Question:

Step 1: Convert the given percentages into the number of families. Let $N$ be the total number of families, which is 10,000. We denote the number of families buying newspaper A, B, C, and their combinations using set notation. $$ |A| = 40\% \times 10,000 = 4000 $$ $$ |B| = 20\% \times 10,000 = 2000 $$ $$ |C| = 10\% \times 10,000 = 1000 $$ $$ |A \cap B| = 5\% \times 10,000 = 500 $$ $$ |B \cap C| = 3\% \times 10,000 = 300 $$ $$ |A \cap C| = 4\% \times 10,000 = 400 $$ $$ |A \cap B \cap C| = 2\% \times 10,000 = 200 $$ Step 2: State the formula for finding the number of families that buy newspaper A only. The number of families that buy newspaper A only can be found using the principle of inclusion-exclusion for three sets, specifically adapted to find elements belonging to only one set. $$ \text{Families buying A only} = |A| - |A \cap B| - |A \cap C| + |A \cap B \cap C| $$ This formula works because when we subtract $|A \cap B|$ and $|A \cap C|$ from $|A|$, the families that buy all three newspapers (i.e., $|A \cap B \cap C|$) are subtracted twice. Once within $|A \cap B|$ and once within $|A \cap C|$. To correct for this double subtraction, we add $|A \cap B \cap C|$ back once. Step 3: Calculate the number of families that buy newspaper A only. Substitute the values obtained in Step 1 into the formula from Step 2. $$ \text{Families buying A only} = 4000 - 500 - 400 + 200 $$ $$ \text{Families buying A only} = 3500 - 400 + 200 $$ $$ \text{Families buying A only} = 3100 + 200 $$ $$ \text{Families buying A only} = 3300 $$ The number of families that buy newspaper A only is 3300. The final answer is $\boxed{\text{3300}}$ families, which corresponds to Option 2.