Job

In a certain college of the 100 students, 60% students took Finance course, 45% took Marketing course and 45 students took Operation Management course. If 10 students took all the courses and 10 students none of the three courses, how many students took exactly 2 of these 3 courses?

(Set)

Created: 2 years ago | Updated: 1 year ago
Updated: 1 year ago
Ans :

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  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

1 year ago

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