Two students appeared for an examination. One of them scored 21 marks more than the other, and his marks were 80 percent of the sum of both their marks. What are the marks obtained by the two students?

Introduction / Context:
This algebra problem involves forming and solving simultaneous equations from a word description. The relationship between the two students scores includes both a difference (21 marks) and a percentage condition (80 percent of their sum). Translating this correctly into equations is crucial in many exam situations and real problems involving comparisons and percentages.

Given Data / Assumptions:

• Let the higher mark be x and the lower mark be y.
• One student scored 21 marks more than the other, so x = y + 21.
• The higher score x is 80 percent of the sum of both scores.
• So x = 0.8(x + y).

Concept / Approach:
We use algebraic substitution. Starting from the 80 percent relation, we derive an equation in x and y. Then we combine this with the simple difference relation x = y + 21. Solving the system gives unique values for both x and y, which can be checked against the options.

Step-by-Step Solution:
Step 1: From x = 0.8(x + y), expand the right side to get x = 0.8x + 0.8y.Step 2: Rearrange: x - 0.8x = 0.8y, so 0.2x = 0.8y.Step 3: Divide both sides by 0.2 to get x = 4y.Step 4: From the difference relation, x = y + 21.Step 5: Substitute x = 4y into x = y + 21 to get 4y = y + 21.Step 6: Solve 4y - y = 21, giving 3y = 21, so y = 7 and x = 28.

Verification / Alternative check:
Check the conditions. The difference is 28 - 7 = 21, which agrees with the problem statement. The sum is 28 + 7 = 35, and 80 percent of 35 is 0.8 * 35 = 28, which equals the higher score. Both conditions are satisfied, confirming the solution is consistent.

Why Other Options Are Wrong:
Pairs like 88 and 67 or 98 and 77 have a difference of 21 but do not satisfy the 80 percent condition when checking x = 0.8(x + y). The pair 89 and 68 does not even maintain the stated difference exactly. The pair 49 and 28 has a difference of 21 but would not make the higher score equal to 80 percent of the sum of the two scores.

Common Pitfalls:

• Misreading the 80 percent condition as the higher score being 80 percent of the lower score.
• Forgetting to use both relationships and trying to guess from one condition only.
• Arithmetic errors in handling decimal coefficients like 0.8 and 0.2.

Final Answer:
The two students scored 28 and 7 marks respectively.