you have to take 1.52 as constant for calculating concrete
mix ,let me explain it this 1.52 is the quqntity of total
dry aggregate which is required.so when ever u require to
know the quantity for any given mix .u need to add up the
ratio of mix and divide 1.54 by the sum of mix ratio u ll
get the quantity of cement for one cum
eg: suppose m20 mix u need to calculate the concrete then
simply add the ratio of m20 i.e 1:1.5:3 so the total is 5.5
and divide 1.52 by 5.5 u ll get 0.27 cum of cement which ll
b required for 1 cum of concrete and if u multiply it(0.27)
by 30 u ll get the no. of cement bags .follow the same rule
for diff type of mix

1. Calculate the volume of concrete needed.
2. Estimate the total volume of dry material by
multiplying the required volume of concrete by 1.65 to get
the total volume of dry loose material needed (this includes
10% extra to compensate for losses).
3. Add the numbers in the volumetric proportion that you
will use to get a relative total. This will allow you later
to compute fractions of the total needed for each
ingredient. (i.e. 1:2:4 = 7).
4. Determine the required volume of cement, sand and
gravel by multiplying the total volume of dry material (Step
2) by each components fraction of the total mix volume (Step
3) i.e. the total amount cement needed = volume of dry
materials * 1/7.
5. Calculate the number of bags of concrete by dividing
the required volume of cement by the unit volume per bag of
cement (0.0332 m3 per 50 kg bag of cement or 1 ft3 per 94 lb
bag).

For example, for a 2 m x 2 m x 10 cm thick pump pad:

1. Required volume of concrete = 0.40 m3
2. Estimated volume of dry material = 0.4 x 1.65 = 0.66 m3
3. Mix totals = 1+2+4 = 7 (1:2:4 cement:sand:gravel)
4. Ingredient Volumes: 0.66 x 1/7 = .094 m3 cement
0.66 x 2/7 = .188 m3 sand
0.66 x 4/7 = .378 m3 gravel
5. # Bags of cement: 0.094 m3 cement / .0332 m3 per 50 Kg bag
= 2.83 bags of cement (use three bags)

Ok lets go to the calculaton part
Now we are going to calculate the quantity of materials required for 1m3 Concrete (M15)

Quantity of Cement required for 1m3

[1/(1+2+4)] * 1.52 = 0.217m3
0.217 * 1440 = 312.48 kg/m3
To calculate in bags
312.48/50 = 6.24 bags/m3
Here 1.52 is the dry co-efficient and 1440 is the unit weignt of cement.

Quantity of Fine aggregate( sand) for 1m3

[2/(1+2+4)] * 1.52 = 0.434m3

Quantity of course aggregate for 1m3

[4/(1+2+4)] * 1.52 = 0.868m3

So this is the Quantity for 1m3. For example, If you want 6m3 concrete of M15 grade

for 6m3
cement = 312.48 * 6 = 1874.88 kg for 6m3 (37.5 bags )
fine aggregate = 0.434 * 6 = 2.604 m3 ( 2.604 m3 sand required for 6m3 concrete )
course aggregate = 0.868 * 6 = 5.208 m3 ( 5.208 m3 aggregate required for 6m3 concrete)

The same procedures should be followed for M20, M25 concrete
For example M20
[1/(1+1.5+2)]* 1.52
For M25
[1/(1+1+2)] * 1.52

1+2+4= 7
cement
1/7+40%(compaction)+5%(waste)
0.142+0.0572+0.00715=0.20735m3
density of cement = 1440kg/cm3
1m3 of cement = 1440kg/cm3
xm3 contains 50kg = 50/1440 = 0.035m3
no of bags of cement =0.20735/0.035= 5.92bags

Sand
2/7 + 40% + 5%
0286 + 0.1144 + 0.143=0.4147m3
3.33m3 of lorry = 5cubic yard
no of load = 0.414/3.33 =0.125load/m3
Granite
4/7 + 40 + 5
0.572 + 0.229 + 0.0286 = 0.8296m3
3.33m3 of lorry = 5cubic yard
no of load = 0.8296/3.33 =0.25load/m3