i) Ratio of fine to total aggregate = 0.315

ii) Water Content = $190 kg/m^3$

iii) Amount of entrapped air = 2$\%$

iv) Specific gravity of cement = 3.15

v) Specific gravity of F.A = 2.68

vi) Specific gravity of C.A = 2.85

vii) Water Cement ratio = 0.50

Determine quantity of cement FA and CA per $m^3$ of concrete.

Aggregate content can be determined from following equation

$V = [W + \frac{C}{SC} + \frac{1}{P} \frac{Fa}{SFa} ] \frac{1}{1000}$

Determination of water content

Water / Cement ratio =0.5

Water = $190 kg/m^3$

Cement = $\frac{190}{0.50} = 380 kg/m^3$

V= Absolute volume of fresh concrete, which is equal to gross volume $(m^3)$ minus the volume of entrapped air

V =$1(m^3) -2 \%$

$\quad 1-0.2$

$\quad 0.98 m^3$

$V = [W + \frac{C}{SC} + \frac{1}{P} \frac{Fa}{SFa} ] \frac{1}{1000}$

$0.98 = [190 + \frac{380}{3.15} + \frac{1}{0.315} \frac{Fa}{2.08} ] \frac{1}{1000}$

Fa = $565 kg/m^3$

$Ca = \frac{1-P}{P} \times Fa \frac{S C a}{F C a}$

$Ca = \frac{1-0.315}{0.315} \times 565 \frac{2.85}{2068}$

$\quad 1306.58 kg/m^3$

The mix portion then becomes