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A 75 kg mixture of copper and tin contains copper and tin in the ratio 3 ∶ 2. 25 kg of the mixture is removed and replaced with copper and the operation is repeated twice. At the end of the two removals and replacements, what is the ratio of tin and copper in the resultant mixture?

A 75 kg mixture of copper and tin contains copper and tin in the ratio 3 ∶ 2. 25 kg of the mixture is removed and replaced with copper and the operation is repeated twice. At the end of the two removals and replacements, what is the ratio of tin and copper in the resultant mixture? Correct Answer 8 : 37

Amount of tin in the original mixture = 75 × 2/5 = 30 kg

⇒ Amount of tin after repeating operation twice = 30 × (1 – 25/75)2 = 30 × (2/3)2 = 40/3 kg

It means final mixture will have 40/3 kg of tin and (75 – 40/3 = 185/3) kg of copper 

∴ Required ratio = 40/3 ∶ 185/3 = 8 : 37

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