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        riddles >> medium >> Exam Questions
        
(Message started by: THUDandBLUNDER on Apr 13^th, 2003, 7:46pm)

Title: Exam Questions
Post by THUDandBLUNDER on Apr 13^th, 2003, 7:46pm

In an exam three questions (A, B, and C) were set.

1) Among the entrants there were 25 who correctly answered at least one question each.

2) Of all the entrants who did not correctly answer Question A, the number who correctly answered Question B was twice the number who correctly answered Question C.

3) The number of entrants who correctly answered only Question A was one more than the number who correctly answered Question A and at least one other question.

4) Of all students who correctly answered just one question, half did not correctly answer Question A.

How many students correctly answered only Question B?

Title: Re: Exam Questions
Post by aero_guy on Apr 13^th, 2003, 11:36pm

Nifty, here is my method:

[hide]1)Draw out those overlapping logic circles, I don't remember what they are called. There is one circel for A, one for B, and one for C. They overlap in such a way as you have every possible combination represented (being on the interior of a circle means you got that question right.

2)Label each of the seven sectiond of the circles and translate the four clues into equations with these labels.

3)Reduce them until you get an equation with just "only A right" and "only B right" left. (Cannot be solved explicitly as there are too many variables.)
4)Note that they must be positive whole numbers less than 26, so there are only 6 possible solutions.

5)Use the equation taken from the part 2 clue to identify the only solution that will give a positive number of students for "B and C"

6)The answer is 6. "A only"-8, "C only"-2, "B and C"-2, "All others (undiferentiable)"-7[/hide]

Title: Re: Exam Questions
Post by THUDandBLUNDER on Apr 14^th, 2003, 3:05am

Nifty?? Are you referring to me, my puzzle, your method, or your search engine? ;)

Title: Re: Exam Questions
Post by aero_guy on Apr 14^th, 2003, 5:39am

Surely the puzzle... and don't call me Shirley.