Class ranks and pass–fail count – Malay is 13th from the front and 17th from the end in his class. Among students who passed, he is 8th from the front and 13th from the end. How many students failed?

Introduction / Context:
This question combines whole-class rank with rank within a subset (the pass list). We convert “kth from front” and “mth from end” into totals using the standard identity: total = front + end − 1. Then we subtract to find failures.

Given Data / Assumptions:

• Overall class ranks: 13th from front, 17th from end.
• Pass-list ranks: 8th from front, 13th from end.
• All ranks are consistent and unique.

Concept / Approach:
For any line of N people, if a person is a from the front and b from the back, then N = a + b − 1. Apply this to the full class to get class size T, and to the pass list to get P. Then failures F = T − P.

Step-by-Step Solution:

Verification / Alternative check:
Visualize the full class of 29 with pass marks on 20 positions; Malay is in both counts consistently. The arithmetic identity ensures correctness.

Why Other Options Are Wrong:

• 7, 8: Underestimate failures; contradict arithmetic.
• Cannot be determined: The totals are uniquely determined by the given ranks.
• 10: Overestimates failures; also contradicts arithmetic.

Common Pitfalls:
Forgetting to subtract 1 in the identity; mixing class and pass ranks; or assuming overlapping ambiguity. The identity T = front + end − 1 resolves the totals directly.

Final Answer:
9