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?
Initially, the second child receives the house and pays $93,750. The first child receives $30,000, the third child receives $25,000 and the fourth child receives $22,500. The first child receives the farm and pays $60,000. The second child receives $15,000, the third child receives $17,500, and the fourth child receives 18,750. The fourth child receives the property and pays $33,750. The first child receives $10,000, the second child receives $7,500, and the third child receives $6,250. The three surpluses are $16,250, $8,750 and $10,000, totaling $35,000. The added shares are $8,750. Summarizing, the first child receives the farm and pays $60,000 less $40,000 plus $8,750, the second child receives the house and pays $93,750 less $22,500 plus $8,750, the third child receives $57,500 and the fourth child receives the property and pays $33,750 and receives $41,250 plus $8,750.
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 bottomup 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 bottomup 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 5050, 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 divideandchoose 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

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

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 HillHuntington 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 HillHuntington method of apportionment. I switched to HillHuntington 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

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 HillHuntington 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)