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
30%
40%
50%
60%.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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.

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%.
(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%.

Vishvas said:   8 years ago
X = % coarse 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.

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%.

Gopal said:   6 years ago
Yes, you are right, Thanks @Naveen Kallan.

Rajesh said:   10 years ago
Fineness Modulus-FM.


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