My friend translated this problem from his native language, and I am unsure of the accuracy of the translation. If necessary, please edit to correct any faults.
A mineral substance, to be sent to a richening factory, contains $9\%$ metal. After maintenance, the substance contains $42\%$ metal, and the leftover contains $2\%$ metal. What percent of the substance becomes leftover?
How shall one proceed in order to solve this?
On
It sounds, that you have a mineral substance and you try to extract as much metal as you can. At the end of the process the total amount is still the same, but the two amounts have different proportions of metal.
Thus the equation is $x\cdot 0.02+(1-x)\cdot 0.42=0.09$
x is the fraction of leftover and mineral substance.
On
Let $X$ be the fraction of the useful substance & $(1-X)$ be the fraction of left over substance after maintenance then we have
$$\text{amount of the total metal before maintenance }=\text{amount of the total metal after maintenance} $$
$$0.09(1)=0.42X+0.02(1-X)$$ $$0.4X=0.07$$$$\implies X=0.175$$
Since, $2\text{%}$ of $(1-X)$ is the metal then rest $98\text{%}$ of $(1-X)$ is the left over substance then
the left over substance $$=0.98(1-X)=0.98(1-0.175)=0.825$$
Hence, $\color{red}{82.5\ \text{%}}$ of the substance is leftover
Suppose you started with $100$ units of mineral, which must then contain $9$ units of metal and $91$ units of non-metal. At the end you have $X$ units of substance and $100-X$ units of leftover. We know that $$.42X +.02(100-X)=9\;\Rightarrow\;.4X=7\;\Rightarrow\;X=17.5$$
Thus $17.5\%$ of the original sample remains in the "substance", so $82.5\%$ has become leftover.