SOLUTION: A candy store sells boxes of candy containing caramels and cremes. Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost $0.25
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A candy store sells boxes of candy containing caramels and cremes. Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost $0.25
Log On
Question 372702: A candy store sells boxes of candy containing caramels and cremes. Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size). If the caramels cost $0.25 to produce and the cremes cost $0.45, how many of each should be in a box to make a profit of $3? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A candy store sells boxes of candy containing caramels and cremes.
Each box sells for $12.50 and holds 30 pieces of candy (all pieces are the same size).
If the caramels cost $0.25 to produce and the cremes cost $0.45,
how many of each should be in a box to make a profit of $3?
:
Let c = no. of caramels
Let m = no. of cremes
:
Total candy equation
c + m = 30
c = (30-m); this form for substitution
:
Cost equation
.25c + .45m = 12.50 - 3.00
.25c + .45m = 9.50
Substitute (30-m) for c
.25(30-m) + .45m = 9.50
7.50 - .25m + .45m = 9.50
-.25m + .45m = 9.50 -7.50
.20m = 2.00
m =
m = 10 creams in each box
then
20 - 10 = 20 caramels in each box
:
:
you can check our solutions by substitution in the original equation
