How many received medals in exactly two of the three sports ?

Concept:

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

Explanation:

Given: x students (x < 6) got medals in all three sports and the medals went to a total of 15x students. It awarded 5x medals in basketball, (4x + 15) medals in football, and (x + 25) medals in volleyball.

Let A = students who got medals in basketball

B = students who got medals in football

C = students who got medals in volleyball

From given data,

⇒ n(A) = 5x, n(B) = 4x + 15, n(C) = x + 25, n(A ∪ B ∪ C) = 15x, n(A ∩ B ∩ C) = x

Putting in the formula,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

⇒ 15x = 5x + 4x + 15 + x + 25 - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + x

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 11x - 15x + 40 

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = 40 - 4x  __(i)

From Venn diagram,

n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = (p + s) + (r + s) + (q + r)

⇒ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) = p + q + r + 3s

where (p + q + r) is the number of students who received medals in exactly two of the three sports.

and s = the number of students who received medals in all the three sports

⇒ The number of students who received medals in exactly two of the three sports

= n(A ∩ B) + n(B ∩ C) + n(C ∩ A) - 3x

Putting value from (i),

⇒ The number of students who received medals in exactly two of the three sports = 40 - 4x - 3x = 40 - 7x

∴ The correct option is (3).