Question

A COLLEG AWARDED 38 MEDALS IN FOOTBALL,15 IN BASKET BALL AND 20 IN Cricket .if these medals went to a total of 58 men and only 3 men got medals in all the three sports hw many received medals in exactly two of the sports? plz answer wid explaination

Answer 1

use concept of set
n (AuBuC)=n (A)+n (B)+n (c)-n (AuB)-n(BuC)-n (CuA)+n (AnBnC)
58 = 38 + 15 + 20 - {n (AuB)+n (BuC)+n (CuA)}+3
58-(38+15+20+3)=-{n (AuB)+n (BuC)+n (CuA)}

men received in exactly two of the sports =18

Answer 2

Solution :

Let,

V = set of students getting medals in volleyball

F = set of students getting medals in football

B = set of students getting medals in basket ball

We are given the following:

n ( V ) = 38,

n ( F ) = 15

n ( B ) = 20

n ( V U F U B ) = 58

n ( V ∩ F ∩ B ) = 3

Let,

a = Number of students getting medals in volleyball and football but not the basketball

b = Number of students getting medals in football and basketball but not volleyball

c = Number of students getting medals in basketball and volleyball but not football

So,

n ( V U F U B ) is as follows :

⇒ [ 38 - ( a + c + 3 ) ] + [ 15 - ( a + b + 3 ) ] + [ 20 - ( b + c + 3 ) ] + a + b + c = 58

⇒ 67 - ( a + b + c ) = 58

⇒ a + b + c = 9

∴ Number of players who received medals in exactly two of the three sports,

⇒ a + b + c

⇒ 9

∴ Number of players who received medals in exactly two of the three sports is 9 respectively.