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Each person attending a fundraising party for a certain club was
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23 Sep 2010, 12:15
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Each person attending a fundraising 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 Attachment:
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Re: Each person attending a fundraising party for a certain club was
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23 Sep 2010, 13:02
Each person attending a fund raising party for a certain club was charged the same admission fee. How many people attended the party?Let the original admission fee be \(x\) and the original # of people be \(n\). Question: \(n=?\) (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 the admission fee \(xn=(x0.75)(n+100)\) > \(100x0.75n=75\). Two unknowns, one equation, not sufficient to calculate \(n\). (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 the admission fee \(xn=(x1.5)(n100)\) > \(100x1.5n=150\). Two unknowns, one equation, not sufficient to calculate \(n\). (1)+(2) \(100x0.75n=75\) and \(100x1.5n=150\) > we have two distinct linear equation with two unknowns, hence we can solve for \(n\). Sufficient. Answer: C.
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Re: Each person attending a fundraising party for a certain club was
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24 Sep 2010, 12:13
let assume x=admin fee,y=no.of. people x*y=100. 1) (x.75)* (y+100)= 100 (statement 1 mentioned same amount,so i assume the amount as 100) from question stem x*y=100 so x=100/y, sufficient to know x value.
2) (x+1.50) * (y100) =100 x*y =100 x=100/y,sufficient to know x value.
Bunnel, plz explain why should we solve the problem like this.



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Re: Each person attending a fundraising party for a certain club was
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24 Sep 2010, 15:17
TomB, how can you assign a random value (100) to xy and claim that x can be solved??
Let a = admission fee and n = number of people attended
(1) If the admission fee had been $0.75 less and 100 more people attended, the club would have received the same amount in admission fees.
an = (a  0.75)(n + 100) an = an  0.75n + 100a 75 0.75n  100a = 75
Insufficient
(2) If the admission fee had been $1.50 more and 100 fewer people had atteneded, the club would have received the same amount in admission fees.
an = (a + 1.50)(n  100) an = an + 1.5n  100a  150 1.5n  100a = 150
Insufficent
Subtrace (1) from (2) 1.5n  100a = 150 0.75n  100a = 75
0.75n = 225 >> n = 300 & a can be solved



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Re: Each person attending a fundraising party for a certain club was
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Updated on: 09 Nov 2012, 07:32
Each person attending a party was charged the same admission fee. How many people attended the party?
(1) If the fee had been 0.75$ less and 100 more people had attended, the club would have made the same amount of money. (2) If the fee had been 1.50$ more and 100 fewer people had attended, the club would have made the same amount of money.
I have a specific question regarding this problem :
Obviously Statement 1 and 2 are insufficient alone, however once we find out equation 1: xy = (x3/4)(y+100) from statement 1 and xy = (x+3/2)(y100) from statement 2, why can't we simply equate them? (x3/4)(y+100) = (x+3/2)(y100)
I'd appreciate any help! Thanks
Originally posted by mockney on 09 Nov 2012, 07:28.
Last edited by Bunuel on 09 Nov 2012, 07:32, edited 1 time in total.
Renamed the topic and edited the question.



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Re: Each person attending a fundraising party for a certain club was
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09 Nov 2012, 07:43
mockney wrote: Each person attending a party was charged the same admission fee. How many people attended the party?
(1) If the fee had been 0.75$ less and 100 more people had attended, the club would have made the same amount of money. (2) If the fee had been 1.50$ more and 100 fewer people had attended, the club would have made the same amount of money.
I have a specific question regarding this problem :
Obviously Statement 1 and 2 are insufficient alone, however once we find out equation 1: xy = (x3/4)(y+100) from statement 1 and xy = (x+3/2)(y100) from statement 2, why can't we simply equate them? (x3/4)(y+100) = (x+3/2)(y100)
I'd appreciate any help! Thanks Well, actually you can: (x3/4)(y+100) = (x+3/2)(y100) > 9y=800x+300. From xy = (x3/4)(y+100) we'll have that 3y=400x300, so 9y=1200x900. Equating again: 800x+300=1200x900 > x=3 > y=300. Though you can directly simplify xy = (x3/4)(y+100) to get 3y=400x300 and xy = (x+3/2)(y100) to get 3y=200x+300 > 400x300=200x+300 > x=3 > y=300. Hope it's clear.
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Re: Each person attending a fundraising party for a certain club was
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09 Nov 2012, 07:54
Thank you Bunuel for your help! I was just expecting to find a value for either x or y directly without needing to isolate for y from one of the 2 equations!



