Khối 6 của một trường có 4 lớp . Số học sinh của lớp 6A bằng 3/8 tổng số học sinh của 3 lớp còn lại . Số học sinh lớp 6B bàng 5/22 tổng số học sinh khối 6 . Số học sinh lớp 6C bằng 1/4 tổng số học sinh...

To solve this problem, we can use algebraic equations to represent the given information and then solve for the number of students in classes 6A, 6B, and 6C. Let's denote: - The total number of students in grade 6 as T - The number of students in class 6A as A - The number of students in class 6B as B - The number of students in class 6C as C From the problem statement, we have the following information: 1. The number of students in class 6A is equal to 3/8 of the total number of students in the other 3 classes: A = (3/8)(T - A - B - C) 2. The number of students in class 6B is equal to 5/22 of the total number of students in grade 6: B = (5/22)T 3. The number of students in class 6C is equal to 1/4 of the total number of students in grade 6: C = (1/4)T 4. The number of students in class 6D is given as D = 33 We need to solve for A, B, and C. First, let's rewrite equation (1) by multiplying both sides by 8 to clear the fraction: $$ Next, substitute equations (2), (3), and (4) into equation (1): $$ $$ $$ $$ Now we can solve for A: $$ $$ So, the number of students in class 6A is $$. Similarly, from equations (2) and (3), we have: The number of students in class 6B: $$ or $$ The number of students in class 6C: $$ or $$ Therefore, using algebraic equations and solving for each variable allows us to find that: The number of students in class 6A: $$ The number of students in class 6B: $$ The number of students in class 6C: $$