X and Y are two alloys which are prepared by mixing zinc and aluminium in the ratio of 1 : 4 and 11 : 8 respectively. If equal quantities of alloys are melted to form a third alloy Z then what is the ratio of zinc and aluminium in alloy Z ?

Ratio of zinc and aluminium in alloy X = 1 : 4

Ratio of zinc and aluminium in alloy Y = 11 : 8

Let quantity of both alloys taken = L.C.M. (5,9) = 95 units

=> Zinc in alloy X = $$\frac{1}{(1+4)}\times95=19$$ units

Aluminium present in alloy X = $$95-19=76$$ units

Similarly, zinc in alloy Y = $$55$$ units and aluminium = $$40$$ units

$$\therefore$$ Ratio of zinc and aluminium in alloy Z = $$\frac{(19+55)}{(76+40)}$$

= $$\frac{74}{116}=\frac{37}{58}$$

=> Ans - (B)