phew.. !

Ramit88 wrote:A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?

(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.

(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.



unable to understand from previous threads..

m -> classroom
n -> students
3<m<13<n

question : Is n/m integer ?

Statement 1:
3n/m = integer = i
=> n/m = i/3

n/m can be (m=3, n = 15) or can't be (m=3, n=14) integer based on this. So Insufficient.

Statement 2:
13n/m = integer = i
=> since, 13 is prime, 13 can't be factor of m,
so n has to be a factor of m, => n/m = integer -----> Sufficient.

Hence, B

thnx ashu.. man u r amazing.. awesome.. if u dont mind ..wats ur gmat score buddy

Ramit88 wrote:thnx ashu.. man u r amazing.. awesome.. if u dont mind ..wats ur gmat score buddy

Glad it helped ! I don't know yet

Edited my earlier post. pls ignore. Made some changes.

IMO should be D

n students, Let Each student be n1,n2, n3 , n4 & so on

m classes . let each class me m1.m2.m3,m4 & so on

3<m<13.

13< n

The question is basically asking is it possible that n/m an be an integer ? lets check

1) 3n students. Lets say n=14 ( as n>13) . 3n =3*14=42 . now this can be assigned to each of the classrooms such that it has the same no of students in it.

m=4 ,5,6,7,8,9,10,11,12, Yes possible. as 42 is divisible by 6 as well as 7. So its possible for n/m to be an integer.

2) 13n = 13* 14 ( took n as 14) = 13 * 2 * 7

Now m=4 ,5,6,7,8,9,10,11,12, . 13*2*7 is definitely divisible by 7.But since m cant be 26. So its not possible for n/m to be an integer in this case.Yes . Sufficient. As it definitely says that n/m cannot be an integer.

D.

Thanks anshumishra for the wowest explanation . You rock !!!

anshumishra wrote:

Ramit88 wrote:A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?

(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.

(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.



unable to understand from previous threads..

m -> classroom
n -> students
3<m<13<n

question : Is n/m integer ?

Statement 1:
3n/m = integer = i
=> n/m = i/3

n/m can be (m=3, n = 6) or can't be (m=3, n=5) integer based on this. So Insufficient.

Statement 2:
13n/m = integer = i
=> since, 13 is prime, 13 can't be factor of m,
so n has to be a factor of m, => n/m = integer -----> Sufficient.

Hence, B

Hi Anshu,

Isn't there a problem with your Statement 1 explanation. You have taken n=6 and n=5 which is invalid. As per premise,
n>13. If you use n>13, statement one is sufficient too. If you think otherwise, please explain.

Thanks.

DarkKnight wrote:

anshumishra wrote:

Ramit88 wrote:A school administrator will assign each student in a group of n students to one of m classrooms. If 3 < m < 13 < n, is it possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students assigned to it?

(1) It is possible to assign each of 3n students to one of m classrooms so that each classroom has the same number of students assigned to it.

(2) It is possible to assign each of 13n students to one of m classrooms so that each classroom has the same number of students assigned to it.



unable to understand from previous threads..

m -> classroom
n -> students
3<m<13<n

question : Is n/m integer ?

Statement 1:
3n/m = integer = i
=> n/m = i/3

n/m can be (m=3, n = 6) or can't be (m=3, n=5) integer based on this. So Insufficient.

Statement 2:
13n/m = integer = i
=> since, 13 is prime, 13 can't be factor of m,
so n has to be a factor of m, => n/m = integer -----> Sufficient.

Hence, B

Hi Anshu,

Isn't there a problem with your Statement 1 explanation. You have taken n=6 and n=5 which is invalid. As per premise,
n>13. If you use n>13, statement one is sufficient too. If you think otherwise, please explain.

Thanks.

You are absolutely right. I'll edit it with proper examples. Please note , even with n>13 statement 1 is insufficient as shown in my first post edited.
Thanks

but the ans in OG is B.... and they have taken the same no as ashu has.. but darknight logic is aalso true...anyone ??

Ramit88 wrote:but the ans in OG is B.... and they have taken the same no as ashu has.. but darknight logic is aalso true...anyone ??

Hey Ramit,

Answer will be "B" only. As you can check my first post of the solution that statement 1 is insufficient (My example of n=5 and 6 was not right as per the question stem n>13, BUT even with n=14 and 15 the statement 1 is insufficient as shown in my first post.)

So, actually I was agreeing on the first part of Dark night's response, that my example was not right within the constraint (13<n).
Hope that clears !

thanx ashu... 1 week left for d day..i guess m freakin out..lol..