Two students appeared for an examination. One of them scored 9 marks more than the other, and his marks were 56 percent of the sum of their marks. What are the marks obtained by the two students?

Introduction / Context:
This question is similar to other two variable word problems that connect differences in marks with a percentage of their combined total. It reinforces algebraic thinking, where real life statements about marks are converted into equations that can be solved systematically.

Given Data / Assumptions:
Let marks of the first student be A. Let marks of the second student be B. A is 9 marks more than B, so A - B = 9. A is 56 percent of their total, so A = 56 percent of (A + B). Marks are assumed to be whole numbers.

Concept / Approach:
The percentage relation is converted into a linear equation involving A and B. Together with the simple difference equation A - B = 9, we obtain a system of two equations. Solving these equations gives the exact marks for both students. This method is typical in quantitative aptitude and helps avoid guesswork.

Step-by-Step Solution:
From the first statement: A - B = 9. From the second statement: A = 56/100 * (A + B) = 0.56 * (A + B). Write A = 0.56A + 0.56B. Bring terms together: A - 0.56A = 0.56B. This gives 0.44A = 0.56B. So A / B = 0.56 / 0.44 = 56 / 44 = 14 / 11. Thus A : B = 14 : 11. Difference in ratio parts = 14 - 11 = 3 corresponds to actual difference 9. So 3 parts = 9 marks, hence 1 part = 3 marks. Therefore A = 14 * 3 = 42 and B = 11 * 3 = 33.

Verification / Alternative check:
Check the conditions. Difference: 42 minus 33 equals 9, matching the first condition. Sum: 42 plus 33 equals 75. Now 56 percent of 75 is 0.56 * 75 = 42. Both statements are satisfied exactly, confirming that 42 and 33 are the correct marks.

Why Other Options Are Wrong:
For 40 and 31, the difference is 9 but 56 percent of 71 is 39.76, not 40. For 72 and 63, the difference is 9 but 56 percent of 135 is 75.6. For 68 and 59, 56 percent of 127 is 71.12. For 56 and 47, 56 percent of 103 is 57.68. None of these satisfy both the difference and percentage conditions. Only 42 and 33 fit both.

Common Pitfalls:
Students sometimes misread the phrase 56 percent of the sum and instead apply the percentage to the second student only. Others may try random substitution from options. Writing the equations clearly and solving through ratios is faster and more reliable in exam conditions.

Final Answer:
The two students obtained 42 and 33 marks respectively.