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Q28,set 9-15testset

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Q28,set 9-15testset

by tanviet » Sat Oct 17, 2009 6:33 pm
pls, help with this. how to solve this kind fast.

A school administer will assign each student in a group of n student to one of m classroom. If 3<m<n<, is it possible to assign each of the n students to one of the m classroom 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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Source: — Data Sufficiency |
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Re: Q28,set 9-15testset

by Talkativetree » Sun Oct 18, 2009 4:34 pm
duongthang wrote:pls, help with this. how to solve this kind fast.

A school administer will assign each student in a group of n student to one of m classroom. If 3<m<n<, is it possible to assign each of the n students to one of the m classroom 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 do this problem quickly, you need to have a good understanding of multiples, factors, and division.

(1) says basically that if you diving 3n by m, or 3n/m, that it will divide evenly, or to say it differently, that there will be no remainder.

Okay, so from this, you can infer several things. 1, M is a factor of N, or to say it differently, that N is a multiple of M. this means that M will divide evenly into N, so this means that the question is sufficient. Now, the 3 can be ignored because n>m>3, so we don't have to worry about whether or not m is a factor or n or 3, because m>3 and 1 is a shared factor in all numbers.

(2) looking at this, it's basically the same thing, but this time 13 is greater than 3, so m may be a factor of n, OR a factor of 13. (remember, students are whole units, so they must be positive integers). This means (i believe) that (2) is insufficient, so the Answer would be A.

So to illustrate this point, for (2)

N could be 12 and the factors of 12 are 1,2,3,4,6, & 12. However, because in (2) 13>3, unlike (1), m could = 13, where as n could be any number greater than m.

So the answer, imo, is A
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by boy141 » Tue Oct 27, 2009 1:41 pm
OA is B


But the reasons they give it is really cryptic.. For 1), why did they use n=12 ? N cannot be 12 because N has to be more than 12.

2)Their reasoning is a whole lot different than 1. Why couldn't they use the same reasoning to do both.

BTW, it's DS: #128 in the 12th Edition.
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by tanviet » Thu Oct 29, 2009 1:00 am
sory,

in the question, the condition is 3<m<13<n
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by tanviet » Thu Oct 29, 2009 1:34 am
picking the number is solution to yes-no question of number property, hardest question.

a, if n=16,m=12, n can not be divided by m

if n=24, m=12, n can be divided by m

because m<13, m can be multiple of 3 such as 3,6,9,12 or 1*3,2*3,3*3,4*3 so 3n/m is interger, so n can be multiple of 1,2,3,4 and greater than 13, pluging the number, we have result above.

not sufficient

b,13n/m is interger, m<13, so 13 and m have no common factor, n must be multiple of m

sufficient.

answer is B
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