Set theory puzzles - chess players and mathematicians

Solution 1:

Yes, it’s correct. If $M$ is the set of mathematicians, and $C$ is the set of chess players, you’re looking rankings of the members of $M\cap C$. If for $x\in M\cap C$ we let $m(x)$ be $x$’s ranking among mathematicians, $c(x)$ be $x$’s ranking among chess players, and $a(x)$ be $x$’s age, then there is a unique $x_a\in M\cap C$ such that $$a(x_a)=\max\{a(x):x\in M\cap C\}\;,$$ but there can certainly be distinct $x_m,x_c\in M\cap C$ such that $$m(x_m)=\max\{m(x):x\in M\cap C\}$$ and $$c(x_c)=\max\{c(x):x\in M\cap C\}\;.$$

All of which just says what you said, but a bit more formally.

Solution 2:

(1) Think of it in terms of sets. Let $M$ be the set of mathematicians, $C$ the set of chess players. Both are asking for the oldest person in $C\cap M$.

(2) Absolutely fantastic reasoning, though perhaps less simply set-theoretically described.