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Sagot :
Let b represent the number of boys and g represent the number of girls initially in detention. We are given that there are 27 boys and 27 girls, so b = 27 and g = 27 .
Let x denote the number of students allowed to leave early. According to the problem, these x students will be chosen such that the ratio of boys to girls remaining in detention becomes 6:5.
Initially, the ratio of boys to girls is:
27/27 = 1
After x students (some boys and some girls) leave, the ratio of boys to girls remaining becomes 6:5:
27 - b'/27 - g'= 6/5
where b' and g' are the numbers of boys and girls remaining after x students leave.
From the ratio 6/5, we can set up the equation:
27 - b'/27 - g' = 6/5
Cross-multiplying gives:
5(27 - b') = 6(27 - g')
Expanding and simplifying:
135 - 5b' = 162 - 6g'
6g' - 5b' = 27
To find the least number of students x that can leave early, we need to solve this equation for integer solutions b' and g' that satisfy b' + g' = x .
Testing values, we find:
If b' = 18 and g' = 15 ,
6 times 15 - 5 times 18 = 90 - 90 = 0
This satisfies the equation 6g' - 5b' = 27 . Therefore, b' = 18 and g' = 15 are valid, and x = b' + g' = 18 + 15 = 33 .
Thus, the least number of students that can be allowed to leave early so that the ratio of boys to girls remaining in the classroom becomes 6 to 5 is boxed 33.
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