Let $x = |A\cap B\cap C|$ be the number of students that got all As.

We use the inclusion-exclusion formula,

$$ |A| + |B| + |C| - |A\cap B| - |B\cap C|-|C\cap A| + |A\cap B\cap C| = |A\cup B\cup C|,$$

to get

$$x = |A\cup B\cup C| - |A| - |B| - |C| + |A\cap B| + |B\cap C|+|C\cap A|$$

To find a bound, we need to bound the intersections: using $$|A\cup B \cup C| \geq |A\cup B| = |A|+|B| - |A\cap B|,$$ we can rearrange to obtain $$|A\cap B| \geq |A| + |B| - |A\cup B \cup C|.$$ Plugging that (and analogous expressions) back into the equality for $x$ gives

$$x \geq |A\cup B \cup C| - |A|-|B|-|C|\\+ (|A|+|B|-|A\cup B \cup C| ) \\+ (|B|+|C|-|A\cup B \cup C| ) \\+ (|C|+|A|-|A\cup B \cup C| ) \\= |A|+|B|+|C| - 2|A\cup B\cup C|.$$

Thus $x \geq |A|+|B|+|C| - 2\cdot40 = 20$, because $40 \geq |A\cup B\cup C|$.