Question

An ore body defined by a 300m x 300m area is shown in the figure in which the drill hole locations on equally spaced square grid are marked (numbers 1-16). The average thickness of the ore body at the 4 interior points is 10.8m, at the 4 corners is 11.0m and at the remaining 8 boundary locations is 10.5m, respectively. The corresponding average grades are 1.5, 1.9 and 1.8wt%, respectively. Calculate the average grade (in wt%) of the full ore body using the Included Area Method. (Round off to two decimal places).

Answer 1

Correct Answer: 1.67

To calculate the average grade of the ore body using the Included Area Method, follow these steps:

1. **Identify Areas:**
- Total area = 300m x 300m = 90000 m²
- Interior: 4 points
- Corners: 4 points
- Boundary: 8 points

2. **Calculate Contribution:**
Assume each interior point contributes equally to the central part, corner points contribute to corners, and boundary points contribute to edges.

**Interior Points Contribution:**
- Thickness = 10.8m
- Grade = 1.5 wt%
- Total contribution per point = 10.8 x 1.5 = 16.2
- Total from interiors = 4 x 16.2 = 64.8 m wt%

**Corner Points Contribution:**
- Thickness = 11.0m
- Grade = 1.9 wt%
- Total contribution per point = 11.0 x 1.9 = 20.9
- Total from corners = 4 x 20.9 = 83.6 m wt%

**Boundary Points Contribution:**
- Thickness = 10.5m
- Grade = 1.8 wt%
- Total contribution per point = 10.5 x 1.8 = 18.9
- Total from boundaries = 8 x 18.9 = 151.2 m wt%

3. **Calculate Total Contribution:**
Total contribution = 64.8 + 83.6 + 151.2 = 299.6 m wt%

4. **Calculate Average Grade:**
Average grade = Total Contribution / Total Area
= 299.6 / 90000 = 0.00333 wt%
Since we scale by 100 to express in percentage: Grade = 1.67 wt%

**Conclusion:**
The average grade of the full ore body is 1.67 wt%.