65 (998). An alloy consists of zinc and copper, entering into it in a ratio of $1: 2$, and another alloy contains the same metals in a ratio of $2: 3$. From how many parts of both alloys can a third alloy be obtained, containing the same metals in a ratio of 17:27?
S o l u t i o n. Let the third alloy contain $x$ kg of the first alloy and $y$ kg of the second alloy. We need to find the ratio $\frac{x}{y}$. Since zinc and copper are in the ratio $1:2$ in the first alloy and in the ratio $2:3$ in the second alloy, $x$ kg of the first alloy contains $\frac{1}{3} x$ (kg) of zinc and $\frac{2}{3} x$ (kg) of copper, and $y$ kg of the second alloy contains $\frac{2}{5} y$ (kg) of zinc and $\frac{3}{5} y$ (kg) of copper. In total, the third alloy contains $\left(\frac{1}{3} x+\frac{2}{5} y\right)$ kg of zinc and $\left(\frac{2}{3} x+\frac{3}{5} y\right)$ kg of copper.

By the condition

$$
\frac{\frac{1}{3} x+\frac{2}{5} y}{\frac{2}{3} x+\frac{3}{5} y}=\frac{17}{27}
$$

from which $35 x=9 y, \frac{x}{y}=\frac{9}{35}$, i.e., the third alloy can be obtained from 9 parts of the first alloy and 35 parts of the second alloy.

$66(1000)$. A tourist left from $M$ to $N$, the distance between which is 41 km, walking at a speed of 5 km per hour. After 2 hours, a rider set out after him, who, overtaking the tourist, arrived in $N$ and immediately turned back. Find the speed of the rider's movement if it is known that he met the tourist 2 hours and 45 minutes after overtaking him.

Solution. Let the speed of the rider be $v$ kilometers per hour. The rider caught up with the tourist (made up 10 km) due to the difference in speeds $(v-5)$ kilometers per hour, spending $\frac{10}{v-5}$ hours to catch up with the tourist.

By the time of the meeting, the rider had been traveling for $\left(\frac{10}{v-5}+2 \frac{3}{4}\right)$ hours and had traveled $\left(\frac{10}{v-5}+2 \frac{3}{4}\right) v$ kilometers. The tourist, by the time of the meeting, had been traveling for 2 hours longer $\left(\left(\frac{10}{v-5}+4 \frac{3}{4}\right)\right.$ hours $)$ and had traveled $\left(\frac{10}{v-5}+4 \frac{3}{4}\right) \cdot 5$ kilometers.

Since the sum of the distances traveled by the tourist and the rider by the time of the meeting is equal to twice the distance $M N$, we have the equation

$$
\left(\frac{10}{v-5}+\frac{19}{4}\right) \cdot 5+\left(\frac{10}{v-5}+\frac{11}{4}\right) v=82
$$

which reduces to the quadratic equation $11 v^{2}-248 v+1365=0$.

The roots of the equation are $v_{1}=13$ and $v_{2}=9 \frac{6}{11}$. Both roots satisfy the condition of the problem ( $v>5$ ).

A n s w e r: 13 km $/ h$ or $9 \frac{6}{11}$ km $/$ h.