Discussion :: Concrete Technology - Section 2 (Q.No.35)
| Pratik Basak said: (Jan 3, 2015) | |
| Formula is (X-Z)/(Z-Y)*100. Here X - 6.8, Y - 2.6, Z - 5.4. |
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| Agar said: (Dec 8, 2015) | |
| FM mean? | |
| Rajesh said: (Jan 26, 2016) | |
| Fineness Modulus-FM. | |
| Nabin Chandra Das From Gimt said: (Jan 27, 2016) | |
| {(6.8-5.4)/(5.4-2.6)}*100 = 50% x = 6.8, y = 2.6, z = 5.4. C is correct answer. |
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| Amrit Raj said: (Mar 18, 2017) | |
| How can we determine the value of X, Y and Z? | |
| Vishvas said: (Jun 3, 2017) | |
| X = % coarse aggregates. Y = % fine aggregates. Z = % combined aggregates. |
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| Pritam said: (Sep 28, 2017) | |
| When it is given "to be combined", the solution is given as (6.8-5.4)/(6.8-2.6). | |
| Naveen Kallan said: (Oct 9, 2017) | |
| If it is the ratio of fine to coarse then only ans is 50%. If it is percentage then ans is X-Z/X-Y = 33.33%. |
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| Rabindra said: (Apr 18, 2018) | |
| Nice @Pratik. | |
| Vikrantsinghchauhan said: (Sep 24, 2018) | |
| I am not understanding. Please explain the correct answer of this question. | |
| Vivek Yadav said: (Oct 8, 2018) | |
| 50% is correct. | |
| Mrityunjoy Mete said: (Nov 16, 2018) | |
| the fine aggregate of finesse modulus 2.6 is mix with a coarse aggregate of finesse modulus 6.8 for obtaining the aggregate of finesse modulus 5.4 then find our the percentage of coarse aggregate in the mix? Can anybody answer this? |
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| Gopal said: (Jun 11, 2019) | |
| Yes, you are right, Thanks @Naveen Kallan. | |
| Chintu said: (Jul 21, 2019) | |
| Agree @Naveen Kallan. The Answer is 33.33%. |
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| Chintu said: (Jul 21, 2019) | |
| Let the proportion of FA is =p. Then, proportion of CA will be =1-p , as sum of proportion is always 1. So, 2.6p+6.8(1-p)=5.4 , by solving it we get, p=1/3=0.33, i.e in percentage it is 33.33% . But in option appropriate answer will be 30%. |
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| Anoms said: (Aug 15, 2019) | |
| (coarse -Combine)/(Combine-Fine) * 100. So, (6.8-5.4)/(5.4-2.6)*100 = 50%. |
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| Anand Ratan said: (Dec 12, 2019) | |
| X = course aggregates. Y = fine aggregates. Z = combined aggregates. Formula , Fine modules ( % )= ( x - z / z - y ) x 100 =( 6.8 - 5.4 / 5.2 - 2.6 ) x 100 = 50% Correct answer is C . |
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| Khongthaw said: (Aug 17, 2020) | |
| Let Fine aggregates be P1. Course aggregates be P2. Combined aggregates be P. Given: P1=2.6, P2=6.8 and P=5.4. Fineness modulus = (P2 - P/P-P1) * 100. = 50%. |
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| Rahul said: (Dec 7, 2020) | |
| @Mrityunjoy mete. First, find the percentage of fine aggregate by FMc-FMcom/FMcm-FMf. Then the percentage of coarse aggregate will be 100- the percentage of fine aggregate. |
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| Dawood Khan said: (Jan 25, 2021) | |
| 5O is the correct answer. FM= x-z/z-y x100. X= coarse agg which is 6.8. Y = fine agg which is 2.6. Z= combined or obtained agg which is= 5.4. Now put the values in the above formula, = 6.8-5.4/5.4-2.6x100. =1.4/2.8x100, = 0.5x100, = 50. |
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| Khaled Mahmoud said: (Jul 11, 2021) | |
| Can anyone give me feedback about the previous percent 50%, Is this percent is relative to the total mix volume or related to the total volume of aggregates (coarse and fine)? Please explain me. | |
| Pankaj Nagar said: (Jul 14, 2021) | |
| Here, The percentage of the aggregate of F.M. 2.6 to be combined. i.e If the total amount of F.M 2.6 is 100% now then how much % of Aggregate will be required to obtain an F.M 5.4 with 100% amount of F. M. 6.8 aggregates. |
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