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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:

49 : 221
39:231
94:181
None of these

Correct Answer: 49 : 221

Explanation:

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

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