$31 \cdot 2$ In a certain store, the unit prices of three types of goods, A, B, and C, are 2 yuan, 3 yuan, and 5 yuan, respectively. A person buys several pieces of these three types of goods, paying a total of 20 yuan. This person then realizes that they bought too much of one type of good, and returns two pieces of this good, but the cashier only has 10 yuan bills and no change. As a result, this person adjusts the quantities of the other two types of goods (and each type of good is still bought at least one piece), keeping the total price unchanged. At this point, the number of items of type B that this person has bought is
(A) 1.
(B) 2.
(C) 3.
(D) 4.
(China Jiangxi Nanchang Junior High School Mathematics Competition, 1990)
[Solution] It is impossible to return two items of type C, because giving back 10 yuan is feasible, and it is also impossible to return two items of type A, because two items of type A are worth 4 yuan, which cannot be replaced with other items to make change. Therefore, only two items of type B can be returned. Replacing with 3 items of type A for change is sufficient.
Suppose someone initially buys $x, y, z$ items of types A, B, and C respectively, then we have the equation
$$
2 x+3 y+5 z=20 \quad(y \geqslant 3) .
$$

The possible positive integer solutions are $x=3, y=3, z=1$.
Thus,
$$
y-2=1 \text {, }
$$

Therefore, the answer is $(A)$.
Note The original problem does not have the requirement of "at least one item of each type after adjustment," in which case $x=2, y=2, z=2$ also satisfies the equation $\circledast$. In this case, the number of items of type B after adjustment would be 0.