3.41 Several students decided to buy an imported tape recorder priced between 170 and 195 dollars. However, at the last moment, two of them refused to participate in the purchase, so each of the remaining had to contribute 1 dollar more. What was the price of the tape recorder?

## Ratios between natural numbers
3.41 Let $x$ be the original number of students, and $p$ be the cost of the tape recorder. Then $\frac{p}{x}$ is the amount each student was originally supposed to contribute, and $\frac{p}{x-2}$ is the amount each of the tape recorder buyers contributed. Using one of the conditions of the problem, we get the equation

$$
\frac{p}{x-2}-\frac{p}{x}=1
$$

$$
p\left(\frac{1}{x-2}-\frac{1}{x}\right)=1 ; p \cdot \frac{2}{x(x-2)}=1 ; p=\frac{x(x-2)}{2}
$$

Now let's use the other condition:

$170<\frac{x(x-2)}{2}<195 ; 340<x^{2}-2 x<390 ;$

$341<(x-1)^{2}<391 ; 18.5<x-1<19.8$.

Since the number $x-1$ must be an integer, then $x-1=19$,

i.e., $x=20$. Therefore, the tape recorder cost $\frac{20 \cdot 18}{2}=180$ dollars.

Answer: 180 dollars.