Question

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 took none of the three courses, how many students took exactly 2 of these 3 courses?​

Answer 1

Answer:

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Answer 2

Answer:

The number of students in the college that took exactly 2 of the 3 courses is found to be 70.

Step-by-step explanation:

The total number of students in the college, n(U), is given as: 100

Number of students that took Finance course, n(A): $\frac{60}{100}\times 100$ = 60

Number of students that took Marketing course, n(B): $\frac{45}{100}\times 100$ = 45

Number of students that took Operation management course, n(C): 45

Number of students that took none of the courses, n[(A∪B∪C)']: 10

Number of students that took all of the three courses, n(A∩B∩C): 10

We have to find the number of students that took exactly 2 of the 3 courses, [i.e., n(A∪B) + n(B∪C) + n(A∪C)]

The number of students that took either of the three courses is as follows:

n(A∪B∪C) = n(U) - n[(A∪B∪C)']

n(A∪B∪C) = 100 - 10

n(A∪B∪C) = 90

We know that the expression of the number of elements in the union of three sets is as follows:

n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∪B) - n(B∪C) - n(A∪C) + n(A∩B∩C)

Substituting the known information in this expression, we get:

90 = 60 + 45 + 45 - n(A∪B) - n(B∪C) - n(A∪C) + 10

Seperating the unknowns on one side of the sign of equality, we get:

n(A∪B) + n(B∪C) + n(A∪C) = 60 + 45 + 45 + 10 - 90

Simplifying it, we get:

n(A∪B) + n(B∪C) + n(A∪C) = 70

Thus, the number of students that took exactly 2 of the 3 courses is found to be 70.