Average marks for section A is 32. Average marks for section B is 60. Let the number of students in section A be \(x\), and in section B be \(y\). It is given that \(x = y - 10\).
Let the average marks of all students be \(a\), where \(32 < a < 60\). Using the weighted average formula: \[ \frac{32x + 60y}{x + y} = a \] Substituting \(x = y - 10\) yields: \[ \frac{32(y - 10) + 60y}{(y - 10) + y} = a \Rightarrow \frac{32y - 320 + 60y}{2y - 10} = a \Rightarrow \frac{92y - 320}{2y - 10} = a \]
Alternatively, using the shortcut provided: Total students = \(x + y = 2y - 10\). The average of the combined group is: \[ a = \frac{32x + 60y}{x + y} = \frac{32(y - 10) + 60y}{2y - 10} = \frac{92y - 320}{2y - 10} \] Simplifying this equation: \[ a = y - 5 \Rightarrow y = a + 5 \Rightarrow x = a - 5 \]
For \(a = 47\): The ratio of students in A to B is \( (60 - a) : (a - 32) = 13 : 15 \). The difference in proportional parts is \(15x - 13x = 10\), so \(2x = 10\), which means \(x = 5\). The number of students in Section A is \(13x = 65\).
For \(a = 56\): The ratio of students in A to B is \( (60 - a) : (a - 32) = 4 : 24 = 1 : 6 \). The difference in proportional parts is \(6x - x = 10\), so \(5x = 10\), which means \(x = 2\). The number of students in Section A is \(1x = 2\).
✅ The difference between the maximum and minimum number of students in Section A is \( 65 - 2 = \boxed{63} \).
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is