A total of 100 members are writing exams. Of these, 48 members are writing the first exam, 45 members are writing the second exam and 38 members are writing the third exam. Exactly 5 members are writing all three exams. Assuming every member writes at least one exam, how many members are writing exactly two exams?

Introduction / Context:
This is a set theory and counting problem that uses the principle of inclusion and exclusion. We have three exams and counts of how many members are taking each exam, plus the number who take all three. The question asks how many members take exactly two exams, under the natural assumption that everyone takes at least one exam. Such questions are common in quantitative and logical reasoning sections, and they test your ability to work with overlapping sets.

Given Data / Assumptions:

• Total number of members is 100.
• |A| = 48 members write the first exam.
• |B| = 45 members write the second exam.
• |C| = 38 members write the third exam.
• |A ∩ B ∩ C| = 5 members write all three exams.
• Assume every member writes at least one exam (so no one is outside A ∪ B ∪ C).
• We must find the number of members who write exactly two exams.

Concept / Approach:
Let x be the number of members who write exactly two exams and y be the number of members who write exactly one exam. The members who write all three exams are already given as 5. Using set theory, the sum of the individual set sizes |A| + |B| + |C| counts members as many times as the number of exams they write. That means it equals 1 * (members in exactly one exam) + 2 * (members in exactly two exams) + 3 * (members in all three exams). We also know that the total number of members is the sum of those in exactly one exam, exactly two exams, and exactly three exams. Using these facts, we can form equations and solve for x, the number writing exactly two exams.

Step-by-Step Solution:
Step 1: Let y be the number of members who write exactly one exam and x be the number who write exactly two exams. We are given that 5 members write all three exams. Step 2: Compute the sum of the set sizes: |A| + |B| + |C| = 48 + 45 + 38 = 131. Step 3: Count by membership. Each member in exactly one exam contributes 1 to this total, each in exactly two exams contributes 2, and each in all three contributes 3. So we have 131 = y + 2x + 3 * 5. Step 4: Simplify: 3 * 5 = 15, so y + 2x + 15 = 131, which gives y + 2x = 116. Call this Equation (1). Step 5: Since every member writes at least one exam, the total 100 members equals y (exactly one) + x (exactly two) + 5 (all three). So 100 = y + x + 5, or y + x = 95. Call this Equation (2). Step 6: Subtract Equation (2) from Equation (1): (y + 2x) − (y + x) = 116 − 95, giving x = 21.

Verification / Alternative check:
From x = 21, use Equation (2) to find y: y + 21 = 95, so y = 74. Now check the count into the sum of set sizes: y + 2x + 3 * 5 = 74 + 2 * 21 + 15 = 74 + 42 + 15 = 131, which matches |A| + |B| + |C|. Also check total members: 74 (exactly one exam) + 21 (exactly two) + 5 (all three) = 100, matching the total number of members. Everything is consistent, so x = 21 is correct.

Why Other Options Are Wrong:
Option 26, 23, 27 and 24: Substituting any of these values for x into the equations y + 2x = 116 and y + x = 95 will make the equations inconsistent, so they cannot represent the actual count of members writing exactly two exams.

Common Pitfalls:
One common mistake is to forget the assumption that every member writes at least one exam, which is needed to set the total members equal to y + x + 5. Another error is to misinterpret “writing 2 exams” as “at least two exams” instead of “exactly two exams”, which changes the counting. Always define variables clearly for “exactly one”, “exactly two” and “exactly three” membership, and use the inclusion–exclusion counting carefully to avoid double counting or undercounting.

Final Answer:
The number of members who are writing exactly two exams is 21.