Answer / kiran basude

Total qty is 0.576 m3
Total mix is =1+4+8= 13

Cement (0.576*1)%13= 0.0443 m3
Sand (0.576*4)%13= 0.1772 m3
aggregate (0.576*8)%13= 0.3545 m3

Answer / kiran basude

Total qty is 0.576 m3
Total ratio is =1+4+8= 13

Cement (0.576*1)%13= 0.0443 m3
0.0443%0.347=0.128 bags
Sand 0.128*4= 0.512 m3
aggregate 0.128*8= 1.024 m3

Answer / dinesh singh rawat

+.600X.600X1.600=.576 CUM
ONE CUM =1.52/13=.117
Cement .117*.576*30= 2.02bags=101 kG
Sand 0.117*4*.576= 0..269 m3
aggregate 0.117*8*.576= .538 m3

Answer / n.kurma rao (kurmi) srkr

volume of concrete quantity is
V = L * B * H.
V = .6*.6*.16
V = .576 CUM

Dry Volume required is 35% = .778 CUM.

in that Cement is .778 * 1 /13 = .0598 CUM of Cement
Cement in terms of bags in .0598/.0347 = 1.724 bags
Sand required is .778 * 4 /13 = .239 CUM of Sand
course aggregae is .778 * 8 /13 = .479 CUM of aggregate.

Answer / charlesroshan

cement requirement for 1:4:8 mix is 3.4 bags per cm3.

For (.6*.6*1.6 )section = 0.576m3

therefore for 0.576m3 cement reuired is 3.4*0.576=1.95 bags.

Answer / manoj kumar

Cement (0.576*1)%13= 0.0443 m3
0.0443x1440/50=1.275 bags or
0.0443/0.0347=1.275 bags
Sand 0.128*4= 0.512 m3
aggregate 0.128*8= 1.024 m3

Answer / sasikumar kalisetti

total qty : 0.4 cum

cement = 0.4*165.6kgs
sand = 0.4*0.46 cum
metal(coarse aggregate) = 0.4*0.92 cum

Answer / sasikumar kalisetti

165.6 kgs of Cement is required.

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