1. **Stating the problem:** We have a survey about preferences for Newari food (set $N$) and Thakali food (set $T$). Given: 70% like Newari food, 60% like Thakali food, 20% like neither, and 550 people like both. 2. **Understanding cardinality notation:** $n(N \cup T)$ represents the number of people who like Newari food or Thakali food or both. 3. **Given data:** - $n(N) = 0.7n(U)$ where $n(U)$ is total people - $n(T) = 0.6n(U)$ - $n(N^c \cap T^c) = 0.2n(U)$ (people who like neither) - $n(N \cap T) = 550$ 4. **Find total number of people $n(U)$:** Since 20% like neither, 80% like at least one type: $$n(N \cup T) = 0.8n(U)$$ Using the formula for union: $$n(N \cup T) = n(N) + n(T) - n(N \cap T)$$ Substitute values: $$0.8n(U) = 0.7n(U) + 0.6n(U) - 550$$ Simplify: $$0.8n(U) = 1.3n(U) - 550$$ Rearranged: $$0.8n(U) - 1.3n(U) = -550$$ $$\cancel{0.8}n(U) - \cancel{1.3}n(U) = -550$$ $$-0.5n(U) = -550$$ Divide both sides by $-0.5$: $$n(U) = \frac{-550}{-0.5} = 1100$$ 5. **Calculate number who like Newari only:** $$n(N \text{ only}) = n(N) - n(N \cap T) = 0.7 \times 1100 - 550 = 770 - 550 = 220$$ 6. **Calculate number who like Thakali only:** $$n(T \text{ only}) = n(T) - n(N \cap T) = 0.6 \times 1100 - 550 = 660 - 550 = 110$$ 7. **Number who don’t like both types:** Given as 20% of total: $$n(N^c \cap T^c) = 0.2 \times 1100 = 220$$ 8. **Comparison:** Number who like Newari only (220) is twice the number who like Thakali only (110). **Final answers:** - (a) $n(N \cup T)$ is the number of people who like Newari or Thakali or both. - (c) Number who don’t like both types is 220. - (d) Newari only: 220, Thakali only: 110; Newari only is twice Thakali only.