3 students are taking an examination of N subjects.
Points are awarded according to the positions in each subject they took.
The person who gets 1st will get the highest points, followed by 2nd and lastly 3rd. (So let 1st=X points, 2nd=Y points, 3rd=Z points.)
Student B obtained X points in English.
Student A obtained 22 points total.
Students B and C obtained 9 points total each.
How many subjects did they take in the examination, and what are the values of X, Y, Z?
First assumption: no ties. (The problem doesn't even mention how they would be scored)
Second assumption: X>Y>Z>0
Third assumption: X, Y and Z are integers
Note that X<9 because of B, in fact since Z≥1, X≤ 91*(N1) = 8N
The students scored 40 points between them so there are 40 points available among the N subjects. (X+Y+Z)*N = 40
Call the point total X+Y+Z = P
If P*N = 40 then (P,N) could only be one of
(1,40) < can't have 3 different values of X,Y,Z
(2,20) < can't have 3 different values of X,Y,Z
(4,10) < can't have 3 different values of X,Y,Z
(5,8) < can't have 3 different values of X,Y,Z
(8,5)
(10,4)
(20,2) < X > 6.6 > 82
(40,1) < X > 13.3 > 81
If there are 4 subjects with 10 point total the values of (X,Y,Z) could only be one of
(5,3,2) < no set of 4 adds to 9
(5,4,1) < no set of 4 adds to 9
(6,3,1) < although 6+1+1+1 = 9, no set of 4 adds to 22
(7,2,1) < no set of 4 adds to 9
So there must be 5 subjects with 8 point totals. (X,Y,Z) could only be one of
(4,3,1) < no set of 5 adds to 9
(5,2,1)
This is the solution
A scored 2+5+5+5+5 = 22
B scored 5+1+1+1+1 = 9 (that first place in English)
C scored 1+2+2+2+2 = 9

Posted by Jer
on 20131120 12:11:51 