$\frac{1}{2}$ 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.

17 : 37

\(\frac{1}{2}\) of 4 Kg of an alloy is lead and rest is tin.

⇒ Quantity of lead in first alloy = \(\frac{1}{2}\) of 4 Kg = 2 Kg .. (1.)

⇒ Quantity of Tin in first alloy = 4 - 2 = 2 Kg  ..(2.)

⇒ \(\frac{1}{6}\) of 5 Kg of another alloy is lead and the rest of it is tin.

⇒ Quantity of lead in second alloy = \(\frac{1}{6}\) of 5 Kg = \(\frac{5}{6}\)  ..(3.)

⇒ Quantity of Tin in first alloy = 5 - \(\frac{5}{6}\)

⇒\(\frac{30 - 5}{6}\) = \(\frac{25}{6}\)  ..(4.)

The ratio of lead and tin in the mixture of the given quantities of these two alloys.

Eq.(1) + Eq.(2) + Eq(3) + Eq(4)

⇒ 2 + \(\frac{5}{6}\) : 2 = \(\frac{25}{6}\)

⇒ \(\frac{17}{6}\) : \(\frac{37}{6}\)

⇒ 17 : 37