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A school administrator will assign each student in a group

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28 Nov 2020, 15:11

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Video solution from Quant Reasoning:
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My answer was D.

I am not able to understand the reasoning behind statement 1 as insuffucient. The question is asking "is it possible to get integer values of n/m" It is NOT asking if n/m is an integer. As most of the posts here have pointed out that statement 1 also provides at least one possible solution where n/m is an integer (for example: n = 20, m = 10) So, even with statement 1 alone, we can answer that IT IS POSSIBLE.

Can someone please help? Thanks.

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31 Jan 2021, 07:43

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Dvaishnav

Yeah, good question. A couple of points to make here:

1. No DS question will ever ask whether something is possible, so if you interpret a question that way you're certainly mistaken. To illustrate, consider the following example:
Is it possible that x is positive?
(1) x<0
(2) x<3
Each statement is sufficient on its own: (1) gives a definite NO and (2) gives a "definite" YES. [side note: statements can't lead to different definite answers since that would mean one of them is false, which is not a thing on the GMAT]
Another example:
Is it possible that x is positive?
(1) x>0
(2) x>-3
Again each statement is sufficient on its own. Can you create a statement that wouldn't be sufficient to answer the question "Is it possible that x is positive?" - such a statement doesn't exist.
Side note: this isn't just for YES/NO questions - same goes for value questions (think about it).

2. Why is it wrong to rephrase the original question to "is it possible that n/m is an integer?"
Under what circumstances would it be 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 would ONLY be possible if n/m is an integer. So, if n/m is an integer the answer is YES (it's possible), and if n/m isn't an integer then the answer is NO (it's not possible).
Therefore, the correct rephrase is: is n/m an integer?

By the way, the second section of my book is dedicated exclusively to DS strategy, and you can read it for free at quantreasoning.com

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Aaah.. I looked at the problem as a remainder one and inferred we need remainder =0 when N/M (or 3N & 13N when divide by M leave Remainder=0) and spent time tacking that; only to realise that we can assign all students to one class (remainder = 0) but that won't help us achieve equal students in the class.

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29 Apr 2021, 10:48

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gwiz87

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 can rephrase this question to be is N divisible by M without a remainder?

Statement 1:
3N/M = Integer without remainder. Now let's consider all possible numbers of M which is 3<m<13, so 4,5,6,7,8,9,10,11,12. Let's consider the scenario where we have 3N/6, which can be factored as 3N/(3)(2) and then N/2.

So this means that if N is even then M can divide into N without a remainder, but if N is odd then there will be a remainder of 1. We can identify two different answers so not sufficient.

Statement 2:
13N/M = integer without remainder. This means that M will either divide into 13 or N, but since 13 is prime and M<13 that means that for 13N/M to be an integer M MUST divide into N evenly, which means N/M = integer without a remainder.

Note that without M being restricted to M<13 we could not answer the question; say for example M were = 26, then 13N/13(2), which would have again left us with N/2. So since M IS restricted to <13 then this statement alone is sufficient to answer the question.

B is the answer.

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01 May 2021, 08:25

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how i solved this Q

we basically need to know whether N/M gives R = 0 or not

now st1 :
3N is div by M

M could be div by 3 or M
and we don't know which?
its possible that M is a multiple of 3 (eg 9) but that wouldn't divide N. Therefore not sufficient

st2:
13N is div by M
here we know M <13
so M cant be div by 13
so it has no option but be a factor of N

So st2 is enough

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02 Jun 2021, 19:36

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gwiz87

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.

Hi Bunuel
In case the question had been "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 number of students in increasing order/decreasing order assigned to each classroom?" -- could you please let me know what would have been the approach to solving the problem. Sorry I could not create the whole problem as I do not know the approach myself to solve such problem

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This is a very good question which tests your knowledge of Number properties. Let us break down the question statement and the question stem:

There are N students and M classrooms. Each of these N students will be assigned to sit in one of the M classrooms.

3<M<13<N – the minimum number of rooms = 4 and maximum number of rooms = 12 and the minimum number of students = 14. So, 4≤M≤12 and N≥14.

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 N divisible by M could be the simplified version of the question statement.

From statement I alone, we can infer that 3N is divisible by M. This is not sufficient to find out if N is divisible by M.

If N = 15 and M = 5, 3N = 45 is divisible by M = 5; also, N =15 divisible by M = 5, so we answer the question with a YES.
If N = 21 and M = 9, 3N = 62 is divisible by M = 9; but N = 21 is not divisible by M = 9, so we answer the question with a NO.

Statement I alone is insufficient to obtain a definite YES or NO. Answer options A and D can be eliminated. Possible answer options are B, C or E.

From statement II alone, we can infer that 13N is divisible by M. This is where we need to be more careful and just not replicate what we did with statement I.

Note that 13 is a prime number and is only divisible only by 1 and 13. Therefore, 13 is not divisible by M since M is neither of these two numbers.
Therefore, the only way in which 13N is perfectly divisible by M is because N is divisible by M.

Statement II alone is sufficient to answer the question with a YES. Answer options C and E can be eliminated.

The correct answer option is B.

