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Mary persuaded n friends to donate $500 each to her election

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Re: Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 30 Mar 2015, 20:07
Bunuel wrote:
Awli wrote:
Mary persuaded n friends to donate $500 each to her election campaign, an then each of these n friends persuaded n more people to donate $500 each to Mary's campaign. If no one donated more than once and if there were no other donations, what was the value of n?

(1) The first n people donated \frac{1}{16} of the total amount donated.

(2) The total amount donated was $120,000


Merging topics. Please refer to the discussion on page 1.


[color=#0000ff]not sure, if my approach is right or wrong. I just took it like:
1. Mary persuaded n friends to donate $500 = n*500
2. then each of these n friends persuaded n more people.= n^(n+1) * 500

Statement 1. first n donated 1/16 of the total. remains > need total amount.
Statement 2. n^(n+1)*500 = 120,000
n*n^n = 240 .. Looks insufficient

1+2

n = 1/16 * 120000 /500 = 15

hence C
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Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 31 Mar 2015, 03:14
Ans: D
1st amount = 500n ( 500 people pay n amount)
2nd amount = 500n^2(n people employ n people which makes the number of people n^2 and each one of them pays 500)
total = 500n + 500 n^2

1st statment
500n = 1/16 * (500n + 500n^2)
n = 15 or 0 (n = 0 not possible)
hence sufficient

2nd statement
500n + 500n^2 = 120,000
=> n + n^2 = 240
=> n = 15 or -16 ( n = -16 not possible for obvious reasons :wink: )
hence sufficient
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Re: Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 31 Mar 2015, 04:44
Original Donations = 500n
Friends' Friends =500n^2
Total = 500n+500n^2

Statement 1 :The first n people donated \frac{1}{16} of the total amount donated.

500n = 1/16 (500n+500n^2)
16(500n) = 500n+500n^2
n = 15
stmt 1 is sufficient

Statement 2: The total amount donated was $120,000

500n+500n^2=$120,000
can solve for n
n=15
stmt 2 is sufficient

Answer is D
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Re: Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 10 Aug 2015, 02:55
aalriy wrote:
I have understood the approach GT took to solve the problem its very similar to mine... but i cannot make out how can the first stmt give a solution for n as 0 or a -ve value.

Could someone explain this?



On the GMAT you ll not be asked a value based DS question if at all there is no such value.That is why n can not be zero.One more thing that U can understand that as there are n people first to donate $500 each and those n people refer n people each .So if U consider that there are 16 portions total money is donated by all then 1 portion is by the first n people.
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Re: Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 22 May 2017, 09:57
Here we have given:

n*500 – number of people who initially donated
n(n*500) – persuaded number of people to donate.

so the total T = 500n + 500nsrd = 500(n + n sqrd)


(1) n*500 – (initial number of people donated) = 1/16, so here we have a fraction and we can solve for n => sufficient.
(2) We have total amount so we can again solve for n => sufficient.

The answer is D.
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Re: Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 12 Nov 2017, 09:07
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seofah wrote:
Mary persuaded n friends to donate $500 each to her election campaign, and then each of these n friends persuaded n more people to donate $500 each to Mary’s campaign. If no one donated more than once and if there were no other donations, what was the value of n?

(1) The first n people donated 1/16 of the total amount donated.
(2) The total amount donated was $120,000.


Target question: What was the value of n?

When I scan the two statements, it seems that statement 2 is easier, so I'll start with that one first...

Statement 2: The total amount donated was $120,000
Let's summarize the given information....

First round: n friends donate 500 dollars.
This gives us a total of 500n dollars in this round

Second round: n friends persuade n friends each to donate
So, each of the n friends gets n more people to donate.
The total number of donors in this round = n²
This gives us a total of 500(n²) dollars in this round

TOTAL DONATIONS = 500n dollars + 500(n²) dollars
We can rewrite this: 500n² + 500n dollars

