Question

In an examination,the average marks of students in sections A and B are 32 and 60,respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer,then the difference between the maximum and minimum possible number of students in section A is

Answer 1

Correct Answer: 63

Average marks of section A: 32
Average marks of section B: 60
Let the number of students in A = \(x\), and in B = \(y\)
Given: \(x = y - 10\)

Let the average marks of all students = \(a\), where \(32 < a < 60\).
Using the formula for weighted average: \[ \frac{32x + 60y}{x + y} = a \] Substituting \(x = y - 10\), we get: \[ \frac{32(y - 10) + 60y}{(y - 10) + y} = a \Rightarrow \frac{32y - 320 + 60y}{2y - 10} = a \Rightarrow \frac{92y - 320}{2y - 10} = a \]

Alternatively, using the shortcut given in the problem:
\[ x = y - 10 \Rightarrow \text{Total students} = x + y = 2y - 10 \] The average of the combined group: \[ a = \frac{32x + 60y}{x + y} = \frac{32(y - 10) + 60y}{2y - 10} = \frac{92y - 320}{2y - 10} \] Let’s simplify: \[ a = y - 5 \Rightarrow y = a + 5 \Rightarrow x = a - 5 \]

Extreme Cases:

When \(a = 47\):
Ratio of A to B = \( (60 - a) : (a - 32) = 13 : 15 \)
Difference = \(15x - 13x = 10 \Rightarrow 2x = 10 \Rightarrow x = 5\)
Students in Section A = \(13x = 65\)

When \(a = 56\):
Ratio of A to B = \( (60 - a) : (a - 32) = 4 : 24 = 1 : 6 \)
Difference = \(6x - x = 10 \Rightarrow 5x = 10 \Rightarrow x = 2\)
Students in Section A = \(1x = 2\)

✅ Difference between maximum and minimum number of students in Section A: \[ 65 - 2 = \boxed{63} \]