A survey shows that $63\%$ of kids like brand $A$ shoes and $76\%$ like brand $B$ shoes. brand $B$. Each boy in the group surveyed expresses at least one preference. Calculate the percentage of the boys who like both brand $A$ and $B$ and the percentage of those who like only the $A$.
My attemps: I have thought that the total number $n_b$ of boys cannot obviously exceed $100\%$; but I have instead
$$|A\cup B|=63\%+76\%=\color{red}{139\%}$$
Subtracting the percentage of boys who like both brand $A$ and $B$ that is $100 \%$ I will have the percentage $n'$:
$$n'=\color{red}{139\%}-100 \%=39\%$$
After the percentage of those who like only the $A$ is:
$$|A - B|=|A|-|A\cap B|=63 \%-39 \%=24\%$$
Question: I have used the logic: surely the percentage of kids can't be more than $100\%$. Is there another motivation or is that the only one?
You have $$|A\cup B|\equiv100\%,$$ since everyone must choose at least one option. The left-hand side equals $$|A|+|B|-|A\cap B|.$$ Isolating $|A\cap B|$ then yields $$|A\cap B|=|A|+|B|-|A\cup B|,$$ so $$|A\cap B|\equiv63\%+76\%-100\%=39\%.$$ The way you computed $|A|$ is fine.