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

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25 Aug 2010, 19:35

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A

B

C

D

E

Difficulty:

25% (medium)

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70% (02:05) correct
30% (00:23) wrong based on 66 sessions

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Hello, I got issues with some other problems in OG11. Please help explain how to solve these. Thanks a lot. #DS133:

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.

For this, IMO answer must be A, but OA is D.

#DS145: Is 1/p > r / (r^2+2) ?

(1) p=r (2) r>0

My solution is A again but OA is C

Thanks.

PS: To Bunuel: sorry if I've posted something already asked (I tried to search last few days but it took me too much time / I couldnt shorten the search by question numbers or sources.If possible please guide me on this search. I'm new to the forum. Thanks again.)

Re: help me on some problems on OG11 (DS #133- #145) [#permalink]

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25 Aug 2010, 20:28

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Quote:

#DS133: 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. For this, IMO answer must be A, but OA is D.

Statement 1: This is straightforward. Divide 60/4 - 15 members.

Statement 2: When 60$ is divided equally, there could be potentially 2,3,4,5,6,10,12,15,20,30 members. Given that when 5 members fail to contribute, hence we know that the difference between the initial and final members is 5. Numbers 5, 10, 15, 20 could be the number of members.

When the count of the members fall by 5 and share of each member increases by $2. We can infer this from the numbers 15 (initial) and 10 (final members).

Hence statement 2 is also sufficient.

Alternatively we can probably take advantage of the fact that statement 1 and 2 will never contradict in GMAT. Hence we observe from statement 1 that the number of initial members is 15, hence the number of final members (after 5 fail to contribute) is 10. And the condition -- "the share of each contributing member will increase by $2" holds good.

Hence answer should be D.
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Last edited by ezhilkumarank on 25 Aug 2010, 20:37, edited 2 times in total.

Hello, I got issues with some other problems in OG11. Please help explain how to solve these. Thanks a lot. #DS133: 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. For this, IMO answer must be A, but OA is D.

#DS145: Is 1/p > r / (r^2+2) ? (1) p=r (2) r>0 My solution is A again but OA is C

Thanks. PS: To Bunuel: sorry if I've posted something already asked (I tried to search last few days but it took me too much time / I couldnt shorten the search by question numbers or sources.If possible please guide me on this search. I'm new to the forum. Thanks again.)

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.

Answer: D.

Is \(\frac{1}{p}>\frac{r}{r^2 + 2}\)?

(1) \(p=r\) --> is \(\frac{1}{r}>\frac{r}{r^2+2}\)? --> as \(r^2+2\) is always positive, we can safely multiply both parts of the inequality by this expression, we'll get: is \(\frac{r^2+2}{r}>r\)? --> is \(r+\frac{2}{r}>r?\)? --> is \(\frac{2}{r}>0\)? This inequality is true when \(r>0\) and not true when \(r<0\). Not sufficient.

Side note: we can not cross multiply for \(\frac{1}{r}>\frac{r}{r^2+2}\) as we don't know the sign of \(r\), thus don't know whether we should flip sign of inequality (>).

(2) \(r>0\). Not sufficient by itself.

(1)+(2) As \(r>0\) then \(\frac{2}{r}>0\) is true. Sufficient.

Answer: C.

As for search: I'd search the words "volunteers" and "certificate" in DS subfroum, don't thin there will be many different questions with theses words. The second question is much harder to search (you can try by tags list by forum: inequalities), so it's no problem at all to post such questions without a search, if I recognize the question I'll merge it with previous discussion myself.

P.S. Please post one question per topic. _________________

Re: help me on some problems on OG11 (DS #133- #145) [#permalink]

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18 Nov 2011, 08:45

Quote:

(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.

I would like to ask our math gurus. The way Bunuel solved the question is clear. I guess there are other ways to do that, but it's not that important. What I want to ask is this:

\(\frac{60}{n}=\frac{60}{n-5}-2\)

The fraction above obviously leads to a quadratic equation, which could have two positive solutions. If it indeed had two positive solutions, (2) would be insufficient. The only way to know how many positive solutions it has is to solve it. But it is time consuming.

Is there a quick way to know whether a quadratic equation has one or two positive solutions without actually solving it?

Re: help me on some problems on OG11 (DS #133- #145) [#permalink]

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29 Nov 2011, 12:53

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nonameee wrote:

\(\frac{60}{n}=\frac{60}{n-5}-2\)

The fraction above obviously leads to a quadratic equation, which could have two positive solutions. If it indeed had two positive solutions, (2) would be insufficient. The only way to know how many positive solutions it has is to solve it. But it is time consuming.

Is there a quick way to know whether a quadratic equation has one or two positive solutions without actually solving it?

Thanks.

My 2 cents- It is better to solve the quadratic equation. or atleast manipulate the equation to point of \(n^2-5n-150 = 0\). because with out getting to this point it will be dangerous to think about it intuitively. Once we have the equation we can quickly see that of the two possible solutions one is negative.

Also, In DS, statement 1 and 2 never contradict each other. We can use this fact to our advantage. So from statement 1 we got n= 15. We shouldnt have to carry this information but when looking at the quadratic equation we should know that one of the solution should be 15.

Re: help me on some problems on OG11 (DS #133- #145) [#permalink]

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21 Apr 2012, 17:45

nonameee wrote:

I would like to ask our math gurus. The way Bunuel solved the question is clear. I guess there are other ways to do that, but it's not that important. What I want to ask is this:

\(\frac{60}{n}=\frac{60}{n-5}-2\)

The fraction above obviously leads to a quadratic equation, which could have two positive solutions. If it indeed had two positive solutions, (2) would be insufficient. The only way to know how many positive solutions it has is to solve it. But it is time consuming.

Is there a quick way to know whether a quadratic equation has one or two positive solutions without actually solving it?

Thanks.

To be sure that the quadratic equations yield only one non-negative answer, I would take advantage of the following:

Having a quadratic equation \(ax^2+bx+c=0\)

The product of the two roots is \(\frac{c}{a}\) The sum of the two roots is \(\frac{-b}{a}\)

Considering this, you just need to manipulate the equation from \(\frac{60}{n}=\frac{60}{n-5}-2\) to \(2x^2-10x-300=0\) (you do not even need to divide by 2). In this case, the product of the two roots (solutions) of the equation is \(\frac{c}{a}=\frac{-300}{2}\) and the sum is \(\frac{-b}{a}=\frac{10}{2}\). Therefore, the roots must have different signs (one positive and one negative) for the product to be negative (-150).

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