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Let's solve the problem step by step.

Let:
- \( g \) represent the number of green pens in the box.
- \( b \) represent the number of blue pens in the box.

### Given Information:
1. There are only green pens and blue pens.
2. There are three more blue pens than green pens:
   \[
   b = g + 3
   \]
3. There are more than 12 pens in total:
   \[
   g + b > 12
   \]
4. Simon takes two pens at random, and the probability that both are of the same color is \( \frac{27}{55} \).

### Probability Analysis:
The total number of ways to pick any 2 pens from the box is:
\[
\text{Total ways} = \binom{g + b}{2} = \frac{(g + b)(g + b - 1)}{2}
\]

The ways to pick 2 green pens are:
\[
\binom{g}{2} = \frac{g(g - 1)}{2}
\]

The ways to pick 2 blue pens are:
\[
\binom{b}{2} = \frac{b(b - 1)}{2}
\]

The total number of favorable outcomes (picking 2 pens of the same color) is:
\[
\text{Favorable ways} = \frac{g(g - 1)}{2} + \frac{b(b - 1)}{2}
\]

The probability that Simon picks two pens of the same color is:
\[
\frac{\frac{g(g - 1)}{2} + \frac{b(b - 1)}{2}}{\frac{(g + b)(g + b - 1)}{2}} = \frac{27}{55}
\]

### Simplifying the Equation:
Multiply both the numerator and the denominator by 2 to clear the fractions:
\[
\frac{g(g - 1) + b(b - 1)}{(g + b)(g + b - 1)} = \frac{27}{55}
\]

Substitute \( b = g + 3 \) into the equation:
\[
\frac{g(g - 1) + (g + 3)(g + 2)}{(2g + 3)(2g + 2)} = \frac{27}{55}
\]

### Solving for \( g \):
Expand and simplify:
\[
g(g - 1) + (g + 3)(g + 2) = g^2 - g + g^2 + 5g + 6 = 2g^2 + 4g + 6
\]
\[
(2g + 3)(2g + 2) = 4g^2 + 10g + 6
\]
So:
\[
\frac{2g^2 + 4g + 6}{4g^2 + 10g + 6} = \frac{27}{55}
\]

Cross-multiply:
\[
55(2g^2 + 4g + 6) = 27(4g^2 + 10g + 6)
\]
\[
110g^2 + 220g + 330 = 108g^2 + 270g + 162
\]
Simplify:
\[
2g^2 - 50g + 168 = 0
\]
Divide by 2:
\[
g^2 - 25g + 84 = 0
\]

### Factoring:
\[
(g - 21)(g - 4) = 0
\]

Thus, \( g = 21 \) or \( g = 4 \). Since the total number of pens must be more than 12, we discard \( g = 4 \).

### Solution:
The number of green pens in the box is:
\[
\boxed{21}
\]

### Verification:
If \( g = 21 \), then \( b = 21 + 3 = 24 \), and the total number of pens is \( 45 \). The total ways to pick any 2 pens are:
\[
\binom{45}{2} = \frac{45 \times 44}{2} = 990
\]

The favorable ways:
\[
\binom{21}{2} + \binom{24}{2} = \frac{21 \times 20}{2} + \frac{24 \times 23}{2} = 210 + 276 = 486
\]

The probability:
\[
\frac{486}{990} = \frac{27}{55}
\]

The solution is correct.

Would you like more details on any part of this solution or have any other questions?

### Related Questions:
1. How would the problem change if there were fewer than 12 pens in the box?
2. What would the calculation be if the probability were different, say \( \frac{1}{2} \)?
3. Can you explain how to derive the probability formulas used in this problem?
4. What would happen if the number of blue pens was not a fixed amount more than the green pens?
5. How would you approach the problem if there were three colors of pens?

### Tip:
When dealing with probability problems involving combinations, always double-check the total number of outcomes and favorable outcomes to ensure accuracy.

There are only green pens and blue pens in a box. There are three more blue pens than green pens in the box. There are more than 12 pens in the box. Simon is going to take at random two pens from the box. The probability that Simon will take two pens of the same colour is 27/55. Work out the number of green pens in the box.

Solve a probability problem involving a box with green and blue pens, using combinatorics and quadratic equations. Learn how to calculate the number of green pens when given the probability of selecting two pens of the same color. This problem involves key concepts in probability and algebra, suitable for students in Grades 9-11.

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