A tractor driver couple working at a farm made a job offer on March 12th for the remaining 20 working days starting from April 4th, where the husband would increase his performance by $1 \%$ of the previous day's performance every day, and the wife would increase her performance by $0.5 \%$ of her previous day's performance every day. On April 4th, it turned out that the couple had collectively achieved a $240 \%$ average daily performance over the last 20 working days. What percentage did the husband and the wife achieve on the first day, if the wife's performance on the twentieth day was $101.2 \%$?
Let's denote the first day's performance of the husband and wife as $x$ and $y$ percent, respectively. According to the problem, using the formula for the sum of a geometric series, we have:

$$
x \frac{1.01^{20}-1}{1.01-1} + y \frac{1.005^{20}-1}{1.005-1} = 4800
$$

and

$$
y \cdot 1.005^{19} = 101.2
$$

From (2) (using a four-digit logarithm table for calculations),

$$
y = \frac{101.2}{1.005^{19}} = 91.92 \%
$$

Substituting this value of $y$ into (1),

$$
\begin{aligned}
x \frac{1.01^{20}-1}{0.01-1} = 4800 - \frac{91.92 \left(1.005^{20}-1\right)}{0.005-1} & = 4800 - 18384 \left(1.005-1\right) = \\
4800 - 18384 \cdot 0.107 & = 2833
\end{aligned}
$$

From which,

$$
x = \frac{2833 \cdot 0.01}{1.01^{20}-1} = \frac{28.33}{0.219} = 129.4 \%
$$

Gábor Vass (Bp., V., Piarist g. III. o. t.)