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 120g (Given pure gold is of 24 carat.)?

Answer

80 g, 40 g

Explanation

Let x g of 18- carat gld be mixed with y g of 12-carat gold to get 120 g of 16- carat gold.
Then, ` x + y = 120 " "` … (i)
Gold % in 18 -carat gold = `((18)/(24) xx 100) % = 75% `
Gold % in 12-carat gold = ` ((12)/(24) xx 100 )% = 50 % `
Gold % in 16 - carat gold = ` ((16)/(24) xx 100)% = (200)/(3)%`
` therefore 75% of x + 50 % of y = (200)/(3) % of 120 `
` rArr (75x )/(100) + ( 50 y )/(100) = (200xx 120)/(3 xx 100) rArr 3x + 2y = 320 " " `... (ii)
Now, solve (i) and (ii).