6. There are three types of stocks, the sum of the shares of the first two equals the number of shares of the third; the total value of the second type of stock is four times that of the first type; the total value of the first and second types of stocks equals the total value of the third type of stock; the price per share of the second type of stock is 1.6 to 2 yuan higher than that of the first type, and the value per share of the third type of stock is no less than 4.2 yuan and no more than 6 yuan. Find the maximum and minimum percentages of the first type of stock in the total number of shares.
6. Let the number of shares of the first, second, and third types of stocks be $k$, $m$, and $n$, respectively, and the value per share be $x$, $y$, and $z$ yuan, respectively. The problem is to find $f=\frac{k}{k+m+n} \times 100 \%$. Given $k+m=n$, $k x+m y=n z$, $4 k x=m y$, $1.6 \leqslant y-x \leqslant 2$, and $4.2 \leqslant z \leqslant 6$. Let $p=\frac{m}{k}$, then $f=\frac{k}{m+n+k} \times 100 \%=\frac{k}{2(k+m)} \times 100 \%=\frac{50}{1+p} \%$, $x=\frac{p}{4} y$, $z=\frac{k x+m y}{k+m}=\frac{x+p y}{1+p}=\frac{\frac{p y}{4}+p y}{1+p}=\frac{5 p y}{4(1+p)}$. Therefore, $1.6 \leqslant y-\frac{p y}{4} \leqslant 2$, which leads to $\frac{6.4}{4-p} \leqslant y \leqslant \frac{8}{4-p}$. Also, $4.2 \leqslant \frac{5 p y}{4(1+p)} \leqslant 6$, so $\frac{16.8(1+p)}{5 p} \leqslant y \leqslant \frac{24(1+p)}{5 p}$. Since for any $a$, $b$, $c$, and $d$ we have $\left\{\begin{array}{l}a \leqslant y \leqslant b \\ c \leqslant y \leqslant d\end{array} \Leftrightarrow\left\{\begin{array}{l}a \leqslant y \leqslant d \\ c \leqslant y \leqslant b,\end{array}\right.\right.$ it follows that $\frac{6.4}{4-p} \leqslant \frac{24(1+p)}{5 p}$ and $\frac{16.8(1+p)}{5 p}<\frac{8}{4-p}$, leading to $\frac{7}{3} \leqslant p \leqslant 3$. Thus, $12.5 \% \leqslant f \leqslant 15 \%$.