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

Question

Accepted Answer

Let the number of students in section B be $$x$$. Then, the number of students in section A will be $$x - 10$$.

1. **Combined Average Formula:**  
   The total marks for section A = $$32(x - 10)$$, and for section B = $$60x$$.  
   The combined average marks across both sections is:  
   $$
   	ext{Combined Average} = \frac{32(x - 10) + 60x}{(x - 10) + x}
   $$  
   Simplify the denominator:  
   $$
   	ext{Combined Average} = \frac{92x - 320}{2x - 10}
   $$

2. **Integer Constraint:**  
   For the combined average to be an integer, $$(92x - 320)$$ must be divisible by $$(2x - 10)$$. Simplify further:  
   Let $$y = x - 5$$, so $$2x - 10 = 2y$$. Rewrite the numerator as:  
   $$
   \frac{46y + 70}{y}
   $$  
   For this expression to be an integer, $$70/y$$ must also be an integer. Hence, $$y$$ must be a factor of 70.

3. **Factors of 70:**  
   Possible values for $$y$$ are: $$1, 2, 5, 7, 10, 14, 35, 70$$. Since $$y = x - 5$$, we calculate $$x = y + 5$$. Additionally, $$x > 10$$ (as $$x - 10 > 0$$). Valid values for $$x$$: $$12, 15, 20, ...,$$.

4. **Calculate Values for Section A ($$x - 10$$):**  
   Using valid values for $$x$$, calculate the corresponding values for section A ($$x - 10$$):  
   - Maximum value: $$65$$ (when $$x = 75$$).  
   - Minimum value: $$2$$ (when $$x = 12$$).

5. **Difference Between Maximum and Minimum Values:**  
   Difference = Maximum value - Minimum value = $$65 - 2 = 63$$.

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