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Each person attending a fund-raising party for a certain clu

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Each person attending a fund-raising party for a certain clu [#permalink] New post 11 Dec 2007, 14:48
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A
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Question Stats:

66% (01:55) correct 34% (01:06) wrong based on 27 sessions
Each person attending a fund-raising party for a certain club was charged the same admission fee. How many people attended the party?

(1) If the admission fee had been 0.75$ less and 100 more people had attended, the club would have received the same amount in admission fee
(2) If the admission fee had been 1.50$ less and 100 fewer people had attended, the club would have received the same amount in admission fee

OPEN DISCUSSION OF THIS QUESTION IS HERE: each-person-attending-a-fund-raising-party-for-a-certain-clu-101572.html
[Reveal] Spoiler: OA
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 11 Dec 2007, 14:55
wow...this one is deceptively tricky for me. I'm not sure anymore
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 11 Dec 2007, 15:30
A- original admission fee
P- original number of people

1) AP=(A-0.75)(P+100)
AP=AP+100A-0.75P -75
0=100A-0.75P-75
1 equation, 2 unknowns

2) AP=(A+1.5)(P-100)
AP=AP-100A+1.5P-150
0=-100A+1.5P-150
1 equation, 2 unknowns

1) + 2) 0=0+0.75P-225=> P=225*4/3=300
answer choice C
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 14 Dec 2007, 10:51
the OA is C. Anyone else wanna try this?
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 14 Dec 2007, 11:06
alexperi wrote:
A- original admission fee
P- original number of people

1) AP=(A-0.75)(P+100)
AP=AP+100A-0.75P -75
0=100A-0.75P-75
1 equation, 2 unknowns

2) AP=(A+1.5)(P-100)
AP=AP-100A+1.5P-150
0=-100A+1.5P-150
1 equation, 2 unknowns

1) + 2) 0=0+0.75P-225=> P=225*4/3=300
answer choice C


I completely agree with this, but it's a little too much. The moment you have the first equation, you should know you can't solve, and the same with the second one. Together, you've got two equations and two variables, so you should be able to solve.

The only risk here is that sometimes, with two equations, you can't solve. But if you learn the times when you can't, you can do a quick check for them and move on. They are:

1) When the equations are the same but look different
Example:
y = 3x + 6
6x = 2y - 12
These are actually the same equation.

2) When there are powers in one or both equation
Example:
y^2 + x = 9
x + 2y = 6

3) When one of the variables actually disappears, and you should have picked A or B
Example:
3x + 2y = 18
x + y = 4x + y + 12 <-- the y's cancel here so there's actually only 1 variable.

So look at the two equations and make sure quickly that you're not in any of those three situations (a very fast test when you get used to looking for them) and then the answer is C right away.
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 14 Dec 2007, 11:27
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tarek99 wrote:
Each person attending a fund-raising party for a certain club was charged the same admission fee. How many people attended the party?

(1) If the admission fee had been $0.75 less and 100 more people had attended, the club would have received the same amount in admission fees.

(2) If the admission fee had been $1.50 more and 100 fewer people had attended, the club would have received the same amount in admission fees.

Please explain your answer.


C.
1: xy = (x-0.75) (y+100)
100x -0.75y = 75
100x = 0.75y + 75

2: xy = (x+1.50) (y-100)
xy = xy - 100x + 1.50y - 150
1.50y - 100x = 150
1.50y - 150 = 100x

togather, solve for 1 and 2:

0.75y + 75 = 1.50y - 150
0.75y = 225
y = 300
x = 2.25
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 11 Aug 2014, 17:19
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 11 Aug 2014, 20:44
tarek99 wrote:
Each person attending a fund-raising party for a certain club was charged the same admission fee. How many people attended the party?

(1) If the admission fee had been $0.75 less and 100 more people had attended, the club would have received the same amount in admission fees.

(2) If the admission fee had been $1.50 more and 100 fewer people had attended, the club would have received the same amount in admission fees.


Please explain your answer.


C is also my answer. If someone thinks D is answer, could you please give an explanation?
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Re: Each person attending a fund-raising party for a certain clu [#permalink] New post 12 Aug 2014, 07:37
Expert's post
tarek99 wrote:
Each person attending a fund-raising party for a certain club was charged the same admission fee. How many people attended the party?

(1) If the admission fee had been 0.75$ less and 100 more people had attended, the club would have received the same amount in admission fee
(2) If the admission fee had been 1.50$ less and 100 fewer people had attended, the club would have received the same amount in admission fee




OPEN DISCUSSION OF THIS QUESTION IS HERE: each-person-attending-a-fund-raising-party-for-a-certain-clu-101572.html
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Re: Each person attending a fund-raising party for a certain clu   [#permalink] 12 Aug 2014, 07:37
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