A, B, C, D and E appeared in the same examination. A gets more marks than B but fewer marks than C. D gets fewer marks than E but more marks than A. If C gets fewer marks than D, then who among A, B, C, D and E obtains the highest marks?

Introduction / Context:
This comparison problem involves ranking five students by their marks using given inequalities. Each statement compares two or three students, and you must combine these inequalities to obtain a complete order from lowest to highest marks. Once that order is known, identifying the highest scorer is straightforward. These types of questions develop comfort with inequality chaining and relational reasoning.

Given Data / Assumptions:

• A scores more marks than B but fewer marks than C, so B < A < C.
• D scores fewer marks than E but more marks than A, so A < D < E.
• We are additionally told that C scores fewer marks than D, so C < D.
• We assume all marks are different and can be arranged in a strict order.

Concept / Approach:
The idea is to combine the inequalities step by step. Starting from B < A < C, and A < D < E, we can place A in a chain where B is below and C is above. Then we insert D and E accordingly. The extra condition C < D tells us how C and D compare, allowing us to fit all five students into a single sequence. The person at the top of this sequence has the highest marks.

Step-by-Step Solution:
From “A gets more marks than B but fewer than C”, we write B < A < C. From “D gets fewer marks than E but more than A”, we write A < D < E. Combine these two chains initially as B < A < C and A < D < E. At this stage, the relative ordering of C and D is not fixed. We are then told that “C gets fewer marks than D”, which is C < D. Now we can merge everything: B is below A, A is below C, C is below D, and D is below E. Thus the full order from lowest to highest is B < A < C < D < E. From this chain, E is clearly at the top, so E obtains the highest marks.

Verification / Alternative check:
We can verify by assigning sample numerical scores that respect the inequalities. For example, take B = 10, A = 20, C = 30. Then choose D = 40 and E = 50. These values satisfy B < A < C, A < D < E, and C < D. With these values, E clearly has the highest marks. Any other choice of numbers respecting the same inequality structure still places E at the top, confirming that E must be the highest scorer.

Why Other Options Are Wrong:
C cannot be the highest because C is explicitly less than D, and D is less than E. D cannot be the highest because D is still less than E. A is less than C and D and therefore cannot be highest. B is even lower, as B is less than A. Only E stands higher than everyone else in the combined chain.

Common Pitfalls:
A frequent error is to overlook the additional condition C < D and conclude that C is highest based only on the first comparison B < A < C. Another pitfall is to mix up “less than” and “more than” when translating the verbal statements into inequalities. Writing each inequality separately and then merging them carefully helps avoid confusion.

Final Answer:
The student who obtains the highest marks among A, B, C, D and E is E.