\section*{Task 1 - V11021}

The industrial production of the Soviet Union increased by 10\% in 1960. In the following years, production is also expected to grow annually by approximately this magnitude.

Calculate how many times the industrial production of the USSR will increase:

a) in the next 10 years,

b) in the next 20 years,

c) by the year 2000,

assuming an annual growth rate of \(10 \%\)!
}

Let the initial production be \(A\). Then, with an annual growth rate of \(10\%\):

\begin{tabular}{llll}
1959 & \(A\) & & \\
1960 & \(A+\frac{1}{10} A\) & \(=A\left(1+\frac{1}{10}\right)\) & \(A \cdot 1.1\) \\
1961 & \(A \cdot 1.1+\frac{1}{10} A \cdot 1.1\) & \(=A \cdot 1.1\left(1+\frac{1}{10}\right)\) & \(A \cdot 1.1^{2}\) \\
1962 & \(A \cdot 1.1^{2}+\frac{1}{10} A \cdot 1.1^{2}\) & \(=A \cdot 1.1^{2}\left(1+\frac{1}{10}\right)\) & \(A \cdot 1.1^{3}\)
\end{tabular}

After \(n\) years of growth, \(A \cdot 1.1^{n}\) results, because each time a factor of 1.1 is added through factoring. Therefore, the formula is:

\[
x=A \cdot 1.1^{n}
\]

Since the problem is not concerned with the production itself but with the multiplication, \(A\) can be set to 1; thus, we get:

for a) \(x=1.1^{10}=2.59 \approx 2.6\),

for b) \(x=1.1^{20}=6.73 \approx 6.8\),

for c) \(x=1.1^{40}=45.26 \approx 45.3\).

The industrial production of the USSR will grow to 2.6 times its original value in the next 10 years, to 6.8 times in the next 20 years, ..., if an annual growth rate of \(10\%\) is assumed.

Adapted from \([6]\)