7.6. Along the shore of a round lake, apple trees grow. Petya and Vasya start walking from point $A$ on the shore in opposite directions along the shore and count all the apple trees they encounter, as well as all the apples growing on the trees. Meeting at some point $B$, they compared their results. It turned out that Petya counted twice as many apple trees as Vasya, and seven times as many apples as Vasya. They were surprised by this result and decided to repeat the experiment. They set off from point $B$ in the same directions as initially, and met again at point $C$. It turned out that on the way from $B$ to $C$, Petya again counted twice as many apple trees as Vasya, and seven times as many apples as Vasya. Their surprise grew even more, and they decided to repeat the experiment again. Setting off from $C$ in the same directions, they met at point $D$. It turned out that Petya again counted twice as many apple trees as Vasya. Who counted more apples on the way from $C$ to $D$ and by how many times?

Answer. On the way from $C$ to $D$, Vasya counted 3 times more apples than Petya.
Solution. Let there be $n$ apple trees growing along the shore in total. On the way from $A$ to $B$, Vasya counted twice as few apple trees as Petya, and together they counted all the apple trees growing on the shore. Therefore, Vasya counted $\frac{n}{3}$, and Petya - $\frac{2 n}{3}$ apple trees. Similarly, on the way from $B$ to $C$, Vasya also counted $\frac{n}{3}$, and from $C$ to $D$ - also $\frac{n}{3}$ apple trees. Thus, in total, he counted exactly $\frac{n}{3}+\frac{n}{3}+\frac{n}{3}=n$ apple trees. Therefore, by traveling the path from $A$ to $D$, Vasya met all the apple trees growing on the shore exactly once.

Next, let there be $m$ apples on the apple trees in total. On the way from $A$ to $B$, Vasya counted seven times fewer apples than Petya, and together they counted $m$ apples. Therefore, Vasya counted $\frac{m}{8}$, and Petya - $\frac{7 m}{8}$ apples. Similarly, on the way from $B$ to $C$, Vasya also counted $\frac{m}{8}$ apples. But in total, on the way from $A$ to $D$, Vasya counted $m$ apples (since he met all the apple trees exactly once). Therefore, on the way from $C$ to $D$, he counted $m-\frac{m}{8}-\frac{m}{8}=\frac{3 m}{4}$ apples. Then Petya on the way from $C$ to $D$ counted $m-\frac{3 m}{4}=\frac{m}{4}$ apples, i.e., three times fewer than Vasya.

Criteria.

(-.) It is proven that Vasya counted the same number of apples on the segments $A B, B C, C D$.

(-/+) The correct answer, but the justification is absent or very incomplete.

(+/-) The problem is solved under the assumption that the boys' speeds are constant, or that the trees grow at equal distances.

(+.) Minor flaws in the correct solution (for example, it is stated that points $A$ and $D$ necessarily coincide, or the location of points $A, B, C, D$ is indicated without justification).