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.
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)
SHASHANK said:
4 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)
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)
Amrit Raj said:
8 years ago
How can we determine the value of X, Y and Z?
(1)
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.
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 .
Anoms said:
6 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%.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers