You are volunteering to help with the soccer team's Valentine's Day fundraiser. Each 16-ounce, bag of nuts the team sold must include at least $60 \%$ chocolate-covered nuts inside.

However, instead of receiving a shipment of separated plain nuts and chocolate nuts, they delivered two large containers of mixed nuts. The first says it is $50 / 50$ plain and chocolate covered. The second contains $80 \%$ chocolate-covered nuts.
The team is dismayed, but you come up with a solution. You suggest combining ounces* of the $50 / 50$ nuts with ounces* of the $80 \%$ chocolate-covered nuts, to create the $60 \%$ mixture required for each bag.
*estimate
Then Bob, another student on the team says, "Wait, we promised at lêast $60 \%$. So if we just do half and half, won't we be giving them at least $60 \%$ ?"
Bob $\hat{\imath}$ correct.

Step 1 ：Bob suggests mixing equal amounts of the 50/50 nuts and the 80% chocolate-covered nuts. This would result in a mixture that is 65% chocolate-covered nuts (the average of 50% and 80%). This is indeed more than the 60% promised, so Bob is correct.

Step 2 ：However, this may not be the most cost-effective solution. If the 80% chocolate-covered nuts are more expensive, it might be more economical to use more of the 50/50 nuts and less of the 80% nuts, while still meeting the 60% requirement.

Step 3 ：To find the most cost-effective solution, we can set up a system of equations. Let x be the amount of 50/50 nuts and y be the amount of 80% nuts. We want to solve for x and y such that: \(0.5x + 0.8y = 0.6(x + y)\). This equation represents the requirement that the final mixture is 60% chocolate-covered nuts.

Step 4 ：The solution to the equation is \(y = 0.5x\). This means that for every ounce of the 50/50 nuts, we should use half an ounce of the 80% chocolate-covered nuts to create a mixture that is 60% chocolate-covered nuts. This is the most cost-effective solution that still meets the 60% requirement.

Step 5 ：Final Answer: For every ounce of the 50/50 nuts, we should use half an ounce of the 80% chocolate-covered nuts to create a mixture that is 60% chocolate-covered nuts. So, \(\boxed{y = 0.5x}\).