Question:medium

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

Updated On: Jan 15, 2026
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Correct Answer: 63

Solution and Explanation

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: \[ a = y - 5 \Rightarrow y = a + 5 \Rightarrow x = a - 5 \]

Extreme Cases:

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 \( 65 - 2 = \boxed{63} \).

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