A2. In a mathematics competition, the contestants solved 3 problems. The first problem was solved correctly by $90 \%$ of the contestants, the second by $80 \%$, and the third by $70 \%$. What is the minimum percentage of contestants who solved all three problems correctly?
(A) 30
(B) 35
(C) 40
(D) 50
(E) 70
A2. Denote by $n$ the number of contestants. If $A$ and $B$ are two sets of contestants, then $n \geq |A \cup B| = |A| + |B| - |A \cap B|$, or equivalently, $|A \cap B| \geq |A| + |B| - n$. Since the first problem was solved correctly by $\frac{9}{10} n$ contestants, and the second problem by $\frac{8}{10} n$ contestants, by the above estimate, at least $\frac{9}{10} n + \frac{8}{10} n - n = \frac{7}{10} n$ contestants solved the first two problems correctly, and at least $\frac{7}{10} n + \frac{7}{10} n - n = \frac{4}{10} n$ contestants solved all three problems correctly, which is $40\%$ of the contestants.