Arthur, Bert and Charles took three exams. In all exams, the first got X points, the second Y points, and the third Z points, X, Y and Z being positive integers such that X>Y>Z.
Summing the three results, Albert got 20 points; Bert, 10; and Charles, 9. Albert was 2nd in Algebra. Who was 2nd in Geometry?
(In reply to
re(2): Solution assuming that Albert is Arthur's real name... by e.g.)
Ahh, I see what you mean. If X, Y, and Z are the same on each test, then the score totals on each test is the same, ie. 13. So the scores work out as
Exam 1: B(8), A(4), C(1)
Exam 2: A(8), C(4), B(1)
Exam 3: A(8), C(4), B(1)
Which would give the same X, Y, and Z and the required point totals. I remember an (8,4,1) in Charlie's solution description. I assume this is what he got.
re(3): Solution assuming that Albert is Arthur's real name...
