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Each person attending a fund raising event was charged the

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Each person attending a fund raising event was charged the [#permalink] New post 19 Dec 2005, 19:18
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C
D
E

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Each person attending a fund raising event was charged the same admission fee. How many people attended the fund raising event?

(1) If the admission fee had been $.75 less and 100 more people had attended, they would have received the same amount in revenue.

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

A) Statement 1 sufficient...
B) Statement 2 sufficient...
C) Both statements together...
D) Each statement alone
E) Neither statement ...

The book answer is C. How can the math for this be done? It seems like you've got three variables here with only 2 equations...

Jim
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Re: DS [#permalink] New post 19 Dec 2005, 19:33
ellisje22 wrote:
Each person attending a fund raising event was charged the same admission fee. How many people attended the fund raising event?

(1) If the admission fee had been $.75 less and 100 more people had attended, they would have received the same amount in revenue.

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

A) Statement 1 sufficient...
B) Statement 2 sufficient...
C) Both statements together...
D) Each statement alone
E) Neither statement ...

The book answer is C. How can the math for this be done? It seems like you've got three variables here with only 2 equations...

Jim


You need 3 equations to yield the value of 3 variables.
n = number of people
f = fee per person
r = total revenue

Stem tells u:
fn = r

1) gives:
(f-.75)(n+100) = r

2) gives:
(f+1.50)(n-100) = r

This gives us 3 distinct equations, so 1) and 2) together are sufficient.
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 [#permalink] New post 19 Dec 2005, 19:34
Assume initial is initial addmission fee is x and the number of people who attended is y.

(1) (x-0.75) (y+100) = xy. We do not know x and y, so this statement is not sufficient.

(2) (x+1.5) (y-100) = xy. Same argument as (1); this statement is not sufficient.

Using both, we can solve for x and y and so the answer is C.
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Re: DS [#permalink] New post 20 Dec 2005, 02:55
From stmt 1 we get xy = (x+100)(y-.75)
Solving we get xy = xy + 100y - .75x - 75
or .75x - 100y = -75 ---> equn 1

From stmt 2 we get xy = (x-100)(y+1.5)
solving we get xy = xy + 1.5x - 100y - 150
1.5x - 100y = 150 ---> equn 2

Solving equn 1 and equn 2 we get
.75x = 225 ==> X = 300
y = 3

Hence C.
Re: DS   [#permalink] 20 Dec 2005, 02:55
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