A maths exam has two papers, each with at least one question and 28 questions in total. Each pupil attempted 7 questions. Each pair of questions was attempted by just two pupils. Show that one pupil attempted either nil or at least 4 questions in the first paper.
Solution
Each pupil attempts 7 questions and hence 21 pairs of questions. There are 28·27/2 = 378 pairs of questions in total and each is attempted by 2 pupils. So there must be 378·2/21 = 36 pupils. Suppose n pupils solved question 1. Each solved 6 pairs involving question 1, so there must be 3n pairs involving question 1. But there are 27 pairs involving question 1, so n = 9. The same applies to any other question. So each question was solved by 9 pupils.
Suppose the result is false. Suppose there are m questions in the first paper, that the number of pupils solving 1, 2, 3 questions in the first paper is a, b, c respectively. So a + b + c = 36, a + 2b + 3c = 9m. Now consider pairs of problems in the first paper. There are m(m-1)/2 such pairs. Pupils solving just 1 solve no pairs, those solving 2 solve 1 pair and those solving 3 solve 3 pairs, so we have b + 3c = m(m-1). Solving for b we get b = - 2m2 + 29m - 108 = -2(m - 29/4)2 - 23/8 < 0. Contradiction. So the result must be true.
© John Scholes
jscholes@kalva.demon.co.uk
24 Aug 2002