An alloy of silver and gold weighs 30 g in air and 28 g in water. Assuming that silver looses one-tength of its weight in water and that gold looses one-ninghteenth of its weight, find the weight of each metal in the alloy

Dear Student,
Please find below the solution to the asked query:
Let the weight of silver be x gram.
Therefore, weight of gold is (30 − x) grams.
According to question,
$$\begin{gathered} x\left(1-\frac{1}{10}\right)+\left(30-x\right)\left(1-\frac{1}{90}\right)=28 \\ \Rightarrow x\left(\frac{10-1}{10}\right)+\left(30-x\right)\left(\frac{90-1}{90}\right)=28 \\ \Rightarrow \frac{9x}{10}+\frac{89\left(30-x\right)}{90}=28 \\ \Rightarrow \frac{81x+89\left(30-x\right)}{90}=28 \\ \Rightarrow 81x+89\left(30-x\right)=2520 \\ \Rightarrow 81x+2670-89x=2520 \\ \Rightarrow -8x=-150 \\ \Rightarrow x=\frac{-150}{-8}=18.75 \\ \mathrm{And} 30-x=30-18.75=11.25 \\ \mathrm{Hence}, \mathrm{amount} \mathrm{of} \mathrm{silver} \mathrm{is} 18.75 \mathrm{g} \mathrm{and} \mathrm{amount} \mathrm{of} \mathrm{gold} \mathrm{is} 11.25 \mathrm{g}. \end{gathered}$$
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