Mongolian Mathematical Olympiad

First we will prove a lemma which is a generalized form of the given problem. Let $|\Omega |$ be the universal set.

Lemma: If $a_1\le |A|\le a_2$; $b_1\le |B|\le b_2$ then the double inequality $$\max \{0,a_1+a_2-|\Omega |\}\le |A\cap B|\le \min \{a_2,b_2\}$$ holds.

Proof of lemma:

a) Obviously $|A\cap B|\le \min \{|A|,|B|\}$. Equality holds when either $A\subset B$ or $B\subset A$.

b) Let us prove that $\max \{|A|+|B|-|\Omega |,0\}\le |A\cap B|$. Consider the case $|A|+|B|-|\Omega |<0$. If $A\cap B=\varnothing$ then $|A\cap B|=0$.

Consider the case $|A|+|B|-|\Omega |\ge 0$. Then by inclusion-exclusion principle $$|A\cap B|=|A|+|B|-|A\cup B|\ge |A|+|B|-|\Omega |$$ and we have done.

Now let us apply the lemma to the given problem. Let $|A|,|B|,|C|$ be percent of students who solved I, II, III problems respectively. Then our task is to find minimal and maximal value of $|A\cap B\cap C|$.

First, consider $|A|=98\%$, $|B|=90\%$. By the aforementioned lemma: $$\max \{0,98\%+90\%-100\%\}\le |A\cap B|\le \min \{98\%,90\%\}$$ So $88\%\le |A\cap B|\le 90\%$.

Now, $|A\cap B|$ and $|C|=85\%$. Apply the lemma again: $$\max \{0,88\%+85\%-100\%\}\le |A\cap B\cap C|\le \min \{90\%,85\%\}$$ So $73\%\le |A\cap B\cap C|\le 85\%$.

Thus, the least percentage of students that solved all three problems is $73\%$, and the most is $85\%$.