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 6 of 6.
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.
Anne said:
5 years ago
Water required = (5% of aggregate + 30% of cement ) * w/c.
Ratio given = 1:3:6 ( c:s:a).
1 bag cement = 50kg.
Sand = 3*50 = 150kg (3 parts of sand).
Therefore,
Water required = (5/100*150) + (30/100*50) * 0.6.
= 13.5 = 14 kg.
Ratio given = 1:3:6 ( c:s:a).
1 bag cement = 50kg.
Sand = 3*50 = 150kg (3 parts of sand).
Therefore,
Water required = (5/100*150) + (30/100*50) * 0.6.
= 13.5 = 14 kg.
(4)
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).
Shubham Angre said:
4 years ago
28% of water is required for the concrete mix and 1 bag of cement = 50 kg.
So, 50*28% = 14 kg.
So, 50*28% = 14 kg.
(9)
Ramchandra chaulagain said:
4 years ago
We know, W/C=weight of water in kg/weight of cement in kg.
so, the weight of water = weight of cement * w/c.
= 50 * 0.6.
= 30 kg.
That's why Ans is 30kg.
so, the weight of water = weight of cement * w/c.
= 50 * 0.6.
= 30 kg.
That's why Ans is 30kg.
(4)
Abdur Rehman said:
4 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.
(30)
Tanmoy Ghosh said:
3 years ago
Here, 0.6 * 50 = 30.
(9)
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