To solve this problem, we need to find the amount of the first alloy in a mixture that ultimately consists of 100 kg of material containing 90% silver. We are given two silver-copper alloys:

• The first alloy contains \(93\tfrac{1}{3}\%\) silver.
• The second alloy contains \(86\tfrac{2}{3}\%\) silver.

Let's denote:

• \(x\) as the amount of the first alloy in kg.
• \(y\) as the amount of the second alloy in kg.

We know that the total weight of the mixture is 100 kg, so:

\(x + y = 100\)

Now, let's express the amounts of silver contained in each alloy:

• The first alloy's silver content is \(93\tfrac{1}{3}\%\), which can be expressed as a fraction: \(\frac{280}{3}\%\).
• The second alloy's silver content is \(86\tfrac{2}{3}\%\), which can be expressed as a fraction: \(\frac{260}{3}\%\).

The weight of the silver in the first alloy is:

\(\frac{280}{3} \cdot \frac{x}{100} = \frac{280x}{300}\)

The weight of the silver in the second alloy is:

\(\frac{260}{3} \cdot \frac{y}{100} = \frac{260y}{300}\)

For the entire mixture to contain 90% silver, the equation using the silver contents is:

\(\frac{280x}{300} + \frac{260y}{300} = \frac{90 \cdot 100}{100}\)

Simplifying this equation gives us:

\(\frac{280x}{300} + \frac{260y}{300} = 90\)

Removing the denominator by multiplying through by 300:

\(280x + 260y = 27000\)

We now have the following system of equations:

• \(x + y = 100\) (Equation 1)
• \(280x + 260y = 27000\) (Equation 2)

We solve the system of linear equations simultaneously. From Equation 1:

\(y = 100 - x\)

Substitute \(y\) in Equation 2:

\(280x + 260(100 - x) = 27000\)

Expand and simplify:

\(280x + 26000 - 260x = 27000\) \(20x = 1000\)

Divide both sides by 20:

\(x = 50\)

Therefore, the amount of the first alloy that should be mixed is 50 kg. The correct answer is:

50 kg