Answer:
150 kg
Step-by-step explanation:
Given that
cement = 310 kg
Fine aggregate = 750 kg
coarse aggregate = 1080 kg
mass of water = 180 kg
given that aggregate in field is 4% excess water
Determine the revised quantity of water
mass of concrete per m^3 = mass of cement + mass of water + mass of fine aggregate + mass of coarse aggregate
therefore mass of concrete = 310 + 180 + 750 + 1080 = 2320 kg
given that fine aggregate in field has 4% excess water this simply means it was void of air in the concrete the new weight = (1.04 * 750) = 780
Hence the revised quantity of water ( x )
= 310 + x + 780 + 1080 = 2320
therefore x = 150 kg