So, statement 2 tells us that 500n² + 500n = 120,000
This is a quadratic equation, so let's set it equal to zero to get: 500n² + 500n - 120,000 = 0
Factor out the 500 to get: 500(n² + n - 240) = 0
Factor more to get: 500(n + 16)(n - 15) = 0
So, EITHER n = -16 OR n = 15
Since n cannot be negative, it must be the case that n = 15
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Statement 1: The first n people donated 1/16 of the total amount donated.
First round donations = 500n
TOTAL donations = 500n² + 500n
So, we can write: 500n = (1/16)[500n² + 500n]
Multiply both sides by 16 to get: 8000n = 500n² + 500n
Set this quadratic equation equal to zero to get: 500n² - 7500n = 0
Factor to get: 500n(n - 15) = 0
Do, EITHER n = 0 OR n = 15
Since n cannot be zero, it must be the case that n = 15
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Answer:

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Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 22 Jun 2018, 23:07
[/quote]

You can solve it using the formula for quadratics, though it's better to use another approach:

\(500n+500n^2=120,000\) --> \(n+n^2=240\) --> \(n(n+1)=240\). Since \(n\) is an integer then we have that the product of two consecutive integers is 240, now it's easy to find that \(n=15\).

Hope it's clear.[/quote]

Within context of GMAT DS question, the moment I manage to set up such relationship n(n+1) = 240, will it be safe to say there is 1 solution for n without trying to find a pair of factors that fit? This would save some time. Whenever I get to this point, I always try to find a pair just to make sure it will not be the case of a) having no solution for n or b) having 2 solutions for n.
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Re: Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 22 Jun 2018, 23:23
gmatcrash wrote:
Within context of GMAT DS question, the moment I manage to set up such relationship n(n+1) = 240, will it be safe to say there is 1 solution for n without trying to find a pair of factors that fit? This would save some time. Whenever I get to this point, I always try to find a pair just to make sure it will not be the case of a) having no solution for n or b) having 2 solutions for n.


n(n + 1) = (positive number) will always have two solutions, one negative and one positive but not always these solutions will be integers.

For example:

n(n + 1) = 2 --> n = -2 or n = 1;

n(n + 1) = 2 --> \(n = -\frac{1}{2}-\frac{\sqrt{13}}{2}\) or \(n = -\frac{1}{2}+\frac{\sqrt{13}}{2}\)
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Re: Mary persuaded n friends to donate $500 each to her election  [#permalink]

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New post 04 Nov 2018, 12:47
seofah wrote:
Mary persuaded n friends to donate $500 each to her election campaign, and then each of these n friends persuaded n more people to donate $500 each to Mary’s campaign. If no one donated more than once and if there were no other donations, what was the value of n?

(1) The first n people donated 1/16 of the total amount donated.
(2) The total amount donated was $120,000.

\({\rm{Total}}\,\, = \,\,500 \cdot n + 500 \cdot n \cdot n\,\,\,\,\,\,\left[ \$ \right]\)

\(? = n\)

\(\left( 1 \right)\,\,\,500 \cdot n = {1 \over {16}} \cdot 500 \cdot n \cdot \left( {1 + n} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,\,\left( {500\,n} \right)\,\,\,\left[ {\,n\, \ne \,0\,} \right]} \,\,\,1 = {1 \over {16}} \cdot \left( {1 + n} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,n\,\,{\rm{unique}}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)

\(\left( 2 \right)\,\,\,500 \cdot n\left( {1 + n} \right) = 120000\,\,\,\,\,\mathop \Rightarrow \limits^{:\,\,500} \,\,\,\,n\left( {1 + n} \right) = 240\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,n\,\, > 0\,\,\,\,{\rm{unique}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.\)

\(\left( * \right)\,\,15 \cdot 16 = 240\,\,\, \Rightarrow \,\,\,\left\{ \matrix{
\,n\left( {n + 1} \right) < 240\,\,{\rm{for}}\,\,0 < n < 15 \hfill \cr
\,n\left( {n + 1} \right) > 240\,\,{\rm{for}}\,\,n \ge 16 \hfill \cr} \right.\,\,\,\,\,\,\left( {{\rm{Now}}\,\,{\rm{rethink}}\,\,{\rm{without}}\,\,{\rm{knowing}}\,\,{\rm{that}}\,\,n = 15...} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Mary persuaded n friends to donate $500 each to her election   [#permalink] 04 Nov 2018, 12:47

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