2. (3 points) Given that the concentration of salt water in bottle A is $8 \%$, and the concentration of salt water in bottle B is $5 \%$, after mixing, the concentration is $6.2 \%$. Then, the concentration of one-fourth of the salt water from bottle A mixed with one-sixth of the salt water from bottle B is ( )
A. $7.5 \%$
B. $5.5 \%$
C. $6 \%$
D. $6.5 \%$
【Analysis】We set the mass of the saltwater in bottle A as $a$, and the mass of the saltwater in bottle B as $b$. Based on the concentration of the mixture being $6.2 \%$, we find the quantitative relationship between $a$ and $b$, and then further calculate the concentration of the mixture when one quarter of the saltwater from bottle A and one sixth of the saltwater from bottle B are mixed.

【Solution】Let the mass of the saltwater in bottle A be $a$, and the mass of the saltwater in bottle B be $b$.
$$
(8 \% a+5 \% b) \div(a+b)=6.2 \% \text {, }
$$

Solving this, we get: $a=\frac{2}{3} b$;
$$
\begin{aligned}
& \left(\frac{1}{4} a \times 8 \%+\frac{1}{6} b \times 5 \%\right) \div\left(\frac{1}{4} a+\frac{1}{6} b\right), \\
= & \left(\frac{1}{50} a+\frac{1}{120} b\right) \div\left(\frac{1}{4} \times \frac{2}{3} b+\frac{1}{6} b\right), \\
= & \left(\frac{1}{50} \times \frac{2}{3} b+\frac{1}{120} b\right) \div\left(\frac{1}{3} b\right), \\
= & \frac{13}{600} b \times \frac{3}{b}, \\
= & 6.5 \%
\end{aligned}
$$

Therefore, the answer is: D.