 .77 seats$

Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down. Find the sum, too.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Fictionville	$\frac{71,000 people}{43, 000 people per seat} = 1 .65 seats$	1
Pretendstead	$\frac{117,000 people}{43, 000 people per seat} = 2 .72 seats$	2
Illusionham	$\frac{211,000 people}{43, 000 people per seat} = 4 .91 seats$	4
Mythbury	$\frac{1,194,000 people}{43, 000 people per seat} = 27 .77 seats$	27

Because the sum (34) is still lower than the number of seats (35), reduce the standard divisor again.

Third Round:

Find each state’s standard quota using 42,500 people per seat.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{42, 500 people per seat} = 1 .67 seats$
Pretendstead	$\frac{117,000 people}{42, 500 people per seat} = 2 .75 seats$
Illusionham	$\frac{211,000 people}{42, 500 people per seat} = 4 .96 seats$
Mythbury	$\frac{1,194,000 people}{42, 500 people per seat} = 28 .09 seats$

Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down. Find the sum, too.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Fictionville	$\frac{71,000 people}{42, 500 people per seat} = 1 .67 seats$	1
Pretendstead	$\frac{117,000 people}{42, 500 people per seat} = 2 .75 seats$	2
Illusionham	$\frac{211,000 people}{42, 500 people per seat} = 4 .96 seats$	4
Mythbury	$\frac{1,194,000 people}{42, 500 people per seat} = 28 .09 seats$	28

Now the sum of the lower quotas is equal to the number of seats.

Step 6: The new sum of the lower quotas equals the house size. The apportionment is complete.

$1 + 2 + 4 + 28 = 35$

Fictionville: 1

Pretendstead: 2

Illusionham: 4

Mythbury: 28

The modified divisor you chose may be different, but the final apportionment seat numbers will be the same for everyone who uses Jefferson’s method.

Each state would receive the following seats: Fictionville 1, Pretendstead 2, Illusionham 4, and Mythbury 28.

Your Turn 11.38 §11.4 source ↗

There are four states in Imaginaria: Fictionville, Pretendstead, Illusionham, and Mythbury. Assume there will be 35 seats in the legislature of Imaginaria. Use Adams's method of apportionment to determine the number of seats in each state based on the populations in the table below. How many seats would each state receive?

	Fictionville	Pretendstead	Illusionham	Mythbury	Total
Population	71,000	117,000	211,000	1,194,000	1,593,000

Answer

Use Adams’s method.

Step 1: Find the standard divisor.

$Standard Divisor = \frac{Total Population}{House Size} = \frac{1,593,000}{35 seats} \approx 45, 514.28571 people per seat$

Step 2: Find each state’s standard quota.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$

Step 3: Give each state the state’s upper quota (with each state receiving at least 1 seat). The upper quota always rounds up. Find the sum, too.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Upper Quota
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$	2
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$	3
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$	5
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$	27

Add up the upper quotas: $2 + 3 + 5 + 27 = 37$

Step 4: If the sum of the upper quotas equals the number of seats, you are done. If the sum of the upper quotas is less than the number of seats, reduce the standard divisor. If the sum is greater, increase the standard divisor. Return to Step 2 using the modified divisor.

Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, upper quota, and the sum of the upper quotas based on the modified divisor.

Second Round:

Find each state’s standard quota using 48,000 people per seat.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{48, 000 people per seat} = 1 .48 seats$
Pretendstead	$\frac{117,000 people}{48, 000 people per seat} = 2 .44 seats$
Illusionham	$\frac{211,000 people}{48, 000 people per seat} = 4 .40 seats$
Mythbury	$\frac{1,194,000 people}{48, 000 people per seat} = 24 .88 seats$

Give each state the state’s upper quota (with each state receiving at least 1 seat). The upper quota always rounds up. Find the sum, too.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Upper Quota
Fictionville	$\frac{71,000 people}{48, 000 people per seat} = 1 .48 seats$	2
Pretendstead	$\frac{117,000 people}{48, 000 people per seat} = 2 .44 seats$	3
Illusionham	$\frac{211,000 people}{48, 000 people per seat} = 4 .40 seats$	5
Mythbury	$\frac{1,194,000 people}{48, 000 people per seat} = 24 .88 seats$	25

Now the sum of the upper quotas is equal to the number of seats.

Step 6: The new sum of the upper quotas equals the house size. The apportionment is complete.

$2 + 3 + 5 + 25 = 35$

Fictionville: 2

Pretendstead: 3

Illusionham: 5

Mythbury: 25

The modified divisor you chose may be different, but the final apportionment seat numbers will be the same for everyone who uses Adams’s method.

Fictionville 2, Pretendstead 3, Illusionham 5, and Mythbury 25.

