Question 64

A and 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

Solution

Let quantity of each be L.C.M.(9,18) = 18 units

=> Gold in alloy A = 14 units and copper in alloy A = 4 units

Similarly, gold in alloy B = 7 units and copper in alloy B = 11 units

$$\therefore$$ Ratio of gold and copper in alloy C = $$\frac{(14+7)}{(4+11)}$$

= $$\frac{21}{15}=7:5$$

=> Ans - (C)

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