A project requires 2,400 cubic yards of fill material to be compacted to 95% standard Proctor density. If the borrow material has a shrinkage factor of 15%, how many cubic yards of material must be excavated from the borrow pit?
Correct Answer
B) 2,824 cubic yards
With 15% shrinkage, you need 15% more material than the final compacted volume. Required material = 2,400 ÷ 0.85 = 2,824 cubic yards to account for the volume loss during compaction.
Why This Is the Correct Answer
When material has a 15% shrinkage factor, it means the final compacted volume will be 85% of the original excavated volume. To find the required excavation amount, divide the needed compacted volume by the remaining percentage: 2,400 ÷ 0.85 = 2,824 cubic yards. This accounts for the 15% volume loss that occurs during compaction to 95% standard Proctor density.
Why the Other Options Are Wrong
Option A: 2,400 cubic yards
This assumes no shrinkage occurs, which ignores the 15% shrinkage factor. Using 2,400 cubic yards would result in only 2,040 cubic yards of compacted fill (2,400 × 0.85), falling short of the required 2,400 cubic yards.
Option C: 2,760 cubic yards
This incorrectly adds 15% to the required volume (2,400 × 1.15 = 2,760). This approach fails to account for the fact that shrinkage is calculated from the original excavated volume, not added to the final compacted volume.
Option D: 2,040 cubic yards
This represents what you would get if you excavated 2,400 cubic yards with 15% shrinkage (2,400 × 0.85 = 2,040). This is the result of insufficient excavation, not the amount needed to be excavated.
Memory Technique
Remember 'Shrink and Divide': When material shrinks, divide the needed amount by what remains (100% - shrinkage%). For 15% shrinkage, divide by 0.85.
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