Since we have at least 2 classes in each of the 32 schools, this means that there will be a total of \(32 \times 2 = 64\) classes in all.

Given that there are 37 teachers, for the first 37 classes, one unique teacher can be assigned to each of the first batch of classes.

This means that if only one teacher is assigned per class, there will be \(64 – 37 = 27\) classes left. So, some of the 37 teachers will have to teach another class[/i][/color] (and possibly more than one).

It’s possible that 27 of the 37 teachers will each teach an additional class. Thus, the least possible number of teachers who will have to teach 3 classes is zero.

To find the maxim number of teachers that could teach 3 classes, we have to add 2 classes to each teacher who is already teaching 1 class. With 27 classes remaining, we have 13 sets of two additional classes, with one left-over class. Thus, the maximum possible number of teachers who could teach 3 classes is 13. This would leave 1 teacher teaching two classes and 23 teachers teaching one class each.

Therefore, the final answer is .

An easier way to solve this problem:

1.2 classes in each of 32 schools means total of 2*32=64 classes.
2.Total teachers = 37

Now, MINIMUM value of n(teachers who took three classes)

Suppose all of them 2 classes then 37*2 =74 classes which is more than 64..so we can imagine few taking one classes say 25 people took 2 classes and rest 12 took 1 classes..This means total classe is 64 and value of n=0

Now since we know value of N is 0, eliminate option C,D and E

Now, MAXIMUM value of n(teachers who took three classes)

Focus only on options A and B for which max value is 13 and 14

If you plugin 14 as value of n means 14(teacher)*3=42 classes...classes remaining =64-42=22 and teachers remaning(37-14=23)

So even if one teacher takes one classe we will have an empty class..Hence B is out..

Directly chose A.

a+2b+3c=32*2=64
a+b+c=37

SO b+2c=27
cmin=0，cmax=13