Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 19 (Q.No. 47)
47.
Given that Plasticity Index (PI) of local soil = 15 and PI of sand = zero, for a desired PJ of 6, the percentage of sand in the mix should be
Discussion:
7 comments Page 1 of 1.
Samyak said:
5 years ago
Let soil and sand be in 1:x Proportion
Now,
Ip(mix)= (1*15+x*0)/(1+x)
6=(1*15+x*0)/(1+x)
x=3/2.
Now,
%of sand in the mix = (x)/(1+x) *100.
= (3/2)/(1+3/2)*100.
= 60%.
Now,
Ip(mix)= (1*15+x*0)/(1+x)
6=(1*15+x*0)/(1+x)
x=3/2.
Now,
%of sand in the mix = (x)/(1+x) *100.
= (3/2)/(1+3/2)*100.
= 60%.
Aditya Tiwari said:
6 years ago
(Ip)mix= (Ip1*X1 + Ip2*X2)/(X1+X2).
(2)
Garima said:
7 years ago
What is pj here? Please explain.
Manish yadav said:
7 years ago
Please elaborate the solution.
Xiyoz said:
7 years ago
How? Can someone elaborate?
Manali rathee said:
8 years ago
X*0 +(1-x)*15=6.
X = 60%.
X = 60%.
Khushhal said:
8 years ago
Please explain it.
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