SOLUTION: 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

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Question 125815: 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?
Found 2 solutions by checkley71, ankor@dixie-net.com:
Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
(500/X)+5=500/(X-5)
(500+5X)/X=500/(X-5) CROSS MULTIPLY.
(X-5)(500+5X)=500X
500X-2500+5X^2-25X=500X
5X^2-25X-2500=0
5(X^2-5X-500)=0
5(X-25)(X=20=0
X-25=0
X=25 STUDENTS WERE THE ORIGINAL NUMBER.
SO 25-5=20 IS THE ACTUAL NUMBER WHO ATTENDED.
PROOF
500/25+5=500/(25-5)
20+5=500/20
25=25

Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
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?
:
Let x = no. of students that attended
then
(x+5) = no. of students anticipated to attend
:
The original cost per student for food:
Actual cost per student:
:
The equation
Actual cost - original cost = $5
- = 5
:
Multiply each term by x(x+5) to get rid of the denominators. cancel them & you have:
500(x+5) - 500x = 5*x(x+5)
:
500x + 2500 - 500x = 5x^2 + 25x
:
5x^2 + 25x - 2500 = 0; 500x's cancel, arrange as a quadratic equation
:
Simplify divide eq by 5:
x^2 + 5x - 500 = 0
:
Factors to:
(x + 25) (x - 20) = 0
:
Positive solution is what we want:
x = +20 students actually attended
;
:
Check solution by finding the cost per student
original 500/25 = 20
actual: 500/20 = 25
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difference cost = 5