Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Connect with professionals on our platform to receive accurate answers to your questions quickly and efficiently. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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.
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
