GMAT Word Problem and Inequality

Expert replies

abhi75: Master | Next Rank: 500 Posts; Posts: 105; Joined: Tue Oct 16, 2007 5:57 am; Thanked: 3 times

by abhi75 » Mon Jun 23, 2008 6:39 pm

A school administrator will assign each student in a group of n students to one of the 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 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 the m classrooms so that each classroom has the same number of students assigned to it.

Source: Beat The GMAT — Data Sufficiency | Mon Jun 23, 2008 6:39 pm

beeparoo: Master | Next Rank: 500 Posts; Posts: 200; Joined: Sun Jun 17, 2007 10:46 am; Location: Canada; Thanked: 9 times

by beeparoo » Thu Jun 26, 2008 10:07 am

Here's my crack at it.

Question is asking: Is n/m is an integer? Yes or No?

Statement (1): 3n/m = integer. Here, m can be a multiple of 3 or be one of n's factors.
If m is 6, while n is say, 14, then n/m will not result in an integer. Ans is NO.
But if m is 7, while n is 14, then indeed, the answer is YES.
INSUFFICIENT

Statement (1): 13n/m = integer.
Since m < 13, m is not a multiple of this prime number. Therefore, m must be a factor of n, and n/m must be an integer. SUFFICIENT

I choose B. What is the OA?

navalpike: Master | Next Rank: 500 Posts; Posts: 103; Joined: Mon May 04, 2009 11:53 am; Thanked: 6 times

by navalpike » Thu Jun 25, 2009 7:18 pm

good solution beeparoo.

Does anyone also have an algebraic solution? Never hurts to know multiple ways to solve these problems.

ketkoag: Legendary Member; Posts: 876; Joined: Thu Apr 10, 2008 8:14 am; Thanked: 13 times

by ketkoag » Sun Jun 28, 2009 8:05 am

Yeah, B it is.. this question has been posted many times on this forum..

aj5105: Legendary Member; Posts: 1169; Joined: Sun Jul 06, 2008 2:34 amFollowed by: 1 member; Thanked: 25 times

by aj5105 » Sun Jun 28, 2009 9:22 pm

Very nice solution beeparoo.

berserk: Newbie | Next Rank: 10 Posts; Posts: 6; Joined: Mon Mar 23, 2009 2:28 pm

by berserk » Sat Nov 28, 2009 8:17 pm

very good explanation!
I hv gotten struggled for this Q for an hour.
Thanks...

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