A piece of an alloy of gold and silver weighs `25gxxg` in air and `23gxxg` in water. Find out the amount of gold and silver in the piece of alloy. [Specific gravity of silver is 10.8 and that of gold is 19.3]

Explanation

Let the amount of gold in the allow be x g.
`therefore` Amount of silver = (25 - x)g
Volume of gold = `x/19.3cm^(3)`,
volume of silver = `(25-x)/10.8cm^(3)`
`therefore` Volume of the alloy = `(x/19.3+(25-x)/10.8)cm^(3)`
Again, weight of water displaced by the alloy
`=(25-23)xxg=2gxxg`
i.e., mass of this displaced water = 2g
`therefore` Volume of displaced water = `2cm^(3)`
`therefore" "x/19.3+(25-x)/10.8=2`
or, `10.8x+19.3xx25-19.3x=2xx19.3xx10.8`
or, x = 7.72
`therefore` Amount of gold = 7.72g and amount of silver = 25 - 7.72 = 17.28g.