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.
SHASHANK said:
3 years ago
X=P2-P/P-P1*100.
P2=6.8.
P1= 2.8.
P=5.4.
X= 6.8-5.4/5.4 - 2.8 * 100.
= 50%.
P2=6.8.
P1= 2.8.
P=5.4.
X= 6.8-5.4/5.4 - 2.8 * 100.
= 50%.
(3)
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.
Dawood khan said:
4 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)
Rahul said:
4 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:
4 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)
Anand Ratan said:
5 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 .
Anoms said:
5 years ago
(coarse -Combine)/(Combine-Fine) * 100.
So, (6.8-5.4)/(5.4-2.6)*100 = 50%.
So, (6.8-5.4)/(5.4-2.6)*100 = 50%.
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%.
Chintu said:
6 years ago
Agree @Naveen Kallan.
The Answer is 33.33%.
The Answer is 33.33%.
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