Let the alloy contain $a \; \text{kg}$ silver and $b \; \text{kg}$ copper.
Now, when mixed with $3 \; \text{kg}$ pure silver.
$\Rightarrow \frac{a+3}{a+b+3} = \frac{90}{100}$
$\Rightarrow \frac{a+3}{a+b+3} = \frac{9}{10}$
$\Rightarrow 10a + 30 = 9a + 9b + 27$
$\Rightarrow a – 9b = \;– 3 \quad \longrightarrow (1)$
Now, silver in $2^{\text{nd}}$ alloy $ = 2 \left( \frac{90}{100} \right) = \frac{9}{5} = 1.8 \; \text{kg}$
And, $\frac{a+1.8}{a+b+2} = \frac{84}{100}$
$\Rightarrow \frac{a+1.8}{a+b+2} = \frac{21}{25}$
$\Rightarrow 25a + 45 = 21a + 21b + 42$
$\Rightarrow 4a – 21b =\; – 3 \quad \longrightarrow (2)$
From equation $(1)\; \&\; (2).$
$\begin{array} {} (a – 9b = -3) \times 4 \\ (4a – 21b = - 3) \times 1 \\\hline 4a – 36b = -12 \\ 4a – 21b = -3 \\ \; – \quad + \qquad \; +\\\hline \end{array}$
$\Rightarrow -15b = -9$
$\Rightarrow \boxed{b = \frac{3}{5} = 0.6 \; \text{kg}}$
Put the value of $b$ is equal in $(1),$ we get.
$\Rightarrow a – 9 \left(\frac{3}{5} \right) = -3$
$\Rightarrow 5a – 27 = -15$
$\Rightarrow 5a = 12$
$\Rightarrow a = \frac{12}{5}$
$\Rightarrow \boxed{a = 2.4 \; \text{kg}}$
$\therefore$ The weight of the initial alloy $ = a+b = 0.6 + 2.4 = 3 \; \text{kg}.$
Correct Answer $: \text{C}$