Modified Divisor Validation:

Target range: \(63.82 < d' \le 64.13\). Example \(d' = 64.00\):
\(\lfloor 2501/64 \rfloor = 39\), \(\lfloor 3638/64 \rfloor = 56\), \(\lfloor 5781/64 \rfloor = 90\). Sum = 185.

Version B Final Key:

Group A = 39 | Group B = 56 | Group C = 90

Hamilton & Jefferson allocations coincide

Apportionment Quiz Test Bank: Version B (185 Seats | Ideal Ratio: 64.43) Page 1 of 2

Quiz Variant Test Bank | Version C

Council Apportionment: Quiz Variant C

Parallel assessment problem with 2-decimal rounded ideal ratio

Council Seats: 195

Population: 12,340

Student Quiz Copy Block Version C Prompt

There are 12,340 people divided into 3 groups of sizes 2,704, 3,827, and 5,809. The council board will have 195 seats.

1. Hamilton Method: How many seats are apportioned to each group using the Hamilton method (highest decimal of quota)?
2. Jefferson Method: How many seats are apportioned to each group using the Jefferson method (find the Jefferson adjusted ratios for each group and find the closest to the ideal ratio)?

TEACHER KEY & DERIVATIONS Ideal Ratio (Standard Divisor): \(d = \frac{12,340}{195} \approx \mathbf{63.28}\) (exact: \(63.2821\))

1. Hamilton Method Calculation (2 Surplus Seats) Lower Quotas Sum = 193 (\(195 - 193 = 2\))

2. Jefferson Adjusted Ratios & Distance to Ideal Ratio (\(63.28\)) Formula: \(\text{Adjusted Ratio} = \frac{p_i}{L_i + 1}\)

Modified Divisor Validation:

Target range: \(62.74 < d' \le 62.88\). Example \(d' = 62.80\):
\(\lfloor 2704/62.8 \rfloor = 43\), \(\lfloor 3827/62.8 \rfloor = 60\), \(\lfloor 5809/62.8 \rfloor = 92\). Sum = 195.

Version C Final Key:

Group A = 43 | Group B = 60 | Group C = 92

Hamilton & Jefferson allocations coincide

Apportionment Quiz Test Bank: Version C (195 Seats | Ideal Ratio: 63.28) Page 2 of 2

Webster & Hill Apportionment: Base Problem (320, 215, 195 | 45 Seats) Page 1 of 2

Quiz Test Bank | Parallel Problem Variants

Webster & Hill: Quiz Variants B & C

Different council seat counts (48 & 52 seats) with identical "just right" outcomes

Variant B: 48 Seats

Variant C: 52 Seats

Quiz Variant B (Populations: 328, 226, 206 | Seats: 48) Pop = 760 | \(d = 760/48 \approx \mathbf{15.83}\)

Prompt: A council will allocate 48 seats among 3 groups of sizes 328, 226, and 206. Using Webster and Hill methods, determine if all 48 seats are allocated, or if there are too few or too many seats given.

Quiz Variant C (Populations: 359, 240, 221 | Seats: 52) Pop = 820 | \(d = 820/52 \approx \mathbf{15.77}\)

Prompt: A council will allocate 52 seats among 3 groups of sizes 359, 240, and 221. Using Webster and Hill methods, determine if all 52 seats are allocated, or if there are too few or too many seats given.

Teacher Assessment Key for All 3 Versions:

Base Problem (45 seats): Group A = 20, B = 13, C = 12 → Exactly 45 seats ("Just Right").
Variant B (48 seats): Group A = 21, B = 14, C = 13 → Exactly 48 seats ("Just Right").
Variant C (52 seats): Group A = 23, B = 15, C = 14 → Exactly 52 seats ("Just Right").

Webster & Hill Apportionment: Variant B (48 Seats) & Variant C (52 Seats) Page 2 of 2