Your Turn 11.39 §11.4 source ↗

If you use Webster’s method to apportion 35 legislative seats to the 4 states of Imaginaria, Fictionville, Pretendstead, Illusionham, and Mythbury, with the populations given in the table below, what is the resulting apportionment?

	Fictionville	Pretendstead	Illusionham	Mythbury	Total
Population	71,000	117,000	211,000	1,194,000	1,593,000

Answer

Use Webster’s method.

Step 1: Find the standard divisor.

$Standard Divisor = \frac{Total Population}{House Size} = \frac{1,593,000}{35 seats} \approx 45, 514.28571 people per seat$

Step 2: Find each state’s standard quota.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$

Step 3: Round each state’s quota to the nearest whole number (with each state receiving at least 1 seat). Use traditional rounding. Find the sum, too.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Rounded Quota
Fictionville	$\frac{71,000 people}{45, 514.28571 people per seat} = 1 .56 seats$	2
Pretendstead	$\frac{117,000 people}{45, 514.28571 people per seat} = 2 .57 seats$	3
Illusionham	$\frac{211,000 people}{45, 514.28571 people per seat} = 4 .64 seats$	5
Mythbury	$\frac{1,194,000 people}{45, 514.28571 people per seat} = 26 .23 seats$	26

Add up the rounded quotas: $2 + 3 + 5 + 26 = 36$ .

Step 4: If the sum of the rounded quotas equals the number of seats, you are done. If the sum of the rounded quotas is less than the number of seats, reduce the standard divisor. If the sum is greater, increase the standard divisor. Return to Step 2 using the modified divisor.

Step 5: Repeat the process of finding the quotas. Find each state’s modified quota, rounded quota, and the sum of the rounded quotas based on the modified divisor.

Second Round:

Find each state’s standard quota using 46,820 people per seat.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$
Fictionville	$\frac{71,000 people}{46, 820 people per seat} = 1 .516 seats$
Pretendstead	$\frac{117,000 people}{46, 820 people per seat} = 2 .499 seats$
Illusionham	$\frac{211,000 people}{46, 820 people per seat} = 4 .507 seats$
Mythbury	$\frac{1,194,000 people}{46, 820 people per seat} = 25 .502 seats$

Give each state the state’s rounded quota (with each state receiving at least 1 seat). Use traditional rounding. Find the sum, too.

Town	$Standard Quota = \frac{State Population}{Standard Divisor}$	Rounded Quota
Fictionville	$\frac{71,000 people}{46, 820 people per seat} = 1 .516 seats$	2
Pretendstead	$\frac{117,000 people}{46, 820 people per seat} = 2 .499 seats$	2
Illusionham	$\frac{211,000 people}{46, 820 people per seat} = 4 .507 seats$	5
Mythbury	$\frac{1,194,000 people}{46, 820 people per seat} = 25 .502 seats$	26

Now the sum of the rounded quotas is equal to the number of seats.

Step 6: The new sum of the rounded quotas equals the house size. The apportionment is complete.

$2 + 2 + 5 + 26 = 35$

Fictionville: 2

Pretendstead: 2

Illusionham: 5

Mythbury: 26

The modified divisor you chose may be different, but the final apportionment seat numbers will be the same for everyone who uses Webster’s method.

The apportionment is Fictionville 2, Pretendstead 2, Illusionham 5, and Mythbury 26.

Your Turn 11.40 §11.4 source ↗

In Your Turn 11.34, 11.37, 11.38, and 11.39, you apportioned 35 legislative seats among the four states of Imaginaria using the Hamilton, Jefferson, Adams, and Webster methods of apportionment. To understand how the differences in the apportionments might affect Imaginarians, answer these questions.

Which apportionment method would the citizens of the largest state likely favor most and least? Justify your answer.

Answer

Town	Jefferson’s Method	Adams’s Method	Webster’s Method
Fictionville	1	2	2
Pretendstead	2	3	2
Illusionham	4	5	5
Mythbury	28	25	26

The largest state is Mythbury. Their citizens would likely favor Jefferson’s method of apportionment most because they received the most seats by that method. They would likely favor Adams’s and Webster’s methods the least because they received the least number of seats by those methods.

The largest state is Mythbury. The citizens would likely favor the Jefferson method of apportionment most since they received the most seats by that method. They would likely favor the Adams and Webster methods of apportionment least because they received the least number of seats by those method.

Your Turn 11.40 (b) §11.4 source ↗

In Your Turn 11.34, 11.37, 11.38, and 11.39, you apportioned 35 legislative seats among the four states of Imaginaria using the Hamilton, Jefferson, Adams, and Webster methods of apportionment. To understand how the differences in the apportionments might affect Imaginarians, answer these questions.

