An exam has 5 questions. Of all candidates, 5% answered all 5 questions and 5% answered none. Among the remaining candidates, 25% answered only 1 question and 20% answered 4 questions. If 396 candidates answered either 2 or 3 questions, how many candidates appeared for the exam in total?

Introduction / Context:
This percentage distribution problem involves conditioning: some percentages apply to the entire group, while others apply only to the remaining candidates after removing initial categories. Careful bookkeeping converts the verbal description into algebra, yielding the total population size.

Given Data / Assumptions:

• Let total candidates = N.
• 5% answered all questions → 0.05N; 5% answered none → 0.05N.
• Remaining group = 0.90N.
• Within the remaining: 25% answered only 1 question → 0.25 * 0.90N = 0.225N.
• Within the remaining: 20% answered 4 questions → 0.20 * 0.90N = 0.18N.
• Given: those who answered 2 or 3 questions together = 396.

Concept / Approach:
First identify how many of the “remaining” candidates are already accounted for by the “only 1” and “4 questions” groups. The balance of the remaining must be exactly those who answered 2 or 3 questions. Equate this balance to 396 and solve for N.

Step-by-Step Solution:

Verification / Alternative check:
Compute counts: all = 40, none = 40, only 1 = 180, four = 144, and 2-or-3 = 396. Sum = 40 + 40 + 180 + 144 + 396 = 800, confirming the total.

Why Other Options Are Wrong:
1000, 900, 850, and 720 do not satisfy the equation 0.495N = 396 and yield inconsistent category counts.

Common Pitfalls:
Treating 25% and 20% as percentages of the total N instead of the remaining 90%, or forgetting to subtract both the “only 1” and “4 questions” groups from the remaining pool.

Final Answer:
800