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Question:

In the final semester, an engineering college offers three elective courses and one mandatory course. A student has to register for exactly three courses: two electives and the mandatory course. The registration in three of the four courses is: 45, 55 and 70. What will be the number of students in the elective with the lowest registration?

Explanation:

There are two cases to be considered.

Since there is only one mandatory course, all students will register for this course.

Since each student registers for exactly 3 courses, it means that sum of registrations for all four courses will be = 3 × (number of students).

Case 1: The mandatory course is one of the courses with registrations 45, 55 or 70.

The only possibility is that the mandatory course has 70 registrations, i.e., total number of students is 70.

∴ The total number of students = 70 and the number of student-course combinations = 3 × 70 = 210

∴ 70 + 55 + 45 + X = 210

∴ X = 40

Therefore, the minimum registrations for a course = 40.

Case 2: The mandatory course is not one with registrations 45, 55 or 70.

If the total number of students is X, the mandatory course will have X registrations.

Now, X will definitely be greater than or equal to 70.

∴ Minimum number of registrations in this case = 45

Hence, option (e).