Civil Engineering - Concrete Technology - Discussion

35. 

The percentage of the aggregate of F.M. 2.6 to be combined with coarse aggregate of F.M. 6.8 for obtaining the aggregates of F.M. 5.4, is

[A]. 30%
[B]. 40%
[C]. 50%
[D]. 60%.

Answer: Option C

Explanation:

No answer description available for this question.

Pratik Basak said: (Jan 3, 2015)  
Formula is (X-Z)/(Z-Y)*100.
Here X - 6.8, Y - 2.6, Z - 5.4.

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.

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.

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%.

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?

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%.

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%.

Anoms said: (Aug 15, 2019)  
(coarse -Combine)/(Combine-Fine) * 100.
So, (6.8-5.4)/(5.4-2.6)*100 = 50%.

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 .

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%.

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.

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.

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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