Question 1153: 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 khwang(438)   (Show Source):
You can put this solution on YOUR website! Sol: Assume there were x studented attended,so their average was 500/x dollars.
Since there were x + 5 students planned (note 5 were absent), the average
among the planned was 500/(x+5).
We know the difference between these two averages is 5, we have
500/x = 500/(x+5) + 5,
Dividing eqch term by 5:
100/x = 100/(x+5) + 1,
Multiplying by x(x+5) to cancel the denominator:
100(x+5) = 100x + x^2 + 5x,
Simplify: x^2 + 5x -500 = 0,
Factoring: (x-20)(x + 25) = 0,
So,x= 20 or -25(negative, invalid answer)
Answer : there were 20 students attended the get-together.
|
|
|
