Here, to solve the problem, we need to use some laws from the primary concept of sets.
Let, n(F) = 60; n(M) = 45 n(Om)=45; n(T) = 100
All courses taken, n(F M Om )=10
None taken, n(F M Om )^ =10
We know, z(F M Om )=n(T)-n(F M Om)^
= n(F M Om)= 100 - 10 = 90
= So, we can write, n (F M Om)
= n(F)+n(M)+n(Om)-n(FM)-n(MOm)-n(OmF)+n(FMOm)
= 90=60+45+45-{n(FM)+n(MOm)+n(OmF}+10
=90=160-ln(F M)+n(M Om )+n(Om F)
n(F M)+n(M Om )+n(Om F)=70
The number of students who took 2 of these 3 courses is
=70-3(F M Om)= 70 - (3 10) = 70 - 30 = 40
Related Question
View AllLet, x number of students have more than 5 less than 9 marbles
Now, required condition is 5 10 students have less than 6 marbles: means at least 10 students have 5 marbles 4 students have more than 8 marbles: means 4 students have at least 9 marbles. Therefore, number of students have more 5 less than 9 marbles=25-(10+4)= 11 ans.
Ler, total number of travelers
Number of traveler who had been to Spain, n(A) = 55
Number of traveler who had been to France, n(B) = 53
Number of traveler who had been to Germany, n(C) =79
Number of traveler who had been to Spain and France, (A∩B) = 18
Number of traveler who had been to Spain and Germany, n (A ∩ C)=17
Number of traveler who had been to France and Germany, n( B ∩ C )=25
Number of traveler who had been to all three countries, n (A ∩ B ∩ C)=10
We know, n(∪)= n(A)+n(B)+n(C)-n(A ∩ B)-n(B ∩ C)-n(A ∩ C)+n(A ∩ B ∩ C) = 55 + 53 + 79 - 18 - 17 - 25 + 10 = 197-60 = 137 ans.
Repeat: Rupali Bank Ltd. (Officer - Cash) 2018
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