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)
Show Hint
First average the grade inside each borehole by intercept thickness, then combine the three boreholes using area times thickness as the weight.
This question is really about combining three separate ore estimates into one overall grade for the deposit. The trick is to weight each borehole by how much ore it actually represents, not just treat the three grades as equal.
Step 1: Treat each borehole polygon as its own small ore block. Take an arbitrary constant density $d$ for the iron ore, the same for all three blocks since it is one deposit. The tonnage each block represents is area times thickness times density.
Block D1: area $= 0.02$ km2, thickness $= 12$ m, so tonnage $\propto 0.02 \times 12 = 0.24$ units.
Block D2: area $= 0.03$ km2, thickness $= 10$ m, so tonnage $\propto 0.03 \times 10 = 0.30$ units.
Block D3: area $= 0.04$ km2, thickness $= 15$ m, so tonnage $\propto 0.04 \times 15 = 0.60$ units.
Step 2: Find the grade inside each block first. Each borehole passes through two bands, so average them by thickness.
D1: $(6 \times 63 + 6 \times 66)/12 = 64.50\%$
D2: $(6 \times 60 + 4 \times 64)/10 = 61.60\%$
D3: $(8 \times 64 + 7 \times 62)/15 = 63.07\%$
Step 3: Find the metal content carried by each block, then add them up. Metal content is tonnage multiplied by grade.
D1: $0.24 \times 64.50 = 15.48$
D2: $0.30 \times 61.60 = 18.48$
D3: $0.60 \times 63.07 = 37.84$
Total metal $= 15.48 + 18.48 + 37.84 = 71.80$, and total tonnage $= 0.24 + 0.30 + 0.60 = 1.14$.
The deposit's overall grade is the total metal divided by the total tonnage: $71.80 / 1.14 = 62.98\%$. Grade is never simply averaged across boreholes, it is always tonnage weighted, the same way a mine planner blends ore coming from different faces.
Let's summarize:
Average each borehole's own grade by its intercept thickness first.
Weight the three boreholes by tonnage (area times thickness), not by area alone.
The deposit's average grade works out to 62.98%.
So the average grade of the iron ore deposit is 62.98%.
