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# A school administrator will assign each student in a group

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13 Feb 2012, 22:10
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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.
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Re: A school admin will assign each student in a group of N [#permalink]

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13 Feb 2012, 22:15
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gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.

Hope its' clear.
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Re: A school administrator will assign each student in a group [#permalink]

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21 Aug 2014, 19:40
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For me, it is more simple to understand the rule as follows:
(A×b)÷c , b is divisible by c or not?
+ case 1: c and a share a common factor => b is not necessarily divisible by c
+ case 2: c and a do not share any common factor => b is surely divisible by c
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Re: A school administrator will assign each student in a group [#permalink]

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05 Dec 2012, 03:02
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thangvietnam wrote:
pls, explain why from 13n/m= interger we can have n/m= interger. only by picking numbers we have this conclusion.

why from 13/m=no integer, we can have n/m=integer.

why b is correct. I can get B correct by picking numbers but the reasoning must be clear.

For (2) we have that $$\frac{13n}{m}=integer$$ and $$3<m<13$$. Now, 13 is a prime number thus it has only 2 factors 1 and 13, which means that $$m$$ cannot be a factor of 13 (since $$3<m<13$$, then it's neither 1 nor 13). Therefore in order $$\frac{13n}{m}$$ to be an integer, $$m$$ must be a factor of $$n$$.

Hope it's clear.
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Re: A school administrator will assign each student in a group [#permalink]

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19 Dec 2012, 16:21
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I actually worked this problem out

I picked numbers for statement 1 and found it to be insufficient where:

N=15, M=5

3n/m = 45/5=15 ---> 15/5=3 -----> n is divisible by m or n/m is an integer

and N=15, M=9

3n/m = 45/9=5 ---> 15/9=1.3333 -----> n is NOT divisible by m or n/m is NOT an integer

For statement 2 I picked numbers:

N=14

13N = 182

M=4 no M=5 no M=6 no

I got lazy at that point and said 182 is an even number and isn't divisible by 4 or 6, probably nothing will work and chose answer E although unsure.

Upon re-examining it, I see 7 would have worked. However I agree if you can express this algebraically as you did and the OG's answers did, it does become much simpler than picking these vast arrays of numbers.

My problem with your logic (and the OG's) is that finding statement 2 sufficient is based solely on the fact that 13 is a prime number, is 3 not also a prime? What condition allows you to draw this conclusion for the statement that 13n/m = integer that you can't for 3n/m = integer?
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Re: A school administrator will assign each student in a group [#permalink]

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19 Dec 2012, 21:56
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I actually worked this problem out

I picked numbers for statement 1 and found it to be insufficient where:

N=15, M=5

3n/m = 45/5=15 ---> 15/5=3 -----> n is divisible by m or n/m is an integer

and N=15, M=9

3n/m = 45/9=5 ---> 15/9=1.3333 -----> n is NOT divisible by m or n/m is NOT an integer

For statement 2 I picked numbers:

N=14

13N = 182

M=4 no M=5 no M=6 no

I got lazy at that point and said 182 is an even number and isn't divisible by 4 or 6, probably nothing will work and chose answer E although unsure.

Upon re-examining it, I see 7 would have worked. However I agree if you can express this algebraically as you did and the OG's answers did, it does become much simpler than picking these vast arrays of numbers.

My problem with your logic (and the OG's) is that finding statement 2 sufficient is based solely on the fact that 13 is a prime number, is 3 not also a prime? What condition allows you to draw this conclusion for the statement that 13n/m = integer that you can't for 3n/m = integer?

For St 2 , we are given that 13N/M= Integer and that 3<M<13<N. Now since M, N are integers, we get Value of M can be from (4,5,6...12) and N can be any no greater than 13.Out of the possible values for M, not a single number can divide 13 because it is prime and its only divisor will be 1 and 13. Then for 13N/M to be an integer N has to be a multiple of M or else 13N/M cannot be an Integer which will contradict the given statement itself.

for St1, we are given that 3N/M =Integer and from Q. stem we get 3<M<13<N. Now In possible values of M that can make 3/M in fraction form will be 6,9 and 12 which will result in 3/M value as 1/2,1/3 and 1/4 respectively.

