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The organizers of a fair projected a 25 percent increase in
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14 Jan 2014, 00:08
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The Official Guide For GMAT® Quantitative Review, 2ND EditionThe organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this year actually decreased by 20 percent. What percent of the projected attendance was the actual attendance? (A) 45% (B) 56% (C) 64% (D) 75% (E) 80% Problem Solving Question: 38 Category: Arithmetic Percents Page: 66 Difficulty: 600 GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition  Quantitative Questions ProjectEach week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution. We'll be glad if you participate in development of this project: 1. Please provide your solutions to the questions; 2. Please vote for the best solutions by pressing Kudos button; 3. Please vote for the questions themselves by pressing Kudos button; 4. Please share your views on difficulty level of the questions, so that we have most precise evaluation. Thank you!
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Re: The organizers of a fair projected a 25 percent increase in
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Re: The organizers of a fair projected a 25 percent increase in
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14 Jan 2014, 00:40
C.. Assume last year attendence ..100====expected increased of 25%========> 125 100..actuall decreased of 20% ==================================>80 actuall/expected... 80/125=====================================16/25==64%... 64% is the ans
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Re: The organizers of a fair projected a 25 percent increase in
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14 Jan 2014, 08:23
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionThe organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this year actually decreased by 20 percent. What percent of the projected attendance was the actual attendance? (A) 45% (B) 56% (C) 64% (D) 75% (E) 80% Let "X" be the last year attendance . Projected attendance = 1.25X Actual attendance = 0.8X So .8/1.25 = (4/5)/(5/4) = 16/25 hence 64%  Option C) [% can be calculated faster if we notice 25*4 = 100 so 16/25 = 16*4/25*4 = 64/100 = 64%]
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Re: The organizers of a fair projected a 25 percent increase in
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14 Jan 2014, 21:22
Let us assume the last year attendance as 100(easiest number to pick, since all are in percent). Projected attendance = 25% increase(of 100) = 125 Actual attendance = 20% decrease(of 100) = 80
Actual attendance/Projected attendance = (80/125) = (16/25) = (16*4)/(25*4) = 64/100 = 64%
Ans is (C).



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Re: The organizers of a fair projected a 25 percent increase in
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18 Jan 2014, 22:34
sanjoo wrote: C..
Assume last year attendence ..100====expected increased of 25%========> 125
100..actuall decreased of 20% ==================================>80
actuall/expected... 80/125=====================================16/25==64%...
64% is the ans I took \(\frac{80}{125}\) and reduced it to \(\frac{16}{25}\) and couldn't figure it out past that soooooI made it \(\frac{15}{25}\) and reduced it into \(\frac{3}{5}\) and then multiplied it by two to get \(\frac{6}{10}=60%\) and since \(16>15\) \(\frac{16}{25}\)>\(60%\) Answer C was the only logical answer



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Re: The organizers of a fair projected a 25 percent increase in
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18 Jan 2014, 22:46
The organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this year actually decreased by 20 percent. What percent of the projected attendance was the actual attendance?
(A) 45% (B) 56% (C) 64% (D) 75% (E) 80%
Projected attendance: \(25 %\) increase \(= \frac{5}{4}\)
Actual attendance : decreased by \(20% = \frac{4}{5}\)
\(\frac{Actual}{Projected} = \frac{4}{5}/\frac{5}{4} = \frac{16}{25} =\frac{64}{100}\) or \(64%\)
Answer: (C)



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Re: The organizers of a fair projected a 25 percent increase in
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18 Jan 2014, 23:28
arunspanda wrote: The organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this year actually decreased by 20 percent. What percent of the projected attendance was the actual attendance?
(A) 45% (B) 56% (C) 64% (D) 75% (E) 80%
Projected attendance: \(25 %\) increase \(= \frac{5}{4}\)
Actual attendance : decreased by \(20% = \frac{4}{5}\)
\(\frac{Actual}{Projected} = \frac{4}{5}/\frac{5}{4} = \frac{16}{25} =\frac{64}{100}\) or \(64%\)
Answer: (C) Dude this was awesome Fast, Quick, Clean and thee correct way to do it...awesome job again....



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Re: The organizers of a fair projected a 25 percent increase in
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23 Apr 2014, 02:58
Let Last year attendance be X; Then Projected Attendance will be 1.25x ; Actual attendance will be 0.8x; Question is asking us 0.8x is what percent of 1.25x ?
0.8x/1.25x = Y/100;
y=64 %;



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Re: The organizers of a fair projected a 25 percent increase in
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20 Jun 2014, 00:27
Let's pick smart numbers:
Last year's attendance: 100 projected attendance this year: 125 (+25%) actual attendance this year : 80 (20 %) 80/125 = 16/25 = 64/100 = 64 %. C.



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Re: The organizers of a fair projected a 25 percent increase in
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28 Sep 2016, 01:25
we can use a smart number to solve this problem quickly. let's say the actual attendance was 100 at the beginning. it easy to find that the projected attendence was 125, whereas the actual attendance was 80. \(\frac{80}{125}\)=64% answer C
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The organizers of a fair projected a 25 percent increase in
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08 Jan 2017, 05:24
From MGMAT Questions with "What is X percent of Y" translates to "What" translates to " Unknown " "is" translates to " = " "X" translates to " X " "percent" translates to " /100 " "of" translates to multiplication sign " * " "Y" translates to "Y" Hence Unknown = (X/100) * Y Use the same logic and Smart Numbers in order to decode the question and generate the formula. You should end up with Actual = X/100 Projected ... Plug the numbers and solve with respect to X. The answer should be C



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Re: The organizers of a fair projected a 25 percent increase in
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13 Apr 2017, 22:50
Bunuel wrote: The Official Guide For GMAT® Quantitative Review, 2ND EditionThe organizers of a fair projected a 25 percent increase in attendance this year over that of last year, but attendance this year actually decreased by 20 percent. What percent of the projected attendance was the actual attendance? (A) 45% (B) 56% (C) 64% (D) 75% (E) 80% Say '\(x\)' is your total attendance last year. Projected attendance: \(\cfrac { 5 }{ 4 } x\) Actual attendance: \(\cfrac { 4 }{ 5 } x\) '\(z\)' is the 'What percent of the projected attendance the actual attendance was' \(\cfrac { z }{ 100 } \left( \cfrac { 5 }{ 4 } \right) x=\cfrac { 4 }{ 5 } x\\ \left( \cfrac { z }{ 100 } \left( \cfrac { 5 }{ 4 } \right) x=\cfrac { 4 }{ 5 } x \right) \div x\\ \left( \cfrac { z }{ 100 } \left( \cfrac { 5 }{ 4 } \right) =\cfrac { 4 }{ 5 } \right) \ast \cfrac { 100\ast 4 }{ 5 } \\ z=\cfrac { 16 }{ 25 } \ast 100\\ z=64%\)
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Re: The organizers of a fair projected a 25 percent increase in
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07 Oct 2017, 02:19
last year = 75 people this year = 60 people (20% decrease from last year)
actual= 60 projected=100 (25% increase from 75 is 100)
60/100 = 60
closest answer is 64%; they wanted you to use 100, 125, but, 75 and 100 still work
therefore, ans is (C) 64%



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Re: The organizers of a fair projected a 25 percent increase in
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Re: The organizers of a fair projected a 25 percent increase in
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