A math competition featured four problems. 85 percent of the participants solved the first problem, 80 percent solved the second, 75 percent solved the third, and 70 percent solved the fourth. What is the minimum percentage of participants who solved all four problems?
Solution. 15 percent of the participants did not solve the first problem, 20 percent did not solve the second, 25 percent did not solve the third, and 30 percent did not solve the fourth. Thus, at most $15+20+25+30=90$ percent of the participants failed to solve at least one problem. This precisely means that at least $100-90=10$ percent solved all the problems.

We also need to check if it is possible that only 10 percent of the participants solved all four problems. The previous line of thought shows that this is the case exactly when there is no contestant who failed to solve more than one problem. Based on this, it is not difficult to provide an appropriate example: if 100 people participated in the competition, out of which 30 solved the first three problems (but not the fourth), 25 solved the first, second, and fourth problems, 20 solved the first, third, and fourth problems, 15 solved all except the first problem, and 10 solved all four, then this meets the conditions, and exactly 10 percent of the contestants solved all four problems.