As a group, which apportionment method would the citizens of the other three states likely favor most and least? Justify your answer.

Answer

As a group the other three states received nine seats by either the Hamilton method, seven seats by the Jefferson Method, and ten seats by either the Adams method or the Webster method. They would likely favor the Adams method and Webster method most and favor the Jefferson methods least.

Your Turn 11.41 §11.4 source ↗

The following table displays the effect of increasing the size of the divisor. Observe the effect this has on the modified quotas of smaller states versus larger states and use the table to answer each question.

		Modified Quotas
State	Population	Divisor: 11,500	Divisor: 12,000	Divisor: 12,500
A	10,000	0.87	0.83	0.8
B	100,000	8.7	8.33	8
C	1,000,000	86.96	83.33	80

When the divisor increases from 11,500 to 12,000, how many representatives are lost by each state based on the upper quota?

Answer

	Modified Quota (Divisor: 11,500)	Upper Quota	Modified Quota (Divisor: 12,000)	Upper Quota
A	0.87	1	0.83	1
B	8.7	9	8.33	9
C	86.96	87	83.33	84

States A and B had no change.

State C lost 3.

State A loses 0, seats State B loses 0, and State C loses 3.

Your Turn 11.41 (b) §11.4 source ↗

The following table displays the effect of increasing the size of the divisor. Observe the effect this has on the modified quotas of smaller states versus larger states and use the table to answer each question.

		Modified Quotas
State	Population	Divisor: 11,500	Divisor: 12,000	Divisor: 12,500
A	10,000	0.87	0.83	0.8
B	100,000	8.7	8.33	8
C	1,000,000	86.96	83.33	80

When the divisor increases from 12,000 to 12,500, how many representatives are lost by each state based on the upper quota?

Answer

State A loses 0 seats, State B loses 1, and State C loses 4.

	Modified Quota (Divisor: 12,000)	Upper Quota	Modified Quota (Divisor: 12,500)	Upper Quota
A	0.83	1	0.8	1
B	8.33	9	8	8
C	83.33	84	80	80

State A lost nothing.

State B lost 1.

State C lost 4.

Your Turn 11.41 (c) §11.4 source ↗

The following table displays the effect of increasing the size of the divisor. Observe the effect this has on the modified quotas of smaller states versus larger states and use the table to answer each question.

		Modified Quotas
State	Population	Divisor: 11,500	Divisor: 12,000	Divisor: 12,500
A	10,000	0.87	0.83	0.8
B	100,000	8.7	8.33	8
C	1,000,000	86.96	83.33	80

Which state loses the most representatives each time the divisor is increased?

Answer

State C, the largest state, loses the most representatives each time the divisor is increased.

	Modified Quota (Divisor: 11,500)	Upper Quota	Modified Quota (Divisor: 12,000)	Upper Quota	Modified Quota (Divisor: 12,500)	Upper Quota
A	0.87	1	0.83	1	0.8	1
B	8.7	9	8.33	9	8	8
C	86.96	87	83.33	84	80	80

The switch from 11,500 to 12,000:

States A and B had no change.

State C lost 3.

The switch from 12,000 to 12,500:

State A lost nothing.

State B lost 1.

State C lost 4.

Overall, state C had the greatest losses each time the divisor was increased.

Check Your Understanding 11.4.25 §11.4 source ↗

Which of the four apportionment methods discussed in this section does not use a modified divisor?

Answer

The Hamilton method.

Check Your Understanding 11.4.26 §11.4 source ↗

Which of the four apportionment methods discussed in this section satisfies the quota rule?

Answer

The Hamilton method.

Check Your Understanding 11.4.27 §11.4 source ↗

Which of the four apportionment methods discussed in this section is biased toward states with larger populations?

Answer

The Jefferson method.

Check Your Understanding 11.4.28 §11.4 source ↗

Which of the four apportionment methods discussed in this section is biased toward states with smaller populations?

Answer

The Adams method.

Adams’s method favors smaller states. One of the Your Turn exercises showed this tendency. Here is a reminder:

	Standard Divisor: 11,500		Divisor: 12,000		Divisor: 12,500
	Modified Quota	Upper Quota	Modified Quota	Upper Quota	Modified Quota	Upper Quota
A	0.87	1	0.83	1	0.8	1
B	8.7	9	8.33	9	8	8
C	86.96	87	83.33	84	80	80

The first change from 11,500 to 12,000:

States A and B had no change.

State C lost 3.

The second change from 12,000 to 12,500:

State A lost nothing.

State B lost 1.

State C lost 4.

Overall, state C had the greatest losses each time the divisor was increased.

Check Your Understanding 11.4.29 §11.4 source ↗

Which of the four apportionment methods discussed in this section begin the apportionment with a state’s upper quota and adjust down?

