Civil Engineering - Concrete Technology - Discussion
Discussion Forum : Concrete Technology - Section 2 (Q.No. 35)
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
Discussion:
23 comments Page 1 of 3.
Anand Ratan said:
6 years ago
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 .
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 .
Chintu said:
6 years ago
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%.
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%.
Dawood khan said:
5 years ago
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.
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.
(5)
Mrityunjoy Mete said:
7 years ago
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?
Can anybody answer this?
Pankaj Nagar said:
4 years ago
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.
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.
Khaled Mahmoud said:
4 years ago
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.
Rahul said:
5 years ago
@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.
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.
Khongthaw said:
5 years ago
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%.
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%.
(2)
Naveen Kallan said:
8 years ago
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%.
If it is percentage then ans is X-Z/X-Y = 33.33%.
Pritam said:
8 years ago
When it is given "to be combined", the solution is given as (6.8-5.4)/(6.8-2.6).
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