Now if N=15, and M=3 we get 3N/M =15 as integer and N/M also as Integer(Y).
If N=16 and M=3, we get 3N/M=16 as Integer but N/M is not an Integer(N) and hence not sufficient.

Therefore from St 1 we can get 2 ans and hence not sufficient.

Thanks
Mridul
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Re: A school administrator will assign each student in a group [#permalink]

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27 Feb 2013, 04:01
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Explained perfectly by Bunuel. Still, those who have a problem visualizing it, just try to plugin numbers for N and M. Given that 3<M<13<N.

Let N=15, M=5. Thus "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" NOW means is it possible to assign each of the 15 students to one of the classrooms(out of 5), the no of students being the same in each class. Thus, as we see that we can send 3 students to each of the 5 classes, we know that YES it is possible. Thus to prove whether the given generalized statement is possible, we have to prove that M is a factor of N.

From F.S 1, we know M is a factor of 3N. We have to find out whether M is a factor for N also. We can not say this with confidence. Assume M=6, and N =16. Here, M is not a factor of N. Yet,M(6) is a factor of 3N(48). Not sufficient.

From F.S 2, we know that M is a factor of 13N. Now, as 3<M<13, we know that M is co-prime to 13. Thus, it can only be a factor to N. Hence sufficient.

B.
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Re: A school admin will assign each student in a group of N [#permalink]

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03 Jul 2013, 14:47
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mitmat wrote:
Bunuel wrote:
gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.
.

Hi Bunuel, I could not understand how you inferred whether \frac{n}{m}=integer[/m] from this question stem?

I tried to plug in few values to the question stem to help me understand your line of thought, but I could not. As always, thank you for your great help and support.

The question asks whether we can divide n students into m classes so that each classroom has the same number of students. For example, if there are 20 students and 10 classrooms, then we can assign 2 students to each of the 10 classrooms but if there are 20 students and say 9 classrooms, then we cannot assign (divide) 20 students to 9 classrooms so that each classroom to have the same number of students.

So, we need to determine whether n will be divisible by m, or which is the same whether n/m=integer.

Hope it's clear.
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Re: A school administrator will assign each student in a group [#permalink]

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25 Jun 2014, 05:05
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Bunuel wrote:
gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.

Hope its' clear.

Dear Bunuel,

I have concern here.

To explain for (2) 13n/m to be integer, you said that since 13 is prime and 3<m<13 --> 13/m = non-integer --> n/m must be integer, then 13n/m = integer.
So why the same reasoning cannot be applied for (1) as following:
--> (1): 3n/m = integer. Since 3<m<13 and 3 is prime --> 3/m = non-integer --> n/m must be integer, then 3n/m = integer --> sufficient

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Re: A school administrator will assign each student in a group [#permalink]

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25 Jun 2014, 07:34
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LucyDang wrote:
Bunuel wrote:
gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.

Hope its' clear.

Dear Bunuel,

I have concern here.

To explain for (2) 13n/m to be integer, you said that since 13 is prime and 3<m<13 --> 13/m = non-integer --> n/m must be integer, then 13n/m = integer.
So why the same reasoning cannot be applied for (1) as following:
--> (1): 3n/m = integer. Since 3<m<13 and 3 is prime --> 3/m = non-integer --> n/m must be integer, then 3n/m = integer --> sufficient

Because for (2), m cannot have any factor of 13, thus for $$\frac{13n}{m}$$ to be an integer n must be divisible by m.

