In two alloys copper and zinc are in the ratio of 1:4 and 3:1 respecti...
Answer – c) 68/3 kg Explanation : Copper = 1/5*20 + 3/4*32 = 28kg zinc = 4/5*20 + 1/4*32 = 24kg now x kg of zinc is added, so [28/24 + x] = 3/5. X = 68/3 kg
View all questions of this test
In two alloys copper and zinc are in the ratio of 1:4 and 3:1 respecti...
Given:
- Ratio of copper and zinc in alloy 1 = 1:4
- Ratio of copper and zinc in alloy 2 = 3:1
- Quantity of alloy 1 = 20 kg
- Quantity of alloy 2 = 32 kg
- Ratio of copper and zinc in final alloy = 3:5
To find:
- Quantity of pure zinc melted
Solution:
1. Let's first calculate the amount of copper and zinc in the two alloys:
- Alloy 1: Copper = (1/5)*20 kg = 4 kg, Zinc = (4/5)*20 kg = 16 kg
- Alloy 2: Copper = (3/4)*32 kg = 24 kg, Zinc = (1/4)*32 kg = 8 kg
2. Let's assume that x kg of pure zinc was melted.
3. Now, let's calculate the amount of copper and zinc in the final alloy:
- Let the quantity of final alloy be y kg
- Copper in final alloy = (3/8)*y + 4 + 24 = (3/8)*y + 28
- Zinc in final alloy = (5/8)*y + 16 + 8 + x = (5/8)*y + 24 + x
4. We know that the ratio of copper and zinc in the final alloy is 3:5. Therefore:
- (Copper in final alloy) / (Zinc in final alloy) = 3/5
- ((3/8)*y + 28) / ((5/8)*y + 24 + x) = 3/5
5. Solving for y, we get:
- y = (40/3) + (8/3)*x
6. We also know that the total quantity of alloy and pure zinc melted is 20 + 32 + x = (52 + x) kg. Therefore:
- y + x = 52 + x
- y = 52
7. Substituting the value of y in equation (5), we get:
- (3/8)*y + 28 = (5/8)*y + 24 + x
- (3/8)*52 + 28 = (5/8)*52 + 24 + x
- 27 = 13 + x
- x = 14
Therefore, the quantity of pure zinc melted is 14 kg.
Answer: Option (c) 68/3 kg.