How would I set up this math problem?

Question

In a certain county, the average number of telephone calls per day between any two cities is directly proportional to th eproduct of their populations and inversely proportional to the square of the distance between them. Cities A and B are 25 miles apart and have populations of 10,000 and 5000, respectively. Telephone records show indicate an average of 2000 calls per day between the two cities. Estimate the average number of calls per day between city A and another city of 15,000 people that is 100 miles away.
How would I about solving this?

Answer 1

2000=x
a=15000 :- 100

Answer 2

n = k p1 p2/d^2, where k is a constant, p1 and p2 represent the population in two cities respectively.

2000 = k (10000)(5000)/25^2......(1) plug in the data for the first case

x = k(10000)(15000)/100^2......(2), plug in the data for the second case. x is the average number of calls to be calculated.

(2)/(1) and solve for x,
x = 3(2000)/16 = 375 calls

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Actually, this problem can be easily done by mental math. Since the population at another city is 3 times the population of city B and the distance to another city is 4 times the distance to city B, the average number of calls should be 2000x3/4^2 = 375 calls.

Answer 3

The ratio from the first city is (10000)(5000)K/(25^2) = 2000

You can either solve for K and use it with the second city to find the average number of calls or just set up the ratios without finding K as follows:

(10000)(5000)/((25^2)(2000)) = (10000)(15000)/((100^2)X)

Now rearrange and solve for X the number of calls

Answer 4

let let calls per day be C and
the product of population be p
the distance between the cities be d
and k be the constant number.
C=k.p/(d*d)

Between the Cities A and B:
C=k.p/(d*d)
p=10000+5000=15000
2000=k*15000/(25*25)
k=83.333....

Between the city A and another city:
C=k.p/(d*d)
p=10000+15000=25000
C=83.333....*25000/(100*100)
C=208
So the average number of calls per day between the city A and another city is 208.