Answer
Using Adams's method, each clinic is apportioned the following number of doctors:
Clinic A is apportioned 3 doctors.
Clinic B is apportioned 3 doctors.
Clinic C is apportioned 4 doctors.
Work Step by Step
We can find the total patient load.
total load = 119 + 165 + 216 = 500
We can find the standard divisor.
$standard~divisor = \frac{total~load}{number~of~ doctors}$
$standard~divisor = \frac{500}{10}$
$standard~divisor = 50$
The standard divisor is 50.
We can find the standard quota for each clinic.
Clinic A:
$standard~quota = \frac{patient~load}{standard~divisor}$
$standard~quota = \frac{119}{50}$
$standard~quota = 2.38$
Clinic B:
$standard~quota = \frac{patient~load}{standard~divisor}$
$standard~quota = \frac{165}{50}$
$standard~quota = 3.30$
Clinic C:
$standard~quota = \frac{patient~load}{standard~divisor}$
$standard~quota = \frac{216}{50}$
$standard~quota = 4.32$
If each clinic is apportioned its upper quota, the number of doctors apportioned is 3 + 4 + 5, which is 12 doctors. However there is only a total of 10 doctors available. To obtain a sum of 10 doctors when we use Adams's method, we need to find a modified divisor that is slightly more than the standard divisor.
Let's choose a modified divisor of 56. Note that it may require a bit of trial-and-error to find a modified divisor that works. We can find the modified quota for each clinic.
Clinic A:
$modified~quota = \frac{patient~load}{modified~divisor}$
$modified~quota = \frac{119}{56}$
$modified~quota = 2.13$
Clinic B:
$modified~quota = \frac{patient~load}{modified~divisor}$
$modified~quota = \frac{165}{56}$
$modified~quota = 2.95$
Clinic C:
$modified~quota = \frac{patient~load}{modified~divisor}$
$modified~quota = \frac{216}{56}$
$modified~quota = 3.86$
Using Adams's method, the modified quota is rounded up to the nearest whole number. Each clinic is apportioned the following number of doctors:
Clinic A is apportioned 3 doctors.
Clinic B is apportioned 3 doctors.
Clinic C is apportioned 4 doctors.
Note that the total number of doctors apportioned is 10, so using a modified divisor of 56 is acceptable.