Answer

The Adams method.

Check Your Understanding 11.4.30 §11.4 source ↗

Which of the four apportionment methods discussed in this section begin the apportionment with a state’s lower quota and adjust up?

Answer

The Hamilton and Jefferson methods.

Check Your Understanding 11.4.31 §11.4 source ↗

Which of the four apportionment methods discussed in this section use traditional rounding?

Answer

Webster’s method.

Check Your Understanding 11.4.32 §11.4 source ↗

Does the change from a standard divisor to a modified divisor tend to change the number of seats for larger or smaller states more?

Answer

Larger

Your Turn 11.42 §11.5 source ↗

The 1900 census recorded the population of Colorado as 539,700 and that of the U.S. as 76,212,168.

Calculate the standard divisor and standard quota for the State of Colorado based on a house size of 356.

Answer

$Standard Divisor = \frac{Total Population}{House Size} = \frac{76,212,168}{356 seats} \approx 214, 079.1236 people per seat$

$Standard Quota = \frac{State Population}{Standard Divisor} = \frac{539,700 people}{214, 079.1236 people per seat} = 2 .521 seats$

The standard divisor is 214,079.1236 citizens per seat. The standard quota for Colorado is 2.5210 seats.

Your Turn 11.42 (b) §11.5 source ↗

The 1900 census recorded the population of Colorado as 539,700 and that of the U.S. as 76,212,168.

Calculate the standard divisor and standard quota for the State of Colorado based on a house size of 357.

Answer

$Standard Divisor = \frac{Total Population}{House Size} = \frac{76,212,168}{357 seats} \approx 213, 479.4622 people per seat$

$Standard Quota = \frac{State Population}{Standard Divisor} = \frac{539,700 people}{213, 479.4622 people per seat} = 2 .528 seats$

The standard divisor is 213,479.4622 citizens per seat. The standard quota for Colorado is 2.5281 seats.

Your Turn 11.42 (c) §11.5 source ↗

The 1900 census recorded the population of Colorado as 539,700 and that of the U.S. as 76,212,168.

Did the standard quota increase or decrease when the house size increased?

Answer

The house size increased from 356 to 357 seats. The standard quota increased from 2.521 seats to 2.528 seats. The standard quota increased.

Your Turn 11.42 (d) §11.5 source ↗

The 1900 census recorded the population of Colorado as 539,700 and that of the U.S. as 76,212,168.

Consider the Hamilton method of apportionment. Explain how Colorado’s final quota could be smaller with a larger standard quota.

Answer

It must have been the case that either the fractional part 0.5281 ranked lower amongst the other fractional parts of the state quotas than the fractional part 0.5210 did, or there were fewer remaining seats to be distributed, or both.

Your Turn 11.43 §11.5 source ↗

Suppose that the founders of Imaginaria decide to have a parliament that apportions seats to four political parties based on the portion of the vote each party has earned. Also, suppose that Party A has 56.7 thousand votes, Party B has 38.5 thousand votes, Party C has 4.2 thousand votes, and Party D has 0.6 thousand votes.

Use the Hamilton method to apportion 323 seats.

Answer

Use Hamilton’s method for 323 seats.

Step 1: Find the standard divisor.

Total population: $56,700 + 38,500 + 4,200 + 600 = 100,000$

$Standard Divisor = \frac{Total Population}{House Size} = \frac{100,000}{323 seats} \approx 309.5975 votes per seat$

Step 2: Find each state’s standard quota.

Party	$Standard Quota = \frac{State Population}{Standard Divisor}$
Party A	$\frac{56,700 votes}{309.5975 votes per seat} = 183 .141 seats$
Party B	$\frac{38,500 votes}{309.5975 votes per seat} = 124 .355 seats$
Party C	$\frac{4,200 votes}{309.5975 votes per seat} = 13 .566 seats$
Party D	$\frac{600 votes}{309.5975 votes per seat} = 1 .938 seats$

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

Party	$Standard Quota = \frac{State Population}{Standard Divisor}$	Modified Quota
Party A	$\frac{56,700 votes}{309.5975 votes per seat} = 183 .141 seats$	183
Party B	$\frac{38,500 votes}{309.5975 votes per seat} = 124 .355 seats$	124
Party C	$\frac{4,200 votes}{309.5975 votes per seat} = 13 .566 seats$	$13 + 1 = 14$
Party D	$\frac{600 votes}{309.5975 votes per seat} = 1 .938 seats$	$1 + 1 = 2$

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $183 + 124 + 13 + 1 = 321$ .

Remaining seats to allot: 2

The largest fractional part is 0.938 for Party D and 0.566 for C. Give the seats to Party C and D.

