Answer
Using Webster's method, 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.
Work Step by Step
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.
