Civil Engineering - Concrete Technology - Discussion
Discussion Forum : Concrete Technology - Section 5 (Q.No. 29)
29.
For a concrete mix 1:3:6 and water cement ratio 0.6 both by weight, the quantity of water required per bag, is
Discussion:
57 comments Page 3 of 6.
Wani saqib said:
7 years ago
You are right @Nitu Bharti. Thanks.
Manoj Balakrishnan said:
7 years ago
Here the question is "the quantity of water required per bag and W/C ratio is 0.6" then the water required for 1 bag of cement is 30 Kg.
Deepak sonu said:
8 years ago
Water required = (5% of fine aggregate+30%of cement)*water cement ratio.
Solve: 1:3:6(c:s:a).
Since one bag cement = 50 kg.
Fine aggregate = 50*3 = 150.
So(7.5+15)*0.6=13.5~14 kg.
Solve: 1:3:6(c:s:a).
Since one bag cement = 50 kg.
Fine aggregate = 50*3 = 150.
So(7.5+15)*0.6=13.5~14 kg.
Lalit Pathade said:
6 years ago
Right answer:
Simple hai - assume 10m^3 , Dry mortar volume =30x...*10=3m^3.
Cement = (3) /(1+3+6) = 0.3m^3.
No.of bags - 0.3/(0.034) =8.8235 bags.
Quantity of water per bag = 8.8235/(0.6) = 14.70 = 14kg.
Simple hai - assume 10m^3 , Dry mortar volume =30x...*10=3m^3.
Cement = (3) /(1+3+6) = 0.3m^3.
No.of bags - 0.3/(0.034) =8.8235 bags.
Quantity of water per bag = 8.8235/(0.6) = 14.70 = 14kg.
Abhi said:
10 years ago
Its wrong answer is 30 kg.
How can we take water for aggregate if nothing is mentioned?
How can we take water for aggregate if nothing is mentioned?
S. kumar said:
10 years ago
3 times of cement is fine sand, aggregate is 6 times actually. How you can take aggregate 3 times? Explain please?
Kamran said:
10 years ago
Why we take aggregate 150 kg?
3 times of cement? Explain please?
3 times of cement? Explain please?
Aman said:
5 years ago
37 kg right answer.
5/100*9*(50)+30/100*1*(50).
22.5+15= 37.5 litre water required.
5 mean 5 per cent of aggregate+ 30 per cent of cement= wt of cement.
9 = 6+3 = aggregate ratio.
1= cement ratio.
5/100*9*(50)+30/100*1*(50).
22.5+15= 37.5 litre water required.
5 mean 5 per cent of aggregate+ 30 per cent of cement= wt of cement.
9 = 6+3 = aggregate ratio.
1= cement ratio.
Nitu bharti said:
10 years ago
Water required = (5% of aggregate + 30% of cement)*w/c.
Cement per bag = 50 kg.
Aggregate = 50*3 = 150 kg.
5% of aggregate = 7.5 kg.
30% of cement= 15 kg.
Water required = (15+7.5)*0.6 = 13.5 kg which may be approximately 14 kg.
Cement per bag = 50 kg.
Aggregate = 50*3 = 150 kg.
5% of aggregate = 7.5 kg.
30% of cement= 15 kg.
Water required = (15+7.5)*0.6 = 13.5 kg which may be approximately 14 kg.
Sudhansu said:
5 years ago
1:3:6 = 7.
1.52/7= 0.217 cm Cement.
1 cm Cement = 1440 kg cement.
0.217 cm Cement = 1440 x 0.217 = 312.48 kg.
For 312.48 kg Cement water required = 312.48 x 0.6 = 187.2 kg.
1 bag of Cement =50 kg So, 312.48 / 50 = 6.25 bags.
For1 bag, cement required = 187.2/6.25 = 30 kg (approx).
1.52/7= 0.217 cm Cement.
1 cm Cement = 1440 kg cement.
0.217 cm Cement = 1440 x 0.217 = 312.48 kg.
For 312.48 kg Cement water required = 312.48 x 0.6 = 187.2 kg.
1 bag of Cement =50 kg So, 312.48 / 50 = 6.25 bags.
For1 bag, cement required = 187.2/6.25 = 30 kg (approx).
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