OG16 - DS 149

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OG16 - DS 149

by amina.shaikh309 » Sun Jun 26, 2016 1:49 am

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by GMATGuruNY » Sun Jun 26, 2016 5:11 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.
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 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 (3n) is divisible by the number of classrooms (m).
Implication:
(3n)/m = 3(n/m) = integer.

Case 1: n/m = integer
It's possible that n=16 and m=4, with the result that n/m = 16/4 = 4.
Here, 3(n/m) = 3(16/4) = 12.

Case 2: n/m = k/3, where k is not a multiple of 3
In this case, since n/m = k/3, m must be a multiple of 3.
It's possible that n=14 and m=6, with the result that n/m = 14/6 = 7/3.
Here, 3(n/m) = 3(7/3) = 7.

Since n/m is an integer in Case 1 but not in Case 2, 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 (13n) is divisible by the number of classrooms (m).
Implication:
(13n)/m = 13(n/m) = integer.

Case 3: n/m = integer
It's possible that n=16 and m=4, since 16/4 = 4.
Here, 13(n/m) = 13(16/4) = 52.

Case 4: n/m = k/13, where k is not a multiple of 13
In this case, since n/m = k/13, m must be a multiple of 13.
Not possible, since 3 < m < 13.

Since only Case 3 is possible, n/m = integer.
SUFFICIENT.

The correct answer is B.
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by Brent@GMATPrepNow » Sun Jun 26, 2016 5:19 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.
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- ... cy?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. In fact, we can't even reduce the 13/m to simpler terms. 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

For even more information on rephrasing the target question, you can read this article I wrote for BTG: https://www.beatthegmat.com/mba/2014/06/ ... t-question

Cheers,
Brent
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by 800_or_bust » Sun Jun 26, 2016 11:16 am
amina.shaikh309 wrote:Image
(1) Not sufficient. Consider the possibility that m = 6. So we know 3n must be divisible by 6. It could be the case that n itself is divisible by 6; however, it could also be the case that n is merely divisible by 2. So that when we multiply it by 3, the resulting product is divisible by 6.

(2) Sufficient. The number of classrooms, m, is less than 13. 13 is not independently divisible by any integer between 4 and 12, inclusive. Further, note that since 13 is prime, it cannot be factored into anything smaller than 13. Thus, in order for 13n to be divisible by some integer in the range from 4 to 12, inclusive, it must be the case that n itself is divisible by that integer m. Hence, this is sufficient.
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