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# At a certain conference, 72% of the attendees registered at

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Director
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At a certain conference, 72% of the attendees registered at least two [#permalink]

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30 Jan 2011, 07:43
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85% (hard)

Question Stats:

50% (01:59) correct 50% (02:05) wrong based on 88 sessions

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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%

hi I tried to use this using Venn Diagram and got A as the answer can some one explain this to me using Venn
[Reveal] Spoiler: OA

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Director
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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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30 Jan 2011, 07:58
Hi,
I do not think that Venn will be helpful here. A table will make more sense. But here is my approach.

72% regestered at least 2 weeks and paid full fee.

10% paid full fee and did not registered at least 2 weeks in advance. Then 90% paid full fee and registered at least 2 weeks before.

90% *X=72% where X is the number of people who registerd 2 weeks in advance and paid full fee.

Regards

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Director
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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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31 Jan 2011, 08:39
whats wrong with using Venn diagram here ? 1 more thing, , using the given data , we get 80% of the total attendees (as those who paid full fee ) , not the % of those who registered two weeks in advance .
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Manager
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At a certain conference, 72% of the attendees registered at least two [#permalink]

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14 May 2011, 22:49
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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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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15 May 2011, 00:52
1
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attendees = 100
people who paid the fees in full = x

0.1x = who paid the fee in full but did not do 2 weeks in advance.
0.9x = people who paid the fee in full and paid atleast 2 weeks in advance = 72

x= 72/0.9 = 80

thus people who paid fee and not 2 weeks in advance = 0.1*80 = 8
so

Fee !Fee
2 weeks advance 72 20 92
! 2 week advance 8 0 8

total = 100

so 92%
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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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15 May 2011, 01:09
fanatico 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
(A) 18.0%
(B) 62.0%
(C) 79.2%
(D) 80.0%
(E) 82.0%

Looks like this question is not complete. The required percent cannot be determined with given data. The answer can range from 72% to 92%. Could you please confirm?
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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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15 May 2011, 02:51
My take :

0.72A -> At least 2 weeks in advance and paid conference fee in full.

FA -> Paid the conference fee in full

0.10FA -> Not 2 weeks in advance paid but conference fee in full

=> 0.90FA -> 2 Weeks in advance and paid the conference fee in full

0.90FA = 0.72A

FA/A = 72/90 * 100

= 80%

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Manager
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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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15 May 2011, 08:16
I got the same answer as amit2k9.

@subhashgosh
FA/A is not wat is asked, I think.

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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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17 May 2011, 11:39
fluke wrote:
fanatico 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
(A) 18.0%
(B) 62.0%
(C) 79.2%
(D) 80.0%
(E) 82.0%

Looks like this question is not complete. The required percent cannot be determined with given data. The answer can range from 72% to 92%. Could you please confirm?

Exactly
we dont have any bifurcation of people who did not pay the full fee (how many registered in 2 week advance and how many did not)

without knowing that we can't reach the desired answer that how many registered in 2 weeks advance
it could be 72% +0% (minimum) to 72% + 20% (maximum)

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Director
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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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17 May 2011, 17:09
Something wrong with the question. we need additional data to solve the problem.

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Director
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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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17 May 2011, 17:12
@amit2k9,

how can you say that people who didnt register two weeks in advance and not paid in full =0 ? Did you assume that?

amit2k9 wrote:
attendees = 100
people who paid the fees in full = x

0.1x = who paid the fee in full but did not do 2 weeks in advance.
0.9x = people who paid the fee in full and paid atleast 2 weeks in advance = 72

x= 72/0.9 = 80

thus people who paid fee and not 2 weeks in advance = 0.1*80 = 8
so

Fee !Fee
2 weeks advance 72 20 92
! 2 week advance 8 0 8

total = 100

so 92%

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Re: At a certain conference, 72% of the attendees registered at least two [#permalink]

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17 May 2011, 17:27
yes between 72 and 92%, so answer could be any of C, D and E.

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At a certain conference, 72% of the attendees registered at [#permalink]

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17 Sep 2012, 00:34
1
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2
This post was
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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%

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
3
KUDOS
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
1
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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
Attachments

72 % attendees.docx [14.39 KiB]

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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] 12 Apr 2013, 23:05

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