22.2. Let the number of nuts the first and second boys have be denoted by $x$ and $y$ respectively. Then $70 \leqslant x+y \leqslant 80$. According to the problem, we have the equation:

$$
\frac{1}{5}\left(\frac{1}{4} x+y\right)+\frac{3}{4} x=x+1
$$

from which $y=x+5$, i.e., $x+y=2 x+5$. Using this inequality for $x+y$, we get:

$$
70 \leqslant 2 x+5 \leqslant 80 \text {, i.e., } 32.5 \leqslant x \leqslant 37.5 \text {. }
$$

Considering that $x$ is divisible by 4, we conclude that $x=36$. Then $y=x+5=41$.
22.2. Two boys together have no less than seven dozen and no more than eight dozen nuts. If the first boy gave the second boy a quarter of his nuts, and then the second boy gave the first boy a fifth of all the nuts he ended up with, the first boy would have one more nut than he originally had. How many nuts does each boy have?

$$
(7-8 \text { grade })
$$