SOLUTION: A busload of campers stopped at a dairy stand for ice cream. They ordered 77 cones, some soft-serve at $2.00 each and the rest hard pack at $1.25 each. If the bill was $116.50, how

Algebra ->  College  -> Linear Algebra -> SOLUTION: A busload of campers stopped at a dairy stand for ice cream. They ordered 77 cones, some soft-serve at $2.00 each and the rest hard pack at $1.25 each. If the bill was $116.50, how      Log On


   



Question 704033: A busload of campers stopped at a dairy stand for ice cream. They ordered 77 cones, some soft-serve at $2.00 each and the rest hard pack at $1.25 each. If the bill was $116.50, how many of each type of cone were ordered?
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Just like a mixture problem. x is count of soft serve, y is count of hard pack.
x+y=77 cones.

The money: x*2.00+y*1.25=116.50, dollars.
Try using x=77-y and substitute into the dollars equation. Then solve for y.

(77-y)*2.00+1.25*y=116.50
154-2y+1.25y=116.5
-0.75y=116.5-154
y=(154-116.5)/0.75
highlight%28y=50%29
...

Further Note:
The cones themselves are hard, cooked like a cookie and then later used for holding ice cream. The forms of ice cream that these campers are buying are "soft serve" and "hard pack". Neither of these are cones; they are ice cream. The "y=50" is for how many cones fit with hard pack ice cream were bought by the campers. Therefore, x=27 is how many cones fit with soft serve were bought by these campers. You do not want to get confused between cones and any type of icecream that may be placed into them.