Question
Question: A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to...
A college awarded 38 medals in football, 15 in basketball 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, how many received medals in exactly two of the three sports?
Solution
Hint: In these types of questions remember to use the basic formula of the sets for example n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C) and solve the given question.
Complete step-by-step answer:
According to the given information
Let set A be the medals awarded in football n (A) = 38
Set B be the medals awarded in basketball n (B) = 15
Set C be the medals awarded in cricket n (C) = 20
And the total number of medals be U = 58 or the total medals awarded to men n(A∪B∪C)=58
Now let the 3 men got the medals in all the sports n(A∩B∩C)=3
Using the formula of n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C) ----- (equation 1)
Putting the values in the equation 1
⇒ n(A∩B)+n(B∩C)+n(A∩C)=38−58+15+20+3
⇒ n(A∩B)+n(B∩C)+n(A∩C)=18 ---(Equation 2)
Only for (A∩B)= n(A∩B)−n(A∩B∩C) -- (equation 3)
Only for(B∩C)=n(B∩C)−n(A∩B∩C) ---(equation 4)
Only for(A∩C)=n(A∩C)−n(A∩B∩C) ---(equation 5)
For medals received by exactly two of the three sports adding equation 3, 4 and 5
⇒ n(A∩B)+n(B∩C)+n(A∩C)−3n(A∩B∩C) --(Equation 6)
Putting the values in equation 6
⇒18 – 3(3) = 9
Therefore medals received by exactly two of the three sports are 9 medals.
Note: In these types of questions first use the given information to find the given values then use the basic formulas of sets like n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C) and then use this formula to find the values of unknown factors like n(A∩B)+n(B∩C)+n(A∩C)and then make the equation for each case of sports and then apply the addition operation in all the three equations the coming result is the answer of the question