We evaluate each option using standard qualitative inorganic chemistry facts.

Option A: From a mixed precipitate of $\mathrm{AgCl}$ and $\mathrm{AgI}$, ammonia solution dissolves only $\mathrm{AgCl}$.

• $\mathrm{AgCl}$ dissolves in aqueous ammonia due to complex formation: $\mathrm{AgCl + 2NH_3 \rightarrow [Ag(NH_3)_2]^+ + Cl^-}$
• $\mathrm{AgI}$ does not dissolve in ammonia because it is much less soluble and does not form the complex sufficiently.

Therefore, A is correct.

Option B: Ferric ions give a deep green precipitate on adding potassium ferrocyanide solution.

• $\mathrm{Fe^{3+}}$ with potassium ferrocyanide gives Prussian blue coloration/precipitate, not deep green.
• The product is ferric ferrocyanide.

Therefore, B is incorrect.

Option C: On boiling a solution having $\mathrm{K^+}$, $\mathrm{Ca^{2+}}$ and $\mathrm{HCO_3^-}$ ions we get a precipitate of $\mathrm{K_2Ca(CO_3)_2}$.

• On boiling bicarbonate solution, bicarbonate decomposes: $2\mathrm{HCO_3^-} \rightarrow \mathrm{CO_3^{2-} + CO_2 + H_2O}$
• Then $\mathrm{Ca^{2+}}$ forms insoluble calcium carbonate: $\mathrm{Ca^{2+} + CO_3^{2-} \rightarrow CaCO_3\downarrow}$
• Potassium salts remain soluble; no such precipitate $\mathrm{K_2Ca(CO_3)_2}$ is formed.

Therefore, C is incorrect.

Option D: Manganese salts give a violet borax bead test in the reducing flame.

• Manganese gives an amethyst/violet bead in oxidizing flame.
• In reducing flame, the color changes and is not violet as stated.

Therefore, D is incorrect.

Hence, the only correct statement is: $\boxed{\text{A}}$