HONORS DISCRETE CHAPTER 4 TEST REVIEW GUIDE:



HONORS DISCRETE 4.1 – 4.5 REVIEW WORKSHEET

Use a separate piece of paper.

For #1 – 4, Identify the STATES, SEATS, and POPULATIONS.

1. Mr. Hughes is going to split the leftover Halloween candy between his 4 children. He plans to use the hours of chores that each child completed the past week to help decide how to apportion the candy.

STATES = CHILDREN, SEATS = CANDY, POPULATIONS = HOURS OF CHORES

2. The ACC is planning to apportion tickets for the ACC Basketball Tournament to each ACC school based upon the number of wins each school had in conference play.

STATES = ACC SCHOOLS, SEATS = TICKETS, POPULATIONS = CONFERENCE WINS

3. A summer camp plans to apportion the number of this summer’s campers for different counties based on the financial support that is has received in the past year from each county.

STATES = COUNTIES, SEATS = CAMPERS, POPULATIONS = FINANCIAL SUPPORT

4. A hospital is planning to apportion the 200 nurses on staff to the 4 shifts of the day based on the average number of patients during each shift.

STATES = SHIFTS, SEATS = NURSES, POPULATIONS = PATIENTS

Problems 5 - 8: There are 200 seats in a nation made up of 4 states.

|States |A |B |C |D |

|Populations |1100 |1750 |2500 |2150 |

5. What is the standard divisor?

6. What are the standard quotas for each state?

7. What are the lower quotas for each state?

8. What are the upper quotas for each state?

|Standard Divisor |Total Population: 1100+1750+2500+2150 = 7500; SD = 7500/200 = 37.5 |

|Standard Quota |1100/37.5 |1750/37.5 |2500/37.5 |2150/37.5 |

| |= 29.33 |= 46.66 |= 66.66 |= 57.33 |

|LOWER Quota |29 |46 |66 |57 |

|UPPER Quota |30 |47 |66 |58 |

9. For the following table, assume there are 155 seats for 5 states.

|States |#1 |#2 |#3 |#4 |#5 |

|Populations |10,300 |8,750 |7,342 |9,978 |4,395 |

a. What is the standard divisor for the problem?

b. What is the standard quota for each state?

c. Find the apportionment using HAMILTON’S METHOD?

|Standard Divisor |SD = 40765/155 = 263 |

|Standard Quota |10,300/263 |8,750/ 263 |7,342/263 |9,978/263 |4,395/263 |

| |= 39.16 |= 33.27 |= 27.92 |= 37.94 |= 16.71 |

|Lower Quotas |39 |33 |27 (+1) |37 (+1) |16 (+1) |

|Apportionment |39 |33 |28 |38 |17 | |

|Populations |123 |257 |234 |196 |135 |165 |

d. What is the standard divisor for the problem?

e. What is the standard quota for each state?

f. Find the apportionment using JEFFERSON’S METHOD?

i. Provide the modified divisor .

|Standard Divisor |1100/150 = 7.4 |

|Standard Quota |123/7.4 = |257/7.4 = |234/7.4 = |196/7.4 = |135/7.4 = |165/7.4 = |

| |16.62 |34.73 |31.62 |26.49 |18.24 |22.30 |

|Divisor |LOWER Quotas |

|7.4 (below) |16 |34 |31 |26 |18 |22 |

|7.35 (below) |16 |34 |31 |26 |18 |22 |

|7.3 (below) |16 |35 |32 |26 |18 |22 |

|7.25 |16 |35 |32 |27 |18 |

|Populations |830 |475 |1132 |689 |1014 |

g. What is the standard divisor for the problem?

h. What is the standard quota for each state?

i. Find the apportionment using ADAM’S METHOD?

i. Provide the modified divisor .

|Standard Divisor |4140/75 = 55.2 |

|Standard Quota |830/55.2 = |475/55.2 = |1132/55.2 = |689/55.2 = |1014/55.2 = |

| |15.03 |8.61 |20.51 |12.48 |18.37 |

|Modified Divisor |UPPER Quotas |

|55.2 (above) |16 |9 |21 |13 |19 |

|56 (above) |15 |9 |21 |13 |19 |

|58 (below) |15 |9 |20 |12 |18 |

|57 |15 |9 |20 |13 |

|Courses |57 |83 |75 |65 |

j. What is the standard divisor for the problem?

k. What is the standard quota for each state?

l. Find the apportionment using all 3 methods: HAMILTON’S, ADAM’S, and JEFFERSON’S METHOD?

i. Based on your apportionments, do you think there is a best answer?

|Standard Divisor |280/100 = 2.8 |

|Standard Quotas |57/2.8 |83/2.8 |75/2.8 |65/2.8 |

| |20.36 |29.64 |26.79 |23.21 |

|Hamilton’s |20 |29 + 1 = 30 |26 + 1 = 27 |23 |

|Jefferson’s: |20 |30 |27 |23 |

|MD = 2.75 | | | | |

|Adam’s: |20 |30 |27 |23 |

|MD = 2.85 | | | | |

In this problem, all 3 apportionments gave the same answer for the number of seats for each state.

10. There are 4 students (Joe, Josh, Jeremy, and Jeffrey). The students are going to split 20 pieces of candy that Mr. Smith has given them. According to their grades, Joe has a standard quota of 6.73 and the students plan to use Hamilton’s method to split the candy.

a. Is it possible for all 4 students to receive 5 pieces of candy, explain?

Joe must receive either 6 or 7 pieces of candy because Hamilton’s Method satisfies the Quota Rule. Therefore it is impossible for all 4 students to receive 5 pieces.

b. Is it possible for Joe to receive 8 pieces of candy, explain?

Joe must receive either 6 or 7 pieces of candy because Hamilton’s Method satisfies the Quota Rule. Therefore it is impossible for Joe to receive 8 pieces.

c. Is it possible for Jeremy to receive 14 pieces of candy, explain?

Jeremy’s largest standard quota is 13.17 because Joe’s is 6.73. Therefore Joe has a larger decimal than Jeremy and Jeremy will not be able to get a (+1) from the surplus because Joe will receive it.

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