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Every member of Meg’s immediate family agrees to share equal

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Manager
Joined: 11 Aug 2012
Posts: 124

Kudos [?]: 135 [0], given: 16

Schools: HBS '16, Stanford '16
Every member of Meg’s immediate family agrees to share equal [#permalink]

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05 Jan 2013, 08:05
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Every member of Meg’s immediate family agrees to share equally the cost of her wedding, which is $18,000 in total. How many people are in Meg’s immediate family? (1) Everyone in Meg’s immediate family will pay$1,500.
(2) If 4 immediate family members do not contribute their share, each of the other family members will have to contribute $750 more than if everyone had contributed. I agree with the OA. However, the statement # 2 requires to solve a quadratic equation, which requires more time. Is there a way to know that I will get just one valid answer in that quadratic equation? That would save time instead of solving the equation. Thanks! [Reveal] Spoiler: OA Kudos [?]: 135 [0], given: 16 Board of Directors Joined: 01 Sep 2010 Posts: 3372 Kudos [?]: 9285 [0], given: 1168 Re: Every member of Meg’s immediate family agrees to share equal [#permalink] Show Tags 05 Jan 2013, 08:13 danzig wrote: Every member of Meg’s immediate family agrees to share equally the cost of her wedding, which is$18,000 in total. How many people are in Meg’s immediate family?
(1) Everyone in Meg’s immediate family will pay $1,500. (2) If 4 immediate family members do not contribute their share, each of the other family members will have to contribute$750 more than if everyone had contributed.

I agree with the OA. However, the statement # 2 requires to solve a quadratic equation, which requires more time. Is there a way to know that I will get just one valid answer in that quadratic equation? That would save time instead of solving the equation. Thanks!

can you please elaborate the 2 statement ??

Thanks
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Kudos [?]: 9285 [0], given: 1168

Manager
Joined: 11 Aug 2012
Posts: 124

Kudos [?]: 135 [0], given: 16

Schools: HBS '16, Stanford '16
Re: Every member of Meg’s immediate family agrees to share equal [#permalink]

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05 Jan 2013, 08:26
carcass wrote:
danzig wrote:
Every member of Meg’s immediate family agrees to share equally the cost of her wedding, which is $18,000 in total. How many people are in Meg’s immediate family? (1) Everyone in Meg’s immediate family will pay$1,500.
(2) If 4 immediate family members do not contribute their share, each of the other family members will have to contribute $750 more than if everyone had contributed. I agree with the OA. However, the statement # 2 requires to solve a quadratic equation, which requires more time. Is there a way to know that I will get just one valid answer in that quadratic equation? That would save time instead of solving the equation. Thanks! can you please elaborate the 2 statement ?? Thanks It is the clue # 2 of the question: x = number of people who will pay for the wedding. $$\frac{18000}{(x - 4)} = \frac{18000}{x} + 750$$ Solving, we get: $$x^2 - 4x - 96 = 0$$ $$(x - 12)(x+8) = 0$$ $$x = 12$$ Sufficient. But that requires a lot of time! Kudos [?]: 135 [0], given: 16 Board of Directors Joined: 01 Sep 2010 Posts: 3372 Kudos [?]: 9285 [0], given: 1168 Re: Every member of Meg’s immediate family agrees to share equal [#permalink] Show Tags 05 Jan 2013, 08:36 mmmm I read the second statement with careless, now infact the translation into math is clear. I disagree It takes time simply because once you setup the quadratic equation you have TWO minus, so whatever wuould be the solution one must be positive and one MUST be negative (is not importanto to solve) A negative solution is not possible because we are talking about people so we must have ONE positive solution to consider. seen and considered we are in the DS land, we have a solution so is sufficient. _________________ Kudos [?]: 9285 [0], given: 1168 MBA Section Director Status: Back to work... Affiliations: GMAT Club Joined: 22 Feb 2012 Posts: 4774 Kudos [?]: 3787 [0], given: 2445 Location: India City: Pune GMAT 1: 680 Q49 V34 GPA: 3.4 WE: Business Development (Manufacturing) Re: Every member of Meg’s immediate family agrees to share equal [#permalink] Show Tags 05 Jan 2013, 11:36 danzig wrote: Every member of Meg’s immediate family agrees to share equally the cost of her wedding, which is$18,000 in total. How many people are in Meg’s immediate family?
(1) Everyone in Meg’s immediate family will pay $1,500. (2) If 4 immediate family members do not contribute their share, each of the other family members will have to contribute$750 more than if everyone had contributed.

I agree with the OA. However, the statement # 2 requires to solve a quadratic equation, which requires more time. Is there a way to know that I will get just one valid answer in that quadratic equation? That would save time instead of solving the equation. Thanks!

Wedding cost = 18000
Number of family members = N
Share of each member=S

(18000/N)=S
Question is What is N? So we need value of either S or N.

S1) S=1500 so N=12 Sufficient.
S2) (18000/(N-4))-750=18000/N ------> 3000N-750N(sq)+72000=0 ------> N(sq)-4N-96=0 ------> (N-12)(N+8)=0 ------> N=12 (N can not be negative) sufficient.

as for your query about Is there a way to know that I will get just one valid answer in that quadratic equation? consider the following characteristics of quadratic equation.
if b(sq)-4ac=0 roots are real and equal.
if b(sq)-4ac>0 roots are real and distinct.
if b(sq)-4ac<0 roots are imaginary.
further if the sign of C is negative then the roots will have opposite signs.
so in our equation N(sq)-4N-96=0 C is negative and b(sq)-4ac>0 so you can conclude at this point - without solving the equation - that you will get real and distinct roots of opposite signs which ultimately means N has one positive & one negative value. Now here negative value has no any meaning because number of family members can not be negative. So with positive value of N we can reach to definite answer and hence statement is sufficient.
_________________

Kudos [?]: 3787 [0], given: 2445

Re: Every member of Meg’s immediate family agrees to share equal   [#permalink] 05 Jan 2013, 11:36
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