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

Let's define our variables clearly:

Let $x$ = number of students in section A

Then $(x + 10)$ = number of students in section B (since A has 10 less students than B)

We're told:

Average marks in section A = 32

Average marks in section B = 60

Combined average of both sections = some integer value

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When we combine two groups with different averages, we need to use a weighted average formula. This is because each group contributes to the total based on how many students it has.

Weighted Average Formula:

$\text{Combined Average} = \frac{\text{(Average of A) × (Students in A) + (Average of B) × (Students in B)}}{\text{Total Students}}$

Substituting our values:

$\text{Combined Average} = \frac{32x + 60(x + 10)}{x + (x + 10)}$

$= \frac{32x + 60x + 600}{2x + 10}$

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$\frac{32x + 60x + 600}{2x + 10} = \frac{92x + 600}{2x + 10}$

We can factor out 2 from both numerator and denominator:

$\frac{92x + 600}{2x + 10} = \frac{2(46x + 300)}{2(x + 5)} = \frac{46x + 300}{x + 5}$

Now we rewrite the numerator in terms of $(x + 5)$:

$46x + 300 = 46(x + 5) + 300 - 46 \times 5

= 46(x + 5) + 300 - 230

= 46(x + 5) + 70$

So our expression becomes:

$\frac{46(x + 5) + 70}{x + 5} = \frac{46(x + 5)}{x + 5} + \frac{70}{x + 5} = 46 + \frac{70}{x + 5}$

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For the combined average to be an integer, $46 + \frac{70}{x + 5}$ must be an integer.

Since 46 is already an integer, we need $\frac{70}{x + 5}$ to be an integer.

This means $(x + 5)$ must be a divisor of 70.

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$70 = 2 \times 5 \times 7$

The factors of 70 are: 1, 2, 5, 7, 10, 14, 35, 70

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Since $x + 5$ can be any of these factors:

If $x + 5 = 1$, then $x = -4$ (invalid - can't have negative students)

If $x + 5 = 2$, then $x = -3$ (invalid - can't have negative students)

If $x + 5 = 5$, then $x = 0$ (invalid - can't have zero students)

If $x + 5 = 7$, then $x = 2$ ✓

If $x + 5 = 10$, then $x = 5$ ✓

If $x + 5 = 14$, then $x = 9$ ✓

If $x + 5 = 35$, then $x = 30$ ✓

If $x + 5 = 70$, then $x = 65$ ✓

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Valid values of $x$ are: 2, 5, 9, 30, 65

Maximum possible students in section A = 65

Minimum possible students in section A = 2

Required difference = $65 - 2 = 63$

Therefore, the difference between the maximum and minimum possible number of students in section A is 63.