Solved till the average part but got confused while finding the range of values. Can someone help?

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

V Ritik · Accepted Answer

It&rsquo;s easy to get stuck after the average, but once the ratio idea clicks, it becomes clear.

Average of section A = 32, B = 60
Let total average = a , it lies between 46 and 60
Use ratio of students = (60 &ndash; a) : (a &ndash; 32)
And given: section B = section A + 10
Try a = 47 &rarr; ratio = 13 : 15,  gives A = 65
Try a = 56 &rarr; ratio = 1 : 6, gives A = 2

The final step is just subtracting: 65 &ndash; 2 = 63. So the answer is 63. Try plugging in values of a from 47 to 59 and match the ratio gap with 10. Once you spot the boundaries, it&rsquo;s just about checking the endpoints.

D. Aarav · Answer

This one gets tricky after the average part, but it becomes manageable once the logic of ratios is clear. Let the overall average be a. We&rsquo;re told section A&rsquo;s average is 32, section B&rsquo;s average is 60, and B has 10 more students than A. Using the weighted average concept, the ratio of the number of students in A to B becomes (60 minus a) to (a minus 32).
When both sections have the same number of students, the overall average would be the average of 60 and 32, which is 46. But since section B has 10 more students than A, the overall average will be higher than 46. So the possible values for a lie between 46 and 60, and since the final average must be an integer, we test integer values like 47 and 56 to find the extremes.
If a is 47, the ratio becomes 13 to 15. Using this ratio, the difference in the number of students between B and A becomes 2x, which equals 10. Solving this gives x equal to 5. So, section A has 13 times 5, which is 65 students.
If a is 56, the ratio becomes 1 to 6. The difference becomes 5x, which equals 10. That gives x equal to 2, and section A has 2 students in this case.
Now take the difference between the maximum and minimum possible values of students in section A. That&rsquo;s 65 minus 2, which is 63.