According to the rank-size rule, which of the following is MOST LIKELY the population of City 3?

A. 200,000
B. 400,000
C. 600,000
D. 80,000
E. 1 million

[tex]\[
\begin{array}{|l|l|}
\hline
\text{City} & \text{Population Size} \\
\hline
\text{City 1} & 1.2 \text{ million} \\
\hline
\text{City 2} & 600,000 \\
\hline
\text{City 3} & \\
\hline
\text{City 4} & 300,000 \\
\hline
\text{City 5} & 240,000 \\
\hline
\end{array}
\][/tex]



Answer :

Alright class, let's solve this problem step-by-step using the data provided in the question and the rank-size rule.

The rank-size rule states that the population of a city is inversely proportional to its rank. This means that if you have the rank and population of one city, you can estimate the population of other cities based on their rank.

Given data:
- City 1's rank is 1, with a population of 1.2 million.
- City 2's rank is 2, with a population of 600,000.
- City 4's rank is 4, with a population of 300,000.
- City 5's rank is 5, with a population of 240,000.

We need to find the population of City 3, which has a rank of 3.

Based on the rank-size rule:
[tex]\[ \text{Population of City 2} = \frac{\text{Population of City 1}}{\text{Rank of City 2}} \][/tex]

So applying the rank-size rule:
[tex]\[ \text{Population of City 2} = \frac{1,200,000}{2} = 600,000 \][/tex]

Now, for City 3:
[tex]\[ \text{Population of City 3} = \frac{\text{Population of City 1}}{\text{Rank of City 3}} \][/tex]
[tex]\[ \text{Population of City 3} = \frac{1,200,000}{3} = 400,000 \][/tex]

Therefore, the population of City 3 is most likely 400,000.

So, among the options given:
a) 200,000
b) 400,000
c) 600,000
d) 800,000
e) 1 million

The correct answer is:
b) 400,000