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Re: Each person attending a fundraising party for a certain club was
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15 Nov 2012, 03:41
WE NEed to show that the two equatin is not the same. but I do not know how to do. pls, help
we can avoid much computation.



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Re: Each person attending a fundraising party for a certain club was
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25 Dec 2013, 23:18
If everyone was charged the same fee, how many people came to the club?
a. If the fee had been %0.75 less and 100 more people came the club would have received the same amount. b. If fee had been $1.50 more and 100 fewer people came, the club would have received the same amount



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Re: Each person attending a fundraising party for a certain club was
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25 Dec 2013, 23:21
aja1991 wrote: If everyone was charged the same fee, how many people came to the club?
a. If the fee had been %0.75 less and 100 more people came the club would have received the same amount. b. If fee had been $1.50 more and 100 fewer people came, the club would have received the same amount The official answer is C, but the solution is very complicated. So, please, can somebody over there help me to handle this problem?



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Re: Each person attending a fundraising party for a certain club was
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26 Dec 2013, 03:05



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Re: Each person attending a fundraising party for a certain club was
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26 Dec 2013, 03:42
Bunuel wrote: aja1991 wrote: If everyone was charged the same fee, how many people came to the club?
a. If the fee had been %0.75 less and 100 more people came the club would have received the same amount. b. If fee had been $1.50 more and 100 fewer people came, the club would have received the same amount Merging similar topics. Please refer to the solutions above. Thanks. I found out why the problem was so confusing. There is a big difference between %0.75 and $0.75. I thought that 0.75 was a percentage increase which was wrong.



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Re: Each person attending a fundraising party for a certain club was
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02 Dec 2014, 11:12
Just to clarify, I just bumped into this question in my practice test the other day, and the second statement was slightly different in mine.
It reads as follows: 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.
I wrote the second equation as: (n100)(p+1.5), so I was surprised when I saw your equations until I spotted the little difference.



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Re: Each person attending a fundraising party for a certain club was
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24 Oct 2015, 05:31
Let Original admission fee= x Original number of people = n 1 xn =(x.75)(n+100) =>100x  .75 n = 75 1 Not sufficient . 2.The second statement should have been 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 fee xn= (x+1.5)(n100) => 100x  1.5 n = 150 2 Not sufficient . Combining 1 and 2 , we get 2 distinct linear equations with 2 unknowns n=300 x=3 Answer C 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 Then, xn= (x1.5)(n100) => 100x +1.5n = 150 3 On solving 1 and 3 , we get x=1 n=33.33 We can even see that statement 2 has an issue by analyzing that for a product xn to remain same if one of the values increase , the other should decrease . The product won't be same if both the values decrease or increase .
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Re: Each person attending a fundraising party for a certain club was
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12 Jan 2017, 19:04
Let attend fee be x, number of person be y: Form 1, (x0.75)(y+100)=xy100x0.75y75=0 From 2, (x1.5)(y100)=xy 100x1.5y150=0 Combine 1 and 2, we can get specific value of x and y. Answer is C
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Re: Each person attending a fundraising party for a certain club was
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30 Jul 2017, 09:00
admission fee = a person attended = p total amount = t t = a*p (as each person will pay some amount)
1) => (a+0.75) (p+100) = t => Insufficient, three variable two equation 2) => (a1.5) (p  100) = t => Insufficient, three variable two equation
1+2 => sufficient, three variable three equation
Ans => C



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