There are 391 boys and 323 girls are present in the school. Therefore, Total 391 + 323 = 714 students are present in the school. Now, 714 students are divided into the largest possible equal classes and each class has an equal number of boys and girls. 714 students can be divided into the largest possible equal classes only when each class consists of a minimum number of boys and girls. Therefore, the minimum number which completely divides 391 is 17. and the minimum number which completely divides 323 is also 17. but, we can't put 17 boys and 17 girls in each class because the number of boys are more than the number of girls. Now, try the combination of boys and girls from the option. 1) 23 girls’ and 19 boys’ This combination is not valid because we can't make a class of 19 boys for a total of 391 boys. 391 is not completely divisible by 19. 2) 23 boys’ and 19 girls’ We can make a class of 23 boys and 19 girls. As 391 is divisible by 23 and 323 is divisible by 19. ⇒ 23 + 19 = 42 students in each class. Now, we have to check 714 is divisible by 42 or not. ⇒ 714 ÷ 42 = 17 classes. Therefore, we can make a class of 23 boys and 19 girls and there will be a total of 17 classes in the school. 3) 17 boys’ and 23 girls’ This combination is also not valid because the total number of boys are more than the total number of girls is already given. 4) 23 boys’ and 17 girls’ We can make a class of 23 boys and 17 girls. As 391 is divisible by 23 and 323 is divisible by 17. ⇒ 23 + 17 = 40 students in each class. Now, we have to check 714 is divisible by 40 or not. ⇒ 714 ÷ 40 = 17.85 714 is not completely divisible by 40, hence this combination is also not valid. Hence, the correct answer is 23 boys’ classes, 19 girls’ classes.