EXAMPLE 2 The High School Mathematics Teacher
A high school has one mathematics teacher who teaches all geometry, precalculus, and calculus classes. She has time to teach a total of five sections, and 100 students are enrolled as follows: 52 for geometry, 33 for precalculus, and 15 for calculus. How many sections of each course should be scheduled?
This is an apportionment problem because the number of sections is specified (5), and the number of sections allotted to each course must be a whole number. The three courses correspond to the states; the number of students enrolled in each course corresponds to each state’s population; and the total number of sections to be taught, 5, is the house size. Thus, the populations of geometry, precalculus, and calculus are 52, 33, and 15, respectively. The total population is 100, so the standard divisor is . Table 14.3 displays the calculations of the quotas.
| Course | Population | Quota | Rounded |
|---|---|---|---|
| Geometry | 52 | 3 | |
| Precalculus | 33 | 2 | |
| Calculus | 15 | 1 | |
| Totals | 100 | 5.00 | 6 |
The sum of the quotas is the house size (in this case, 5). it is tempting to round each quota to the nearest whole number, as in the right column of the table, but this makes 6 sections in all—too many! The purpose of an apportionment method is to find an equitable way to round a set of numbers such as these quotas without increasing or decreasing the original sum.