Party	Modified Quota
Party A	183
Party B	124
Party C	$13 + 1 = 14$
Party D	$1 + 1 = 2$

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$184 + 125 + 14 = 323$

The final apportionment is: A 183, B 124, C 14, and D 2, which sums to 323.

Your Turn 11.43 (b) §11.5 source ↗

Suppose that the founders of Imaginaria decide to have a parliament that apportions seats to four political parties based on the portion of the vote each party has earned. Also, suppose that Party A has 56.7 thousand votes, Party B has 38.5 thousand votes, Party C has 4.2 thousand votes, and Party D has 0.6 thousand votes.

Use the Hamilton method to apportion 324 seats.

Answer

Use Hamilton’s method for 323 seats.

Step 1: Find the standard divisor.

Total population: $56,700 + 38,500 + 4,200 + 600 = 100,000$

$Standard Divisor = \frac{Total Population}{House Size} = \frac{100,000}{324 seats} \approx 308.6420 votes per seat$

Step 2: Find each state’s standard quota.

Party	$Standard Quota = \frac{State Population}{Standard Divisor}$
Party A	$\frac{56,700 votes}{308.6420 votes per seat} = 183 .708 seats$
Party B	$\frac{38,500 votes}{308.6420 votes per seat} = 124 .740 seats$
Party C	$\frac{4,200 votes}{308.6420 votes per seat} = 13 .608 seats$
Party D	$\frac{600 votes}{308.6420 votes per seat} = 1 .944 seats$

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

Party	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Party A	$\frac{56,700 votes}{308.6420 votes per seat} = 183 .708 seats$	$183 + 1 = 184$
Party B	$\frac{38,500 votes}{308.6420 votes per seat} = 124 .740 seats$	$124 + 1 = 125$
Party C	$\frac{4,200 votes}{308.6420 votes per seat} = 13 .608 seats$	13
Party D	$\frac{600 votes}{308.6420 votes per seat} = 1 .944 seats$	$1 + 1 = 2$

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $183 + 124 + 13 + 1 = 321$ .

Remaining seats to allot: 3

The largest fractional part is 0.944 for D, 0.740 for B, 0.708 for A. Give A, B, and D a seat.

Party	Lower Quota
Party A	184
Party B	125
Party C	13
Party D	2

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$184 + 125 + 13 + 2 = 324$

The final apportionment is: A 184, B 125, C 13, and D 2, which sums to 324.

Your Turn 11.43 (c) §11.5 source ↗

Suppose that the founders of Imaginaria decide to have a parliament that apportions seats to four political parties based on the portion of the vote each party has earned. Also, suppose that Party A has 56.7 thousand votes, Party B has 38.5 thousand votes, Party C has 4.2 thousand votes, and Party D has 0.6 thousand votes.

Does this example demonstrate the Alabama paradox? If so, how?

Answer

Party	Seat Size of 323	Seat Size of 324	Difference
Party A	183	184	1
Party B	124	125	1
Party C	14	13	–1
Party D	2	2	0

The Alabama paradox occurs when an increase in the house size reduces a state’s quota. Party C lost a seat despite an increase in the house size.

Yes, this demonstrates the Alabama paradox because State C receives 14 seats if the house size is 323, but only 13 seats if the house size is 324.

Your Turn 11.44 §11.5 source ↗

Suppose that 18 respirators are to be apportioned to three hospitals based on their capacities. The respirators are allocated based on the Hamilton method in 2020, then reallocated based on new capacities in 2021. The results are shown in the table below. How does this demonstrate the population paradox?

Hospital	Capacity in 2020	Respirators in 2020	Change in Capacity	Growth Rate $\frac{Change in Capacity}{Capacity in 2020}$	Capacity in 2021	Respirators in 2021
A	237	5	6	$6 \div 237 = 0.0253 = 2.53 %$	243	6
B	889	21	69	$69 \div 889 \approx 0.0776 = 7.76 %$	958	21
C	674	16	18	$18 \div 674 \approx 0.0267 = 2.67 %$	692	15

Answer

Hospital C lost a respirator while hospital A gained a seat, but hospital C has a higher growth rate than hospital A.

The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats. More generally, the population paradox occurs when a state with a higher population growth rate loses seats while a state with a lower population growth rate retains or gains seats.

Hospital C lost a respirator while hospital A gained a seat, but hospital C has a higher growth rate than hospital A.

Your Turn 11.45 §11.5 source ↗

The country of Elbonia has three states: Mudston with a population of 866,000; WallaWalla with a population of 626,000; and Dilberta with a population of 256,000. There are 38 seats in the Congress of Elbonia. The apportionment of the seats is determined by Hamilton’s method to be: 19 for Mudston, 14 for WallaWalla, and 5 for Dilberta. A census reveals that the population has grown and the seats must be reapportioned. If Mudston now has 921,000 residents, WallaWalla now has 640,000 residents, and Dilberta now has 260,000 residents, how many seats will each state receive upon reapportionment? How is this an example of the population paradox?

