In two alloys copper and zinc are in the ratio of 1:3 and 4:1 respecti...
Answer – b) 4.4 Explanation : In 1 alloy copper = (1/4)*20 = 5kg and zinc = (3/4)*20 = 15kg in 2 alloy copper = (4/5)*35 = 28kg and zinc = (1/5)*35 = 7kg So, 33/(22+x) = 5/4 (X is the amount of pure zinc added)
View all questions of this test
In two alloys copper and zinc are in the ratio of 1:3 and 4:1 respecti...
To solve this problem, we can use a system of equations. Let's assume that the amount of pure copper in the final alloy is x kg and the amount of pure zinc melted is y kg.
1. Set up the equations:
- For copper: (amount of copper in the first alloy) + (amount of copper in the second alloy) = (amount of copper in the final alloy)
- For zinc: (amount of zinc in the first alloy) + (amount of zinc in the second alloy) + (amount of pure zinc melted) = (amount of zinc in the final alloy)
2. Calculate the amount of copper and zinc in each alloy:
- In the first alloy, the ratio of copper to zinc is 1:3. So, the amount of copper in the first alloy is (1/4) * 20 kg = 5 kg, and the amount of zinc is (3/4) * 20 kg = 15 kg.
- In the second alloy, the ratio of copper to zinc is 4:1. So, the amount of copper in the second alloy is (4/5) * 35 kg = 28 kg, and the amount of zinc is (1/5) * 35 kg = 7 kg.
3. Substitute the values into the equations:
- For copper: 5 kg + 28 kg = x kg
- For zinc: 15 kg + 7 kg + y kg = 4/9 * (x kg + y kg)
4. Simplify the equations:
- For copper: 33 kg = x kg
- For zinc: 22 kg + y kg = (4/9) * (33 kg + y kg)
5. Solve for y:
- Multiply both sides of the second equation by 9 to eliminate the fraction: 198 kg + 9y kg = 4(33 kg + y kg)
- Simplify: 198 kg + 9y kg = 132 kg + 4y kg
- Combine like terms: 5y kg = -66 kg
- Divide both sides by 5: y kg = -66 kg / 5 = -13.2 kg
6. Since we cannot have a negative amount of zinc, we discard this solution. Therefore, there is no amount of pure zinc that can be melted to achieve the desired ratio.
Therefore, the correct answer is e) None of these.