A school administrator will assign each student in a group : GMAT Data Sufficiency (DS)
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# A school administrator will assign each student in a group

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

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07 Oct 2005, 13:46
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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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-school-administrator-will-assign-each-student-in-a-group-127509.html
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Re: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink]

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31 Oct 2009, 10:11
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Hi,

you can solve it by Picking Numbers strategy, but I suggest you another, which for me is less confusing.

"...so that each classroom has the same number of students assigned to it" means n is divisible by m. Therefore, m is a factor of n.

The stem asks "If 3<m<13<n, is n divisible by m?"

To be divisible by any of 3<m<13, the number should be at least a LCM of all 3<m<13. To find LCM write all possible m by prime factors...

4=2*2*1
5=5*1
6=3*2*1
7=7*1
8=2*2*2*1
9=3*3*1
10=5*2*1
11=11*1
12=3*2*2*1

So, the LCM of all 3<m<13 is 2*2*2*3*3*5*7*11.

Let's test the statements.

1) 3n is divisible by m.
From the expression of LCM you can see that if 3n is at least a multiple of 2*2*2*3*3*5*7*11, n is at least a multiple of 2*2*2*3*5*7*11, which could or could not be divisible by 3<m<13. So, (1) is NOT SUFFICIENT.

2) 13n is divisible by m.
Since expression of LCM of all 3<m<13 doesn't contain 13 it is necessary for n to be at least a multiple of 2*2*2*3*3*5*7*11. Therefore, n is divisible by m. So, (2) is SUFFICIENT.

The answer is B.

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Re: "A school administrator..." [#permalink]

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11 Jun 2011, 17:24
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Here is a video solution to the above problem:

http://www.gmatquantum.com/list-of-vide ... ds128.html

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Re: Students per classroom [#permalink]

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05 Sep 2011, 08:41
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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.

Question: n/m = integer? (i.e. is m a divisor of n?)

From Statement 1
3n/m = integer
factors of 3n: 3n, 3, n, 1
Since m>3 and m<13, m can be divisor of either 3n or n.
Insufficient.

From Statement 2
13n/m = integer
factors of 13n: 13n, 13, n, 1
Since m>3 and m<13, m can only be divisor of n.
therefore, n/m is an integer.
Sufficient!

Answer: B
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Re: Data Sufficiency #128 [#permalink]

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22 Nov 2010, 07:19
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jferes2345 wrote:
Hello everyone,

I need some help with an official question. #128 in the DS section of the 12th ed. official guide:

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.

Correct answer is B.

I understand why (1) is not sufficient however I can not grasp why (2) is sufficient. Can someone please explain?

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.
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05 Feb 2006, 01:19
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willget800 wrote:
B?

If 13n is assigned to m classes... then the number n is surely divisible by m since m cannot be 13 or 1.. So we are sure that m can be divided into n..

Is that right?

OA?

Good explanation.
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a hard one [#permalink]

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10 Sep 2009, 05:18
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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.

[Reveal] Spoiler:
OA is B
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Re: a hard one [#permalink]

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10 Sep 2009, 06:18
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1) insuff. it means,
3n can be divided by m. it means m can divide n or m can divide some of n's factors*3.
e.g.
n=14, m=6
n cannot be divided by m. But 3n can be
or let say
n=15, m=5
not only n but also 3n is divisible by m
2)suff.
it means 13n can be dividen by m. it means m can divide n or m can divide some of n's factors*13.
in first case, n can be divided by m. in latter case, since m is less than 13, m must divide n.
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Divisibility problem - n students in m classes (+/-700 lvl) [#permalink]

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31 Oct 2009, 06:33
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I was hoping somebody could take a crack at this problem and specifically explain how you beat any useful info out of statement (1) - that statment in specific has me LOST.

A school administrator will assigen 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 the 3n students to one of the 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: Divisibility problem - n students in m classes (+/-700 lvl) [#permalink]

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31 Oct 2009, 08:57
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We are given that 3<m<13<n. this means m has to be between 4-12 or must necessarily be lesser than 13. Also, the number of students n can be assigned to each of the classrooms m equally which means n/m is an intger.

Statement 1 says - (1) It is possible to assign each of the 3n students to one of the m classrooms so that each classroom has the same number of students assigned to it. This means the number of students in each classroom after dividing 3n/m is an integer. This possibility states the number n, when multiplied by 3 is divisble by m. since m>3, this data is sufficient to answer the 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?

Stetement 2 says - (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. Going by the same logic mentioned above, the number of students 13n/m. However, n>13 and m<13, the data is not sufficient for the required question. For.e.g 52 students and 4 classroom is divisible into equal numbers, but 169 students can not be divided equally into any number of given m. Hence statement 2 is not sufficient.

I would go with ans A.
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Re: wrong explanations? [#permalink]

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02 Aug 2010, 10:36
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Hi there,

It looks to me that they do indeed use invalid numbers for n in the explanation for statement 1. However, the principle that they demonstrate is still applicable. The idea is that, because you'll end up with 3n students, the number of classrooms must itself be a multiple of 3. So the value of m can only be 6 or 9. The key is that n itself doesn't have to be divisible by 3 just because m is 6 or 9. n could be 18-->3n = 54-->54 is divisible by 6, and so is 18. But n could also be 22-->3n = 66-->66 is divisible by 6, but the original n, 22, is not. In fact, as long as n is even to begin with, it will always be divisible by 6 when you multiply n by 3. The same thing applies if m = 9. n could be divisible by 9 to begin with (like 18), or it could have just one factor of 3 (like 21), so that when you multiply it by 3 to get 3n, it becomes divisible by 9 thanks to the additional factor of 3.

