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 2 of 3.
NABIN CHANDRA DAS FROM GIMT said:
10 years ago
{(6.8-5.4)/(5.4-2.6)}*100 = 50%
x = 6.8, y = 2.6, z = 5.4.
C is correct answer.
x = 6.8, y = 2.6, z = 5.4.
C is correct answer.
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)
Vikrantsinghchauhan said:
7 years ago
I am not understanding. Please explain the correct answer of this question.
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%.
Vishvas said:
8 years ago
X = % coarse aggregates.
Y = % fine aggregates.
Z = % combined aggregates.
Y = % fine aggregates.
Z = % combined aggregates.
PRATIK BASAK said:
1 decade ago
Formula is (X-Z)/(Z-Y)*100.
Here X - 6.8, Y - 2.6, Z - 5.4.
Here X - 6.8, Y - 2.6, Z - 5.4.
Amrit Raj said:
8 years ago
How can we determine the value of X, Y and Z?
(1)
Chintu said:
6 years ago
Agree @Naveen Kallan.
The Answer is 33.33%.
The Answer is 33.33%.
Gopal said:
6 years ago
Yes, you are right, Thanks @Naveen Kallan.
Rajesh said:
10 years ago
Fineness Modulus-FM.
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