Let the marks of the two students be $x$ and $y$, where $x$ is the marks of the first student and $y$ is the marks of the second student.

We know two things from the problem:

1. The first student achieved 9 marks more than the second student, so: $x = y + 9$
2. The marks of the first student were 56% of the sum of both students' marks: $x = 0.56(x + y)$

Step 1: Substituting the first equation into the second equation

Substitute $x = y + 9$ into $x = 0.56(x + y)$: $y + 9 = 0.56((y + 9) + y)$ $y + 9 = 0.56(2y + 9)$ $y + 9 = 1.12y + 5.04$

Step 2: Solving for $y$

Subtract $1.12y$ from both sides: $y - 1.12y + 9 = 5.04$ $-0.12y + 9 = 5.04$ Subtract 9 from both sides: $-0.12y = 5.04 - 9$ $-0.12y = -3.96$

Now, divide both sides by $-0.12$: $y = \frac{-3.96}{-0.12}$ $y = 33$

Step 3: Finding $x$

Now, use $x = y + 9$: $x = 33 + 9 = 42$

Final Answer:

The marks of the first student are $42$, and the marks of the second student are $33$.

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Do you want further details on the process or have any questions?

Here are 5 related questions:

1. What if the percentage had been 60% instead of 56%? How would that affect the results?
2. What is the sum of both students' marks in this case?
3. How would you set up the equations if the first student had 12 more marks than the second?
4. If the marks of the first student were a different percentage of the sum, say 50%, what would the equations look like?
5. Can you solve the problem using a different approach, such as graphically?

Tip: When solving word problems, try to identify relationships between variables early on and translate them into equations.