\section*{Problem 4 - 310914}

a) A school has a total of 825 students. It has been calculated that during a school year, the number of participants in an interest group increased by \(4\%\) of its initial value, and, equivalently, the number of non-participants decreased by \(7\%\) of its initial value.

If this were exactly true, how many non-participants were there at the beginning of the school year, and by how many students did this number decrease by the end of the school year?

b) However, it was subsequently reported that the percentages were only approximate values of \(4.0\%\) and \(7.0\%\), determined according to rounding rules to one decimal place.

Are the numbers sought in a) still uniquely determined in this case? If not, determine all those values for the numbers sought in a) that also lead to the rounded percentages of \(4.0\%\) and \(7.0\%\)!
}

a) At the beginning of the school year, let \(x\) be the number of participants, \(y\) the number of non-participants. Then it holds that \(x+y=825\) (1) and since the increase in the number of participants was as large as the decrease in the number of non-participants, it follows that

\[
\frac{4}{100} \cdot x=\frac{7}{100} \cdot y
\]

From (1) it follows that \(x=825-y\); from (2) it follows that

\[
4 \cdot(825-y)=7 y \quad(3) \quad ; \quad 11 y=3300
\]

Thus, the number of non-participants at the beginning of the school year was \(y=300\) (5); this number decreased by 7% of this value, i.e., by the number \(z=21\).

b) If \(p\) and \(q\) are the true values, approximated by 4.0 and 7.0, respectively, then \(p\) lies between 3.95 and 4.05, and \(q\) lies between 6.95 and 7.05; furthermore, instead of (3), (4), (5), it is only known that with such values \(p, q\)

\[
p \cdot(825-y)=q \cdot y \quad ; \quad y=\frac{p \cdot 825}{p+q}
\]

This leads to the statement that \(y\) lies between

\[
\frac{3.95 \cdot 825}{4.05+7.05}=\frac{3258.75}{11.1} \quad \text { and } \quad \frac{4.05 \cdot 825}{3.95+6.95}=\frac{3341.25}{10.9} \quad \text { and }
\]

thus, certainly between 293.58 and 306.54 and must therefore be one of the numbers

\[
294,295, \ldots, 306
\]

The number \(z=\frac{q}{100} y\) of students by which \(y\) decreased during the school year therefore lies between \(\frac{6.95}{100} \cdot 294=20.433\) and \(\frac{7.05}{100} \cdot 306=21.573\), i.e., it is uniquely determined and is \(z=21\).

Solutions of the 1st Round 1991 adopted from [5]

\title{
6.33.2 2nd Round 1991, Class 9
}

\begin{abstract}