How many people came

[rohit_gmat]

Q. If everyone was charged the same fee, how many people came?
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.

source : 198 700+ Qs list

OA : C
IMO : A

plz provide explanations as well.. thanks !!

[cans]

n people came. fees = f.. fee received = fn
A) (f- f/7500)(n+100) = fn
insufficient.
B) (f+1.5)(n-100)=fn -> 1.5n -100f -150=0.
3n - 200f - 300 =0 insufficient.
combining A,B we have 2 eqn and 2 variable. Thus IMO C

[rohit_gmat]

n people came. fees = f.. fee received = fn
A) (f- f/7500)(n+100) = fn
insufficient.
B) (f+1.5)(n-100)=fn -> 1.5n -100f -150=0.
3n - 200f - 300 =0 insufficient.
combining A,B we have 2 eqn and 2 variable. Thus IMO C

hi cans.. need ur help... for stmnt 1, i did the following :

(f - 0.0075f) (n + 100) = fn
f (1 - 0.0075) (n + 100) = fn
and i cancel f out on both sides...
(1 - 0.0075) (n+100) = n
n + 100 - 0.0075n - 0.75 = n
100 = 0.0075n + 0.75
and thats solvable to n...

what did i do wrong in here?

[cans]

I made a mistake in A.... it is as you mentioned but when you solve for n, its not an integer. But # of people should be integer. so A can't be correct....... I think question is wrong......

[GMATGuruNY]

Q. If everyone was charged the same fee, how many people came?
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.

source : 198 700+ Qs list

OA : C
IMO : A

plz provide explanations as well.. thanks !!

I've corrected a typo in the posted question: Statement 1 should say if the fee had been .75 less.

Let the number of people = p.
Let the fee = f.
Sum of the fees = pf.

Statement 1: If the fee had been 0.75 less and 100 more people came, the club would have received the same amount.
The sum here = (p+100)(f-75).
Since the sum here is the same as the original sum:
(p+100)(f-75) = pf.
pf - 75p + 100f - 7500 = pf.
-75p + 100f = 7500.
No way to solve for p.
Insufficient.

Statement 2: If fee had been $1.50 more and 100 fewer people came, the club would have received the same amount.
The sum of the fees here = (p-100)(f+150)
Since the sum here is the same as the original sum:
(p-100)(f+150) = pf.
pf + 150p - 100f - 15000 = pf.
150p - 100f = 15,000.
No way to solve for p.
Insufficient.

Statements 1 and 2 combined:
Two variables, two distinct linear equations.
Thus, we can solve for p.
Sufficient.

The correct answer is C.

[saketk]

I think if we solve the original question (with typo error) then the answer should be A.

[just_dan]

n people came. fees = f.. fee received = fn
A) (f- f/7500)(n+100) = fn
insufficient.
B) (f+1.5)(n-100)=fn -> 1.5n -100f -150=0.
3n - 200f - 300 =0 insufficient.
combining A,B we have 2 eqn and 2 variable. Thus IMO C

hi cans.. need ur help... for stmnt 1, i did the following :

(f - 0.0075f) (n + 100) = fn
f (1 - 0.0075) (n + 100) = fn
and i cancel f out on both sides...
(1 - 0.0075) (n+100) = n
n + 100 - 0.0075n - 0.75 = n
100 = 0.0075n + 0.75
and thats solvable to n...

what did i do wrong in here?

Yeah, your solution seems correct, thus A shall be the right answer!

[Brent@GMATPrepNow]

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.

Let F = admission fee
Let N = number of attendees
So, the TOTAL revenue = FN

Target question: What is the value of N?

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.
In other words, if the fee had been (F - 0.75) and the number of attendees had been (N + 100) the total revenue would still be FN
We can write: (F - 0.75)(N + 100) = FN
Expand: FN + 100F - 0.75N - 75 = FN
Simplify: 100F - 0.75N - 75 = 0
Since we cannot solve this equation for N, we cannot answer the target question with certainty.
So statement 1 is NOT SUFFICIENT

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.
In other words, if the fee had been (F + 1.50) and the number of attendees had been (N - 100) the total revenue would still be FN
We can write: (F + 1.50)(N - 100) = FN
Expand: FN - 100F + 1.5N - 150 = FN
Simplify: -100F + 1.5N - 150 = 0
Since we cannot solve this equation for N, we cannot answer the target question with certainty.
So statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that 100F - 0.75N - 75 = 0
Statement 2 tells us that -100F + 1.5N - 150 = 0
Here we have two different linear equations involving two variables.
Since we COULD solve this system for N, we could answer the target question with certainty.
So, the combined statements are SUFFICIENT

ASIDE: We don't need to actually solve the system. We need only recognize that we have SUFFICIENT information to do so.

Answer = C

Cheers,
Brent