The cement content in a mix design is 378 kg/m³, water content 170 kg, sand is 30% of total aggregate, entrapped air is 1%, specific gravity of cement, coarse aggregate and fine aggregate are, respectively 3·15, 2·70 and 2·60. The fine aggregate is approximately :
- A510 kg
- B600 kg
- C550 kgCorrect
- D450 kg
Solution
Concept
Absolute volume method of mix proportioning
Formula
Vaggregate = 1 − (Vcement + Vwater + Vair)
Step-by-step solution
1. Cement volume = 378 / (3.15 × 1000) = 0.120 m³. 2. Water volume = 170 / 1000 = 0.170 m³; air = 0.010 m³. 3. Volume of aggregates = 1 − 0.300 = 0.700 m³. 4. Fine aggregate = 30% of 0.700 = 0.210 m³ → mass = 0.210 × 2.60 × 1000 = 546 kg ≈ 550 kg.
Explanation
In the absolute volume method the volumes of all ingredients add up to 1 m³. The aggregates occupy what is left after cement, water and air; sand is 30 percent of that volume.
Why the other options are wrong
510, 600 and 450 kg do not match 546 kg (the mass-basis split gives about 560 kg, still closest to 550).
Shortcut
Aggregate volume = 0.70 m³; sand 0.21 m³ × 2.6 = 546.
Common mistake
Forgetting to subtract the 1% air or dividing by the wrong specific gravity.
Correct answer: C) 550 kg