A student took 5 papers in an exam where the full marks were same for each…

Concept: Quantities that are known only through a ratio a : b : c : … are written as ak, bk, ck, … with one common unit k. A stated aggregate percentage then anchors that unit to the real maximum, because total obtained = (aggregate %) × (total maximum). Once the maximum of a single item is known, that item crosses a given percentage threshold exactly when its own share exceeds that percentage of the single-item maximum, so the count is decided by comparing every ratio share against one threshold value.

Application to this question

1. Write the five scores as 6k, 7k, 8k, 9k and 10k. Their sum is 6k + 7k + 8k + 9k + 10k = 40k.

2. Let the full marks of one paper be F. All five papers carry the same full marks, so the total maximum is 5F.

3. The aggregate is 60% of the total, so 40k = (60/100) × 5F = 3F.

4. Solving for the full marks of one paper: F = 40k/3.

5. The 50% line for a single paper is F/2 = 20k/3 ≈ 6.67k.

6. Compare every share with 6.67k: 6k lies below it, while 7k, 8k, 9k and 10k each lie above it.

7. Four of the five shares exceed the 50% line, so the required count is 4.

Cross-check with concrete numbers

Take k = 3. The scores become 18, 21, 24, 27 and 30, summing to 120. Then F = 40 × 3 / 3 = 40, the total maximum is 5 × 40 = 200, and 120/200 = 60%, which matches the given aggregate. Each paper is now scored out of 40.

Result: the student scored more than 50% in 4 of the 5 papers.