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Re: A school administrator will assign each student in a group
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14 Sep 2014, 03:37
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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20 Sep 2014, 02:39
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. Answer: B. 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 = noninteger > 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 = noninteger > n/m must be integer, then 3n/m = integer > sufficient Please help me to clarify this point. Thank you so much! Hi LucyDang This is how I understood it...hope it helps you as well... For any fraction to result in an integer value, the denominator needs to be less than or equal to the numerator. In case it isn't, then despite the denominator being a multiple of the numerator, you'll still get a fraction in form of 1/x as your answer (eg: 15/5 = 3 = integer, but 15/45 = 1/3 = fraction) Now applying this logic to statement 1... 3n/m= integer > m</=3n. If we apply the information from the question stem, which is 3<m<13, we can have m= 3, 6, 9, 12. As per these values of m, corresponding n values (for 3n/m to be an integer) can be 1, 2, 3, 4. Out of these only n=3 will result in n/m = integer. But n = 1/2/4 can also satisfy Statement 1. Hence, we can't be sure if n/m will be an integer or not. Now if we apply the same logic to statement 2... 13n/m = integer > m</=13n Applying the condition 3<m<13 to this, we know that m cannot be any integer that would divide 13 (since 13 is a prime number). And hence, it'll have to divide n for 13n/m to be an integer, implying n/m=integer. Hope this helped. Good luck! NB



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24 Sep 2014, 16:52
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. Answer: B. 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 = noninteger > 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 = noninteger > n/m must be integer, then 3n/m = integer > sufficient Please help me to clarify this point. Thank you so much! 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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15 Oct 2015, 09:44
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. Answer: B. Hope its' clear. I have to start to think like bunuel, amazing! So easy if you read it...
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Updated on: 16 Nov 2015, 19:36
Q: if 3<M<13. Is n divisible by m?
St1: 3n divisible by m.
Possibility #1: n is a multiple of m only if multiplied by 3. In which case 3n is divisible by m, but n alone isn't divisible by m. e.g. m = 9, n = 6
Possibility #2: n is divisible by m. e.g. m =6, n = 12
INSUFF
St2: 13n is divisible by m. Since m != 13, the n must be divisible by m.
SUFF Insuff.
Originally posted by kham71 on 15 Nov 2015, 16:48.
Last edited by kham71 on 16 Nov 2015, 19:36, edited 1 time in total.



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15 Nov 2015, 20:54
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 itThis 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/gmatdatasufficiency?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. Consider these contradictory cases. 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 itThis 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 Answer = B Cheers, Brent
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05 May 2016, 20:31
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.
Please clarify my doubt.
Regards, Bhamini



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05 May 2016, 20:45
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.
Please clarify my doubt.
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 coprimes 3 and 5.. so BOTH possiblity that N/M is an integer and not an integer exist.. so insuff..
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24 Oct 2016, 01:27
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?



