Math wizards get in here

[Nobody911]

Well duhh. I just realized we have AI now.

ChatGPT:

To solve this, we need to:

1. Convert the metal bar weight from pounds and ounces to just pounds for easier calculation.
2. Determine the amount of tin in the original bar and adjust for the desired alloy composition.
3. 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.

While AI has made significant strides in solving various mathematical problems, certain categories still present challenges. Here is an example:

Undecidable Problems: These are problems for which no algorithm can provide a solution. A classic example is the Halting Problem, which involves determining whether a given program will finish running or continue indefinitely.

The Halting Problem is a fundamental concept in computer science and mathematics, formulated by Alan Turing in 1936 ( this gay genius is a key reason why I am not against LGBTQ+. He saved countless lives from being killed by Hitler ).

It involves determining whether a given program, when run with a specific input, will eventually halt (finish execution) or run indefinitely (loop forever).

Turing proved that a general algorithm to solve the Halting Problem for all possible program-input pairs cannot exist. The proof is based on a technique known as diagonalization, which shows that if we could create such an algorithm, we could construct a program that would lead to a contradiction.The essence of the proof is as follows:

Assume there exists a hypothetical algorithm, called "H," that can determine if any program "P" halts with input "I." We can then create a new program, "D," which behaves as follows: it uses "H" to check if "D" halts when given itself as input. If "H" determines that "D(D)" halts, "D" enters an infinite loop; if "H" determines that "D(D)" does not halt, "D" halts.

This creates a paradox, as "H" cannot correctly predict the behavior of "D."This result has profound implications, showing that there are inherent limitations to computation and that certain problems are undecidable, meaning no algorithm can solve them for all cases.

AI is evidently not a magic wand. It's not the answer to every question.

 

[zaangalewa]

JGalt

I made a miscalculation (never make a calculation when you are in hurry!!!). 92 / 2 is not 41 but 46!!! (What a stupid mistake. I'm an idiot!)

So:

tin = 30% = 0.229631137311 kg.
antimon (3 times more than tin)=0.688893411933 kg.
lead (46 times more than tin)=10.563032316306 kg.
...
10.563032316306 - 0.535805987059 = 10.027226329247 kg
...
Per bar ~10.0 kg lead and ~0.7 kg antimon. (10027 g + 690 g)