$$\frac{1}{2}$$ of 4 kg of an alloy is lead and the rest of it is tin. $$\frac{1}{6}$$ of 5 kg of another alloy is lead and the rest of it is tin. Find the ratio of lead and tin in the mixture of the given quantities of these two alloys.

$$\frac{1}{2}$$ of 4 kg of an alloy is lead and the rest of it is tin.

quantity of lead in the first alloy = $$\frac{1}{2}$$ of 4 kg

= 2 kg    Eq.(i)

quantity of tin in the first alloy = 4-2

= 2 kg    Eq.(ii)

$$\frac{1}{6}$$ of 5 kg of another alloy is lead and the rest of it is tin.

quantity of lead in the second alloy = $$\frac{1}{6}$$ of 5 kg

= $$\frac{5}{6}$$ kg    Eq.(iii)

quantity of tin in the first alloy = $$5-\frac{5}{6}$$

= $$\frac{30-5}{6}$$

= $$\frac{25}{6}$$ kg     Eq.(iv)
Ratio of lead and tin in the mixture of the given quantities of these two alloys = $$Eq.\left(i\right)+Eq.\left(iii\right)\ :\ Eq.\left(ii\right)+Eq.\left(iv\right)$$

= $$2+\frac{5}{6}\ :\ 2+\frac{25}{6}$$

= $$\frac{12+5}{6}\ :\ \frac{12+25}{6}$$

= $$\frac{17}{6}\ :\ \frac{37}{6}$$

= 17 : 37