Question

A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7 : 2 and 7 :11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be

• A. 5 : 7
• B. 5 : 9
• C. 7 : 5
• D. 9 : 5

Answer 1

Answer: (C)

Let 100 gm of alloy A and 100 gm of alloy B
Gold in alloy A=$\frac{7}{9}\times 100$
copper in alloy A=$\frac{2}{9}\times 100$
Gold in alloy B=$\frac{7}{18}\times 100$
Copper in alloy B=$\frac{11}{18}\times 100$
Gold in alloy C=$\frac{700}{9}+\frac{700}{18}=\frac{2100}{18}$
Copper in alloy C=$\frac{200}{9}+\frac{1100}{18}=\frac{1500}{18}$
Ratio of copper and iron in alloy C=$\frac{2100}{18}:\frac{1500}{18}$
$7:5$