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Intern  B
Joined: 15 Sep 2018
Posts: 31
A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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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 .
Intern  S
Joined: 07 May 2015
Posts: 41
Location: India
Schools: Darden '21
GPA: 4
A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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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.
Senior Manager  P
Joined: 09 Jun 2014
Posts: 354
Location: India
Concentration: General Management, Operations
A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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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.
Manager  B
Joined: 03 Aug 2018
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GMAT 1: 590 Q45 V26 GPA: 3.5
Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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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
Intern  Joined: 10 Aug 2019
Posts: 2
Location: United States
Schools: NUS '22
GPA: 3.7
Re: A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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1
a+2b+3c=32*2=64
a+b+c=37

SO b+2c=27
cmin=0，cmax=13
Manager  S
Joined: 09 Nov 2015
Posts: 142
A certain experimental mathematics program was tried out in 2 classes  [#permalink]

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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 A certain experimental mathematics program was tried out in 2 classes   [#permalink] 06 Sep 2019, 23:45

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