The population of a certain city on January 1, 1950, was 117,751. The average annual increase due to migration (in-migration and out-migration) is 640 people, while the natural growth rate is $0.87 \%$ per year. a) What will the population be on January 1, 1960? - b) What is the percentage of annual growth? - c) Under the same conditions, in which year will the city reach a population of 200,000?
a) According to a known formula, the naturally increased population at the end of the 10th year is

$$
A_{10}=A q^{10}, \quad \text{where} \quad A=117751, \quad \text{and} \quad q=1+\frac{p}{100}=1.0087
$$

In addition, there is an increase due to migration, which over 10 years, according to the formula of the arithmetic series, is

$$
a_{10}=a \frac{q^{10}-1}{q-1}, \quad \text{where} \quad a=640
$$

Using a four-digit table,

$$
A_{10}+a_{10}=128500+6694 \approx 135200
$$

b) If the sought percentage is denoted by $p_{10}$, then

$$
117751 q_{10}^{10}=135200
$$

where $q_{10}=1+\frac{p_{10}}{100}$.

Thus,

$$
q=\sqrt[10]{\frac{135200}{117751}} \approx 1.014, \quad \text{from which} \quad p_{10}=1.4 \%
$$

c) Suppose that under the same conditions, the number of inhabitants in the city will be $A_{x}=200,000$ after $x$ years, then

$$
A q^{x}+a \frac{q^{x}-1}{q-1}=A_{x}
$$

or

$$
A(q-1) q^{x}+a q^{x}-a=A_{x}(q-1)
$$

from which

$$
q^{x}=\frac{A_{x}(q-1)+a}{A(q-1)+a}=\frac{200000 \cdot 0.0087+640}{117751 \cdot 0.0087+640} \approx \frac{1740+640}{1024+640} \approx \frac{2380}{1664}
$$

From this,

$$
x=\frac{\lg 2380-\lg 1664}{\lg 1.0087}=\frac{3.3766-3.2211}{0.00378}=\frac{15550}{378} \approx 41.1
$$

Thus, the population of the city will reach 200,000 in the year 1991.

Attila Kerekes (Pécs, Bányaip. techn. III. o. t.)