Mathematics for Liberal Arts 2 Sample Test 1 Solutions

1. After having been roommates for four years at college, Alex and Bob are moving on.  Several items they have accumulated belong jointly to the pair, but now must be divided between the two.  They assign points to the items as follows:

 Object Alex's points Bob's points Bicycle 10 12 Textbooks 20 16 Barbells 5 2 Rowing machine 7 10 Music collection 8 11 Computer 15 17 Novels 20 22 Desk 15 10

Use the adjusted winner procedure to determine a fair division of the property.

SOLUTION:

Initially:

 Alex Bob Textbooks 20 Bicycle 12 Barbells 5 Rowing machine 10 Desk 15 Music collection 11 Computer 17 Novels 22 Total 40 72

The point ratios for the items on Bob’s side are: 12/10; 10/7; 11/8; 17/15; 22/20.  The smallest ratio is 22/20, so we try to move the novels, but that would put Alex at 60 and Bob at 50.  So we say that Bob keeps x part of the novels, and Alex gets 1 – x   Then we need 50 + 22x = 40 + 20(1 – x).  Solving:

22x + 20x = 40 + 20 – 50

42x = 10

x = 10/42 = 5/21

1 – x = 16/21

Alex gets the textbooks, the barbells, the desk, and 16/21 of the novels.  Bob gets the rest.

2. Henry and Lisa inherit a house.  If their monetary bids on the house are \$135,000 and \$114,000, respectively, what is the fair division arrived at by the Knaster inheritance procedure?

SOLUTION: Initially, Henry gets the house and pays \$67,500.   Lisa receives \$57,000.  These are based n their original valuations.  There is left a sum of \$10,500, which they split.  This means that Henry pays \$62,250 and Lisa receives \$62,250.

Henry receives the house and gives Lisa \$62,250.

3. A parent leaves a house, a farm, and a piece of property to be divided equally among four children who submit dollar bids on these objects as follows:

 Children Objects First Second Third Fourth House 120,000 125,000 100,000 90,000 Farm 80,000 60,000 70,000 75,000 Property 40,000 30,000 25,000 45,000

What is the fair division arrived at by the Knaster inheritance procedure?

The first child receives the farm and pays \$11,250.  The second child receives the house and pays \$62,500.  The third child receives \$57,500 and the fourth child receives the piece of property and \$16,250.

4. (a) Suppose Kelly and Jan want to take turns, using the bottom-up strategy, to allocate several textbooks currently held jointly.  Their ranked preferences are listed below:

If Kelly chooses first, how are books allocated?

SOLUTION: Kelly selects Painting, Sculpture, Music.  Jan selects Architecture, Design, Textiles.

(b) Suppose Kelly and Jan want to take turns, using the bottom-up strategy, to allocate several textbooks currently held jointly.  Their ranked preferences are listed below:

If Jan chooses first, how are books allocated?

SOLUTION: Jan selects Architecture, Sculpture, Design.  Kelly selects Painting, Music, Textiles.

5. Suppose Chris and Terry need to make a fair division of a stereo they jointly purchased.  Suppose also, that both agree Chris is entitled to 2/3 of the stereo and Terry to 1/3.  Chris and Terry place monetary bids on the value of the stereo of \$500 and \$450 respectively.  Modify the Knaster inheritance procedure to make a fair division of the stereo.

Chris gets the stereo because of the higher evaluation, and pays 1/3 of the value, \$166.67.  Terry gets 1/3 of \$450, \$150.  This leaves a surplus of \$16.67, which they split, 2/3 (= \$11.11) to Chris and 1/3 (= \$5.56) to Terry.

Chris gets the stereo and pays Terry \$155.56.  If you interpret the procedure as demanding that the surplus be split 50-50, the amount would be \$158.33.  This is acceptable, but I disagree with it.

6.  Tom and Jerry must rake all the leaves in a yard.  They each view the yard as shown below and agree to divide the chore by the divide-and-choose method.  Tom will divide the yard in two parts and Jerry will choose the side he has to rake.  Describe the results of this fair division.

