This problem involves mixing two alloys with different ratios of tin and copper to achieve a desired ratio in the final mixture. We need to find the quantity of the first alloy required.

Alloy Composition Analysis

• First Alloy: Contains tin and copper in the ratio 5:3.
• Second Alloy: Contains tin and copper in the ratio 3:1. We have 20 kg of this alloy.
• Target Mixture: The new alloy should have tin and copper in the ratio 4:2 (which simplifies to 2:1).

Calculating Quantities in the Second Alloy

The second alloy is present in a quantity of 20 kg with a tin to copper ratio of 3:1.

• Total parts in the second alloy ratio = 3 + 1 = 4.
• Quantity of tin in 20 kg of the second alloy = $ \frac{3}{4} \times 20 \text{ kg} = 15 \text{ kg} $.
• Quantity of copper in 20 kg of the second alloy = $ \frac{1}{4} \times 20 \text{ kg} = 5 \text{ kg} $.

Calculating Quantities in the First Alloy

Let the required quantity of the first alloy be $x$ kg. The ratio of tin to copper is 5:3.

• Total parts in the first alloy ratio = 5 + 3 = 8.
• Quantity of tin in $x$ kg of the first alloy = $ \frac{5}{8}x \text{ kg} $.
• Quantity of copper in $x$ kg of the first alloy = $ \frac{3}{8}x \text{ kg} $.

Setting up the Equation for the New Alloy

The new alloy is formed by mixing $x$ kg of the first alloy and 20 kg of the second alloy. The final ratio of tin to copper is 4:2.

• Total quantity of tin in the new alloy = (Tin from first alloy) + (Tin from second alloy) = $ \frac{5}{8}x + 15 $ kg.
• Total quantity of copper in the new alloy = (Copper from first alloy) + (Copper from second alloy) = $ \frac{3}{8}x + 5 $ kg.
• The ratio of total tin to total copper in the new alloy is $ \frac{4}{2} $, which simplifies to $ \frac{2}{1} $.

We can set up the equation based on the final ratio:

Solving for the Quantity 'x'

Now, we solve the equation for $x$:

1. Cross-multiply: $ \frac{5}{8}x + 15 = 2 \left( \frac{3}{8}x + 5 \right) $
2. Distribute the 2: $ \frac{5}{8}x + 15 = \frac{6}{8}x + 10 $
3. Rearrange the terms to isolate $x$: $ 15 - 10 = \frac{6}{8}x - \frac{5}{8}x $
4. Simplify: $ 5 = \frac{1}{8}x $
5. Solve for $x$: $ x = 5 \times 8 $
6. Result: $ x = 40 $ kg.

Therefore, 40 kg of the first alloy must be mixed.