a gols ornament weighing720g contains 90% of pure gold and rest silver.How much of silver(in grams)must be added so that the resultant mixture have 19% silver?

• 80 gm.
  
  Ornament contains gold=720*90/100=648 gm. So silver=720-648=72 gm.
  If 'x' gm. of silver is added to make 19% silver, then
  (72+x)/(720+x)=19/100
  On solving, x=80
• 8 years agoHelpfull: Yes(4) No(0)