11. In the arithmetic expression below, each square represents a digit. Question: What is the sum of the 6 digits in the squares?
$$
\begin{array}{r}
\square \square \square \\
+\square \square \square \\
\hline 1991
\end{array}
$$
11. The total is 47
11.【Solution】Each number in the boxes can only be $0 \sim 9$, so the sum of any two numbers in the boxes cannot exceed 18. Now, let's look at the sum of the digits in the "hundreds" place of the addend and the augend. This sum cannot be less than 18, because no matter what the following two-digit numbers are, their sum will be less than 200, meaning the maximum carry is 1. Therefore, we can conclude that the sum of the digits in the "hundreds" place is 18, and the sum of the following two-digit numbers must carry 1. Similarly, the sum of the digits in the "tens" place is 18, and the sum of the digits in the "units" place must carry 1. Therefore, the sum of the digits in the "units" place must be 11. The total sum of the six digits in the boxes is $18+18+11=47$.

【Another Solution】The addend cannot be greater than 999, so the augend cannot be less than 1991-999=992. Similarly, the addend cannot be less than 992, meaning that both the addend and the augend are numbers no less than 992 and no greater than 999. This determines that the digits in the "hundreds" and "tens" places of both the addend and the augend are 9, and the sum of the digits in the "units" place must be 11.

Thus, the total sum is $9 \times 4+11=47$.