SOLUTION: Hello. I have been workingon this problem all night and I know it's probably something I'm overlooking. Please help. 2. Some students planned for a get-together. The budget

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Hello. I have been workingon this problem all night and I know it's probably something I'm overlooking. Please help. 2. Some students planned for a get-together. The budget       Log On


   

Question 22160: Hello. I have been workingon this problem all night and I know it's probably something I'm overlooking. Please help.
2. Some students planned for a get-together. The budget for food was $500. Five of the students failed to come because of the distance and therefore the cost of food for each member increased by $5. How many students attended the get-together?
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
Let number of student originally = x
Therefore number of students who actually turn up = x-5

Original cost per head = 500/x whereas the actual Cost per head = 500/(x-5)

Now, the actual cost (per head) - original cost (per head) = 5. So,

500%2F%28x-5%29+-+500%2Fx+=+5. Now get rid of the fractions by multiplying every term by x(x-5). Some cancelling happens and we are left with:

500x - 500(x-5) = 5x(x-5).

We need to multiply this out, collect terms and then solve the resulting quadratic:

500x+-+500x+%2B+2500+=+5x%5E2+-+25x%29
2500+=+5x%5E2+-+25x%29
5x%5E2+-+25x+-+2500+=+0%29
x%5E2+-+5x+-+500+=+0%29
(x + 20)(x - 25) = 0

so either x+20 = 0 OR x-25 = 0
so either x = -20 or x = 25

As x is the number of students, we cannot have a negative, so the answer we require is 25.

CHECK:
500/25 --> $20 per head

then down to 20 students, 500/20 --> $25 per head.

So this is correct, the price did increase by $5 per head.

jon.