1. The problem asks to find all numbers in the set $Q \cap R'$ where $Q = \{4, 8, 12, 16, 20, 24\}$ and $R = \{6, 12, 18, 24, 30\}$. Here, $R'$ denotes the complement of $R$ relative to the universal set that contains $Q$ and $R$. 2. First, identify the universal set. Since $Q$ and $R$ contain numbers from 4 to 30, we consider the universal set as all numbers appearing in either $Q$ or $R$, i.e., $U = \{4, 6, 8, 12, 16, 18, 20, 24, 30\}$. 3. The complement of $R$, denoted $R'$, is the set of elements in $U$ that are not in $R$: $$R' = U \setminus R = \{4, 8, 16, 20\}$$ 4. Now, find the intersection $Q \cap R'$, which are elements common to both $Q$ and $R'$: $$Q \cap R' = \{4, 8, 12, 16, 20, 24\} \cap \{4, 8, 16, 20\} = \{4, 8, 16, 20\}$$ 5. Therefore, the numbers in $Q \cap R'$ are $4, 8, 16,$ and $20$. Final answer: $$Q \cap R' = \{4, 8, 16, 20\}$$