Answer

Use Hamilton’s method.

Step 1: Find the standard divisor.

Total population: $866,000 + 626,000 + 256,000 = 1,748,000$

$Standard Divisor = \frac{Total Population}{House Size} = \frac{1,748,000}{38 seats} = 46, 000 people per seat$

Step 2: Find each state’s standard quota.

State	$Standard Quota = \frac{State Population}{Standard Divisor}$
Mudston	$\frac{866,000 people}{46, 000 people per seat} \approx 18 .826 seats$
WallaWalla	$\frac{626,000 people}{46, 000 people per seat} \approx 13 .609 seats$
Dilberta	$\frac{256,000 people}{46, 000 people per seat} \approx 5.565 seats$

Step 3: Give each state the state’s lower quota (with each state receiving at least 1 seat). The lower quota always rounds down.

State	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Mudston	$\frac{866,000 people}{46, 000 people per seat} \approx 18 .826 seats$	18
WallaWalla	$\frac{626,000 people}{46, 000 people per seat} \approx 13 .609 seats$	13
Dilberta	$\frac{256,000 people}{46, 000 people per seat} \approx 5.565 seats$	5

Step 4: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $18 + 13 + 5 = 36$ .

There are two more seats to allot. The largest fractional parts are 0.825 for Mudstown and 0.608 for WallaWalla. Give them the extra seats.

State	Lower Quota
Mudston	$18 + 1 = 19$
WallaWalla	$13 + 1 = 14$
Dilberta	5

Step 5: Check the solution by confirming that the sum of the modified quotas equals the house size.

$19 + 14 + 5 = 38$

Now, redo for the changed population:

Step 1: Find the standard divisor.

Total population: $921,000 + 640,000 + 260,000 = 1,821,000$

$Standard Divisor = \frac{Total Population}{House Size} = \frac{1, 821, 000}{38 seats} \approx 47921.05263 people per seat$

State	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Mudston	$\frac{921,000 people}{47921.05263 people per seat} \approx 19 .219 seats$	19
WallaWalla	$\frac{640,000 people}{47921.05263 people per seat} \approx 13 .355 seats$	13
Dilberta	$\frac{260,000 people}{47921.05263 people per seat} \approx 5.426 seats$	5

Step 2: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $19 + 13 + 5 = 37$ .

There is one more seat to allot. The largest fractional part is 0.426 for Dilberta. Give Dilberta the extra seat.

State	Lower Quota
Mudston	19
WallaWalla	13
Dilberta	$5 + 1 = 6$

Step 3: Check the solution by confirming that the sum of the modified quotas equals the house size.

$19 + 13 + 6 = 3$

State	Original Population	New Population	Difference	Difference	Percentage Difference
Mudston	866,000	921,000	921,000 – 866,000	55,000	6.35%
Walla Walla	626,000	640,000	640,000 – 626,000	14,000	2.24%
Dilberta	256,000	260,000	260,000 – 256,000	4,000	1.56%

With the original population numbers, WallaWalla had 14 seats. Even though WallaWalla had a higher growth rate than Dilberta, Dilberta gained a seat and WallaWalla lost a seat.

State	Original Population	Increased Population
Mudston	$18 + 1 = 19$	19
WallaWalla	$13 + 1 = 14$	13
Dilberta	5	$5 + 1 = 6$

This violates the population paradox.

The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats.

The Hamilton reapportionment is: 19 for Mudston, 13 for WallaWalla, and 6 for Dilberta. This is an example of the population paradox because WallaWalla lost a seat to Dilberta, even though WallaWalla’s population grew by 2.24 percent while Dilberta’s only grew by 1.56 percent.

Your Turn 11.46 §11.5 source ↗

The country of Narnia has grown from two states to three. The house size of the congress has been increased by five and the seats have been reapportioned to accommodate the new state of Chippingford. The constitution mandates the use of the Hamilton method of apportionment. Use this information and the following table to answer the questions.

State	Population (in hundreds)	Original Apportionment	Reapportionment
Beruna	7600	39	40
Beaversdam	9720	51	50
Chippingford	1000	Not Applicable	5

What was the original house size?

Answer

The house size is the number of seats.

$39 + 51 = 90$ seats

The original house size is 90 seats.

90

Your Turn 11.46 (b) §11.5 source ↗

The country of Narnia has grown from two states to three. The house size of the congress has been increased by five and the seats have been reapportioned to accommodate the new state of Chippingford. The constitution mandates the use of the Hamilton method of apportionment. Use this information and the following table to answer the questions.

