Let’s denote the marks obtained by the first student as x and the second student as y.

Since one student secured 9 marks more than the other:

x = y + 9 or y = x + 9 (we’ll consider both cases)

Also, the marks of one student are 56% of the sum of their marks:

x = 0.56(x + y)

Now, substitute y = x + 9 (or x = y + 9) into the equation.

Using y = x + 9:

x = 0.56(x + (x + 9))

x = 0.56(2x + 9)

x = 1.12x + 5.04

-0.12x = 5.04

x = -5.04 / 0.12

x = 42

Now, find y:

y = x + 9

y = 42 + 9

y = 51

However, we should verify the other case (x = y + 9) as well.

Using x = y + 9:

y + 9 = 0.56(y + (y + 9))

y + 9 = 0.56(2y + 9)

y + 9 = 1.12y + 5.04

-0.12y = -3.96

y = 3.96 / 0.12

y = 33

Now, find x:

x = y + 9

x = 33 + 9

x = 42

Both cases yield the same marks for one student (x = 42).

The marks obtained by the two students are 33 and 42.