Well duhh. I just realized we have AI now.
ChatGPT:
To solve this, we need to:
- Convert the metal bar weight from pounds and ounces to just pounds for easier calculation.
- Determine the amount of tin in the original bar and adjust for the desired alloy composition.
- Calculate the amount of lead and antimony to add to the tin from the bar to achieve the target alloy composition.
Step 1: Convert the bar weight to pounds
The bar weighs 1 pound, 11 ounces. There are 16 ounces in a pound, so:
Total weight in pounds=1+1116=1.6875 pounds\text{Total weight in pounds} = 1 + \frac{11}{16} = 1.6875 \text{ pounds}Total weight in pounds=1+1611=1.6875 pounds
Step 2: Determine the amount of tin in the bar
Since the bar is composed of 30% tin and 70% lead, the amount of tin in the bar is:
Tin in the bar=1.6875×0.30=0.50625 pounds of tin\text{Tin in the bar} = 1.6875 \times 0.30 = 0.50625 \text{ pounds of tin}Tin in the bar=1.6875×0.30=0.50625 pounds of tin
Step 3: Determine the target weight for the alloy
The final alloy needs to contain 2% tin, 6% antimony, and 92% lead.
Let’s assume the total final weight of the alloy is WWW pounds. We know:
- 2%2\%2% of WWW is tin.
- 6%6\%6% of WWW is antimony.
- 92%92\%92% of WWW is lead.
Given that we have 0.50625 pounds of tin, and that must represent 2% of the total alloy, we can calculate the total weight WWW of the final alloy:
0.50625=0.02×W0.50625 = 0.02 \times W0.50625=0.02×W
Solving for WWW:
W=0.506250.02=25.3125 poundsW = \frac{0.50625}{0.02} = 25.3125 \text{ pounds}W=0.020.50625=25.3125 pounds
Step 4: Determine the required amounts of lead and antimony
Now that we know the total weight of the alloy is 25.3125 pounds, we can calculate the amounts of lead and antimony needed:
- Lead: 92%92\%92% of 25.3125 pounds is:
Lead required=0.92×25.3125=23.2875 pounds\text{Lead required} = 0.92 \times 25.3125 = 23.2875 \text{ pounds}Lead required=0.92×25.3125=23.2875 pounds
- Antimony: 6%6\%6% of 25.3125 pounds is:
Antimony required=0.06×25.3125=1.51875 pounds\text{Antimony required} = 0.06 \times 25.3125 = 1.51875 \text{ pounds}Antimony required=0.06×25.3125=1.51875 pounds
Step 5: Adjust for the lead already in the bar
The original bar contains 70% lead, so:
Lead in the bar=1.6875×0.70=1.18125 pounds of lead\text{Lead in the bar} = 1.6875 \times 0.70 = 1.18125 \text{ pounds of lead}Lead in the bar=1.6875×0.70=1.18125 pounds of lead
The additional lead required is:
Additional lead needed=23.2875−1.18125=22.10625 pounds\text{Additional lead needed} = 23.2875 - 1.18125 = 22.10625 \text{ pounds}Additional lead needed=23.2875−1.18125=22.10625 pounds
Final Summary:
To create the alloy with the composition of 2% tin, 6% antimony, and 92% lead, using one of your bars:
- You need 1.51875 pounds of antimony.
- You need 22.10625 pounds of additional lead.
Would you like the weight of additional lead and antimony in ounces as well?
Let's convert the amounts of additional lead and antimony into pounds and ounces.
Step 1: Lead
You need 22.10625 pounds of additional lead.
To convert 0.10625 pounds to ounces:
0.10625×16=1.7 ounces0.10625 \times 16 = 1.7 \text{ ounces}0.10625×16=1.7 ounces
So, the total additional lead required is:
22 pounds and 1.7 ounces22 \text{ pounds and } 1.7 \text{ ounces}22 pounds and 1.7 ounces
Step 2: Antimony
You need 1.51875 pounds of antimony.
To convert 0.51875 pounds to ounces:
0.51875×16=8.3 ounces0.51875 \times 16 = 8.3 \text{ ounces}0.51875×16=8.3 ounces
So, the total antimony required is:
1 pound and 8.3 ounces1 \text{ pound and } 8.3 \text{ ounces}1 pound and 8.3 ounces
Final Summary:
- You need 22 pounds, 1.7 ounces of additional lead.
- You need 1 pound, 8.3 ounces of antimony.