32 Schools X 2-Classes each = 64 Classes (in total)
Total no. of teachers = 37

Least no. of teachers who teach 3 classes = 0
[since 2*37=74, can cover up 64 classes]

Greatest no. of teachers who teach 3 classes = 13


Why?
No of Teachers who teach 3 classes = 14
Remaining teachers = 37-14 = 23
No of classes covered = 14*3 = 42
No of remaining classes = 22
Hence we will have a case where one teacher will have no class to teach
Hence, its 13 teachers.

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.

Hi Chetan , thanks for the explanation.

May I ask, why didn't we consider 14 instead of 13 , when your calculation had 13.5 and we aim to increase /look for maximum value ?

Thanks
Abhinav

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

SO b+2c=27
cmin=0，cmax=13

chii has provided a very elegant solution. I am merely elucidating his solution to make it easier to grasp:

The number of teachers teaching one, two and three classes each are 'a', 'b' and 'c' respectively. Thus:

a+b+c=Total # of teachers=37.......(i)
a+2b+3c=Total # of classes=64.....(ii)
Subtracting (i) from (ii), we get:
b+2c=27

Cmin=0 works because, in that case, 27 teachers teach 27*2=54 classes and the other 37-27=10 teachers teach 10*1=10 classes.
For cmax, we have to derive the lowest possible value of 'b'. It can't be 0 because 'c' would then be a fraction. So the least value of 'b' is 1 so the max value of 'c' is 13. So 13 teachers teach 39*3=39 classes, 1 teacher teaches 1*2=2 classes and 23 teachers teach 23*1=23 classes.

ANS:A

Hi chetan2u

I'd like to disagree with the above explanation provided for the equation.
X teachers can teach 3 classes. - As per the solution, we have 13 teachers teaching 3 classes.

X classes can't have 3 teachers.
As per the question,

Each class can only have 1 TEACHER.
If I re-frame the answer according to your explanation, it'd say "13 classes had 3 teachers."

Maybe you meant, X teachers taught 3 classes instead of


Let me know if my interpretation is correct

Yes it is correct., it’s a typo.. Thanks
X teachers take 3 classes and remaining 37-x take one class each.
That is why the total number of classes = 3x+(37-x)=64...x=13.5 or 13
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This is a difficult problem, but if you break it up into small chunks, it begins to easily reveal itself.

1) What is the problem asking us to find? We need to find the (minimum, maximum) values for how many teachers could have taught 3 classes. We are given the following: 64 total classes (2 classes per 32 elementary schools), 37 teachers, and the fact that each teacher can choose between 1, 2, or 3 classes.

2) Looking at the answer choices, there is a much narrower range of potential minimums than maximums, so lets test minimums first to narrow down our answers. It is easier to test than to solve this algebraically.

We know there are 64 total classes. So let's add the first layer of classes. We know that every teacher teaches at least 1 class, so the total number of classes remaining for teachers after all teachers taught their first class is 64 - 37 = 27 classes (left for teachers' 2nd or 3rd class if they teach one).

We have 37 teachers and 27 classes left to cover, so it is very possible that 27 of the remaining 37 teachers choose to teach 2 classes, thus leaving no classes left. This gives us: Teachers with 1 class = 10, Teachers with 2 classes = 27, Teachers with 3 classes = 0.

We know that the minimum is 0.

3) Now, to find the maximum, let's say go back to our previous scenario after every teacher has taught their 1st class only. We have 27 classes remaining for teachers to choose a 2nd or 3rd class if they choose to do so. Let's say that 26 of the remaining classes are taught by 13 teachers, which means each teacher taught 3 classes. Since the max they can teach is 3 classes, a different teacher will need to teach the final 27th class remaining. This gives us: Teachers with 1 class = 23, Teachers with 2 classes = 1, Teachers with 3 classes = 13.

We know that the minimum must be 13.

4) Thus the correct answer is A. There is a minimum of 0 and a maximum of 13.

Another approach

let x,y & z be the number of teachers who take 1,2 & 3 number of classes respectively. So x+y+z=37 (A) and x+2y+3z=64 (B) (as given in question).
(B)-(A) y+2z=27.
z=(27-y)/2. The minimum value of z can be 0 ie none of the teachers taking three classes. For the maximum value of z, y has to be minimum ie y=0.
if y=0, z=27/2=13.5. Since the number of teachers can be integers only so 13.

