SOLUTION: In an indoor games tournament at Sekolah Menengah Tanjung Malim, medals are awarded in three games. 36 medals in chess game, 12 medals in scrabble and 18 medals in sahibba. If t

Let C (chess), B (scrabble) and H (sahiba) be the abbreviations for these games.


Let p is the number of students having medals in the game C only.

Let q is the number of students having medals in the game B only.

Let r is the number of students having medals in the game H only.


Let x is the number of students having medals in exactly two games C and B.

Let y is the number of students having medals in exactly two games C and H.

Let z is the number of students having medals in exactly two games B and H.


    The problem asks about the amount x+y+z.


From the condition, we have these equations

    p + x + y + 4 = 36     (1)

    q + x + z + 4 = 18     (2)

    r + y + z + 4 = 12     (3)


Add equations (1), (2) and (3).   You will get

    (p + q + r) + 2*(x + y + z) + 3*4 = 36 + 18 + 12,   or

    (p + q + r) + 2*(x + y + z) +   4 = 66 - 8,         or

    (p + q + r) + 2*(x + y + z) +   4 = 58.    (4)


From the other side, from the condition, we have another equation 

    (p + q + r) +   (x + y + z) +   4  = 45    (5)


for the total number of students.


Now subtract equation (5) from equation (4).    You will get

    x + y + z = 58 - 45 = 13.


This value  x + y + z = 13  is the desired ANSWER.


ANSWER.  The number of students received medals in exactly two of these games is 13.