To solve this problem, we need to calculate the composition of the new alloy C that is created by mixing equal quantities of two alloys, A and B, and deduce the amount of gold in C based on the given information.

Here are the steps to solve the problem:

1. Alloy A has gold and copper mixed in the ratio 7:2. This means, if the total quantity of alloy A is \(9\) parts, \(7\) parts are gold and \(2\) parts are copper.
2. Alloy B has gold and copper mixed in the ratio 7:11. This means, if the total quantity of alloy B is \(18\) parts, \(7\) parts are gold and \(11\) parts are copper.
3. Alloys A and B are mixed in equal quantities to form alloy C. Let's assume each alloy has a total weight of \(x\) kg.
4. The gold content in alloy A is \(\frac{7}{9}x\) kg and the copper content is \(\frac{2}{9}x\) kg.
5. The gold content in alloy B is \(\frac{7}{18}x\) kg and the copper content is \(\frac{11}{18}x\) kg.
6. When these two alloys are mixed to form alloy C, we add their gold contents and copper contents separately.
7. Gold content in C = \(\frac{7}{9}x + \frac{7}{18}x\).
8. Copper content in C = \(\frac{2}{9}x + \frac{11}{18}x\).

Given that the copper content in C is 10 kg:

• Therefore, \(\frac{2}{9}x + \frac{11}{18}x = 10\).

We solve for \(x\):

• First, find a common denominator for the fractions: \(\frac{4}{18}x + \frac{11}{18}x = 10\).
• \(\frac{15}{18}x = 10\).
• \(\frac{5}{6}x = 10\).
• x = 10 × \(\frac{6}{5}\) = 12 kg.

Now, substituting value of \(x\) back to find gold content in C:

• Gold content in C = \(\frac{7}{9} \times 12 + \frac{7}{18} \times 12\)
• = \( \frac{7 \times 12}{9} + \frac{7 \times 12}{18}\)
• = \(\frac{28}{3} + \frac{14}{3}\)
• = \(\frac{42}{3}\) = 14 kg.

Thus, the amount of gold in alloy C is 14 kg.