Question

A school awarded 42 medals in hockey, 18 in basketball and 23 in cricket. If these medals were bagged by total of 65 students and only 4 student got medals in all the three sports, how many students receive the medals in exactly two of three sports ?​

Answer 1

Answer:22

Solution : Let A,B,C denote the sets of students who bagged medals in hockey, basketball and cricket respectively. ⇒65=(42+18+23+4)-x⇔x=(87-65)=22.

Answer 2

Answer :

Given -

• Hockey = n(A) = 42 medals
• Basketball = n(B) = 18 medals
• Cricket = n(C) = 23 medals
• Total students = n(A U B U C) = 65
• Students who got medal in all three = n(A ∩ B ∩ C) = 4

To Find -

• The number of students that got medal in two of these.

Solution -

We know that,

Number of elements in exactly two of the sets = n(A ∩ B) + n(B ∩ C) + n(C ∩ A) – 3n(A ∩ B ∩ C).

➸ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) – 3*4

➸ n(A ∩ B) + n(B ∩ C) + n(C ∩ A) – 12

We know that,

n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C).

➸ n(A∩B) + n(B∩C) + n(C∩A) = n(A) + n(B) + n(C) + n(A∩B∩C) - n(A U B U C)

➸ n(A∩B) + n(B∩C) + n(C∩A) = 42 + 18 + 23 + 4 - 65

➸ n(A∩B) + n(B∩C) + n(C∩A) = 87 - 65

➸ n(A∩B) + n(B∩C) + n(C∩A) = 22

Hence, the number of students who received the medals in exactly two of three sports is 22.