what is the quantity of cement( in bags),sand and aggregates required for 1 cmt of M 25 concrete
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Answer / harika nukala
M25= 1:1:2
VOLUME = 1+1+2=4
TOTAL VOL OF DRY MIX =1.54
CEMENT = 1/4(1.54*1440)=554.4 Kgs = 554.4/50 = 11 BAGS
SAND= 1/4(1.54)= 0.38 CUM
AGGREGATE = 2/4(1.54)= 0.77 CUM
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / vikas d v
m25=1:1:2
for 1cum add 50% for wet mix so 1.5cum
1.50/4=0.375cum
for 1cum=1440/50=28.8 approximately 30bags
cement=0.375*30=11.25bags
sand=0.375*1=0.375cum
aggregate=0.375*2=0.75cum
| Is This Answer Correct ? | 1 Yes | 0 No |
Answer / harish dutt pandey
it is depends upon mix design .in nominal mix we can used 370 kg.
| Is This Answer Correct ? | 1 Yes | 3 No |
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