Multiple choice

An alloy A contains two elements copper and tin in the ratio of 2 : 3, whereas an alloy B contains the same elements in the ratio of 3 : 4. If 20 kg of alloy A, 28 kg of alloy B and some more pure copper are mixed to form a third alloy C, which now contains copper and tin in the ratio of 6 : 7, then what is the quantity of pure copper mixed in the alloy C?

  1. 3 kg

  2. 4 kg

  3. 5 kg

  4. 7 kg
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Alloy A: 20kg (2:3) -> 8kg Cu, 12kg Sn. Alloy B: 28kg (3:4) -> 12kg Cu, 16kg Sn. Total Cu = 20 + x, Total Sn = 28. Ratio (20+x)/28 = 6/7. 7(20+x) = 168. 140 + 7x = 168. 7x = 28. x = 4.

AI explanation

From 20 kg of alloy A, the copper is (2/5)*20 = 8 kg and the tin is 12 kg, while from 28 kg of alloy B, the copper is (3/7)*28 = 12 kg and the tin is 16 kg. Let x be the mass of pure copper added, so the total copper is 20 + x and the total tin is 28 kg. Since the new ratio of copper to tin is 6 : 7, we set up the equation (20 + x) / 28 = 6 / 7. Solving this gives 20 + x = 24, which means x = 4 kg of pure copper was added.

Keep practicing — related questions