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(C) Let the weight of alloy $A$ be $1 	ext{ kg}$.$	herefore$ Gold in $A = \frac{7}{9} 	ext{ kg}$ and copper in $A = \frac{2}{9} 	ext{ kg}$.In $1 \

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$7:5$

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(C) Let the weight of alloy $A$ be $1 	ext{ kg}$.$	herefore$ Gold in $A = \frac{7}{9} 	ext{ kg}$ and copper in $A = \frac{2}{9} 	ext{ kg}$.In $1 	ext{ kg}$ of alloy $B$,Gold $= \frac{7}{18} 	ext{ kg}$ and copper $= \frac{11}{18} 	ext{ kg}$.Since equal quantities are mixed,the total gold in alloy $C = \frac{7}{9} + \frac{7}{18} = \frac{14+7}{18} = \frac{21}{18} 	ext{ kg}$.The total copper in alloy $C = \frac{2}{9} + \frac{11}{18} = \frac{4+11}{18} = \frac{15}{18} 	ext{ kg}$.$	herefore$ Required ratio of gold to copper in $C = \frac{21}{18} : \frac{15}{18} = 21 : 15 = 7 : 5$.

$A$ 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

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