Question

A jeweller has bars of 18-carat gold and 12-carat gold. How much of each must be melted together to obtain a
bar of 16-carat gold, weighing 120 g?

Answer 1

Let x g and y g be the weight of 18-carat and 12-carat gold respectively.
As per the given condition
$$\begin{gathered} \frac{18x}{24}+\frac{12y}{24}=\frac{120\times 16}{24} \\ \Rightarrow 3x+2y=320.....\left(i\right) \end{gathered}$$
And
$x+y=120.....\left(ii\right)$
Multiplying (ii) by 2 and subtracting from (i), we get
$x=320-240=80$
Now, putting x = 80 in (ii), we have
$80+y=120\Rightarrow y=40$
Hence, the required weight of 18-carat and 12-carat gold bars are 80 g and 40 g respectively.