A and B are two alloys of gold and copper prepared by mixing metals on the ratios 5:3 and 5:11 respectively. Equal quantities of these alloys are melted to form a third alloy C. The ratio of gold and copper in the alloy C is :

• A contains Gold=5/8 Copper=3/8
  B contains Gold=5/16 Copper=11/16
  After mixing A & B , C will have Gold=5/8+5/16=15/16 & Copper=3/8+11/16=17/16
  Therefore ratio of Gold:Copper=15/16 : 17/16 =15:17
• 11 years agoHelpfull: Yes(10) No(2)
• 
  
  suppose 16 units of each alloy is mixed , then
  32 units of final alloy will have A and B in ratio of
  (10+5) : (6+11) or 15:17
• 11 years agoHelpfull: Yes(8) No(3)
• explain it clearly.
• 11 years agoHelpfull: Yes(0) No(3)
• Alloy A : 8x(Gold=5x/8 and copper=3x/8)
  Alloy B : 16y(Gold= 5y/16 and copper= 11y/16)
  now it is written there it should be in equal proportion so
  (5x/8)+(5y/16)=(3x/8)+(11y/16)
  i.e. x/y= 3:2
• 7 years agoHelpfull: Yes(0) No(0)
• 7/5 no explanation
• 6 years agoHelpfull: Yes(0) No(0)