Civil Engineering - Concrete Technology - Discussion

Discussion :: Concrete Technology - Section 4 (Q.No.34)

34.

If 50 kg of fine aggregates and 100 kg of coarse agregates are mixed in a concrete whose water cement ratio is 0.6, the weight of water required for harsh mix, is

 [A]. 8 kg [B]. 10 kg [C]. 12 kg [D]. 14 kg [E]. 15 kg.

Answer: Option C

Explanation:

No answer description available for this question.

 Mohanad said: (Aug 25, 2013) How to get this result?

 M.Pagavandoss said: (Jul 17, 2014) Wt of water = (5% wt of coarse aggregate+30% wt of cement)*w/c ratio.

 Sumit said: (Oct 12, 2014) 100 (wt.of Coarse sand) *5/100 = 5 kg. 50 (wt.of cement)* 30/100 = 15 kg. (5+15)*0.6 = 12 kg.

 Vishal Singh said: (Feb 28, 2015) Its 30% of coarse aggregate.

 Nungsang said: (May 12, 2015) If the mix is 1:3:6, Can we use same method keeping same wc ratio?

 Nabaratna said: (Aug 5, 2015) WC ratio is always 0.6.

 Ruhi Varshney said: (Jan 26, 2016) Water cement ratio are different for different grade of concrete.

 Bahram said: (Oct 4, 2016) @Sumit, 50 kg is the weight of fine aggregates, not cement.

 Baloch said: (Nov 14, 2016) Please read question, it's giving 50kg of FINE aggregates not cement.

 Buet-K said: (Nov 30, 2016) C = 1/7 * 150 = 21 (1;2;4) = 7, W/C = 0.6, W = 0.6 * 21 = 12.

 Apcivilian said: (Dec 2, 2016) .3p+.1F+.01C= w/c xp By using this equation, we can get the answer.

 Concretesukan said: (Dec 11, 2016) (5% Coarse aggregate + 30% of cement (50kg here))* w/c is correct.

 Prem. Vit Jaipur said: (Dec 13, 2016) 0.3 * cement + 0.1FA + 0.01CA = w/c * cement. 0.3 * cement + 0.1 * 50 + .01 * 100 = 0.6*cement. cement = 20. W/c = 0.6 water = 0.6 * 20 = 12kg.

 Bivek Pradhanang said: (Jan 23, 2017) D. w= 5%aggregate + 30%cement and here w/c=0.6. So, here w= .05 * [50+150]+.3 * [w/0.6], so w = 15kg.

 Sanjeev Kr Anad said: (Feb 6, 2017) W/c *p = 0.3p +0.1F +0.01C. 0.6*p = 0.3p + 0.1*50 +0.01*100, 0.3p = 5+1, P = 6 ÷ 0.3, P = 20. Then, water= w/c*P =0.6 * 20 = 12kg ans.

 Ram Soni said: (Jul 22, 2017) Thanks @Apcivilian & @Prem. Vit.

 Pakkirappa said: (Sep 9, 2017) Thank you very much @Sanjeev Kr Anad.

 Qazi said: (Sep 21, 2017) Why water cement ratio always used in weight?

 Ak Jain said: (Oct 19, 2017) Please solve, How to find the water required for concrete 1:2:4 if w/c =.6 and unit w.=2400kg/m3?

 Priya said: (Nov 9, 2017) W/C*P = 0.3P + 0.1Y + 0.01Z.

 Hussain said: (Jan 24, 2018) W= 30%cement +4%of total aggregates. W=. 3W/.6+.04*150 (Given- Water/Cement =0.6). W=. 5W+6. W=12kg.

 Kunal said: (Mar 13, 2018) Thank you so much for explaining it.

 Qasim Afzal said: (Apr 8, 2018) W/C *P = 0.3P + 0.1F.A +.01C.A 0.6*P=0.3P+0.1(50)+0.01(100) 0.6P-0.3P=5+1 P=6/0.3 P=20 => Where P= Cement. Now; W/C=0.6, W=C*0.6, W=20*0.6, W=12Kg.

