Gunpowder contains 75% nitre and 10% sulphur. Find the amount of gunpowder which carries 9 kg nitre. What amount of gunpowder would contain 2.5 kg sulphur?
Let $x$ be the amount of gunpowder.
Amount of nitre $=75 \%$
Let $x \mathrm{~kg}$ be the amount of gunpowder containing $9 \mathrm{~kg}$ of nitre.
i.e., $(75 \%$ of $\mathrm{x})=9 \mathrm{~kg}$
$\Rightarrow\left(x \times \frac{75}{100}\right)=9$
$\Rightarrow \frac{75 x}{100}=9$
$\Rightarrow x=\left(9 \times \frac{100}{75}\right)$
$\Rightarrow x=12 \mathrm{~kg}$
Hence, $12 \mathrm{~kg}$ of gunpowder contains $9 \mathrm{~kg}$ of nitre.
Now, amount of sulphur $=10 \%$
Let $x \mathrm{~kg}$ be the amount of gunpowder containing $2.5 \mathrm{~kg}$ of sulphur.
i.e., $(10 \%$ of $\mathrm{x})=2.5 \mathrm{~kg}$
$\Rightarrow\left(x \times \frac{10}{100}\right)=2.5$
$\Rightarrow \frac{10 x}{100}=2.5$
$\Rightarrow \frac{x}{10}=2.5$
$\Rightarrow x=(2.5 \times 10)$
$\Rightarrow x=25 \mathrm{~kg}$
Hence, $25 \mathrm{~kg}$ of gunpowder contains $2.5 \mathrm{~kg}$ of sulphur.