State	Population (in hundreds)	Original Apportionment	Reapportionment
Beruna	7600	39	40
Beaversdam	9720	51	50
Chippingford	1000	Not Applicable	5

What is the new house size?

Answer

The house size is the number of seats.

$40 + 50 + 5 = 95$

The new house size is 95 seats.

95

Your Turn 11.46 (c) §11.5 source ↗

The country of Narnia has grown from two states to three. The house size of the congress has been increased by five and the seats have been reapportioned to accommodate the new state of Chippingford. The constitution mandates the use of the Hamilton method of apportionment. Use this information and the following table to answer the questions.

State	Population (in hundreds)	Original Apportionment	Reapportionment
Beruna	7600	39	40
Beaversdam	9720	51	50
Chippingford	1000	Not Applicable	5

How is this reapportionment an example of the new-states paradox?

Answer

The existing state of Beaversdam lost a seat to the existing state of Beruna when the new state of Chippingford was added.

This is an example of the new-state paradox, which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment.

The original state of Beaversdam lost a seat to the original state of Beruna when the new state of Chippingford was added.

Your Turn 11.47 §11.5 source ↗

The country of Neverland has two states: Neverwood with a population of 760,000 and Mermaids Lagoon with a population of 943,000. The constitution of Neverland requires that the 84 congressional seats be apportioned by Hamilton’s method. Neverwood has received 37 seats while Mermaids Lagoon has received 47 seats. Recently, the island of Marooners Rock with a population of 190,000 has joined the union, becoming a state of Neverland. When Marooners Rock is included, nine additional seats must be apportioned to maintain the same ratio of seats to citizens. Use Hamilton’s method to reapportion the 93 seats to the three states. How is the resulting apportionment an example of the new-states paradox?

Answer

Use Hamilton’s method for the inclusion of the new state.

Step 1: Find the standard divisor.

Total population: $760,000 + 943,000 + 190,000 = 1,893,000$

$Standard Divisor = \frac{Total Population}{House Size} = \frac{1, 893, 000}{93 seats} \approx 20, 354.839 people per seat$

State	$Standard Quota = \frac{State Population}{Standard Divisor}$	Lower Quota
Neverwood	$\frac{760,000 people}{20, 354.839 people per seat} \approx 37 .338 seats$	37
Mermaids Lagoon	$\frac{943,000 people}{20, 354.839 people per seat} \approx 46 .328 seats$	46
Marooners Rock	$\frac{190,000 people}{20, 354.839 people per seat} \approx 9.334 seats$	9

Step 2: Give each remaining seat one at a time to the states with the largest fractional parts of their standard quotas until no seats remain.

Add up the lower quotas: $37 + 46 + 9 = 92$ .

There is one more seat to allot. The largest fractional part is 0.338 for Neverwood. Give Neverwood the extra seat.

State	Lower Quota
Neverwood	$37 + 1 = 38$
Mermaids Lagoon	46
Marooners Rock	9

Step 3: Check the solution by confirming that the sum of the modified quotas equals the house size.

$38 + 46 + 9 = 93$ , the same as the new house size.

State	Original Seats	New Seats
Neverwood	37	38
Mermaids Lagoon	47	46
Marooners Rock	N/A	9

This violates the new state paradox. The existing state of Neverwood lost a seat to the existing state of Mermaids Lagoon when the state of Marooners Rock was added.

This is an example of the new-state paradox, which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment.

The reapportionment gives 38 seats to Neverwood, 46 seats to Mermaids Lagoon, and 9 seats to Marooners Rock. This is an example of the new-states paradox because the original state Mermaids Lagoon lost a seat to the original state Neverwood when the new state was added to the union.

Your Turn 11.48 §11.5 source ↗

New Mexico was admitted as the 47th state on January 6, 1912. Before New Mexico joined the union, the U.S. population was approximately 76,000,000 and the House of Representatives had 391 seats. The new state had a population of approximately 300,000. Use this information to estimate the original standard divisor to the nearest hundred, the new population, the new house size, and the number of seats New Mexico should receive.

Answer

$Original Standard Divisor \approx \frac{76, 000, 000}{391} \approx 194, 400$

$New Population \approx 76, 000, 000 + 300, 000 = 76, 300, 000$

$New House Size \approx \frac{76,300,000}{194, 400} \approx 392$

There are $392 - 391 = 1$ new seats to be apportioned to New Mexico.

$Standard Divisor = \frac{Total Population}{House Size} = \frac{76,000,000}{391 seats} \approx 194, 400 people per seat$

New total population $= 76,000,000 + 300,000 = 76,300,000$ people.

To estimate the new house size, rearrange the equation for the standard divisor.

