We can solve this problem using a weighted average method. Let's define the variables:

• Let $x$ be the pounds of candy selling at 75 cents per pound.
• Let $150 - x$ be the pounds of candy selling at 90 cents per pound (since the total mixture is 150 pounds).

Step 1: Set up the equation for the price.

The total cost of the 150-pound mixture must equal the total cost of the individual parts, weighted by their respective amounts. This gives the equation:

$\text{Total Cost of Mixture} = \text{Cost of 75-cent Candy} + \text{Cost of 90-cent Candy}$

The total cost of the mixture can also be written as:

$150 \times 0.78 = x \times 0.75 + (150 - x) \times 0.90$

Step 2: Simplify the equation.

First, calculate the total value for the mixture:

$150 \times 0.78 = 117$

Now, expand the right side:

$117 = 0.75x + 0.90(150 - x)$ $117 = 0.75x + 135 - 0.90x$

Step 3: Solve for $x$.

Combine like terms:

$117 = 135 - 0.15x$

Subtract 135 from both sides:

$117 - 135 = -0.15x$ $-18 = -0.15x$

Now divide both sides by -0.15:

$x = \frac{-18}{-0.15} = 120$

Step 4: Find the amount of the other candy.

Since $x = 120$, the amount of candy selling at 75 cents per pound is 120 pounds. The amount of candy selling at 90 cents per pound is:

$150 - 120 = 30$

Final Answer:

• 120 pounds of candy at 75 cents per pound.
• 30 pounds of candy at 90 cents per pound.

Would you like more details or have any questions?

Related Questions:

1. What if the mixture had to sell at 80 cents per pound instead of 78 cents?
2. How would the solution change if the total amount of candy were 200 pounds?
3. Can this problem be solved using a different method like a system of equations?
4. How much would the mixture cost per pound if the ratio of candies was reversed?
5. How does this mixture problem relate to the concept of weighted averages?

Tip:

In mixture problems, weighted averages play a crucial role in balancing the costs or properties of different components.