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24 Oct 2016, 10:09
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? No, you absolutely don't. All you need to do with these problems is be able to recognize them. The only reason to recognize them is so you can guess on them and move on quickly, rather than mistaking them for simpler problems and spending too much time trying to solve before giving up.
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25 Oct 2016, 05:19
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? Hi AK125You need to understand whether your strength is Quant or Verbal If your strength is Quant and your verbal section is weak then you must try all ways to get as many questions as possible however if your Strength area is Verbal then such hard questions can be done away with. Being an instructor I would atleast suggest or motivate you to try to understand such question. After giving it a try even if you fail to solve such questions in actual test then atleast it will help you build concepts and confidence in cracking comparatively easier question Getting 650 with aspiration of 600 is not a bad idea anyway...
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25 Oct 2016, 05:27
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 20Question REPHRASED: Is N/M=Integer???Statement 1: 3N/M is an IntegerNow 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 IntegerNow 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 Answer: option B
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21 May 2017, 22:05
Bunuel wrote: gwiz87 wrote: 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. 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. Hi Bunuel, \(3<M<13<N\) Possible value of M = 4, 5, 6, 7, 8, 9, 10, 11, 12 Possible value of N = 14, 15, 16, 17, 18, 19 I could not understand the solution for statement (2). \(\frac{13N}{M} = integer\) What if M=4, N=14? \(\frac{13(14)}{4} ≠ integer\). Could you help to explain? Shall we use the value as below? M= 7___8___9__ N=14__16__18__ \(\frac{13N}{M}=13(\frac{14}{7})=26\)
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21 May 2017, 23:16
hazelnut wrote: Bunuel wrote: gwiz87 wrote: 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. 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. Hi Bunuel, \(3<M<13<N\) Possible value of M = 4, 5, 6, 7, 8, 9, 10, 11, 12 Possible value of N = 14, 15, 16, 17, 18, 19 I could not understand the solution for statement (2). \(\frac{13N}{M} = integer\) What if M=4, N=14? \(\frac{13(14)}{4} ≠ integer\). Could you help to explain? Shall we use the value as below? M= 7___8___9__ N=14__16__18__ \(\frac{13N}{M}=13(\frac{14}{7})=26\) The question asks whether \(\frac{n}{m}=integer\). (i) From stem we are given that \(3<m<13<n\). (ii) From (2) we are given that \(\frac{13n}{m}=integer\). When testing values, we should choose those which satisfy (i) and (ii) and test whether they give definite answer to the question: is \(\frac{n}{m}=integer\). All values that satisfy (i) and (ii) will give an YES answer to the question "is \(\frac{n}{m}=integer\)?". Does this make sense?
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Re: A school administrator will assign each student in a group
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16 Jun 2017, 08:39
manimgoindowndown wrote: Bunuel wrote: 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? It is possible as you demonstrated in your solutions post to have a situation where premise I will make n/m an integer. However it is also possible that there is a situation where it wont be an integer. I guess I focused on the possibility part. Since its possible I would see I. As correct. Hello. I had the same doubt as you to this question. According to me, the question simply asks "is it possible?", which means that are there any values of n and m which can solve this problem. Since for Statement (1), we do have possible values, shouldn't the answer be "D"?



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Re: A school administrator will assign each student in a group
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16 Jun 2017, 09:10



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Re: A school administrator will assign each student in a group
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27 Jun 2017, 19:59
gwiz87 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. Responding to a pm: We want to know if we can divide N evenly by M. We know that 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. This tells us that 3N is divisible by M. M is greater than 3 so it could be 4/5/6/7... etc Now note that M could be 5 in which case it must be a factor of N. Or M could be 6 (a multiple of 3) in which case only 2 needs to be a factor of N. N may not be divisible by 6 in this case. M = 5, N = 20 is possible M = 6, N = 14 is also possible (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. This tells us that 13N is divisible by M. M cannot be 13 and cannot have 13 as a factor since M is less than 13. So if 13N is to be divisible by M, N must be divisible by M. e.g. if 13*14 is divisible by M (which is less than 13), M must be a factor of 14. Hence this statement alone is sufficient. Answer (B)
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Re: A school administrator will assign each student in a group
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28 Jul 2017, 00:23
Here is how I thought about it:
From the stem we learn that we need to divide a certain number of students equally to a certain number of classes. It basically implies that we need n (number of students) to be divisible by m (number of classes). If that wasn't the case, we would have "remainder" students that were to be allocated to some classes (and, thus, we would have uneven number of students in all the classses)
Once we understand that, the problem becomes simple plugandplay:
3 < m < 13 < n
(1) 3N / M  integer
'Yes' case: 3 * 15 / 5 = 9 (thus, 3N is divisible) 15 / 3 = 5 (thus, N is divisible)
'No' case: 3*16 / 6 = 8 (thus, 3N is divisible) 16 / 6 = 2.66666... (thus, N is not divisible)
NOT SUFFICIENT
(2) 13N / M  integer
'No' case: 13 * 14 / 7 = 26 (thus, 13N is divisible) 13/7 = 1.85... (thus, N is not divisible)
'Yes' case: doesn't exist as 13 is not divisible by any number in between 3 and 13 (exluding both ends)
B; second statement is sufficient!



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Re: A school administrator will assign each student in a group
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07 Dec 2017, 04:34
One approach based on reasoning instead of arithmetic and choosing numbers:
Basically, N has to be a multiple of M within the constraint 3<M<13<N.
Per a. 3N is a mutliple of M => M could be a factor of 3 or N. Does not suffice Per b. 13N is a multiple of M => M is a factor of N given the constraint. Suffices
Hope that helps.
 A




Re: A school administrator will assign each student in a group &nbs
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