Alloy A contains iron and copper in the ratio of 4:5. Alloy B contains iron and copper in the ratio of 3:4. If 45 kg of alloy A and 63 kg of Alloy B are mixed together, what will be the final ratio of copper to iron?

Explanation

Given The ratio of iron and copper in alloy A = 4 : 5 The ratio of iron and copper in alloy B = 3 : 4 The quantity of alloy A in the new mixture = 45 kg. The quantity of alloy B in the new mixture = 63 kg. Let the quantity of iron and copper in alloy A be 4x and 5x respectively. $4x + 5x = 45$ $9x = 45$ $x = 5$ The quantity of iron in alloy A = $4x = 4 \times 5 = 20$ kg. ## Page 11 Quantity of copper in Alloy A = 5x = 5 × 5 = 25 kg. Let the quantity of iron and copper in alloy B be 3y and 4y respectively. $$3y + 4y = 63$$ $$7y = 63$$ $$y = 63/7 = 9$$ Alloy B = 3y = 3 × 9 = 27 kg. Quantity of Copper Alloy B = 4y = 4 × 9 = 36 kg. The quantity of iron in the final mixture = 20 + 27 = 47 kg. The quantity of copper in the final mixture = 25 + 36 = 61 kg. The ratio of copper to iron in the final mixture is 61 : 47. ∵ It takes 1 minute to make 100 lock.