A project requires 150 cubic yards of fill material with 95% compaction. The soil has a shrinkage factor of 15%. How many cubic yards of loose material must be ordered?
Correct Answer
D) 186.3 cubic yards
First, account for compaction: 150 ÷ 0.95 = 157.9 cubic yards. Then account for shrinkage: 157.9 ÷ 0.85 = 186.3 cubic yards of loose material needed. Shrinkage factor means the material loses 15% of its volume when compacted.
Why This Is the Correct Answer
Option D correctly applies both the compaction and shrinkage factors in the proper sequence. First, we calculate the uncompacted volume needed by dividing 150 by 0.95 (95% compaction), giving 157.9 cubic yards. Then we account for the 15% shrinkage by dividing by 0.85 (100% - 15%), resulting in 186.3 cubic yards of loose material needed. This two-step process ensures we order enough material to achieve the required compacted volume.
Why the Other Options Are Wrong
Option A: 127.5 cubic yards
This answer only accounts for the shrinkage factor (150 × 0.85 = 127.5) but completely ignores the compaction requirement. This would result in insufficient material to meet the project specifications.
Option B: 157.9 cubic yards
This answer only accounts for the compaction factor (150 ÷ 0.95 = 157.9) but fails to consider the shrinkage factor. This would still leave the project short of the required loose material.
Option C: 172.5 cubic yards
This answer incorrectly applies the shrinkage factor by multiplying instead of dividing (150 × 1.15 = 172.5), and doesn't properly account for the compaction requirement. The calculation method is fundamentally flawed.
Memory Technique
Remember 'CUSH' - Compaction first, Understand shrinkage second, Shrinkage divides (don't multiply), Heavy equipment needs more material than the final volume
Reference Hint
Florida Building Code Chapter 18 - Soils and Foundations, or construction materials reference sections covering soil compaction and volume calculations
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