1. **Problem Statement:** We have three subsets of whole numbers from 1 to 20: - $F = \{4, 6, 8, 10, 12, 14, 18\}$ - $G = \{1, 2, 3, 4, 5, 7, 9, 11, 16, 18\}$ - $H = \{1, 2, 3, 5, 7, 9, 14\}$ We need to find: (i) $F \cap G$ (ii) $(F \cup G)^c$ (iii) $(F \cap H)^c$ --- 2. **Recall Set Operations:** - Intersection ($\cap$): elements common to both sets. - Union ($\cup$): all elements in either set. - Complement ($^c$): elements not in the set, relative to the universal set (here, whole numbers 1 to 20). --- 3. **Calculate (i) $F \cap G$:** Find elements common to both $F$ and $G$. $F = \{4, 6, 8, 10, 12, 14, 18\}$ $G = \{1, 2, 3, 4, 5, 7, 9, 11, 16, 18\}$ Common elements: $4$ and $18$ So, $$F \cap G = \{4, 18\}$$ --- 4. **Calculate (ii) $(F \cup G)^c$:** First find $F \cup G$ (all elements in $F$ or $G$): $F \cup G = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 14, 16, 18\}$ Universal set $U = \{1, 2, 3, ..., 20\}$ Elements not in $F \cup G$ are: $$U \setminus (F \cup G) = \{13, 15, 17, 19, 20\}$$ So, $$(F \cup G)^c = \{13, 15, 17, 19, 20\}$$ --- 5. **Calculate (iii) $(F \cap H)^c$:** Find $F \cap H$ first: $F = \{4, 6, 8, 10, 12, 14, 18\}$ $H = \{1, 2, 3, 5, 7, 9, 14\}$ Common element is $14$ only. So, $$F \cap H = \{14\}$$ Complement relative to $U$: $$(F \cap H)^c = U \setminus \{14\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20\}$$ --- **Final answers:** (i) $F \cap G = \{4, 18\}$ (ii) $(F \cup G)^c = \{13, 15, 17, 19, 20\}$ (iii) $(F \cap H)^c = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 20\}$