 Dilip said: (Mar 6, 2019) How to find out 5% CA and 30% FA?

 Intwali said: (Mar 28, 2019) W/C * P = 0.3P + 0.1F.A +.01C.A. 0.6 * P = 0.3P + 0.1(50) + 0.01(100), 0.6P - 0.3P = 5 + 1, P = 6/0.3 P = 20 => Where P= Cement. Now; W/C = 0.6, W = C * 0.6, W = 20 * 0.6, W = 12Kg.

 Sanjay Kumar Mandal said: (Jun 21, 2019) Thanks all for explaining it.

 Amit Mishra said: (Aug 20, 2019) Water content = 5/100 * (weight of aggregate) + 30/100 * (wt of cement). So the answer is 15 kg.

 Nabi Nyadv said: (Sep 13, 2019) 0.3 * cement + 0.1fine + 0.01coarse = 0.6 * cement. 0.3 * cent + 0.1 * 50 + .0.01 * 100 = 0.6 * cement. 0.3 * cement = 6. Cement = 20, W/c = 0.6, W = 0.6 * 20 = 12kg.

 Muet 12Ce41 said: (Dec 13, 2019) DEAR all, First of all, pick the question, In the question, we are asked for two quantities not for one. We have to fine cement as well as water. Therefore We know that, Weight of water / Weight of cement = Water to cement ratio. And the water to cement ratio is 0.6. Rearrange the above formula, Weight of water = Weight of cement x Water to cement ratio. So, Weight of water = Weight of cement x 0.6.....................(i) Here, We have in weight of water = 0.3 x weight of cement + 0.1 x fine aggregate + 0.1 x coarse aggregate= 0.3 x weight of cement + 0.1 x (50)+ 0.1 x (100) = 0.3 x weight of cement + 5 +1 0.3 x weight of cement + 0.1 x (50)+ 0.1 x (100) = 0.3 x weight of cement +6. So, Weight of water = 0.3 x weight of cement +6...........................(ii) From equation 1, Weight of cement x 0.6 = 0.3 x weight of cement + 6. Let weight of cement be denoted by L L x 0.6 =0.3 x L + 6 0.6 L-0.3 L = 6 0.3 L = 6 L = 6/ 0.3= 20 kg. Where L is the weight of cement. By using formula, Weight of water = weight of cement x water to cement ratio. Weight of water = 20 x 0.6. = 12 kg.

 Madhu Ratan said: (Aug 6, 2020) Weight of water = (5%wt.of coarse aggregate +30%wt. Of fine aggregate)*W/C ratio. = ( 5 * 100/100 + 30 * 50/100) * 0.6. = 12kg.

 Ashish Sharma said: (Sep 2, 2020) W/C *P = 0.3P + 0.1F.A +.01C.A. 0.6*P =0.3P + 0.1(50) + 0.01(100), 0.6P - 0.3P = 5 + 1, P = 6/0.3, P = 20 => Where P= Cement. Now; W/C = 0.6, W = C * 0.6, W = 20 * 0.6, W = 12Kg.

 Charan said: (Sep 7, 2020) Is this (equation) useful for all concrete mix? Please explain.

 Ejaz Ahmad said: (Jan 10, 2021) Wt. Of water = (5% of Coarse aggregate + 30% of fine aggregate) * water cement ratio. Therefore, Wt. Of water = {5 * 100/100 +30 * 50/100} * 0.6, Wt. Of water= 12kg.

 Sameer Khan said: (Apr 13, 2021) Where does this formula comes from? W/C *P = 0.3P + 0.1F.A +.01C.A How do you justify it? Please explain me.

 Mir Imroz said: (May 29, 2021) In harsh mix weight of water =. 30% wt of fine aggregate + .05% wt of coarse aggregate) water-cement ratio.

 Muhtasim said: (Jun 8, 2021) What will be formula for rich mix?

 Er Vikash Saini said: (Jul 9, 2021) Here 50 kg weight of cement: M 25 ratio 1:1:2. (5% weight of aggregate + 30 weight of cement)w/c ratio.

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