So correct answer is A
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This is a difficult question, for me at least.

Total number of classes = 32 schools * 2 classes per school = 64 classes

From the information given in the problem, i know that Every class has to be assigned at least one teacher, hence number of classes that can have a variable distribution will be all besides 37 classes, which is

64 classes - 37 teachers (Every teacher has t be assigned at least one class) = 27

To have the least number of teachers teaching 3 classes i assume the 37 teachers are all teaching 2 classes which makes it = 37*2 = 74 (Hence we dont need any teacher with 3 classes) - 0 teachers with 3 classes required

To have the maximum number of teachers teaching 3 classes - 27 / 2 (additional 2 classes besides the 1 already assigned) = 13.5
The number will be 13. If i round it to 14, it would mean the number of classes are 28 which isnt the case and the number of teacher has to be a whole number.
Hence the range is 0-13

Hope this helps a little

Total number of classes= 64
Total number of Teachers= 37
Minimum 1 class need to be allocated to all teachers.
Maximum 3 classes can be allocated to each teachers.

Case 1: For Minimum number of 3 classes taken by teachers.
1 class allocated to 37 teachers, so 27 classes left for allocation,
If we allocate 1 more round of classes to all teachers then only 27 teachers will have double or 2 classes, leaving 10 teachers with 1 class. No (zero) teacher will be given 3 classes. So minimum value of n=0
Case 1 situation is - 0 teacher 3 classes, 27 teachers 2 classes and 10 teachers 1 class

Case 2: For Maximum number of 3 classes allocated to teachers.
The number 3 classes per teacher we'll be at-least 1 class, then additional 2 classes to any teacher can be allocated, means our last priority will be to allocate 2 class to any teacher-

Allocate 1 class to each teacher- 37 teachers (64 class -37= 27 classes pending)
Allocate additional 2 classes to same 37 teachers, only multiple 2 can be allocated, that is out of 27, only (2 X 13) classes will be given to 13 teachers and pending 1 class to separate one teacher.
So, max value of n=13
Case 2 situation is - 13 teachers 3 classes, 1 teacher 2 classes and 23 teachers 1 class


A is the answer

given - 32 schools & 2 classes in each school. So, 64 schools & teachers are 37.
teachers can take atleast on class but not more than 3 class : 1<= teachers can take class<=3
Min. value of teacher taking 3 class : if each teacher will take 1 class, 37 classes will get covered & rest will be 27 class left. So, among 37 teacher, 27 teacher can take left 27 classes. so, zero (0) teacher will take 3 class.

Max. value of teacher taking 3 classes: in this left 27 class post each teacher has taken 1 class. lets try to make as many teacher to complete 3 class. for that, teachers have to take on 2 class. 27 class /2 class for a teacher = 13.5. it means 13 teacher can complete additional 2 class to complete their 3 class.

So final answer will be (0, 13)

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If this asks for maximum possible value , why does the rounding of take the lower value (13) and not (14) ? Bunuel VeritasKarishma help please

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It has nothing to do with rounding off rules.
If x can be 13.5 at the MOST, can it take the value 14? No.
But if we need the value of x to be an integer, then it means the highest integer value it can take is 13.


Now say there is a completely different question - How many people are needed to complete the work within 3 days?
Say you get x = 13.5. What will be your answer? You cannot have 13.5 people. Will you have 13 people or 14 people? Note that since you need 13.5 people to do the job, 13 will NOT be sufficient to complete it. So you need 14 (even if the job will be completed some time before the end of the 3rd day)
_________________

1) # of schools = 32 ; 2 classes each.
2) 37 teachers
3) Each class had 1 teacher, and each taught at least 1 but no more than 3 classes.

32*2 = 64 ( The total number of classes available)
64-37= 27 ( The total number of classes left to be distributed by the 37 teachers)
27/2 = 13.5 ( We divide by 2 because each of the 37 teachers can only teach 2 more classes MAX)

We have to round down 13.5 to 13.

Ans) A

Total Classes = 32 schools and 2 classes each = 32*2 = 64 classes in total
if each teacher teaches one class then remaining teachers = 64-37 = 27 teachers
divide the remaining by 2 to put two teachers in one class in addition to 1 teacher already in class - 27/2 = 13.5 => integer value = 13 teachers
so 13 classes can have 3 teachers
Minimum = if each class has two teachers then 64