For (1) m could be a multiple of 3, for example, 6, 9, or 12, and in this case n is not necessary to be divisible by m.
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Re: A school admin will assign each student in a group of N [#permalink]

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13 Feb 2012, 22:24
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Thanks! Your explanation was much clearer than the guide.
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Re: A school administrator will assign each student in a group [#permalink]

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20 Dec 2012, 04:08
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My problem with your logic (and the OG's) is that finding statement 2 sufficient is based solely on the fact that 13 is a prime number, is 3 not also a prime? What condition allows you to draw this conclusion for the statement that 13n/m = integer that you can't for 3n/m = integer?

Not so.

(2) implies that $$\frac{13n}{m}=integer$$ but we also know that $$3<m<13$$. If we were not told that, then it would be possible m to be for example 26 and in this case n may or may not be a multiple of m, for example consider n=2 and n=26.

Hope it's clear.
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Re: A school administrator will assign each student in a group [#permalink]

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14 Sep 2014, 04:37
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Hey everyone, I finally managed to explain the solution to myself so I thought I'd share my simplification with everyone- it might help. No need to plug in numbers, as that approach confused me even more.
You are offered two statements,and you have to see if they undeniably help you in figuring out whether n is divisible by m.

The first statement tells you that 3n is divisible by m. Therefore: 3n/m=x. Can this statement assure you about the relationship of n and m, given the limitations on their values? NO. Since 3<m<13, you could pick a number that is divisible by 3, and from that point onwards it won't matter if n is divisible by m or not. You are not able to give a clear answer using this information. This statement is insufficient.

The second statement tells you that 13n is divisible by m; 13n/m=x This time, by looking at the limitations you can be sure that there is no way that m is divisible by 13 because m can neither be 1 nor 13. (3<m<13) So for this statement to be true, n HAS to be divisible by m. This time you are able to give a 100% answer on the relationship of n and m.

There ya go. No numbers needed.
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24 Sep 2014, 17:52
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LucyDang wrote:
Bunuel wrote:
gwiz87 wrote:
Hi,

I'm new to this forum and I'm hoping someone can help me understand the problem below:

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.

Can you please explain this to me?

Welcome to GMAT Club. Below is a solution for your problem. Please don't hesitate to ask in case of any question.

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.

Hope its' clear.

Dear Bunuel,

I have concern here.

To explain for (2) 13n/m to be integer, you said that since 13 is prime and 3<m<13 --> 13/m = non-integer --> n/m must be integer, then 13n/m = integer.
So why the same reasoning cannot be applied for (1) as following:
--> (1): 3n/m = integer. Since 3<m<13 and 3 is prime --> 3/m = non-integer --> n/m must be integer, then 3n/m = integer --> sufficient

Really good question!!! It got me thinking too, however, I noticed that, in statement 2, m has to less than 13 --> as per the condition 3<m<13<n, therefore in the case of statement 2: 13n/m -->13 cannot divide m or in other words "cancel out" m, but, in statement 1: 3n/m, 3 can cancel out m since 3 <m<13.

To understand, lets take an example: Let m =6 classrooms and n= 28 student; this adhere's to the condition given 3<m<13<n, if you notice, in this case n/m (28/6) is not an integer, whereas, as per statement 1: 3n/m (3 x 28 /6)=14--> is an integer (you would cancel out 3 & 6 to give you 2 in the denominator)

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Re: A school administrator will assign each student in a group [#permalink]

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15 Nov 2015, 21:54
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rakeshd347 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.

Target question: 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

This is a great candidate for rephrasing the target question (more info about rephrasing the target question can be found in this free video:
http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1100)

In order to be able to assign the same number of students to each classroom, the number of students (n) must be divisible by the number of classrooms (m). In other words, n/m must be an integer.

REPHRASED target question: Is n/m an integer?

Statement 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 to it.

This statement is telling us that the number of students (3n) is divisible by the number of classrooms (m). In other words, 3n/m is an integer.

Does this mean mean that m/n is an integer? No.
case a: m = 4 and n = 20, in which case n/m is an integer.
case b: m = 6 and n = 20, in which case n/m is not an integer.
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 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 to it
This statement tells us that the number of students (13n) is divisible by the number of classrooms (m). In other words, 13n/m is an integer.

