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# M13-34

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Math Expert
Joined: 02 Sep 2009
Posts: 44293

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16 Sep 2014, 00:50
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Difficulty:

55% (hard)

Question Stats:

57% (01:17) correct 43% (01:16) wrong based on 82 sessions

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Each of the $$N$$ members of a club agreed to contribute $$X$$ dollars to buy a new computer. If the club contained 2 more members, how much less would everyone have to contribute?

A. $$\frac{NX}{(N + 2)}$$
B. $$X(N + \frac{N}{(N + 2)})$$
C. $$\frac{N}{(XN + 2X)}$$
D. $$X(1 - \frac{N}{(N + 2)})$$
E. $$XN - \frac{N}{(N + 2)}$$
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 44293

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16 Sep 2014, 00:50
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KUDOS
Expert's post
Official Solution:

Each of the $$N$$ members of a club agreed to contribute $$X$$ dollars to buy a new computer. If the club contained 2 more members, how much less would everyone have to contribute?

A. $$\frac{NX}{(N + 2)}$$
B. $$X(N + \frac{N}{(N + 2)})$$
C. $$\frac{N}{(XN + 2X)}$$
D. $$X(1 - \frac{N}{(N + 2)})$$
E. $$XN - \frac{N}{(N + 2)}$$

The computer costs $$NX$$ dollars. If the club contained 2 more members, everyone would have to contribute $$Y$$ dollars. Because $$(N + 2)Y = NX, Y = \frac{NX}{(N + 2)}$$. This is by $$X - \frac{NX}{(N + 2)}$$ less than the actual contribution of $$X$$.

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Joined: 02 Jun 2015
Posts: 191
Location: Ghana

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16 Aug 2016, 04:01
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KUDOS
(# of members) * (Contribution per member) = Total contribution ===> Let's N = 8 and X = $5 N * X XN 8 * 5 = 40 (Plus 2 members) * New Contr. per member = Total contribution (N + 2) * (XN/N+2) = XN 10 * 4 = 40 Now everybody pays$1 (i.e., 5 - 4) less after the addition of two more members.

Now backsolve using the answer choices:

Only D results in 1
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Joined: 10 Jul 2017
Posts: 2
Location: United States
Schools: Stanford MSx"19
GMAT 1: 720 Q47 V41
GPA: 3.56

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13 Aug 2017, 06:40
1
KUDOS
Wording on this one is tricky...

I read it as "Each of the N members agreed to (collectively) contribute X dollars" when it is actually asking "Each of the N members agreed to contribute X dollars (each)"
Re: M13-34   [#permalink] 13 Aug 2017, 06:40
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# M13-34

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