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A school administrator will assign each student in a group

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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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Video solution from Quant Reasoning:
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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.

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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:
https://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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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.

To assign the same number of students to each classroom, the number of students (n) must be divisible by the number of classrooms (m).

Question rephrased: Is $$\frac{n}{m}$$ 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 assigned to it.
In other words, the number of students here (3n) is divisible by the number of classrooms (m), implying that $$\frac{3n}{m}$$ is an integer.
Since we need to determine whether m will always divide into n, plug in EXTREME values for m.

m=4:
It's possible that m=4 and n=16, with the result that $$\frac{3n}{m} = \frac{3*16}{4} = 12$$.
In this case, then $$\frac{n}{m} = \frac{16}{4} = 4$$, which is an integer.

m=12:
It's possible that m=12 and n=16, with the result that $$\frac{3n}{m} = \frac{3*16}{12} = 4$$.
In this case, $$\frac{n}{m}= \frac{16}{12} = \frac{4}{3}$$, which is NOT an integer.
INSUFFICIENT.

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 assigned to it.
In other words, the number of students here (13n) is divisible by the number of classrooms (m), implying that $$\frac{13n}{m}$$ is an integer.
It is not possible that m divides into 13, since the only factors of 13 are 1 and 13, and m must be BETWEEN 3 and 13.
Thus, for $$\frac{13n}{m}$$ to be an integer, m must divide into n, implying that $$\frac{n}{m}$$ is an integer.
SUFFICIENT.

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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.

MATH REVOLUTION VIDEO SOLUTION:

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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.

We are given that each student in a group of n students is going to be assigned to one of m classrooms. We are being asked whether it is possible to assign each of the n students to one of the m classrooms so that each classroom has the same number of students.

Thus, we need to determine whether n/m = integer.

Statement One Alone:

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.

Statement one is telling us that 3n is evenly divisible by m. Thus, 3n/m = integer.

However, we still do not have enough information to answer the question. When n = 16 and m = 4, n/m DOES equal an integer; however, when n = 20 and m = 6, n/m DOES NOT equal an integer.

Statement Two Alone:

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 statement is telling us that 13n is divisible by m. Thus, 13n/m = integer.

What is interesting about this statement is that we know that n is greater than 13 and that m is less than 13 and greater than 3. Thus, we know that m could equal any of the following: 4, 5, 6, 7, 8, 9, 10, 11, or 12. We see that none of those values (4 through 12) will divide evenly into 13.

Knowing this, we can say conclusively that m will never divide evenly into 13. Thus, in order for m to divide into 13n, m must divide evenly into n.

General Discussion
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Thanks Bunuel! Your explanation was much clearer than the guide.
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Re: A school administrator will assign each student in a group [#permalink]
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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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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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
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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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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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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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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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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.

Given: 3<M<13<N

Question: Can N/M be an integer?

Statement 1: 3N/M is an integer. Not sufficient as this does not conclusively answer whether N/M is an integer. Either N/M is an integer or M is a multiple of 3. Both are possible. For example when 3N/M = (3*15)/5 then N/M is an integer but when 3N/M = (3*14)/6, M is a multiple of 3 and N/M is not an integer.

Statement 2: 13N/M is an integer. Either N/M has to be an integer or M is a multiple of 13. Because M<13, the latter is ruled out. Therefore N/M is an integer.

So the question can be answered from this statement alone.
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Bunuel 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?

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.
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