In a cinema, the entrance fees for the I., II., and III. sections are 6, 4, and 3 Ft, respectively. In one year, 30%, 50%, and 20% of the total revenue came from tickets for the I., II., and III. sections, respectively. In the following year, these revenues showed an increase of 20%, 30%, and a decrease of 5%, respectively. By how much did the total revenue change, and by how much did the total number of cinema visitors change?
I. solution. Taking the first year's revenue as $100 A$ Ft, the revenue from I., II., III. class tickets was $30 A, 50 A, 20 A$. In the following year, these became $1.20 \cdot 30 A=36 A, 1.30 \cdot 50 A=65 A, (1-0.05) 20 A=91 A$, thus their sum is $120 A$, meaning the total revenue increased by $20 \%$.

The number of tickets sold, i.e., the number of visitors, based on the partial revenues and unit prices, in the 1st year was $(30 A / 6) + (50 A / 4) + (20 A / 3) = 145 A / 6$, and in the 2nd year $(36 A / 6) + (65 A / 4) + (19 A / 3) = 171.5 A / 6$, which is $17150 / 145 = 118.27 \%$ of the former, so the increase in the number of visitors was $18.27 \%$.

Jenő Fekete (Mosonmagyaróvár, Kossuth L. g. II. o. t.)

II. solution. Let the number of visitors to the I., II., III. class in the 1st year be $x_{1}, y_{1}, z_{1}$, and the total revenue $B_{1}$. Considering the unit prices and the percentage distribution simultaneously, the partial revenues are: $6 x_{1} = 0.3 B_{1}, 4 y_{1} = 0.5 B_{1}, 3 z_{1} = 0.2 B_{1}$. From these, $B_{1} = 20 x_{1} = 8 y_{1} = 15 z_{1}$, thus $y_{1} = 5 x_{1} / 2, z_{1} = 4 x_{1} / 3$ and the total number of visitors in the 1st year is $N_{1} = x_{1} + y_{1} + z_{1} = 29 x_{1} / 6$.

Let the corresponding numbers for the 2nd year be $x_{2}, y_{2}, z_{2}, B_{2}, N_{2}$. The given percentage changes, due to the constancy of the ticket prices, also give the increase or decrease in the number of viewers at each location. Thus, $x_{2} = 1.2 x_{1}, y_{2} = 1.3 y_{1} = 3.25 x_{1}, z_{2} = 0.95 z_{1} = 3.8 x_{1} / 3$, and furthermore, on the one hand $N_{2} = x_{2} + y_{2} + z_{2} = 17.15 x_{1} / 3$, and on the other hand $B_{2} = 6 x_{2} + 4 y_{2} + 3 z_{2} = 7.2 x_{1} + 13 x_{1} + 3.8 x_{1} = 24 x_{1}$.

Accordingly, the increase in $B$ is $B_{2} - B_{1} = 4 x$, which is $20 \%$ of $B_{1}$, and the increase in $N$ is $5.3 x_{1} / 6$, which is $530 / 29 = 18.27 \%$ of $N_{1}$.

László Béres D. (Ráckeve, Ady E. g. II. o. t.)