Question
Question: In a beauty contest, half the number of experts voted for Miss A and two thirds voted for Miss B, 10...
In a beauty contest, half the number of experts voted for Miss A and two thirds voted for Miss B, 10 voted for both and 6 did not vote for either. Then how many experts were there in all?
(A). 20
(B). 24
(C). 22
(D). 26
Solution
To find the total number of experts in the contest, we can form an equation from the given information and then we can assign N as the total number of experts. This gives an equation in one variable which when solved gets the total number of experts in the beauty contest.
Complete step-by-step answer :
Given half the number of experts voted for Miss A
And two thirds voted for Miss B
And 10 voted for both Miss A and Miss B.
Let the total number of experts be N.
E is the set of experts who voted for Miss A.
F is the set of experts who voted for Miss B.
Since 6 did not vote for either, n(E∪F)=N-6.
So from the above information given
Half the number of experts voted for Miss A,n(E)=2N
Two thirds number of experts voted for Miss B, n(F)=32N
And 10 voted for both Miss A and Miss B that is, n(E∩F)=10
The intersection of two sets A and B, denoted by A∩B, is the set containing all elements of A that also belong to B (or equivalently, all the elements of B that also belongs to A).
So from the formula of sets, If there are two sets P and Q, n(P∩Q)represents the number of elements present in one of the sets P and Q. n(P∪Q)represents the number of elements present in both the sets P & Q.
n(P∪Q)=n(P) + n(Q) - n(P∩Q)
Therefore, from the above values
⇒n(E∪F)=n(E) + n(F) - n(E∩F)
So, N - 6 = 2N+32N - 10
⇒N - 2N−32N = - 10 + 6
⇒66N - 3N - 4N = - 4
Solving the above equation gives6N=4⇒N = 24
Therefore, total number of experts = 24.
Hence option B is the correct answer.
Note : If we have three sets P, Q, R then to solve we haven(P∪Q∪R)=n(P) + n(Q) + n(R)− n(P∩Q)− n(Q∩R)− n(R∩P)+n(P∩Q∩R). Apart from these union and intersection of sets, we have Complement of sets and Cartesian product of sets.