This problem is testing your understanding of the properties of factors, which is why it doesn't really matter which numbers they use in the explanation for statement 1. That said, you do seem to have found a typo.

Good eyes!
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This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 00:33
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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: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 00:46
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amitjash 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.

If each of the 13n students are assign to one of the m classes => 13n and m have HCF greater than 1.
=> since m <13 and it has hcf>1 with 13n, n and m also have hcf >1. Thus statement 2 is sufficient.

If you apply the same with statement 1, 3n and m have hfc >1. But it is possible that n and m does not have hcf>1 as m could be 3,6,9. Thus not sufficient.

Answer:B

Eg of statement 1: Suppose you have 30 students, you can distribute them in m classes only if 30 and m have hcf>1.
30 = 2*3*5. If m does not have any common factor how can you distribute? Say m = 7. 30 can not be distributed in 7 classes.

Eg of statement 2: say 13n = 13*10 =130 are distributed among m classes and since m<13 => for 13n students to be distributed in m classes, m must be multiple of 5.

Since n and m both have a common factor 5, n students can be distributed among m classes.
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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 00:49
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Mathematically speaking, the question is: is n divisible by m !!

(1) 3n is divisible by m. It doesn't say if n is divisible by m because 3 < m < 13. Indeed, n can be equal to 14 and m = 6. In that case m does neither divide 3 nor n. But with m = 6 and n = 18, m divides n.
INSUFFICIENT

(2) 13n is divisible by n. But 13 is a prime number that m can't divide! Besides 3 < m < 13 so it can't be a combination of 13 and n (like 13* 2) that can be divided by 13.

ANS : B.

Hope it's clear enough...
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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 07:17
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amitjash 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.

I can rephrase the question as :

3<m<13<n. Does m divide n ?
(1) m divides 3n
(2) m divides 13n

(1) : Not sufficient. Eg. m=6, n=14 m does not divide n. But m=6, n=18 m divides n. But each case m divides 3n. So clearly not sufficient

(2) : Since m<13 and 13 is prime. m cannot divide 13, hence m must divide n. So sufficient

Answer is (B)
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Re: This one is tough one... atleast for me [#permalink]

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19 Sep 2010, 15:20
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vigneshpandi wrote:
amitjash 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.

Please help me understand what is wrong in this approach. I just used number.
chice 2: Let n=14..Hence 13X14=182. Now choosing m between 3 < m < 13 < n.
Say m=7. Then 182 is divisible by 7. But when m=5 it is not divisible. So (2) is not sufficient.
Some one please explain what is wrong in choosing numbers to solve this and where I am missing something.

In (2) it is given that m divides 13n. In your example, 7 does divide 13n, ie, 182. But 5 does not, so it is not a valid choice for m
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Re: a hard one [#permalink]

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07 Jun 2011, 02:36
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if you rephrase the question he is asking if n/m = 0

case 1: 3n/m = 0
case 2: 13n/m = 0

in case 1: consider n = 15 3n = 45 and 45 is only divisible by 9 coz 3<n<13 now consider n = 18 3n = 54 now 3n is divisible by 6,9 and both are between 3&12 So Insuff

case 2 : 13n/m = 0 since 13 is prime and m<13 n/m will always be zero.

I hope I could help you.
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Re: a hard one [#permalink]

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09 Jun 2011, 20:56
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If 3 < m < 13, Is n/m an integer?

(1) 3n/m is an integer

(2) 13n/m is an integer

The only way way to ensure that n students can be grouped, without remainder, into m classrooms is that all (m) factors of 3n or 13n were also factors of n itself.

Stat (1)

Where n > 13 and 3n is an integer.  3n and n may or may not have the same factors within the range of m.

This is because 3 has multiples within the range, and these multiples could create factors that n does not have.

Insufficient.

Stat (2)

Where n > 13 and 3 < m < 13,
13n and n will have the same factors within the range of m.  13 has no factors within:
3 < m < 13

n/m is an integer

Last edited by Study1 on 14 Jan 2012, 08:44, edited 2 times in total.
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Re: School Administrator [#permalink]

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10 Jun 2011, 05:26
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To find n/m= int?

a) 3n/m = int.
If a*b/c = int and gcd(a,c) = 1 then b/c = int

Hence 3n/4 = int but 3n/6 is not int

b) 13n/m = int as m<13
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Re: OG 12ed Data Sufficiency question - CONFUSION [#permalink]

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25 Sep 2011, 22:31
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tititalin wrote:

My idea is that, after rephrasing, the question is essentially asking whether "it is possible that n is divisible by m", NOT whether "n IS divisible by m". So given (1) & (2), both statements can lead to the conclusion that it is POSSIBLE that n is divisible by m, not n is ABSOLUTELY divisible by m. So I think the correct answer should be (D) - each statement alone is sufficient.

Does this make sense? Anyone could share some insights on this argument?

Thanks a lot!

Think of it this way:

My question to you: "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?"
The statement will be sufficient if you can answer with a 'YES' or a 'NO'
It will be insufficient if you answer with a 'MAY BE'.

When will you say 'YES'? You will say yes when it will be possible to assign students the way I want. When will it be possible to assign students the way I want? When n will be divisible by m. If you do not know whether n is divisible by m and I ask you, "Can you assign students the way I have asked?" what will you say? You will say, "May be. If n is divisible, then yes, if it is not divisible then no." You will not say yes. You answered my original question with a 'may be'. Statement 1 gives you 'may be it is possible to completely divide n by m'. So your answer is 'may be we can assign the students as you requested' which makes the statement insufficient.
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Re: OG 12ed Data Sufficiency question - CONFUSION   [#permalink] 25 Sep 2011, 22:31

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