Question 120

Logical Reasoning Venn Diagram Easy

A university is offering elective course in Mathematics, Economics and Sociology. Each of its 100 undergraduate students has to opt for at least one of these electives. Course enrollment data showed that 47 students enrolled for Mathematics, 47 students enrolled for Economics and 57 students enrolled for Sociology. If 7 students enrolled for all three courses, how many students enrolled for exactly one course?

(A) 58
(B) 56
(C) Cannot be determined
(D) 60
✓ Correct Answer: Option B Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Solving with Set Theory Formulas

This problem can be solved by using standard formulas from set theory that relate the total number of students to the enrollments in individual and overlapping course categories.

1. Define Variables and List Given Information

  • Total Students = 100
  • Students in Mathematics, n(M) = 47
  • Students in Economics, n(E) = 47
  • Students in Sociology, n(S) = 57
  • Students in all three courses, n(M ∩ E ∩ S) = 7
  • Let Y be the number of students enrolled in exactly one course (this is what we need to find).
  • Let X be the number of students enrolled in exactly two courses.

2. Formulate Equations

There are two key relationships we can use:

  1. The total number of students is the sum of those taking exactly one, exactly two, or all three courses.
    Total = (Exactly 1) + (Exactly 2) + (Exactly 3)
    $100 = Y + X + 7$
    This simplifies to: $Y + X = 93$ (Equation 1)
  2. The sum of the enrollments for each course counts students in multiple courses multiple times. The relationship is:
    n(M) + n(E) + n(S) = (Exactly 1) + 2 × (Exactly 2) + 3 × (Exactly 3)
    $47 + 47 + 57 = Y + 2X + (3 \times 7)$
    $151 = Y + 2X + 21$
    This simplifies to: $Y + 2X = 130$ (Equation 2)

3. Solve the System of Equations

Now we have a system of two linear equations:

  1. $Y + 2X = 130$
  2. $Y + X = 93$

By subtracting Equation 2 from Equation 1, we can solve for X:

(Y + 2X) - (Y + X) = 130 - 93

X = 37

Now, substitute X = 37 back into Equation 2 to find Y:

Y + 37 = 93

Y = 93 - 37

Y = 56

Conclusion

The number of students enrolled for exactly one course is 56.

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