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

Re: A school administrator will assign each student in a group
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15 Nov 2015, 20:54

1

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

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 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

Re: A school administrator will assign each student in a group
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22 Jan 2016, 08:57

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.

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

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05 May 2016, 20:45

1

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 co-primes 3 and 5.. so BOTH possiblity that N/M is an integer and not an integer exist.. so insuff..
_________________

Re: A school administrator will assign each student in a group
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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.
_________________

You 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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Re: A school administrator will assign each student in a group
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25 Oct 2016, 05:27

1

2

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

Answer: option B
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Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

A school administrator will assign each student in a group
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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.

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?

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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.

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 \(\frac{n}{m}=integer\)?".

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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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16 Jun 2017, 09:10

bhavikagoyal2009 wrote:

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"?

The point is that m and n are some specific numbers there. So, what the question means is that CAN we 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?
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27 Jun 2017, 19:59

1

1

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 plug-and-play:

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)

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13 Dec 2017, 06:12

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.

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21 Sep 2018, 12:53

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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06 Jan 2020, 13:31

1

If you simplify this question, it will be one of the easiest questions. One of the most important rephrasing : Whenever you see "EACH something has the SAME/EQUAL number of another thing", remember that it's a DIVISIBILITY question.

In other words, Arrange x entities in y groups in a way that each y has the same/equal number of x= x is divisible by y.

Even if you do not know which is GROUP and which one is entity, it's not a problem. Just make sure that you understand it's a DIVISIBILITY problem. Normally, LARGER value is ENTITY and SMALLER value is GROUP.

In simple words, is the LARGER number divisible by SMALLER number?

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06 Jan 2020, 13:57

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BelalHossain046 wrote:

Back to the question ( Simplified version): If 3<M<13<N, Is it possible that N is DIVISIBLE by M? i.e. is N DIVISIBLE by M?

Statement 1: 3N is DIVISIBLE by M. Statement 2: 13N is DIVISIBLE by M.

I think it's an easy question now.

Statement 1: 3N is DIVISIBLE by M Minimum value N=14, So, 3N=42. 42 is DIVISIBLE by 6 or 7. So, M=6 or M=7. If M=7, N(=14) is DIVISIBLE by M (YES to MAIN QUESTION) and, If M=6, N(=14) is NOT DIVISIBLE by M (NO to MAIN QUESTION) NOT SUFFICIENT.

Statement 2: 13N is DIVISIBLE by M [13*N][/M]=Completely DIVISIBILITY=Integer Here, M is less than 13. So, 13 is NOT DIVISIBLE by M [More importantly, 13 & M has NO DIVISIBILITY or REDUCTION/COMMON FACTOR relationship.] So, N MUST BE DIVISIBLE by M to make [13*N][/M] Completely DIVISIBLE SUFFICIENT.

My Answer: B

gmatclubot

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