Lớp 6A của 1 trường THCS có 45 học sinh ,,cuối kì 1 kết quả học gồm 3 loại tốt,khá,đạt (ko có học sinh nào xếp loại chưa đạt).Số học sinh đạt loại tốt =1/3 số học sinh cả lớp Số học sinh xếp loại khá=8...

To solve this problem, we can use algebraic equations to represent the given information and then solve for the number of students in each category. Let's denote: - Number of excellent students as $x$ - Number of good students as $y$ - Number of satisfactory students as $z$ We are given the following information: 1. Total number of students: 45 2. Number of excellent students = $\frac{1}{3}$ * total number of students 3. Number of good students = $\frac{8}{5}$ * percentage of excellent students 4. The remaining students are satisfactory We can express these conditions in equations: 1. Total number of students: $x + y + z = 45$ 2. Number of excellent students: $x = \frac{1}{3} * (x + y + z)$ 3. Number of good students: $y = \frac{8}{5} * x$ 4. The remaining satisfactory students: $z = 45 - x - y$ Now, we can solve these equations to find the values of $x$, $y$, and $z$. First, we substitute the value of $x$ from equation (2) into equation (3): \[y = \frac{8}{5} * (\frac{1}{3} * (x + y + z))\] Solving for $y$ gives us: \[y = \frac{8}{15} * (x + y + z)\] Substitute the value of $z$ from equation (4) into the above equation: \[y = \frac{8}{15} * (x + y + (45 - x - y))\] \[y = \frac{8}{15} * (45 - x)\] Now, substitute this value back into equation (1) to solve for $x$: \[x + \frac{8}{15}*(45-x) + 45-x= 45\] \[x+24-\frac{8}{15}*x+45-x=45\] \[\frac{7}{15}*x=21\] \[x=15\] Now that we have found the value for x, we can substitute it back into our equations to find the values for y and z. From equation(2), we get: \[y=\frac{8}{5}*15=24\] And from equation(4), we get: \[z=45-15-24=6\] So, the number of excellent students is 15, good students is 24, and satisfactory students is 6. Therefore, the final answer is: Number of excellent students: 15 Number of good students: 24 Number of satisfactory students: 6