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.

For those comfortable with algebra, Brent's solution is the way to go.
For those who prefer to test cases, here's an alternate approach:

Let F = fee per person, P = the number of people, and T = the total collected.
Then:
FP = T.

In each statement, the value of T does not change.
Implication:
F and P are INVERSELY PROPORTIONAL:
If the value of F is multiplied by factor of x/y, then the value of P must be multiplied by a factor of y/x, so that value of T remains constant.
Thus, when the values of F and P change, the following constraint must be satisfied:
If (new F)/F = x/y, then (new P)/P = y/x.

Statement 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.
Here, new P = P+100.

Case 1: F=$1.50
In this case, new F = 1.50 - 0.75 = 0.75.
Thus:
(new F)/F = 0.75/1.50 = 1/2.
(new P)/P = 2/1, implying the following:
(P+100)/P = 2
P + 100 = 2P
P = 100.

Case 2: F=$3.00
In this case, new F = 3.00 - 0.75 = 2.25.
Thus:
(new F)/F = 2.25/3.00 = 3/4.
(new P)/P = 4/3, implying the following:
(P+100)/P = 4/3
3P + 300 = 4P
P = 300.

Since P can be different values, INSUFFICIENT.

Statement 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.
Here, new P = P-100.

Case 3: F=$1.50
In this case, new F = 1.50 + 1.50 = 3.00
Thus:
(new F)/F = 3.00/1.50 = 2/1.
(new P)/P = 1/2, implying the following:
(P-100)/P = 1/2
2P - 200 = P
P = 200.

Case 4: F=$3.00
In this case, new F = 3.00 + 1.50 = 4.50.
Thus:
(new F)/F = 4.50/3.00 = 3/2.
(new P)/P = 2/3, implying the following:
(P-100)/P = 2/3
3P - 300 = 2P
P = 300.

Since P can be different values, INSUFFICIENT.

Statements combined:
The values of F and P are both consistent only in Cases 2 and 4:
In each case, F = $3.00 and P = 300.
Implication:
The only combination that will satisfy both statements is F = $3.00 and P = 300.
SUFFICIENT.

The correct answer is C.