Answer / karim

Quantity: 1cum
Proportion: 1:2:4[M15]
Add 54% extra
[54/100×1]+1=1.54cum
Cement required =1.54/7=0.22cum
We know: 1bag cement=0.0347cum
So, No of bags of cement =0.22/0.0347=6.34bags
Sand required: [1.54/7]×2=0.44cum
1cum=35.347cft
So 0.44cum=15.55cft
Aggregate required =[1.54/7]×4=0.88cum
So,0.88cum=31.18cft
Finally,
[Cement required =6.34bags
Sand=15.55cft
Aggregate =31.18cft]

Answer / seduraman

m15 1:2:4 (7 parts)
cement 1/7 x 1.57 = 0.225
Sand 2/7 x 1.57 = 0.45
Jelly 4/7 x 1.57 = 0.90

Cement : 0.225 x1440 = 324kg
Total nos of bag of cement : 324/50 = 6.48 Bags of cement

Answer / huner

v=1c*2s*4G (1+2+4)=7
so for 1 cu m3 1/7= 0.15

the unit weigh of cement =1400 kg/m3
hence for
cement 0.15*1400 =210 kg
sand 2*0.15=0.3 m3
Gravel 4*.15 =0.16 m3

Answer / g.selvaraj

Cement - 6.38,Sand -2.12cft,Aggregate - 4.59cft

Answer / er. nirakar behera

Ratio of concrete mix is 1:2:4
we have to calaculate only about the dry volume.

dry density of concrete =14.2qntl/cum
volume of concrete =
1.54/(1+2+4)=0.22cum
Hence for weight of mix ingredients=(14.2qntl/cum*0.22cum)
i.e=3.124qntl
required weight of cement=1*3.124qntl=3.124qntl or say
6.2 bags
for sand =2*6.2cft=12.4 cft
for aggregate=4*6.2cft=24.8cft

Answer / a.mohammed yahya

For all Common Grade of Concrete Quantity 1Cum fill the
1.57Cum(for 20mm size Jally)
So For the Ratio = 1: 2: 4 (Total = 1+2+4=8)
( there for 1 - Cement,2- sand,3- 20mm Size Jelly)
Calculate for Cement = 1/8 = 0.125
= 0.125 x 1.57
= 0.196
For Common 1Cum Cement Quantity = 1440kg
there for = 1440 x 0.196
Required Cement = 282kg
Calculate for Sand = 2/8 = 0.25
= 0.25x 1.57=0.393Cum
= 0.393 x 35.32
Required Sand =14 Cft
Calculate For Jelly = 4/8 = 0.5
= 0.5 x 1.57
= 0.79Cum
Required Jally = 0.79 x 35.32 = 28 cft

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