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 $V$. Find the value of $x$.
Answer: 20. Solution. The condition of the problem means that the equation is satisfied

$$
v\left(1-\frac{x}{100}\right)\left(1+\frac{0.5 x}{100}\right)=v\left(1-\frac{0.6 x}{100}\right) \Leftrightarrow\left(1-\frac{x}{100}\right)\left(1+\frac{x}{200}\right)=1-\frac{3 x}{500} \Leftrightarrow \frac{x^{2}}{20000}=\frac{3 x}{500}-\frac{x}{200} .
$$

$$
\Leftrightarrow x=0 ; x=20 \text {. The value } x=0 \text { contradicts the condition. Therefore, } x=20 \text {. }
$$