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# A certain experimental mathematics program was tried out in 2 classes

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Joined: 15 Sep 2018
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A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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20 Sep 2018, 20:29
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 .
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A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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14 Oct 2018, 00:36
Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

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.
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A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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17 Oct 2018, 21:22
1
Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

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.
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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22 Oct 2018, 22:33
chetan2u wrote:
Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

Equation - 64 classes and 37 teachers...

LEAST possible
- all have two so 37*2 = 74, more than 64 classes, so it is possible that NONE of classes had 3 teachers..

GREATEST possible - say x classes have 3 teacher..... so 3x + (37-x) = 64............3x+37-x = 64...............2x = 27................x = 13.5.....
take the INTEGER value as greatest possible = 13

ans 0 and 13..
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
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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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06 Sep 2019, 20:08
1
a+2b+3c=32*2=64
a+b+c=37

SO b+2c=27
cmin=0，cmax=13
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A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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06 Sep 2019, 22:45
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
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Joined: 06 Mar 2018
Posts: 47
Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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15 Nov 2019, 00:14
chetan2u wrote:
GREATEST possible - say x classes have 3 teacher..... so 3x + (37-x) = 64............3x+37-x = 64...............2x = 27................x = 13.5.....
take the INTEGER value as greatest possible = 13

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,
Quote:
Each of the classes had 1 teacher

Each class can only have 1 TEACHER.

Maybe you meant, X teachers taught 3 classes instead of
Quote:
say x classes have 3 teacher.

Let me know if my interpretation is correct
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Posts: 8295
Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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15 Nov 2019, 02:23
jkbk1732 wrote:
chetan2u wrote:
GREATEST possible - say x classes have 3 teacher..... so 3x + (37-x) = 64............3x+37-x = 64...............2x = 27................x = 13.5.....
take the INTEGER value as greatest possible = 13

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,
Quote:
Each of the classes had 1 teacher

Each class can only have 1 TEACHER.

Maybe you meant, X teachers taught 3 classes instead of
Quote:
say x classes have 3 teacher.

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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Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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02 Mar 2020, 08:42
Bunuel wrote:
A certain experimental mathematics program was tried out in 2 classes in each of 32 elementary schools and involved 37 teachers. Each of the classes had 1 teacher and each of the teachers taught at least 1, but not more than 3, of the classes. If the number of teachers who taught 3 classes is n, then the least and greatest possible values of n, respectively, are

A) 0 and 13
B) 0 and 14
C) 1 and 10
D) 1 and 9
E) 2 and 8

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.
Re: A certain experimental mathematics program was tried out in 2 classes   [#permalink] 02 Mar 2020, 08:42

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