SOLUTION: i need help with this uhmm....word problem. Please and Thank You!!! A dealer has some hard candy worth $2.00 a pound and some worth $3.00 a pound. He wishes to make a mixture of 8

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: i need help with this uhmm....word problem. Please and Thank You!!! A dealer has some hard candy worth $2.00 a pound and some worth $3.00 a pound. He wishes to make a mixture of 8      Log On


   

Question 38509: i need help with this uhmm....word problem. Please and Thank You!!!
A dealer has some hard candy worth $2.00 a pound and some worth $3.00 a pound. He wishes to make a mixture of 80 pounds that he can sell for $2.20 a pound. How many pounds of each candy does he need?
thank you so much.....
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A dealer has some hard candy worth $2.00 a pound and some worth $3.00 a pound. He wishes to make a mixture of 80 pounds that he can sell for $2.20 a pound. How many pounds of each candy does he need?
Let amount of $2 hard candy be "x".
Then amount of $3 hard candy is "80-x"
Value of $2 candy is 2x dollars
Value of $3 candy is 3(80-x)=240-3x dollars
Value of the mixed candy is 2.2(80)=176 dollars
EQUATION:
Value of $2 candy + Value of $3 candy = Value of mixed candy
2x + 240-3x=176
-x=-64
x= 64 pounds of $2 hard candy
80-x=16 pounds of $3 hard candy.
Cheers,
Stan H.