1. A car was moving at a speed of $V$. Upon entering the city, the driver reduced the speed by $x \%$, and upon leaving the city, increased it by $0.5 x \%$. It turned out that this new speed was $0.6 x \%$ less than the speed $V$.

A) Can the value of $x$ be found? If so, what is it?

B) Can the value of $V$ be found? If so, what is it?
Answer: a) Yes; 20; b) no. Solutions.

A) The condition of the problem means that the equation is satisfied

$$
\begin{aligned}
& V\left(1-\frac{x}{100}\right)\left(1+\frac{0.5 x}{100}\right)=V\left(1-\frac{0.6 x}{100}\right) \Leftrightarrow\left(1-\frac{x}{100}\right)\left(1+\frac{x}{200}\right)=1-\frac{3 x}{500} \Leftrightarrow \frac{x^{2}}{20000}=\frac{3 x}{500}-\frac{x}{200} . \\
& \Leftrightarrow x=0 ; x=20 . \text { The value } x=0 \text { contradicts the condition. Therefore, } x=20 . \\
& \text { B) Any positive number } V \text { satisfies the condition. }
\end{aligned}
$$