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Hi there,

Great question! Let's set up the constraints and work through both parts.

Define: G = total girls, B = total boys. We know:
- G + B = 50
- G > B, so G >= 26 (since students are whole numbers)
- 20 absent, 30 present
- Among those present, boys > girls

Let g = girls absent. Then boys absent = 20 - g.
- Girls present = G - g
- Boys present = (50 - G) - (20 - g) = 30 - G + g

Key Insight: The key constraint (more boys than girls present) gives us:
30 - G + g > G - g
30 + 2g > 2G
g > G - 15

Part 1: Maximum (max girls in the class):
We need at least one valid value of g to exist. Since g <= 20 (only 20 absent total), we need G - 15 < 20, which means G < 35, so G <= 34.

Check G = 34: g must be > 34 - 15 = 19, so g = 20. That means all 20 absentees are girls. Girls present = 34 - 20 = 14, boys present = 16. Indeed 16 > 14. Works!

Check G = 35: g must be > 20, but g can't exceed 20 (only 20 absent). Impossible!

Therefore, maximum girls in the class = 34.

Part 2: Minimum (min girls absent):
We need g > G - 15, meaning g >= G - 14 (since g must be a whole number). To make g as small as possible, we minimize G. The smallest G can be is 26.

At G = 26: g >= 26 - 14 = 12.

Check g = 12: Girls present = 26 - 12 = 14, boys present = 24 - 8 = 16. Indeed 16 > 14. Works!

Check g = 11: Girls present = 15, boys present = 15. Not more boys than girls. Fails!

Therefore, minimum girls absent = 12.

The correct pair: 34 for Maximum, 12 for Minimum.

Answer: 4A (34), 3B (12)
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