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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 these medals went to a total of 45 students and only 4 students got medals in all three games,
how many students received medals in exactly two of these games?
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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.
Solved.