What is the largest number of students who could have answered exactly 4 of the 5 questions correctly?

Question

---

How do I solve this question? The original question asked for the "smallest number of students". I solved that by taking 75 as the maximum number of students who got all 5 questions correct, then from the remaining 25 students I considered the 5 students who got 1st question wrong, 3 students who got 3rd question wrong, 5 students who got 4th question wrong, 4 students who got 5th question wrong as distinct. Therefore, 17 students got exactly 3 questions correct. Thus, 8 students got exactly 4 questions correct.

I was wondering if a similar logic could be applied to find the "largest number of students who got exactly 4 questions correct".

Answer 1

We can see that Question 2 has the least amount of students that got it correct. Then we can have $95$ students that get all of Q1, 3, 4, 5 but not Q2, but that is a contradiction as Q2 had $75$ students that got it correct, which means $100-75=25$ got it wrong. Then, if $25$ students get all of Q1, 3, 4, 5 but no Q2, it is possible.

If we consider the cases where we need to find the maximum possible of students that only get
Q1, 2, 3, 4; 1, 2, 3, 5; 1, 2, 4, 5; or 2, 3, 4, 5; we can find that the maximum possible is $100-95=5$ students who get only those questions and not the excluded question.

Therefore, $25$ is the final answer for Q1, 3, 4, 5. We can also have an extra $100-95=5$ for Q2, 3, 4, 5, an extra $100-95=5$ for 1, 2, 3, 5, an extra $100-96=4$ for 1, 2, 3, 4, and also an extra $100-97=3$ for 1, 2, 4, 5. So the final answer for the entire question is $25+5+5+4+3=\boldsymbol{42}$ students.

We can see that if there are at least $43$ students that get exactly four questions correct, that is a contradiction, because then the numbers for amount of people getting a question correct will be wrong, because the amount of people getting a certain question right plus the amount of people getting the same question wrong will exceed $100$ in one or more question.

Answer 2

A good starting point is to consider the scenario where every student who answered any one question incorrectly answered the other four questions correctly (to keep track, we can assign them names). We then check if that scenario works; if it does it will be the maximum. If it doesn't work, then we take two students and change them so that one made all their incorrect answers and one who answered all five questions correctly. We then check if it works; if so, we found our solution, if not, repeat.

We choose our order of preference as follows:

1. We cannot choose two students who both answered any of the five questions incorrectly
2. We want one to have his incorrect answer be the question that has the most unique incorrect answers (by which I mean the question with the most students who answered that question incorrectly and all other questions correctly). In our starting case, we start with $25$ students answering $2$ incorrectly and the other four correctly, which is more than for any other question
3. We want the other student to have the most possible incorrect answers without violating either of the previous rules. In the event of a tie, we want his $x$ incorrect answers to be from the $x$ questions with the next most unique incorrect answers, with priority given to the first incorrect answer, then to the second, etc. In event of a tie, choose randomly.

Our starting point in this case gives a solution. We can attain the case where every student either gets a perfect score or answers exactly one question incorrectly, resulting in 42 students who answer exactly four questions correctly.

Answer 3

Given data is:

5 people answered Q1 wrong. (set A)

25 people answered Q2 wrong. (set B)

3 people answered Q3 wrong. (set C)

5 people answered Q4 wrong. (set D)

4 people answered Q5 wrong. (set E)

If A,B,C,D,E are completely disjoint sets (which is possible here), then the total number of people who answered only 4 questions correctly are cardinal of A U B U C U D U E = 5+25+3+5+4 = 42