All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Quiz Results (Posted on 2004-09-02)
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?

 See The Solution Submitted by Brian Smith Rating: 3.2222 (9 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 4 of 29 |

There are 6 possibile scores: 0 - 5, so only one score is missing, and further, one must have gotten 5 or one all wrong. If one had gotten all 5 right, taking the five people's answers in turn shows that if any one person's answers had been all right, there would have been duplicate scores.  So the missing score is 5, and in particular there's someone who got all 5 questions wrong.

As a preliminary, taking the true/false questions alone

If A had been all wrong, they'd be FFF, and the scores for A-E on the T/F questions would be
0, 1, 1, 1, 1.  But with four people getting 1 on the T/F part, and there being only three possible scores for the multiple choice part, there'd have to be at least one duplicate, so A is not the one who got all wrong.

If B had been all wrong, the T/F answers would be FTF, and the scores for A-E would be
1, 0, 2, 2, 2. To differentiate the three scores of 2 on this part, there'd need to be three different scores for C thru E on the multiple choice part, but only one of these could be correct on Q1 and one correct on Q2; to have three different scores, there'd have to be two correct on one of them: one a solo hit, and one as part of a pair of hits. So B was not the one to be all wrong.

If C or D had been all wrong, the T/F answers would be FFT, and the scores for A-E on these parts would be
1, 2, 0, 0, 2.  Offhand this didn't look like an immediate possibility of removal, so I went on to the possibility of E being all wrong.

If E was all wrong, the T/F answers would be TFF, and the scores for A-E on these would be
1, 2, 2, 2, 0. This time B-D have to be differentiated by the multiple choice questions.  In  this case this is possible, as, if either A and B were the correct answers or B and C, then the scores among these three people for the multiple choice questions would have been 0, 1 and 2, giving overall scores of 2, 3 and 4.

However, if AB were the correct multiple choice answers, Alex and Bert would both have had a n overall score of 2. So B and C must have been the correct multiple choice answers.

That means BCTFF is a viable set of correct answers, giving A-E scores of 1, 3, 2, 4 and 0 respectively, and Dave has the highest score: 4.

We didn't check the possibility that C or D could have been all wrong, but we have a solution that works without having proved it's a unique solution.

Edited on September 2, 2004, 9:07 am
 Posted by Charlie on 2004-09-02 08:52:46

 Search: Search body:
Forums (0)