SOLUTION: Tom divides the yard in the middle of the third column.  Jerry chooses the left side to rake.

7. Use the Hamilton method of apportionment to distribute 15 representatives to the three states with the populations shown below.

 State Population A 253,000 B 182,000 C 85,000

SOLUTION: Note that total population is 520,000.  With a house size of 15, this gives a standard divisor of 34,667.

 State Population Quota Apportionment A 253,000 7.298 7 B 182,000 5.250 5 C 85,000 2.452 3 Totals 520,000 15 15

Initially, A gets 7, B gets 5 and C gets 2.  With the largest fractional part, C is increased to 3.

8. Use the Jefferson method of apportionment to distribute 15 representatives to the three states with the populations shown below.

 State Population A 253,000 B 182,000 C 85,000

SOLUTION:

 State Population Quota Apportionment Critical divisor New apportionment A 253,000 7.298 7 253,000/8 = 31,625 8 B 182,000 5.250 5 182,000/6 = 30,333 5 C 85,000 2.452 2 85,000/3 = 28,333 2 Totals 520,000 15 14 15

There was only one seat to add, and state A had the largest critical divisor.

9. Use the Webster method of apportionment to distribute 15 representatives to the three states with the populations shown below.

 State Population A 253,000 B 182,000 C 85,000

SOLUTION:

 State Population Quota Tentative Apportionment Critical divisor New apportionment A 253,000 7.298 7 253,000/7.5 = 33,733 7 B 182,000 5.250 5 182,000/5.5 = 33,090 5 C 85,000 2.452 2 85,000/2.5 = 34,000 3 Totals 520,000 15 14 15

There was only one seat to add, and state C had the largest critical divisor.

10. Use the Hill-Huntington method of apportionment to distribute 15 representatives to the three states with the populations shown below.

 State Population A 253,000 B 182,000 C 85,000

SOLUTION:

 State Population Quota and q* Tentative Apportionment Critical divisor New apportionment A 253,000 7.298  7.483 7 7 B 182,000 5.250  5.477 5 5 C 85,000 2.452  2.449 3 3 Totals 520,000 15 15 15

In this case, all seats were apportioned, and there was no need to calculate critical divisors.

11. A country has three states with the populations shown below.  The House of Representatives for the country is to have 15 members.  Find the district population for each state if the country uses the Hill-Huntington method of apportionment.  I switched to Hill-Huntington for kicks, to see if we would have to calculate a critical divisor, but it didn’t happen.  It still could on the test.

 State Population A 245,000 B 320,000 C 155,000

SOLUTION: The standard divisor is 720,000/15 = 48,000.

 State Population Quota and q* Tentative Apportionment Critical divisor New apportionment A 245,000 5.104  5.477 5 7 B 320,000 6.667  6.481 7 5 C 155,000 3.229  3.464 3 3 Totals 720,000 15 15 15

For A, 245,000/5 = 49000

For B, 320,00/7 = 45714.29

For C, 155,000/3 = 51666.67

12. A town board has 10 members, divided among three regions of the town using the Hamilton method of apportionment.  The regions and their populations are given below.  Find the difference in representative share for the North and South region.

 Region Population North 13,400 Central 21,900 South 8700

SOLUTION: The standard divisor is 2933

 Region Population Quota In. Apportionment Final Apportionment North 13,400 4.569 4 5 Central 21,900 7.467 7 7 South 8700 2.966 2 3 Totals 44,000 15.002 13 15

There were two seats to allocate; South and North had the largest fractional parts.

The representative share for North is 5/13,400 = .0003731; for South, it’s 3/8700 = .0003448.  The difference is

13.  Round the following numbers by the method used in the Hill-Huntington apportionment procedure:  3.67, 9.42, 2.46, and 6.49

SOLUTION: 3.67 rounds to 4 (q* = 3.464)

9.42 rounds to 9 (q* = 9.487)

2.46 rounds to 3 (q* = 2.449

6.49 rounds to 7 (q* = 6.481)