Let $A, B$ and $C$ be three sets such that:

- $|A|=100,|B|=50$ and $|C|=48$,

- the number of elements belonging to exactly one of the three sets is equal to twice the number of elements belonging to exactly two of the sets,
- the number of elements belonging to exactly one of the three sets is equal to three times the number of elements belonging to all the sets.

How many elements belong to all the sets?
Let $x$ be the number of elements that belong to exactly one of the three sets (i.e., belong to $A$ but not to $B$ or $C$, or to $B$ but not to $A$ or $C$, or to $C$ but not to $A$ or $B$). Similarly, let $y$ be the number of elements that belong to exactly two of the three sets, and $z$ be the number of elements that belong to the intersection of the three sets. According to the problem, we have $x=2 y=3 z$. When calculating $|A|+|B|+|C|$, we count twice the elements that belong to two sets, and three times those that belong to all three sets. Therefore: $|A|+|B|+|C|=x+2 y+3 z$, which is $100+50+48=198=x+2 y+3 z$. Thus, we get $198=3 x$. Hence, $x=66$ and $z=x / 3=22$. In conclusion, there are 22 elements that belong to all the sets.