C Roettger, Spring 16
Math 105 – Practice Exam 1 – solutions
Best do these problems only after you are comfortable with the material,
not to learn it. This is to test yourself, so take a stopwatch – you should be
able to do all of these in 60 minutes.
Problem 1
Solution This solution has been written
DrExam
Stines,
MATHby
105:
1 and it’s a beautiful
example of how to show all work in the smallest amount of space possible.
Name: in the last column for the Borda count should be 10, 8, 6,
But the numbers
4, 2.
Number 20
15
7
3
2
1st
A
C
E
B
D
2nd
B
E
D
E
C
3rd
C
B
B
A
B
4th
D
D
C
C
A
a. Plurality.
5th
E
A
A
D
E
b. Borda count
1. Use the preference schedule shown to find the
full ranking of candidates using the following methods
(please show work below).
c. Plurality with elimination
d. Pairwise comparisons
Problem 2 Is there a Condorcet winner for the election in Problem 1? If
so, who? If not, why not?
Solution Yes, there is: B. (not part of this problem: ask yourself why this
is the same candidate as the winner of the H2H-comparison).
Problem 3 A candidate who loses in head-to-head comparison against every
other candidate is called a Condorcet Loser. Is there a Condorcet loser in
the election in Problem 1? If so, who? If not, why not?
Solution Yes, there is: A. Just check all the H2H-comparisons, A loses every
one. (again, not part of the problem: Surprising that this would happen,
when A is the plurality winner! and note how A is the last-ranked candidate
in the H2H system.)
Problem 4 In the preference schedule below, determine the winner using
the Borda Count system.
a) Why is the IIA fairness criterion violated in this example?
b) Why is none of the other three fairness criteria we discussed violated here?
3
Place\ Votes 10 8
1st
B
C A
2nd
A
B C
3rd
C
A B
Solution a) IIA is violated. The Borda ranking is B, C, A. If A withdraws,
C wins instead of B.
b) There is no majority winner. There is also no Condorcet winner (B wins
against A, A wins against C, C wins against B. Monotonicity is never violated
by Borda.
Problem 5 The sequence of squares 1, 4, 9, 16, 25 . . . has colors Onyx, Zinnia, Onyx, Zinnia, . . . (colors determined by the remainder after division by
4). Use this to find the remainder of 201720172 after division by 4.
Solution From the simple pattern Onyx, Zinnia, Onyx, Zinnia, . . . , we see
that all those terms in odd position have color Onyx, which means they have
remainder 1 after division by 4. This applies in particular to 201720172 , since
20172017 is an odd number.
Problem 6 Below are some weighted voting systems in generic description.
a) Circle every weight belonging to a dictator.
b) Underline every weight of a player with veto power.
Solution Could not do circling here, so I drew all weights belonging to any
dictator in blue. Then I put the only weight belonging to a dictator in a box.
Note that any dictator will always have veto power, unless there is anarchy
at the same time.
[19 : 7 , 6 , 6 , 5]
[18 : 7 , 6, 6, 5]
[15 : 7, 6, 6, 5]
[20 : 25 , 11, 3, 2]
[20 : 15 , 11, 3, 2]
[16 : 15, 11, 3, 2]