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A school admin will assign each student in a group of N [#permalink]
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06 May 2006, 21:16
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52% (02:44) correct
48% (01:56) wrong based on 126 sessions
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A school admin 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. OPEN DISCUSSION OF THIS QUESTION IS HERE: aschooladminwillassigneachstudentinagroupofn127509.html
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Last edited by Bunuel on 22 Jun 2012, 17:44, edited 1 time in total.
Edited the question and added the OA. Topic is locked.



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Re: DS  assigning students [#permalink]
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06 May 2006, 22:31
The question stmt asks whether each of the n students can be assigned to the m classroom so that equal number of students are accomodated in the classrooms.
statement 1: 3n students can be accomodated in the m classrooms, which means that 3n is divisible m. This does not necessarily tell us that n should also be divisible by m. Hence the question string can be either yes or no.
statement 2: 13n should be divisible by m, which implies n should be definately divisible by m. hence above stmt can be answered in yes.
Hence 2nd stmt alone is sufficient to answer the question. Hence B



Manager
Joined: 29 Apr 2006
Posts: 85

I think the answer is E.
i) Students can be divided evenly only if 3n/m is a positive integer. If n is odd then 3 n = odd x odd = odd. For m=even nos. statement 1 is not possible but for some m odd nos it is. Hence (i) is insufficient.
ii) Same as (i). 13n will be odd for n odd nos. and maybe possible for some m odd nos. Hence (ii) is also insufficient.
(i) & (ii) will also be insufficient.
Thus E.



SVP
Joined: 01 May 2006
Posts: 1796

I would say B
1) 3xn / m = k with k integer
and we cannot conclude anything at all.
2) 13xn / m = j with j integer
As we know that m < 13, it means that 13 cannot be a prime factor of m
which means all prime factors from m are prime factors of n, in other words n/m is an integer
>>> Sufficient



Manager
Joined: 27 Mar 2006
Posts: 136

Agree with B



Senior Manager
Joined: 29 Jun 2005
Posts: 403

it is B
Statement 1 is insufficient: Though 3n is divisible by m, that doesn’t mean that n is also divisible by m. Consider n=15, m=9 or n=15, m=5.
Statement 2 is sufficient: Because 13 is prime, and 13n is divisible by m, than n must also be divisible by m.



VP
Joined: 29 Apr 2003
Posts: 1403

I agree with B Also!
Nice question... Makes me feel good.



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Re: A school administrator will assign each student in a group [#permalink]
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22 Jun 2012, 17:36
Can any one provide a much elaborate solution, did not understand the above's....
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Re: A school administrator will assign each student in a group [#permalink]
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22 Jun 2012, 17:43
riteshgupta wrote: Can any one provide a much elaborate solution, did not understand the above's.... 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?Basically the question asks whether \(n\) (# of students) is a multiple of \(m\) (# of classrooms), or whether \(\frac{n}{m}=integer\), because if it is then we would be able to assign students to classrooms so that each classroom has the same number of students assigned to it. Given: \(3<m<13<n\). (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 > \(\frac{3n}{m}=integer\), from this we can not say whether \(\frac{n}{m}=integer\). For example \(n\) indeed might be a multiple of \(m\) (\(n=14\) and \(m=7\)) but also it as well might not be (\(n=14\) and \(m=6\)). Not sufficient. (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 > \(\frac{13n}{m}=integer\), now as given that \(3<m<13\) then 13 (prime number) is not a multiple of \(m\), so \(\frac{13n}{m}\) to be an integer the \(n\) must be multiple of \(m\). Sufficient. Answer: B. Hope its' clear. OPEN DISCUSSION OF THIS QUESTION IS HERE: aschooladminwillassigneachstudentinagroupofn127509.html In case of any questions please post there.
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Re: A school administrator will assign each student in a group
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