Civil Engineering - Concrete Technology - Discussion
Discussion Forum : Concrete Technology - Section 1 (Q.No. 29)
29.
If 20 kg of coarse aggregate is sieved through 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron and 150 micron standard sieves and the weights retained are 0 kg, 2 kg, 8 kg, 6 kg, 4 kg respectively, the fineness modulus of the aggregate, is
Discussion:
44 comments Page 4 of 5.
Shailendra said:
7 years ago
FM= Mi/M=148/20=7.4.
Where M1=m1=0 kg.
M2=m1+m2=0+2=2 kg,
M3=m1+m2+m3=2+8=10 kg,
M4=m1+m2+m3+m4=10+6=16 kg,
M5=m1+m2+......+m5=16+4=20 kg,
M6=m1+m2+......+m6=20+0=20 kg,
M7=m1+m2+......+m7=20+0=20 kg,
M8=m1+m2+......+m8=20+0=20 kg,
M9=m1+m2+......+m9=20+0=20 kg,
M10=m1+m2+......+m10=20+0=20 kg.
Where M1=m1=0 kg.
M2=m1+m2=0+2=2 kg,
M3=m1+m2+m3=2+8=10 kg,
M4=m1+m2+m3+m4=10+6=16 kg,
M5=m1+m2+......+m5=16+4=20 kg,
M6=m1+m2+......+m6=20+0=20 kg,
M7=m1+m2+......+m7=20+0=20 kg,
M8=m1+m2+......+m8=20+0=20 kg,
M9=m1+m2+......+m9=20+0=20 kg,
M10=m1+m2+......+m10=20+0=20 kg.
(3)
Ganga said:
7 years ago
Well explained @Shailendra.
Munir said:
7 years ago
Thanks for the answers.
Santosh said:
7 years ago
Can't understand it, please explain it clearly.
Bhavin said:
7 years ago
As per my knowledge the answer is 7.35.
(1)
Ravi said:
6 years ago
Thanks @Shailendra.
(2)
Ajay said:
6 years ago
Thank you all.
Naveen said:
6 years ago
Thanks @Shailendra.
Muhammad Saqib said:
6 years ago
Answer Will be 7.4.
Kaisiki said:
6 years ago
There are 10 serial numbers.
No 7.6 mm sieve is there still its added by someone that's why it's addition is coming wrong
Removing calculation for 7.5 n shifting it to 4.75 you will get correct solution with 740 as total.
Thanks.
No 7.6 mm sieve is there still its added by someone that's why it's addition is coming wrong
Removing calculation for 7.5 n shifting it to 4.75 you will get correct solution with 740 as total.
Thanks.
(1)
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