1 .		Questions 1 through 9 refer to the following: A small country consists of four states. The population of State 1 is 1,251, the population of State 2 is 14,749, the population of State 3 is 5,651, and the population of State 4 is 3,349. The total number of seats in the legislature is 250. The standard divisor is
2 .		The standard quota for State 3 is
3 .		Under Hamilton’s method the states getting the “extra” seats are
4 .		Under Hamilton’s method the apportionments to each state are
5 .		Which of the following modified divisors D will apportion the seats using Jefferson’s method?
6 .		Under Jefferson’s method the apportionments to each state are
7 .		Using a divisor of D = 101 the modified quotas (to 2 decimal places) are
8 .		Under Adams’ method the apportionments to each state are
9 .		Under Websters’ method the apportionments to each state are
10 .		Suppose that D represents the standard divisor. If A is the modified divisor used for Adams’ method, then
11 .		Under a certain apportionment method a state receives an apportionment of 34 seats when the total number of seats in the legislature is 435 but only 33 seats when the total number of seats in the legislature is 436. This is called
12 .		Questions 12 through 15 refer to the following: Consider a country consisting of three states whose populations are given in the table below. There are 50 states in the legislature. State A B C Population (in millions) 42 75 283 The standard divisor is
13 .		Under Jefferson’s method the apportionments to each state are
14 .		Under Adams’ method the apportionments to each state are
15 .		Under Hamilton’s method the apportionments to each state are
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