Question

The borehole data for resource estimation for an iron ore deposit are shown.

The average grade of the deposit, in %, is . (rounded off to two decimal places)

Hint:

First average the grade inside each borehole by intercept thickness, then combine the three boreholes using area times thickness as the weight.

Answer 1

Correct Answer: 62.98

Step 1: Find the thickness-weighted average grade inside each borehole.
Each borehole cuts through two ore bands of different thickness and grade. Within one borehole, the grade must first be averaged over its own two intercepts, weighting each by how thick it is, since a thicker band carries more ore.
Borehole D1 has 6 m at 63% and 6 m at 66%:
\[ g_{D1} = \frac{6(63) + 6(66)}{6+6} = \frac{378+396}{12} = 64.50\% \]
Borehole D2 has 6 m at 60% and 4 m at 64%:
\[ g_{D2} = \frac{6(60) + 4(64)}{6+4} = \frac{360+256}{10} = 61.60\% \]
Borehole D3 has 8 m at 64% and 7 m at 62%:
\[ g_{D3} = \frac{8(64) + 7(62)}{8+7} = \frac{512+434}{15} = 63.07\% \]

Step 2: Weight each borehole's grade by its share of ore volume.
Each borehole represents a polygon of influence with plan area A1 = 0.02 km2 (D1), A2 = 0.03 km2 (D2), and A3 = 0.04 km2 (D3). The tonnage a borehole represents depends on both the area it covers and the thickness of ore it meets, so the correct weight for combining grades is area times thickness, not area alone.
\[ W_{D1} = 0.02 \times 12 = 0.24, \quad W_{D2} = 0.03 \times 10 = 0.30, \quad W_{D3} = 0.04 \times 15 = 0.60 \]

Step 3: Combine into the overall average grade.
The deposit's average grade is the tonnage-weighted mean of the three borehole grades.
\[ \bar{g} = \frac{W_{D1} g_{D1} + W_{D2} g_{D2} + W_{D3} g_{D3}}{W_{D1}+W_{D2}+W_{D3}} \]
\[ \bar{g} = \frac{0.24(64.50) + 0.30(61.60) + 0.60(63.07)}{0.24+0.30+0.60} = \frac{15.48+18.48+37.84}{1.14} = \frac{71.80}{1.14} = 62.98\% \]

Final Answer:
The average grade of the deposit is about 62.98%. \[ \boxed{62.98\%} \]