4 kg of a metal contains 1/5 copper and rest in Zinc. Another 5 kg of metal contains 1/6 copper and rest in Zinc.The ratio of Copper and Zinc into the mixture of these two metals:

Copper in 4 kg = 4/5 kg          and      Zinc in 4 kg = 4 x (4/5) kg

Copper in 5 kg = 5/6 kg          and      Zinc in 5 kg = 5 x (5/6) kg

Therefore, Copper in mixture =  $\frac{4}{5}+\frac{5}{6}=\frac{49}{30}$ kg

and  Zinc in the mixture = $\frac{16}{5}+\frac{25}{6}=\frac{221}{30}$kg

Therefore the required ratio = 49 : 221