Question

Data from a borehole log with collar elevation at 590 mRL are given below. Composite grade is calculated using cores of 5 m above and below the reference bench at 580 mRL. The composite grade, in %, is:

 

Hint:

To compute composite grades, use a weighted average based on core length. Carefully match the elevation range to include appropriate samples above and below the target level.

Answer 1

Correct Answer: 39

Step 1: Determine elevation of each core.

Starting from 590 mRL and moving downward:

Core 1: 590–587 mRL (3 m) → Average elevation = 588.5 m
Core 2: 587–584 mRL (3 m) → Average elevation = 585.5 m
Core 3: 584–582 mRL (2 m) → Average elevation = 583.0 m
Core 4: 582–579 mRL (3 m) → Average elevation = 580.5 m
Core 5: 579–577 mRL (2 m) → Average elevation = 578.0 m
Core 6: 577–574 mRL (3 m) → Average elevation = 575.5 m
Core 7: 574–571 mRL (3 m) → Average elevation = 572.5 m

We consider 5 m above and below the reference bench at 580 mRL:

5 m above: 580 to 585 mRL → Cores 2, 3, and 4
5 m below: 580 to 575 mRL → Cores 4, 5, and 6

Step 2: Identify relevant cores and their valid lengths.

Core 2: Full 3 m (grade 42)
Core 3: Full 2 m (grade 41)
Core 4: Full 3 m (grade 43)
Core 5: Full 2 m (grade 42)
Core 6: Full 3 m (grade 41)

Step 3: Compute composite grade (weighted average).

\[ \text{Total length} = 3 + 2 + 3 + 2 + 3 = 13~\text{m} \]

\[ \text{Weighted grade} = (3 \times 42) + (2 \times 41) + (3 \times 43) + (2 \times 42) + (3 \times 41) \]

\[ = 126 + 82 + 129 + 84 + 123 = 544 \]

\[ \text{Composite grade} = \frac{544}{13} = 41.85 \approx 42.0\% \]

Final Answer:
42.0 %