6. (10 points) A convoy of trucks is delivering supplies to a disaster victim resettlement point. Each truck has a carrying capacity of 10 tons. If each tent is allocated 1.5 tons of supplies, there will be less than one truck's worth of supplies left over. If each tent is allocated 1.6 tons of supplies, there will be a shortage of more than 2 trucks' worth of supplies. How many tents are there at the resettlement point at a minimum?
6. (10 points) A convoy of trucks is delivering supplies to a disaster victim resettlement point. Each truck has a carrying capacity of 10 tons. If 1.5 tons of supplies are allocated to each tent, there will be less than a full truck's worth of supplies left over. If 1.6 tons of supplies are allocated to each tent, there will be a shortage of more than 2 trucks' worth of supplies. How many tents are there at the resettlement point at a minimum?
【Analysis】The number of trucks used and the number of tents at the resettlement point are the same in both allocation plans. Therefore, let's denote them as $X$ and $Y$ respectively, and set up the corresponding system of inequalities based on the problem statement. First, eliminate $Y$ from the system of inequalities, then solve the resulting inequality for $X$. Substitute integer values of $X$ within the solution set back into the original system of inequalities to find the solution set for $Y$.
【Solution】Let the number of trucks be $X$ and the number of tents be $Y$. According to the problem, we have:
$$
\left\{\begin{array}{l}
10 \times(\mathrm{X}-1)<1.5 \mathrm{Y}<10 \mathrm{X} \\
10 \times(\mathrm{X}+2)<1.6 \mathrm{Y}<10 \times(\mathrm{X}+3)
\end{array}\right.
$$

Eliminating $Y$, we get: $\left\{\begin{array}{l}\frac{100 \times(X-1)}{15}<\frac{100 \times(X+3)}{16} \\ \frac{100 \times(X+2)}{16}<\frac{100 \times X}{15}\end{array}\right.$
Solving, we get: $30<X<61$
Since we are looking for the minimum number of tents, the number of trucks should also be the minimum. Therefore, we start with the integer value 31 for $X$ and test until we find a value of $Y$ that satisfies the original system of inequalities.
(1) Substituting $X=31$ into the original system of inequalities, we get:
$$
\left\{\begin{array}{l}
200<\mathrm{Y}<206.7 \\
206.2<\mathrm{Y}<212.2
\end{array}\right.
$$

Thus, there is no solution for $Y$.
(2) Substituting $X=32$ into the original system of inequalities, we get:
$$
\left\{\begin{array}{l}
206.7<\mathrm{Y}<213.3 \\
212.2<\mathrm{Y}<218.7
\end{array}\right.
$$

Thus, the minimum integer value for $Y$ is 213.
Answer: The resettlement point has a minimum of 213 tents.