Alloy A contains iron and copper in ration 4 : 5. Alloy B contains iron and copper in ratio 3 : 4. If 45 kg of alloy A and 63 kg of alloy B are mixed, then what is the final ratio of copper and iron?

Correct option is A

Given:
Ratio of iron and copper in alloy A = 4 : 5
Ratio of iron and copper in alloy B = 3 : 4
Amount of alloy A = 45 kg
Amount of alloy B = 63 kg
Solution:
Let the amount of iron and copper in alloy A be 4x and 5x respectively.
4x + 5x = 45
=> 9x = 45
=> x = 5
Amount of iron in alloy A will be 4x = 4 × 5 = 20 kg
Amount of copper in alloy A will be 5x = 5 × 5 = 25 kg
Let the amount of iron and copper in alloy B be 3y and 4y respectively.
3y + 4y = 63
=> 7y = 63
=> y = 9
Amount of iron in alloy B will be 3y = 3 × 9 = 27 kg
Amount of copper in alloy B will be 4y = 4 × 9 = 36 kg
Amount of iron in final mixture = 20 + 27 = 47 kg
Amount of copper in final mixture = 25 + 36 = 61 kg
Ratio of iron and copper in final mixture = 47 : 61
∴ The final ratio of copper and iron is 61 : 47.