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Every member of a certain club volunteers to contribute equally to the purchase of a $60 gift certificate. How many members does the club have?

Let the # of members be \(n\).

(1) Each member's contribution is to be $4 --> \(n=\frac{60}{4}=15\). Sufficient.

(2) If 5 club members fail to contribute, the share of each contributing member will increase by $2 --> share of each contributing member is \(\frac{60}{n}\), if 5 club members fail to contribute, the share of each contributing member will be \(\frac{60}{n-5}\) and we are told that this values is $2 more than the previous one --> \(\frac{60}{n}=\frac{60}{n-5}-2\) --> \(n=15\). Sufficient.

Re: Every member of a certain club volunteers to contribute equa [#permalink]

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18 Feb 2013, 14:53

1

This post received KUDOS

Hi thangvietnam,

Every member of a certain club volunteers to contribute equally to the purchase of a $60 gift certificate. How many members does the club have?

(1) Each member's contribution is to be $4. (2) If 5 club members fail to contribute, the share of each contributing member will increase by $2.

The problem states that each member contributes equally to the gift certificate. Let the number of people be n and the amount contributed by each member be x. Now, from the question we know that total amount contributed=$60.

Hence, n*x=60

1) This statement tells us that contribution of each member=$4. i.e x=4 Hence, n=60/4=15 SUFFICIENT

2) This tells us that if 5 members fail to contribute every person's contribution share increases by $2. We know n=60/x old n=n new n=n-5 Old contribution=60/n new contribution=60/(n-5)

Now, new contribution=old contribution+2 i.e 60/(n-5)=2+ (60/n)

What remains is 2n^2-10n-300. Take out 2 n^2-5n-150=0

Factorize 150, you get 5*3*2*5 15-10=5 Hence, n^2-15n+10n-150=0 Take out common factors n(n-15)+10(n-15)=0 n=-15 or 10. Since, the number of people cannot be negative it has to be 15.

SUFFICIENT Let me know if I can clarify something else.

Equations such as these: \(\frac{60}{n}=\frac{60}{n-5}-2\) are solved using hit and trial.

Try to look for values of n which give us simple numbers i.e. try to plug in values which are factors of the numerator. Say, if n = 10, you get 60/10 = 60/5 - 2 which is not true.\(\frac{60}{n}\)and \(\frac{60}{n-5}\) need to be much closer to each other so that the difference between them is 2. Next put n = 15. It satisfies. Notice that this equation gives us a quadratic so be careful while working on DS questions. A little manipulation shows us that the product of the roots of this quadratic will be -ve. So one root is positive i.e. 15, hence the other must be -ve and we can ignore it.

Notice that when you see such an equation in PS questions, the options can give you ideas on values on n. Here, you anyway got that n = 15 from the first statement so you know which value you must try.
_________________

Re: Every member of a certain club volunteers to contribute equa [#permalink]

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20 Jun 2013, 09:59

I took the quadradic path as i usually do, but dint solve the equation to see if stmt 2 was sufficient, here's how Given : nx=60 (n: number of ppl, x: contribution of each) Stmt 1 : gives x, can solve for n : sufficient Stmt 2 : (n-5)(x+2)=60 =>(n-5)(60/n+2)=60 Simple rearrangement =>2n^2-10n-300=0 I stopped here knowing this equation will have one positive and one negative root (or solution) here's how : ax^2+bx+c=0 b=sum of roots (or solutions) (or -600 c=product of roots => 2n^2-10n-300=0 will have roots that add to -10 and have roots that multiply to -300 or (-600 to be specific) The second statement that -300 (or -600) will be the product of two roots of the equation implies, one is +ve and other is -ve. and n cant take -ve values, therefore only one solution. Stmt 2 is sufficient.

Could someone cpls verify if this method has any flaw?

I took the quadradic path as i usually do, but dint solve the equation to see if stmt 2 was sufficient, here's how Given : nx=60 (n: number of ppl, x: contribution of each) Stmt 1 : gives x, can solve for n : sufficient Stmt 2 : (n-5)(x+2)=60 =>(n-5)(60/n+2)=60 Simple rearrangement =>2n^2-10n-300=0 I stopped here knowing this equation will have one positive and one negative root (or solution) here's how : ax^2+bx+c=0 b=sum of roots (or solutions) (or -600 c=product of roots => 2n^2-10n-300=0 will have roots that add to -10 and have roots that multiply to -300 or (-600 to be specific) The second statement that -300 (or -600) will be the product of two roots of the equation implies, one is +ve and other is -ve. and n cant take -ve values, therefore only one solution. Stmt 2 is sufficient.

Could someone cpls verify if this method has any flaw?

Re: Every member of a certain club volunteers to contribute equa [#permalink]

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20 Jun 2013, 21:42

VeritasPrepKarishma wrote:

nanz236 wrote:

I took the quadradic path as i usually do, but dint solve the equation to see if stmt 2 was sufficient, here's how Given : nx=60 (n: number of ppl, x: contribution of each) Stmt 1 : gives x, can solve for n : sufficient Stmt 2 : (n-5)(x+2)=60 =>(n-5)(60/n+2)=60 Simple rearrangement =>2n^2-10n-300=0 I stopped here knowing this equation will have one positive and one negative root (or solution) here's how : ax^2+bx+c=0 b=sum of roots (or solutions) (or -600 c=product of roots => 2n^2-10n-300=0 will have roots that add to -10 and have roots that multiply to -300 or (-600 to be specific) The second statement that -300 (or -600) will be the product of two roots of the equation implies, one is +ve and other is -ve. and n cant take -ve values, therefore only one solution. Stmt 2 is sufficient.

Could someone cpls verify if this method has any flaw?

Thanks Karishma. I chanced upon that article and then got deep into the pool of other articles you've published. Really good, would have been great if i wasnt 1 month away from my gmat...better late than after GMAT :D

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06 Jul 2014, 15:17

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Re: Every member of a certain club volunteers to contribute equa [#permalink]

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02 Nov 2015, 13:41

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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