Candy that costs 31 cents per pound is mixed with candy that costs 46 cents per pound to obtain 50 pounds of candy that costs 40 cents per pound. How many pounds of each of the two types of candy are in the mix?
The Correct Answer and Explanation is:
To solve this problem, we will use a system of equations based on the principles of weighted averages and total amount.
Let’s define the variables:
- Let xx be the number of pounds of 31 cents per pound candy.
- Let yy be the number of pounds of 46 cents per pound candy.
We are told that the total mixture is 50 pounds, so: x+y=50(Equation 1)x + y = 50 \quad \text{(Equation 1)}
The mixture costs 40 cents per pound, so the total cost of 50 pounds at 40 cents per pound is: 50×0.40=20 dollars50 \times 0.40 = 20 \text{ dollars}
The cost of xx pounds of 31¢ candy is: 0.31×0.31x
The cost of yy pounds of 46¢ candy is: 0.46y0.46y
So the total cost equation is: 0.31x+0.46y=20(Equation 2)0.31x + 0.46y = 20 \quad \text{(Equation 2)}
Now solve the system:
From Equation 1: y=50−xy = 50 – x
Substitute into Equation 2: 0.31x+0.46(50−x)=200.31x + 0.46(50 – x) = 20
Distribute: 0.31x+23−0.46x=200.31x + 23 – 0.46x = 20
Combine like terms: −0.15x+23=20-0.15x + 23 = 20
Subtract 23 from both sides: −0.15x=−3-0.15x = -3
Divide by -0.15: x=20x = 20
Now plug back in: y=50−x=50−20=30y = 50 – x = 50 – 20 = 30
✅ Final Answer:
- 20 pounds of 31¢ candy
- 30 pounds of 46¢ candy
Explanation (300+ words):
This is a classic mixture problem that involves combining two products with different costs to obtain a specific weight and price. The key to solving such problems is setting up equations that reflect both the quantity and the value.
We first define variables to represent how much of each type of candy is used. The total weight constraint gives one equation: x+y=50x + y = 50, representing that the sum of the two candies must equal the final 50-pound mixture.
Next, we calculate the total cost of the final mixture. Since 50 pounds of candy is sold at 40 cents per pound, the total value is $20. This leads us to the second equation: the cost of the 31¢ candy plus the cost of the 46¢ candy must total $20. Using algebra, we substitute one variable (y) from the first equation into the second, which reduces the system to a single variable.
After simplifying and solving, we find that 20 pounds of the cheaper candy and 30 pounds of the more expensive candy produce a blend that averages out to 40 cents per pound. This makes intuitive sense too: since 40 cents is closer to 46 than 31, there must be more of the more expensive candy.
This problem demonstrates how algebra can be used to solve real-world scenarios involving cost-efficiency and product blending.