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At a certain conference, 72% of the attendees registered at [#permalink]
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17 Sep 2012, 00:34
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At a certain conference, 72% of the attendees registered at least two weeks in advance and paid their conference fee in full. If 10% of the attendees who paid their conference fee in full did not register at least two weeks in advance, what percent of conference attendees registered at least two weeks in advance? (A) 18.0% (B) 62.0% (C) 79.2% (D) 80.0% (E) 82.0%
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Last edited by Bunuel on 01 Jan 2013, 02:28, edited 1 time in total.
Renamed the topic and edited the question.



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Re: At a certain conference, 72% of the attendees [#permalink]
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17 Sep 2012, 02:52
Option D. Number of people who registered early and paid full amount /Total number of people who paid full amount. =(72/82)*100 =80.0 %
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Re: At a certain conference, 72% of the attendees [#permalink]
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17 Sep 2012, 13:14
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saikarthikreddy wrote: Option D.
Number of people who registered early and paid full amount /Total number of people who paid full amount. =(72/82)*100 =80.0 % Can you explain more clearly? I still don't get it. The problem ask what percent of conference attendees registered at least two weeks in advance. I think the formula should be Number of people who registered at least two weeks in advance/Total number of attendees.
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Re: At a certain conference, 72% of the attendees [#permalink]
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23 Sep 2012, 15:46
saikarthikreddy wrote: Option D.
Number of people who registered early and paid full amount /Total number of people who paid full amount. =(72/82)*100 =80.0 % I guess 72/82 is greater than 80 % and 82 % is also an option why did u choose 80% Experts pls comment



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Re: At a certain conference, 72% of the attendees [#permalink]
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08 Nov 2012, 15:47
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Refer to the table in the attachment: Let x= No. of members who have paid in Full 10 % members paid in full and did not register in advance = 0.1x 72 % registerd in advance and paid in full. So if total No. of members = 100, then 72 members paid Full and registered in advance. Hence total members who paid full amount = 0.1x + 72 =x 0.9x =72 Hence x = 80 i.e. 80 out of 100 or 80 % Ans. D



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Re: At a certain conference, 72% of the attendees [#permalink]
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01 Jan 2013, 01:11
But they don't ask for % of attendees who paid full amount, they ask for % of attendees who registered 2 weeks in advanced?



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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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01 Jan 2013, 11:29
Even im not clear on this.Can u share the source plz?



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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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12 Apr 2013, 23:05
Using Double matrix method
Let the attending people be 100
Full pay not full pay 2 weeks 72 8 80 < 2weeks 0.1x 12 20
x=80 20 100
So 0.9x = 72 . therefore X = 80 And the other numbers fill in accordingly



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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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27 Apr 2013, 07:51
I tried doing this by taking 100 as the total no. of attendees and using the the chart approach. in this case both 80% and 82% satisfied the chart. Then I assumed that there were 200 attendees. if you then try to fill up chart the total no. of attendees who booked 2 weeks in advance is 160 which is 80% of 200. therefore choice (D).



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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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29 Apr 2013, 22:36
I'm struggling with this question.
When I read it I assumed 72% + 10 % as the total % who have registered before two weeks and paid in full and naturally 18% was my answer.
Why do we need to take 10% of total and 72 as the sum to total? Why is my above assumption wrong..
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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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25 May 2013, 12:40
Hello Mal. I know this is a late reply, but I hope it helps. You are correct to reach the summation that 18% have registered 2 weeks in advance but it's important to remember that this 18% is clean and excludes the part of the overlap. SO remember that the 72% (common) has a bit related to "registration 2 weeks in advance". If we draw a Venn diagram, we will see that Paid in full =10% The common part related to "registration 2 weeks in advance" = 7210 =62 Total registration 2 weeks in advance = 62+18=80 Hope this helps!
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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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17 Aug 2013, 13:34
Hi Folks, Not clear with the solutions above. Can anyone enlighten me as far as this question is concerned. Rgds, TGC !
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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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17 Aug 2013, 18:41
Shawshank wrote: At a certain conference, 72% of the attendees registered at least two weeks in advance and paid their conference fee in full. If 10% of the attendees who paid their conference fee in full did not register at least two weeks in advance, what percent of conference attendees registered at least two weeks in advance?
(A) 18.0% (B) 62.0% (C) 79.2% (D) 80.0%,
(E) 82.0% Let, R_F be number ofpeople who registered in advance and who paid in full R_NF be the number of people who registed in advance but did not pay in full NR_F be the number of people who did not register in advance but paid in full NR_NF be the number of people who did not register in advance and not pais in full we have, R_F + R_NF + NR_F + NR_NF=100 Also, R_F =72  (1) and NR_F / (NR_F + R_F ) = 10% or NR_F = 1/ 9 R_F NR_F= 72/9 = 8  (2) So from (1) and (2), R_F + NR_F = 80 So we can find out only the number of people who paid in full. The question seems to be wrong.
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Re: At a certain conference, 72% of the attendees registered at [#permalink]
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26 Oct 2013, 13:56
This is from Jeff Sackman Problem Solving Challenge. I keep getting this one wrong as well. I can only solve for the % who paid in full > 80%. From the information given I don't know if there are any people in the "neither" category, i.e. didn't signed up in advance and didn't pay in full. If there is no "neither" category I get 92% for paid in advance, but I am making a big assumption (and 92% isn't an answer choice). Is there a typo and the question should be asking for who paid in full and not who signed up in advance? Perhaps I am missing something.




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