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Earlier I provided an "answer" as a hint (in the form of a question). It was deleted as being more appropriate as a comment ("request for clarification") it was not a request for clarification. It was a hint in the form of a question.
"Hint: How many total A's were there out of how many total grades?"
The point is that there were $30+33+37=100$ total A's out of $3\cdot 40=120$ total grades. Hence there were $20$ grades that were not A's. If these grades were maximally spread around among the $40$ students, there would have been $20$ students who got a non A. (There may have been less such students, as some students could have gotten more than one non A.) But in any case that would mean that there are least $40-20=20$ students who got all A's.
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 $40 \geq |A\cup B\cup C|$.