13.380 Several kilograms of two grades of goods were purchased: the first grade at 45 rubles and the second grade at 20 rubles, with 1 kg more of the first grade being purchased. The cost of 1 kg of the first grade is a rubles higher than the cost of 1 kg of the second grade. How many kilograms of each grade of goods were purchased? Determine the number of solutions depending on the possible values of $a$.
Solution. Let $x$ be the mass (in kg) of goods of the first grade. Then $\frac{45}{x}, \frac{20}{x-1}$ are the costs per 1 kg of goods of the first and second grades, respectively. Hence, $\frac{45}{x}-\frac{20}{x-1}=a \Rightarrow a x^{2}-x(a+25)+45=0 \Rightarrow x=\frac{a+25 \pm \sqrt{D}}{2 a}, D=(a-5)(a-125)$.

By the meaning of the problem, $0 < x < 1$ and $0 < x-1 < 1$. Therefore, $D > 0$ when $0 < a \leq 5$.

Answer: $\left(\frac{25+a+\sqrt{D}}{2 a}, \frac{25-a+\sqrt{D}}{2 a}\right)$ and $\left(\frac{25+a-\sqrt{D}}{2 a}, \frac{25-a-\sqrt{D}}{2 a}\right)$; the problem has one solution when $a=5$, two solutions when $0<a<5$, and no solutions for other values of $a$.