1. A bus, a truck, and a car are traveling in the same direction on the same straight line. At a certain moment, the truck is in the middle, the bus is in front, and the car is behind, and the distances between them are equal. After 10 minutes, the car catches up with the truck; after another 5 minutes, the car catches up with the bus. How many more minutes will it take for the truck to catch up with the bus?
Ni.1.15.
As shown in Figure 4, let the distances between the truck and the bus, and the truck and the car be $S$. The speeds of the car, truck, and bus are approximately $a, b, c$. It is given that the truck catches up with the bus after $x$ minutes. Therefore, we have
$$
\begin{array}{l}
10(a-b)=S, \\
15(a-c)=2S, \\
x(b-c)=S .
\end{array}
$$

From equation (1), we get
$$
a-b=\frac{1}{10} S, \quad a-c=\frac{2}{15} S \text{. }
$$

Subtracting the two equations, we get $b-c=\frac{1}{30} S$, which means $S=30(b-c)$.
Substituting into equation (3), we get $x(b-c)=30(b-c)$.
Since $b-c \neq 0$, we get $x=30.30-10-5=15$ (minutes).