Thinking Mathematically (6th Edition)

We can find the total number of passengers. total passengers = 1087 + 1323 + 1592 + 1596 + 5462 total passengers = 11,060 We can find the standard divisor. $standard ~divisor = \frac{total ~passengers}{buses}$ $standard ~divisor = \frac{11,060}{200}$ $standard ~divisor = 55.3$ If we use the standard divisor and round each standard quota to the nearest whole number, the sum of the apportioned buses will be 201 buses. To obtain a sum of 200 buses, we need to find a modified divisor that is slightly more than the standard divisor. Let's choose a modified divisor of 55.5. Note that it may take some trial-and-error to find a modified divisor that works. We can find the modified quota for each route. Route A: $modified~quota = \frac{passengers}{modified~divisor}$ $modified~quota = \frac{1087}{55.5}$ $modified~quota = 19.59$ Route B: $modified~quota = \frac{passengers}{modified~divisor}$ $modified~quota = \frac{1323}{55.5}$ $modified~quota = 23.84$ Route C: $modified~quota = \frac{passengers}{modified~divisor}$ $modified~quota = \frac{1592}{55.5}$ $modified~quota = 28.68$ Route D: $modified~quota = \frac{passengers}{modified~divisor}$ $modified~quota = \frac{1596}{55.5}$ $modified~quota = 28.76$ Route E: $modified~quota = \frac{passengers}{modified~divisor}$ $modified~quota = \frac{5462}{55.5}$ $modified~quota = 98.41$ Using Webster's method, we need to apportion the buses by rounding the modified quota to the nearest whole number. The buses are apportioned to the routes as follows: Route A is apportioned 20 buses. Route B is apportioned 24 buses. Route C is apportioned 29 buses. Route D is apportioned 29 buses. Route E is apportioned 98 buses. Note that the sum of the apportioned buses is 200 buses, so using 55.5 as a modified divisor is acceptable.