Five students each answered five questions on an quiz consisting of two multiple-choice questions (A, B or C) and three True-False questions. They answered the questions as follows:

Student Q1 Q2 Q3 Q4 Q5 
Alex    A  A  T  T  T 
Bert    B  B  T  F  T 
Carl    A  B  T  T  F 
Dave    B  C  T  T  F 
Eddy    C  A  F  T  T 

No two students got the same number of correct answers. Who got the most correct answers?

Well, the number of correct answers could be from 0 to 5 (6 possibilities). So the person with the most correct answers either got 4 or 5 correct.

Wow, it’s really hard not to go through brute force on this one.

First, I did go through and check that if one person got all 5 right, then the other scores could not follow the "no two students got the same number of correct answers" rule. So we are dealing with scores from 0 to 4.

It seems like there might be something useful in Q3 and Q4, since only one person answered differently, but I can’t figure out how to use that information.

Well, I caved and just went through it. I went through and checked what would happen when one person got all 5 wrong. You can know how many they got wrong and right in the T/F sections for sure, but the multiple choice gets a little iffy. Ex: If Alex got all 5 wrong, then Carl definitely got Q1, Q3 and Q4 wrong, definitely got Q5 right, and might have gotten Q2 right, so Carl could only get either 1 or 2 right.

As it turns out, if Alex gets 0 right, no one can get 4 right, which means that at least two people will have the same number of correct answers. So this is out.

If Bert gets 0, only Eddy can get 4 and only Alex can get 1.
This means that Q1=C, Q2=A, Q3=F, Q4=T, and Q5=F. But that would mean that Alex got 2 right, not 1, so this does not work.

If Carl gets 0, only Eddy can get 4 and then only Bert can get 3.
This means that Q1=C, Q2=A, Q3=F, Q4=F, and Q5=T. But that would mean that Bert got 2 right, not 3, so this does not work.

If Dave gets 0, only Eddy can get 4.
This means that Q1=C, Q2=A, Q3=F, Q4=F, and Q5=T. But that would mean that Carl also got 0, when Dave already got 0, so this does not work.

If Eddy gets 0, only Alex can get 1.
This means that Q1=B, Q2=B or C, Q3=T, Q4=F, and Q5=F. In this case (since we don’t know the correct answer for Q2 yet) Carl can have 2 or 3 correct, and Bert and Dave both could get 3 or 4 correct. Well, someone has to get two, and the only person who can get 2 now is Carl, so Q2 must not be B, Q2=C. This means that Bert gets 3 and Dave gets 4.

So the answer is Dave is the student with the most correct answer.

Is this truly brute force? I "checked" every case, but I didn’t finish every case because I found my contradiction early.

Posted by nikki on 2004-09-02 09:34:20