Suppose a society consists of three cities (A, B , C. ) with populations of 99,0

Question

Suppose a society consists of three cities (A, B , C. ) with populations of 99,000 , 51,000 , 6,000 people respectively. Suppose that the society has 15 doctors total.

a. What cities will have how many doctors?

b. If the number of doctors doubles to 30. How many will live in each city?

My professor didn't answer it when I asked if I had it correct... but his advice : Calculate the number of doctors per capita in each city for the answer you think is correct. If you cannot move doctors from one city to another in a way that allows that doctor to have more patients in the new city than in the old one, then they are in equilibrium.

Explanation / Answer

Yes your professor's advice is correct

Total Population of society = 99,000 + 51,000 + 6,000 = 156,000

percentage of population in A = 99,000/156,000 = 63.46

percentage of population in B = 51,000/156,000 = 32.69

percentage of population in C = 6,000/156,000 = 3.85

So, Doctors should be divided among cities such that per capita doctor in each city is same.

So, doctors shouldd be divided according to the percentage of population in each city

So doctors in City A = 0.63*15 = 9 doctors

  doctors in City B = 5 doctors

doctors in City A = 1 doctors

If the number of doctors doubles to 30

So doctors in City A = 0.63*30 = 18 doctors

  doctors in City B = 10 doctors

doctors in City A = 2 doctors

If you don't understand anything, then comment, I will revert back on the same.

And If you liked the answer then please do review the same. Thanks :)

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