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Problem 128 from OG12

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Problem 128 from OG12

by WannabeGeek » Sun Feb 28, 2010 11:01 am
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.

The OA is: B

Looking at 3<m<13 I would think m can be anything from 4 to 12. So how can we match the factors from (1)3n or (2)13n to m?
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by firdaus117 » Sun Feb 28, 2010 1:14 pm
For each classroom to have the same number of students assigned to it,n should be divisible by 'm'.We need to check if given data is sufficient to prove that.
Now,m=4 to 12 and n=14,15,16,17......
Case 1: 3n is divisible by m.As m>3,3 is not divisible by m.This implies that:
(i) n is divisible by m.
(ii) n is not divisible by m.Even then,3n is divisible by n.Examples are n=14,m=6 or n=16,m=12
Case 2:13n is divisible by m.As 13 is a prime number,it is not divisible by any values of m(4 to 12).This implies that n has to be divisible by m.
Hence,Option B
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by WannabeGeek » Sun Feb 28, 2010 5:01 pm
Thank you for the response Firdaus117. I did understand the question, so I did have what you mentioned.

One of my concerns was that the only way I could derive to this answer was with trial and error, wasn't sure if there was a cleaner and faster way to do it.

Secondly, I understand (1) is insufficient. But with an example, I can prove (2) is insufficient. While n=17, no value of m between 4 and 12 divides with 13*17=221. So how can the answer be B?

PS: I have looked up an old post on this question which does not seem to answer my concern:
https://www.beatthegmat.com/a-school-adm ... 42237.html
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by firdaus117 » Mon Mar 01, 2010 9:09 am
WannabeGeek wrote: Secondly, I understand (1) is insufficient. But with an example, I can prove (2) is insufficient. While n=17, no value of m between 4 and 12 divides with 13*17=221. So how can the answer be B?
You can't take n=17 as statement B says: 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.
For n=17,13n=221 is not divisible by any value of m (4 to 12).
On the other hand pick any value of n such that 13n is divisible by any value of m,you will find n is also divisible by m.
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by GMATSUCKER » Mon Mar 01, 2010 11:47 am
firdaus117 wrote:
WannabeGeek wrote: Secondly, I understand (1) is insufficient. But with an example, I can prove (2) is insufficient. While n=17, no value of m between 4 and 12 divides with 13*17=221. So how can the answer be B?
You can't take n=17 as statement B says: 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.
For n=17,13n=221 is not divisible by any value of m (4 to 12).
On the other hand pick any value of n such that 13n is divisible by any value of m,you will find n is also divisible by m.
Firdaus117 has adopted a good aproach !
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by Stuart@KaplanGMAT » Mon Mar 01, 2010 1:12 pm
WannabeGeek wrote:Thank you for the response Firdaus117. I did understand the question, so I did have what you mentioned.

One of my concerns was that the only way I could derive to this answer was with trial and error, wasn't sure if there was a cleaner and faster way to do it.

Secondly, I understand (1) is insufficient. But with an example, I can prove (2) is insufficient. While n=17, no value of m between 4 and 12 divides with 13*17=221. So how can the answer be B?

PS: I have looked up an old post on this question which does not seem to answer my concern:
https://www.beatthegmat.com/a-school-adm ... 42237.html
Firdaus's reply is excellent, I just want to focus on the big takeaway from this thread.

Here's one of the primary rules of data sufficiency:

The statements are laws of the universe. When picking numbers, we are only allowed to work with numbers that follow those laws.

The error you made was picking an impermissible number. Choosing n=17 doesn't give you a "no" answer to the original question, it merely violates the rule stated in statement (2).

There are two steps for picking numbers in data sufficiency:

1) pick permissible numbers (i.e. numbers that follow the rules in the original question and the statement); and

2) plug those numbers back into the original question.

It's only if the number chosen passes the test in step 1 that we get to the plugging in stage. If you choose a number that violates one of the laws you're given, you have to throw that number out and start again with a new one.

Mistaking an impermissible number for a "no" answer is one of the most common mistakes in data sufficiency.
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by kstv » Mon Mar 01, 2010 7:56 pm
trying to understand Firdaus's approach.
3<m<13 (4, ....12)
n > 13 (14..............

1) 3n is divisible by m
but m > 3 so 3 cannot be divided by m
but it is possible 3 is a factor of m , so possible values of m are 6,9,12 then n is a even no. or multiple of 3.
then 3*14/6 = 7 or 3*21/9
but 14 (n) may not be divisible by 6(m) similary 21 is not divisble by 9 . Insufficient.

2) 13n is divisible by m, 13 a prime no so not divisible by m
also 13 cannot be a factor of m as m < 13
so m has to be a factor of n. Sufficient.

Not the easiest of problems and it is in OG12. Is it a 700+ Q ?
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