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 2 of 6.
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)
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.
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)
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)
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.
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.
NIKUNJ said:
9 years ago
Water required per bag means:
Cement bag = 50 kg.
W/c ratio = 0.6.
Water = 50 x 0.6.
Hence water = 30 kg.
If we assume water absorption in calculation then water require more than 30 kg.
Cement bag = 50 kg.
W/c ratio = 0.6.
Water = 50 x 0.6.
Hence water = 30 kg.
If we assume water absorption in calculation then water require more than 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.
Maheshsinh said:
5 years ago
Dry volume of mix 1.54 cmt.
There fore qty of cement in mix of 1:3:6= 1.54*50/10*0.035.
= 220 kg.
Here w/c =0.6.
Qty of water= 0.6*220.
=13.3.
Rounding 14 kg.
There fore qty of cement in mix of 1:3:6= 1.54*50/10*0.035.
= 220 kg.
Here w/c =0.6.
Qty of water= 0.6*220.
=13.3.
Rounding 14 kg.
(9)
Ajay Agnihotri said:
9 years ago
However, I believe that the density of cement is 1440 Kg/Cum.
Therefore, the volume of a 50 kg cement bag works out to be [ 50/1440 ]*1000 = 34.722 Liters.
Therefore, the volume of a 50 kg cement bag works out to be [ 50/1440 ]*1000 = 34.722 Liters.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers