In an examination, a student has to secure 36% marks to pass. He scores 53 marks but still fails by 37 marks. What are the maximum total marks of the examination?

Introduction / Context:
This problem again uses the idea of passing percentage and total marks. It requires converting the information about how many marks the student falls short by into an equation involving the total marks of the examination.

Given Data / Assumptions:

• Passing percentage = 36 percent.
• Marks obtained by the student = 53.
• The student fails by 37 marks.
• Maximum marks of the examination = unknown (let it be M).
• Passing marks are 36 percent of M.

Concept / Approach:
If a student fails by a certain number of marks, then: Passing marks = Marks obtained + Shortfall. These passing marks are also equal to the required percentage of the maximum marks: Passing marks = 36/100 * M. Thus, we set up an equation: Marks obtained + Shortfall = 0.36 * M. We then solve for M.

Step-by-Step Solution:
Step 1: Shortfall = 37 marks. Step 2: Passing marks = 53 + 37 = 90 marks. Step 3: Passing marks also equal 36 percent of M, so 0.36 * M = 90. Step 4: Solve for M: M = 90 / 0.36. Step 5: Compute 90 / 0.36 = 250. Step 6: Therefore, maximum marks M = 250 marks.

Verification / Alternative check:
Check with M = 250. Passing marks = 36 percent of 250 = 0.36 * 250 = 90. The student has 53 marks, so the shortfall is 90 - 53 = 37 marks, which matches the problem statement. Hence, the solution is consistent.

Why Other Options Are Wrong:
275 marks: 36 percent of 275 is 99, which does not give a shortfall of 37 when compared with 53. 300 marks: 36 percent of 300 is 108, making the shortfall 55 marks, not 37. 325 marks: 36 percent of 325 is 117, again inconsistent with the given shortfall. 225 marks: 36 percent of 225 is 81, which does not match the required difference.

Common Pitfalls:
Some students mistakenly use 36 percent of marks obtained instead of 36 percent of total marks. Others misinterpret "fails by 37 marks" as 37 percent rather than a difference of 37 marks. Always write passing marks both as a sum (obtained plus shortfall) and as a percentage of total, then solve for the total marks from the resulting equation.

Final Answer:
The maximum marks of the examination are 250 marks.