SOLUTION: im having a hard time trying to figure out the formula. boxes of candy containing creams and caramels sells for $8 per box and hold 50 pieces. caramels cost $0.10 to produce and cr
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Question 51687This question is from textbook finite mathematics
: im having a hard time trying to figure out the formula. boxes of candy containing creams and caramels sells for $8 per box and hold 50 pieces. caramels cost $0.10 to produce and creams cost 0.20. how many creams and caramels should be in each box for no profit or loss. i have come up with a formula but it is not working, please help. .10x+.20y=8(50). ive tried different but this seems to be the way other formulas are in but wont work for me.
This question is from textbook finite mathematics
Answer by tutorcecilia(2152) (Show Source): You can put this solution on YOUR website!
Use the standard formula of a line to solve:
Ax+By=C
A=rate of item A
x=amount of item A
B=rate of item
y=amount of item B
C=(rate of C)(Amount of item C)
C = ($8 X 1 box = $8)
C also equals 50 individual pieces of candy
So, x+y=50 or 50-x=y
.
Ax+By=C
Setting it up:
(Rate A)(amount A) + (Rate B)(amount B)=C
.10x+.20y=8 [The equation of the line]
(.10)(x)+(.20)(50-x)=8 [Substitute (50-x)=y]
.10x+10-.20x=8 [Solve for x]
-.10x=-2
x=20
.
Check by plugging all of the values back into the original equation:
Ax+By=C
(.10)(20)+(.20)((50-20)=8
(.10)(20)+(.20)(50-20)=8
2+10-4=8
8=8 [Checks out]
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