A candy store makes a 10-lb mixture of gummy worms, candy corn, and sourballs. The cost of gummy worms is $1.00 per pound, candy corn cost $3.00 per pound, and sourballs cost $1.50 per pound. The mixture calls for three times as many gummy worms as candy corn. The total cost of the mixture is $15.00. How much of each ingredient did the store use?
To solve this problem, we will set up a system of equations to represent the given information and then solve for the unknowns.
Let's denote the amounts of gummy worms, candy corn, and sourballs in pounds as g, c, and s, respectively.
We know that the mixture weighs 10 pounds, so we can write the first equation as:
g + c + s = 10 (equation 1)
We also know that the cost of the gummy worms is $1.00 per pound, so the cost of the gummy worms in dollars is g * $1.00 = $1.00g.
Similarly, the cost of the candy corn is $3.00 per pound, so the cost of the candy corn in dollars is c * $3.00 = $3.00c.
And the cost of the sourballs is $1.50 per pound, so the cost of the sourballs in dollars is s * $1.50 = $1.50s.
The total cost of the mixture is given as $15.00, so we can write the second equation as:
$1.00g + $3.00c + $1.50s = $15.00 (equation 2)
Lastly, it is mentioned that the mixture calls for three times as many gummy worms as candy corn, so we can write the third equation as:
g = 3c (equation 3)
Now, we have three equations to solve simultaneously: equations 1, 2, and 3.
From equation 3, we can substitute g in equation 1 and equation 2:
3c + c + s = 10 (substituting g from equation 3 into equation 1)
$1.00(3c) + $3.00c + $1.50s = $15.00 (substituting g from equation 3 into equation 2)
Simplifying equation 1 and equation 2:
4c + s = 10 (equation 4)
$3.00c + $4.50c + $1.50s = $15.00 (equation 5)
Next, we need to convert equation 5 into dollars by multiplying both sides by 100 (to remove decimals):
300c + 450c + 150s = 1500 (equation 6)
Now, we can solve equations 4, 5, and 6 simultaneously to find the values of c, s, and g.
Subtracting equation 4 from equation 6, we get:
750c + 150s = 1490 (equation 7)
Multiplying equation 4 by 150 and subtracting from equation 7, we get:
750c + 150s - 600c - 150s = 1490 - 600
150c = 890
Dividing both sides of equation 8 by 150, we find:
c = 5.93
Since the number of candies must be a whole number, we can take the nearest whole number, which is 6.
Now, we can substitute the value of c into equation 4 to find s:
4(6) + s = 10
24 + s = 10
s = 10 - 24
s = -14
Since the number of candies cannot be negative, we discard this solution.
Let's go back to equation 6 and substitute the value of c = 6 to find s:
300(6) + 450(6) + 150s = 1500
1800 + 2700 + 150s = 1500
150s = 1500 - 1800 - 2700
150s = -3000
s = -20
Again, we discard this negative solution.
Since we couldn't find a valid solution for sourballs, we will try a different approach.
Going back to equation 3, where g = 3c, we know that the candy corn must be a multiple of 3.
Let's substitute c = 3 in equation 1:
g + 3 + s = 10
g + s = 7 (equation 8)
Now, let's substitute c = 3 in equation 2:
$1.00g + $3.00(3) + $1.50s = $15.00
$1.00g + $9.00 + $1.50s = $15.00
$1.00g + $1.50s = $15.00 - $9.00
$1.00g + $1.50s = $6.00 (equation 9)
From equation 8, we can write g = 7 - s and substitute this in equation 9:
$1.00(7 - s) + $1.50s = $6.00
$7.00 - $1.00s + $1.50s = $6.00
$0.50s = $6.00 - $7.00
$0.50s = -$1.00
s = -$1.00 / $0.50
s = -2
Again, we discard this negative solution.
From all the attempts, we didn't find a valid solution within the given constraints. It is likely that there was an error in the problem statement or the constraints.