Heuristics could reduce trial-and-error to a reasonable search space, but is there a purely deductive way to solve this by hand?
Examine what Mary and Dan have in common besides those 4... Nothing. They disagreed on each of the remaining 6 answers.
We can break this down into three cases, either 1, 2, or 3 of the common answers are correct:
1 - Impossible. Mary gets 6 of the remaining correct, Dan gets 4 of the 6 remaining correct, but they have none in common so this doesn't work.
2 - Impossible. Mary gets 5 of the remaining correct, Dan gets 3 of the remaining correct, but again their answers disagree so they cannot share a common answer out of the 6 remaining.
3. Mary needs 4 of the remaining 6 and Dan needs 2, this is feasible.
At this point you can ignore the 4 common ones other than to say Colin has at least a score of 3.
Since we now know that everything outside those common ones are incorrect for Lisa, cross everything off that Lisa and Colin share of the 6 non-common answers. There are only 3 that disagree with Lisa so they must be correct (because everything Lisa answered here is wrong). This gives Colin a final score of 6, three in common with everyone and three from disagreeing with Lisa.
The reason there are 4 keys in Hillel's solution is actually kind of funny, they don't matter. Those 4 keys only disagree on which of the 4 common answers are incorrect. They agree everywhere else.
EDIT: Presentation and some typos
1 2 3 4 5 6 7 8 9 10
Mary B B A B A B B A B B 7
Dan B A A A B A B A A A 5
Lisa B A A A B B B A B A 3
Colin B B A A A B B A A A ?
1 2 3 4 5 6 7 8 9 10
Mary _ B _ B A B _ _ B B 7
Dan _ A _ A B A _ _ A A 5
Lisa _ A _ A B B _ _ B A 3
Colin _ B _ A A B _ _ A A ?
1 2 3 4 5 6 7 8 9 10
Mary _ B _ B A x _ _ x B 7
Dan _ x _ x x A _ _ A x 5
Lisa _ x _ x x x _ _ x x 3
Colin _ B _ x A x _ _ A x ?
So the answer for Colin would by 6 because he has 7 remaining answers.
This is incorrect. The assumption is true, but that's just a coincidence - you haven't justified it.