In semester I, the number of good students in class 6D was equal to \(\dfrac{2}{7}\) of the remaining students. Into semester II, the number of good students increased by 8 students (the number of students in the class was unchaged), so the number of good students became \(\dfrac{2}{3}\) of the rest. How many students were there in class 6D?
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Lớp tôi có 40 học sinh. Số học sinh khá nhiều hơn số học sinh giỏi là 4 bạn nhưng ít hơn số học sinh trung bình là 8 bạn. Tìm số học sinh mỗi loại. ( Không có học sinh yếu ).
I call the number of students didn't take part in Mathemas Competition in class A & B are a & b respectively
(with condition: a & b \(\in\)N*)
,From the theme, we have: \(\frac{1}{3}b+a=\frac{1}{5}a+b\)(Because rhe number of students in 2 class is the same)
\(\Leftrightarrow\frac{b}{3}-b+a-\frac{a}{5}=0\Leftrightarrow\left(-\frac{2b}{3}\right)+\frac{4a}{5}=0\)
\(\Leftrightarrow\frac{4a}{5}-\frac{2b}{3}=0\Leftrightarrow\frac{12a-10b}{15}=0\Leftrightarrow12a=10b\)
\(\Rightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\)
So the ratio of the number of student didn't take part of Class A & B is \(\frac{5}{6}.\)
The number of students of class 5B is:
8 : (9 - 7) × 7 = 28 (students)
The number of students of class 5A is:
8 : (9 - 7) × 9 = 36 (students)