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26 Sep 2012, 11:34
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A club collected exactly $599 from its members. If each member contributed at least$12, what is the greatest number of members the club could have?

(A) 43
(B) 44
(C) 49
(D) 50
(E) 51
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Sep 2012, 12:08, edited 1 time in total.
Renamed the topic.
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Re: A club collected exactly $599 from its members. If each [#permalink] ### Show Tags 26 Sep 2012, 12:17 3 This post received KUDOS Expert's post 2 This post was BOOKMARKED Archit143 wrote: A club collected exactly$599 from its members. If each member contributed at least $12, what is the greatest number of members the club could have? (A) 43 (B) 44 (C) 49 (D) 50 (E) 51 Obviously club could not have 50 or more members, since$12*50=$600>$599. What about 49? If 48 members contributes $12 ($12*48=$576) and 1 member contributed the remaining$23, then the club would have is 48+1=49.

OR: $$12x\leq{599}$$ --> $$x\leq{49\frac{11}{12}}$$ --> $$x=49$$ (since x must be an integer).

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30 Jun 2014, 22:34
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21 Jun 2016, 10:34
Archit143 wrote:
A club collected exactly $599 from its members. If each member contributed at least$12, what is the greatest number of members the club could have?

(A) 43
(B) 44
(C) 49
(D) 50
(E) 51

Let's first divide= 599 by 12 (the minimum amount each member could contribute) and then use the remainder to finish the problem.

599/12 = 49 R 11

This means that: 49 people x $12 + 1 person x$11 = $599 We see that if 49 members each contribute$12, someone would have to contribute the extra $11. Note that, since each member contributed at least$12, the $11 could not have come from an additional member. Therefore, the extra$11 must have been contributed by one (or more) of the existing 49 members. Regardless of who contributed the extra $11, the maximum number of members the club could have is 49. Answer is C. _________________ Jeffrey Miller Jeffrey Miller Head of GMAT Instruction Re: A club collected exactly$599 from its members. If each   [#permalink] 21 Jun 2016, 10:34
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