in a competition School awarded medal in different categories 36 metal and install medal in Ramayan 18 medal in music release medal in all the three categories how many received medal in exactly two of these categories
Answers
13 Students
Correct Question:
In a competition, a school awarded medals in different categories. 36 medals in dance, 12 medals in dramatics and 18 medals in music. If these medals went to a total of 45 persons and only 4 persons got medals in all the three categories, how many received medals in exactly two of these categories?
Given:
Number of students who got awarded medal in Dance n(A)
= 36 Students
Number of students who got awarded medal in Dramatics n(B)
= 12 Students
Number of students who got awarded medal in Music n(C)
= 18 Students
Total number of students that were there n(A ∪ B ∪ C)
= 45 Students
Number of students that got a medal in all the three categories
n(A ∩ B ∩ C)
= 4 Students
To Find:
The number of students that got medal in exactly two of these categories.
Calculating:
The number of elements belonging to exactly two of the three sets A, B, C :
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3n(A ∩ B ∩ C)
= n(A ∩ B) + n(B ∩ C) + n(A ∩ C) - 3 × 4 ----------Eq(1)
= n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(A ∩ C) + n(A ∩ B ∩ C)
Hence,
n(A ∩ B) + n(B ∩ C) + n(A ∩ C) = n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C)
From Equation(1) the required number is:
We can write it as:
= n(A) + n(B) + n(C) + n(A ∩ B ∩ C) - n(A ∪ B ∪ C) - 12
Substituting all the values known to us in this equation we get:
= 36 + 12 + 18 + 4 - 45 - 12
= 70 - 57
= 13
Therefore, 13 students received medal in exactly two of these categories.
Step-by-step explanation:
answer =7
I hope it will be helpful