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Let's carefully analyze the problem and solve it step by step:

We are told that the marks scored by A, B, C, and D are in geometric progression (GP). Let's denote their marks as:

- \( A = a \)
- \( B = ar \)
- \( C = ar^2 \)
- \( D = ar^3 \)

where \( a \) is the first term and \( r \) is the common ratio of the GP.

### Step 1: Relationship between marks of C and sum of A and B
It is given that the marks scored by C are 30\(\frac{5}{9}\)% less than the sum of the marks scored by A and B. This can be mathematically written as:

\[
C = (A + B) \times \left(1 - \frac{30 \frac{5}{9}}{100}\right)
\]

First, convert \(30 \frac{5}{9}\)% into a decimal:
\[
30 \frac{5}{9}\% = \frac{275}{9} \% = \frac{275}{900} = 0.3055
\]

Thus, the relation becomes:
\[
ar^2 = (a + ar) \times (1 - 0.3055)
\]
\[
ar^2 = (a + ar) \times 0.6945
\]

Simplifying:
\[
ar^2 = 0.6945a(1 + r)
\]
\[
r^2 = 0.6945(1 + r)
\]

### Step 2: Solving the quadratic equation
Expand and rearrange:
\[
r^2 - 0.6945r - 0.6945 = 0
\]

We can solve this quadratic equation using the quadratic formula:
\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 1 \), \( b = -0.6945 \), and \( c = -0.6945 \).

Substitute the values:
\[
r = \frac{-(-0.6945) \pm \sqrt{(-0.6945)^2 - 4(1)(-0.6945)}}{2(1)}
\]
\[
r = \frac{0.6945 \pm \sqrt{0.4823 + 2.778}}{2}
\]
\[
r = \frac{0.6945 \pm \sqrt{3.2603}}{2}
\]
\[
r = \frac{0.6945 \pm 1.8056}{2}
\]

Thus, the two possible values for \( r \) are:
\[
r_1 = \frac{0.6945 + 1.8056}{2} = 1.25
\]
\[
r_2 = \frac{0.6945 - 1.8056}{2} = -0.5555
\]

Since the common ratio in a geometric progression should be positive, we take \( r = 1.25 \).

### Step 3: Percentage comparison of marks between D and B
Now, we want to find what percentage the marks scored by D are more than the marks scored by B. The marks scored by D are \( ar^3 \) and the marks scored by B are \( ar \). The percentage increase is given by:

\[
\text{Percentage Increase} = \frac{D - B}{B} \times 100 = \frac{ar^3 - ar}{ar} \times 100 = \frac{r^3 - r}{r} \times 100 = (r^2 - 1) \times 100
\]

Substitute \( r = 1.25 \):
\[
\text{Percentage Increase} = (1.25^2 - 1) \times 100 = (1.5625 - 1) \times 100 = 0.5625 \times 100 = 56.25\%
\]

Thus, the marks scored by D are **56.25%** more than the marks scored by B.

### Conclusion:
The answer is \( \boxed{56.25\%} \).

Would you like more details on any part of this explanation?

Here are some related questions:
1. How is a geometric progression defined and used in solving problems?
2. Can you explain more about the quadratic formula and its application here?
3. How does the concept of percentage increase work in real-life problems?
4. What are the properties of geometric progressions that help in calculating ratios between terms?
5. How do you handle decimal percentages like \(30 \frac{5}{9}\%\) in problems?

**Tip:** In geometric progression, the ratio between any two consecutive terms remains constant, which makes it easy to model growth or decay situations.

The marks scored in Mathematics by A, B, C and D form a geometric progression in that order. If the marks scored by C were 30 5/9% less than the sum of the marks scored by A and B, then the marks scored by D were what percent more than the marks scored by B?

Learn how to determine marks in a geometric progression with a step-by-step solution, where marks scored by D are compared to B after establishing a quadratic relation. This advanced problem involves geometric sequences, quadratic equations, and percentage calculations. Suitable for high school students.

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