$Standard Divisor = \frac{Total Population}{House Size}$	Multiply.
$(House Size) (Standard Divisor) = Total Population$	Divide.

$New House Size = \frac{New Total Population}{Standard Divisor} = \frac{76, 300, 000}{194, 400 people per seat} \approx 392 seats$

New House Size – Old House Size = 392 – 391 = 1 new seat to be apportioned to New Mexico.

The original standard divisor: $194, 400 people per seat$

The new population: 76,300,000 people

The new house size: 392 seats

The number of seats New Mexico should receive: 1 seat

Your Turn 11.49 §11.5 source ↗

Suppose that in 2016, States A, B, and C had populations of 13 million, 12 million, and 112 million, respectively. In 2020, State A has grown by 1 million residents, State B has lost 1 million residents, and State C has gained 2 million residents. Compare the apportionments in 2016 to 2020 using each method given below. Which of the four methods violate(s) the population paradox in this scenario?

Answer

Only Hamilton’s method violates the population paradox.

Check Your Understanding 11.5.33 §11.5 source ↗

In the following questions, a scenario is given. Determine whether the scenario is an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these.

A city purchased five new firetrucks and apportioned them among the existing fire stations. Although your neighborhood fire station has the same proportion of the city’s firetrucks as before the new ones were purchased, it now has one fewer.

Answer

The Alabama paradox occurs when an increase in the house size reduces a state’s quota. This is a violation of the Alabama paradox because your neighborhood fire station suffered a reduction in fire trucks when the house size (number of fire trucks) increased. The Alabama paradox

Check Your Understanding 11.5.34 §11.5 source ↗

In the following questions, a scenario is given. Determine whether the scenario is an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these.

The school resources officers in a county were reapportioned based on the most recent census. The number of students at Chapel Run Elementary went up while the number of students at Panther Trail Elementary went down. However, Chapel Run now has one fewer resources officer while Panther Trail has one more than it did previously.

Answer

The population paradox occurs when a state with an increasing population loses a seat while a state with a decreasing population either retains or gains seats. More generally, the population paradox occurs when a state with a higher population growth rate loses seats while a state with a lower population growth rate retains or gains seats.

This is a violation of the population paradox because the Chapel Run school numbers went up, but they lost resources.

Population paradox

Check Your Understanding 11.5.35 §11.5 source ↗

In the following questions, a scenario is given. Determine whether the scenario is an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these.

The standard quota for States A, B, and C are 1.19, 2.73, and 5.71 respectively. State A received 1 seat, State B received 3 seats, and State C received 4 seats.

Answer

State C’s standard quota is 5.71. The lower quota is 5 and the upper quota is 6. If a scenario exists in which a particular apportionment allocates a value greater than the upper quota or less than the lower quota, then that apportionment violates the quota rule and the apportionment method that was used violates the quota rule. State C received 4 seats, which is lower than the lower quota. Quota rule violation

Check Your Understanding 11.5.36 §11.5 source ↗

In the following questions, a scenario is given. Determine whether the scenario is an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these.

A corporation that owns several hospitals purchased an additional hospital, causing the doctor to patient ratio to decrease.

Answer

None of these

Check Your Understanding 11.5.37 §11.5 source ↗

In the following questions, a scenario is given. Determine whether the scenario is an example of a quota rule violation, the Alabama paradox, the population paradox, the new-states paradox, or none of these.

When the city of Cocoa annexed an adjacent unincorporated community, the number of seats on the city council was increased to maintain the standard ratio of citizens to seats, but one existing community of Cocoa still lost a seat on the city council to another existing community of Cocoa when the new community was added.

Answer

This is an example of the new-state paradox, which occurs when the addition of a new state is accompanied by an increase in seats to maintain the standard ratio of population to seats, but one of the existing states loses a seat in the resulting reapportionment.

An existing community of Cocoa lost a seat to another existing community when a new community was added.

New-states paradox

Check Your Understanding 11.5.38 §11.5 source ↗

In the following questions, a type of scenario is described. Indicate which paradox could arise in a scenario of this kind.

A reapportionment occurs because the populations of the states change, and the house size remains the same.

Answer

The population paradox could occur.

Check Your Understanding 11.5.39 §11.5 source ↗

In the following questions, a type of scenario is described. Indicate which paradox could arise in a scenario of this kind.

A reapportionment occurs because the house size increases, and the populations of the states remain the same.

Answer

The Alabama paradox could occur.

Check Your Understanding 11.5.40 §11.5 source ↗

In the following questions, a type of scenario is described. Indicate which paradox could arise in a scenario of this kind.

A reapportionment occurs because an additional state is added to the union, the populations of the original states remain the same, and the house size is increased to correspond to the population of the new state.

Answer

The new-state paradox could occur.