## Thinking Mathematically (6th Edition)

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.