Hope that helps!
Aravind B T

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We are asked whether N students are evenly divisible into M groups. N/M = integer value?

(1) 3N are evenly divisible by M, so 3N/M = integer.

For this to be true, 3 and N must account for all the prime factors of M, one by itself or together.

If M = 4, then prime factors are 2*2. In this case, M must be a multipe of 4.
If M = 9, then prime factors are 3*3. In this case, M must only be a multiple of 3 as one of the threes are already accounted for. So M can be 15, which is not evenly divisible by 9, or M can be 18, which is. Therefore, 1 is insufficient.

(2) 13N is evenly divisible by M, so 13N/M = integer. Because M can not have any factors in common with 13 (itself a prime number), N definitely must be a multiple of M. Therefore, 2 is sufficient.

Answer: B

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I do understand the reasoning behind this question.

But I guess I made it harder that it was. So I have a question. How were you all sure that there were no students in the m classes before? I mean, let's say there were 4 classes (m=4) and there are say 2 in class 1, 4 in class 2, 3 in class 3 and 6 in class 4 already. Then you have the n students that need to be adjusted among these 4 (m) classes. Am I the only one who understood it this way?

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ravikirancg

ravikirancg, if we think along those lines, there is no end to what we can and cannot assume. Perhaps, there is a miscount? It's best to stick to what is given.

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I want to reiterate Brunel's post, but with a slight twist.

1) If 3 times the number of students (n) divided by the number of classrooms (m) yields an integer (no remainder), we know this is because either 3 is a common factor of the number of classrooms or because the original number of students (n) has a common factor with the number of classrooms (m) (or both!) (this is because some valid integers for 'm' are 6, 9, or 12, which all have a 3 in their prime factorization). Because both of these scenarios can be true, we cannot yield a definitive 'yes' or 'no' to this statement.

I.E. If 'n' doesn't hold a prime factor of 'm' (say for example n = 17), we could still derive an integer from 3n / m. On the flip side of the coin, if 'n' did hold a prime factor of 'm' (say for example 15), we would also derive an integer from 3n / m.

2) Along the same line of logic, because 13 ISN'T a valid prime factor of 'm' (remember, our only valid prime factors come from our valid range of integer values - namely, 4 through 12), 'n' therefore MUST be have a prime factor of 'm' in it, yielding us an integer value and, therefore, each student will be evenly divided into 'm' classrooms.

Hope this helps! If there's any flaw in my logic, please let me know.

Lee

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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?

In order for N to be equally divided into M, M must be a multiple of N.
Because 3<M<13<N, M possible number would be 4, 5, 6, 7, 8, 9, 10, 11, or 12.

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.

For 3N to be equally distributed to M, M must be a multiple of 3N. In other words, we have three scenarios: M is a multiple of N, of 3, or of 3N.

If M is indeed a multiple of N, then we can safely say that it is possible to assign each of N to M classrooms. However, if M is a multiple of 3, then M might or might not a multiple of N. Consider the following example:

M = 12
N = 40

You can't distribute 40 students equally to 12 classrooms. 3N, which equals to 120, however, can clearly be equally distributed to 12 classrooms.
Therefore, statement 1 is insufficient.

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.

Three scenarios: M is a multiple of N, of 13, or of 13N.

But remember that it is impossible for M to be a multiple of 13, since 3<M<13<N.
Therefore, M must be a multiple of N, and the answer to the question is yes.

Thus, statement 2 is sufficient.

Answer: B

Thank you!

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BrentGMATPrepNow

rakeshd347

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.
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

Answer = B

Cheers,
Brent

Hi Brent BrentGMATPrepNow, in St 1. Assigning m = 4 and n = 20, in which case n/m is an integer. Does n not need to multiply with 3 here like (3 * 20)/4?
To clarify with St 2. is it there's no number of m that is a factor of 13 due to 3 < m < 13 < n , therefore n/m = integer? Thanks Brent

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Statement 1 tells us that the number of students (3n) is divisible by the number of classrooms (m). In other words, 3n/m is an integer.

So, yes, we are multiplying n by 3.
The two pairs of numbers satisfy statement 1.
case a: m = 4 and n = 20, in which case n/m is an integer.
Notice that, in this case, 3n/m = 3(20)/4 = 15, which is an integer.

case b: m = 6 and n = 20, in which case n/m is not an integer.
Notice that, in this case, 3n/m = 3(20)/6 = 10, which is an integer.

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25 Nov 2022, 14:47

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Kimberly77

Thanks BrentGMATPrepNow for confirmation. I presumed my question of St 2 below is correct right since you didn't say is not right

?

with St 2. is it there's no number of m that is a factor of 13 due to 3 < m < 13 < n , therefore n/m = integer? Thanks Brent

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Kimberly77

Thanks BrentGMATPrepNow for confirmation. I presumed my question of St 2 below is correct right since you didn't say is not right

?

with St 2. is it there's no number of m that is a factor of 13 due to 3 < m < 13 < n , therefore n/m = integer? Thanks Brent

"with St 2. is it there's no number of m that is a factor of 13 due to 3 < m < 13 < n , therefore n/m = integer?"
The above is correct

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Brilliant and thanks BrentGMATPrepNow for confirmation

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