The given information tells us that 3 < m < 13 < n. Since m is between 3 and 13, there's no way that 13/m can be an integer. From this, we can conclude that n/m must be an integer.
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Cheers,
Brent
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Re: A school administrator will assign each student in a group [#permalink]

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05 May 2016, 21:45
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bhamini1 wrote:
Dear Bunuel,

I am still not clear with the line of reasoning used to prove if n/m is an integer with the given option statements.
For option 1 we are saying that 3n/m could give us an integer and also could not give us an integer basis values of 3<m<13,such as 3*15/5=Integer ,however,3*15/6 is not an integer. Then ,in the second option statement we are considering that since m is greater than 13 then n/m will definitely give us an integer.Why are we changing the thought process for option 2 statement.Even in this case,then 13* 20/5=Integer and 13*21/5 is not an integer.

Regards,
Bhamini

Hi,
the statement II tells us that 13N/M is an integer...
But M is less than 13 and 13 is a PRIME number, so there are no common factors of 13 and M..
so ONLY possiblity is that N/M is an integer..
so suff..

But statement I tells us that 3N/M is an integer...
and M is between 3 and 13, so there can be common factors between 3 and M like 3 and 6 and they can be co-primes 3 and 5..
so BOTH possiblity that N/M is an integer and not an integer exist..
so insuff..
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Re: A school administrator will assign each student in a group [#permalink]

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25 Oct 2016, 06:27
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AK125 wrote:
Hey guys,

might be a little bit off topic.
But do I have to understand (be able to solve) these very hard questions if I am aiming for a low 600?

Here is an effort to make the calculation of question a little easy for you

Quote:
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.

It will be possible to assign each student in classrooms equally only if No. of Students (N) is a factor of No. of classrooms (M)
i.g. 20 students can be assigned in 5 room because 5 is factor of 20 but 20 students can't be assigned in 6 room because 6 is not a factor of 20

Question REPHRASED: Is N/M=Integer???

Statement 1: 3N/M is an Integer
Now I wish to prove this statement insufficient therefore I take two cases
Case 1: N=20 and M=5, such that 3N/M is an Integer And also N/M is and Integer
Case 2: N=20 and M=6, such that 3N/M is an Integer BUT N/M is NOT and Integer
NOT SUFFICIENT

[P.S. I could prove this statement Insufficient because I could take denominator as multiple of 3 in order to use the coefficient of N given in numerator]

Statement 2: 13N/M is an Integer
Now I wish to prove this statement insufficient BUT since M<13 and 13 is a prime Number so I can't choose the value for denominator which can have any factor of 13 so for 13N/M to be an Integer N/M must also be an Integer
SUFFICIENT

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Re: A school administrator will assign each student in a group [#permalink]

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05 Dec 2012, 02:52
pls, explain why from 13n/m= interger we can have n/m= interger. only by picking numbers we have this conclusion.

why from 13/m=no integer, we can have n/m=integer.

why b is correct. I can get B correct by picking numbers but the reasoning must be clear.
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Re: A school administrator will assign each student in a group [#permalink]

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26 Mar 2013, 11:17
The hardest part of this problem wasn't the math. It was the prompt.

Could someone break it down for me?

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?

"is it possible" to me implied whether we can choose an two values within the problem's given inequality to make n/m an integer, at all which we could for condition 1.
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Re: A school administrator will assign each student in a group [#permalink]

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27 Mar 2013, 06:30
manimgoindowndown wrote:
The hardest part of this problem wasn't the math. It was the prompt.

Could someone break it down for me?

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?

"is it possible" to me implied whether we can choose an two values within the problem's given inequality to make n/m an integer, at all which we could for condition 1.

There are:
m classrooms where 3 < m < 13 < n.
n students where 3 < m < 13 < n.

We are asked to find whether we can divide equally n students in m classrooms, so whether n/m is an integer.

Does this make sense?
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Re: A school administrator will assign each student in a group   [#permalink] 27 Mar 2013, 06:30

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