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A school administrator will assign each student in a group [#permalink]
07 Oct 2005, 13:46

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00:00

A

B

C

D

E

Difficulty:

75% (hard)

Question Stats:

54% (02:19) correct
46% (01:19) wrong based on 71 sessions

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.

Question basically asks whether n/m is an integer.
from A, we can say 3(n/m) is an integer. So, n/m must be an integer.
This will fail if n=1 and m=3 and other lower numbers, but the range of values for n and m exclude this possibility.
Similarly with B.

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.

If 13n is assigned to m classes... then the number n is surely divisible by m since m cannot be 13 or 1.. So we are sure that m can be divided into n..

If 13n is assigned to m classes... then the number n is surely divisible by m since m cannot be 13 or 1.. So we are sure that m can be divided into n..

Is that right?

OA?

Good explanation. _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

using the same reasoning can't I be sufficent also?

3n/m , 3 is prime so n has to be divisible by m?

For n = 14 and m = 6
3n/m is divisible but n/m is not!

For n=15, and m = 5 both 3n/m and n/m are divisible.

Hence 1 is INSUFF.

HTH _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

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 eachclassroom 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 eachclassroom has the same number of students assigned to it.

Hallo,
Think that A is insufficient
From A) 3n=K*m now n-15 then m can be 5 or 9 which makes A insuff
From B) 13n=K*m then m can not be a prime bigger than 13 , n=15 m can be 3 or 5, n=20 m can be 2,4,10
So think that B Is sufficient

Ive just decided to start studying for the gmat, so im a rookie here, but....
The question asks: is it possible to do so. and i think that (D) is correct b/c they both are sufficient to recognizing that it is possible.

Ive just decided to start studying for the gmat, so im a rookie here, but.... The question asks: is it possible to do so. and i think that (D) is correct b/c they both are sufficient to recognizing that it is possible.

This should be B.

In A if 3n = 42 and m = 6 then stem fails but if 3n = 48 and m = 4 then it works. So its INSUFF.

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...