Two students appeared at an examination. One of them secured 6% more marks…

Let the smaller mark be x and the larger mark be y.

Given that the larger mark is 56% of the sum of their marks:

y = 0.56(x + y)

Rearrange: 0.44y = 0.56x, so

y/x = 0.56/0.44 = 56/44 = 14/11.

Thus the marks are in the ratio 14:11. We look for a pair of whole-number marks with this ratio.

Choose k so that the marks are 14k and 11k. For k = 3, the marks are 42 and 33.

• Check: 42/(42 + 33) = 42/75 = 0.56, so the 56% condition is satisfied.

Answer: 42 and 33.

Note: The statement '6% more marks' in the question seems inconsistent with 42 and 33 (because 42 is 27.27% more than 33). The solution above is based on the 56% condition; the intended wording in the question may contain a typographical error.