The problem involves blending a stock of high grade ore with waste to achieve a target grade. Let's analyze the given data: The waste has a grade of 0.25 wt% and weighs 1 million tonnes. The high grade ore has a grade of 1.8 wt%, and we need to blend it to achieve a target grade of 0.5 wt%. Denote x as the mass of high grade ore needed in million tonnes.
Using the formula for the weighted average of grades:
(Grade of Waste)×(Mass of Waste) + (Grade of High Grade Ore)×(Mass of High Grade Ore) = (Target Grade)×(Total Mass)
Substituting the values:
0.25×1 + 1.8×x = 0.5×(1 + x)
Expanding and simplifying this equation:
0.25 + 1.8x = 0.5 + 0.5x
Rearranging terms to isolate x:
1.8x - 0.5x = 0.5 - 0.25
1.3x = 0.25
Solving for x:
x = 0.25 / 1.3 = 0.1923
Rounding to three decimal places, the amount of high grade ore needed is 0.192 million tonnes.
| Column I | Column II | ||
| P. | Malanjkhand | 1. | Uranium ore |
| Q. | Tummalapalle | 2. | Gold ore |
| R. | Bhukia | 3. | Tin ore |
| S. | Tosham | 4. | Copper ore |