Bunuel
A class of boys and girls has more girls than boys. The total number of students is 50, and the number of boys is at least 30% of the number of girls.
On a particular day, 20 students were absent, resulting in the number of boys present being greater than the number of girls present.
Select for
Maximum the maximum possible number of girls that could have been in the original class of 50 students, and select for
Minimum the minimum possible number of girls who could have been absent on that day. These two values could correspond to separate scenarios and are not necessarily consistent with each other. Make only two selections, one in each column.

Official Solution: Given that the total number of students is 50, and the number of boys is at least 30% of the number of girls, we have:
G + B = 50 and B ≥ 0.3G
Substituting B = 50 - G into the second equation:
50 - G ≥ 0.3G
Solving this inequality gives:
G ≤ 38.46
Since the number of students must be an integer and there are more girls than boys, the number of girls can range from 26 to 38, inclusive. Therefore, the maximum possible number of girls in the class is 38.
To minimize the number of girls who could have been absent on the day when the number of boys exceeded the number of girls, we should start with the smallest possible number of girls, which is 26. Since 20 students were absent, 30 students were present in the class that day. To achieve the smallest possible number of absent girls, we need to distribute the 30 students as evenly as possible while ensuring that the number of boys exceeds the number of girls. This approach maximizes the number of girls present, minimizing the difference between 26 (the initial number of girls) and the number of girls present. The closest distribution that meets this condition is 16 boys and 14 girls, meaning the minimum number of girls who could have been absent is 26 - 14 = 12.
Correct answer: Maximum
"38"Minimum
"12"Attachment:
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