$29 \cdot 2$ Among all two-digit numbers composed of two different digits, when each two-digit number is divided by the sum of its two digits, the minimum value of the quotient is
(A) 1.5 .
(B) 1.9 .
(C) 3.25 .
(D) 4 .
(Chinese Jiangsu Province Junior High School Mathematics Competition, 1986)
[Solution 1] Let the two-digit number be $10x + y$ ($x = 1, 2, \cdots, 9; y = 0, 1, 2, \cdots, 9$, and $x \neq y$).
Then $\frac{10x + y}{x + y} = 1 + \frac{9x}{x + y}$. Let $u = \frac{9x}{x + y}$,
we have $\frac{9 - u}{u} = \frac{y}{x}$, and $0 \leq \frac{y}{x} \leq 9$,
so $0 \leq \frac{9 - u}{u} \leq 9$, which means $0 \leq \frac{9}{u} - 1 \leq 9$, or $\frac{9}{10} \leq u \leq 9$. Therefore, $u_{\min} = \frac{9}{10}$, and the minimum value of $\frac{10x + y}{x + y}$ is 1.9. Hence, the answer is $(B)$.
[Solution 2] From the above solution, we know $\frac{10x + y}{x + y} = 1 + \frac{9x}{x + y} = 1 + \frac{9}{1 + \frac{y}{x}}$.
When $1 + \frac{y}{x}$ is maximized, the above value is minimized. But $\left(\frac{y}{x}\right)_{\text{max}} = 9$, then
$$
\left(\frac{10x + y}{x + y}\right)_{\text{min}} = 1 + \frac{9}{1 + 9} = 1.9.
$$

Hence, the answer is $(B)$.