1. **State the problem:** We are given the sizes of complements and unions of sets and need to find $n(A \cap B')$. 2. **Given:** - $n(A') = 70$ - $n(B') = 100$ - $n(A' \cup B') = 120$ - $n(U) = 160$ 3. **Recall important formulas:** - $n(A) = n(U) - n(A')$ - $n(B) = n(U) - n(B')$ - $n(A' \cup B') = n(U) - n(A \cap B)$ (since $A' \cup B' = (A \cap B)'$) 4. **Calculate $n(A)$ and $n(B)$:** $$n(A) = 160 - 70 = 90$$ $$n(B) = 160 - 100 = 60$$ 5. **Find $n(A \cap B)$ using the complement union formula:** $$n(A' \cup B') = 120 = n(U) - n(A \cap B)$$ $$120 = 160 - n(A \cap B)$$ $$n(A \cap B) = 160 - 120 = 40$$ 6. **Use the formula for $n(A)$ to find $n(A \cap B')$:** $$n(A) = n(A \cap B) + n(A \cap B')$$ $$90 = 40 + n(A \cap B')$$ $$n(A \cap B') = 90 - 40 = 50$$ **Final answer:** $$\boxed{50}$$