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 5 of 5.
Rhythm said:
1 decade ago
But the answer is in 7.35 and your answer is 7.4 which is correct?
Maha said:
1 decade ago
2 out of 20 kg retained in that particular sieve, so for 100 how much is retained give % retained hence.
(2/20) * 100 = 10
(8/20) * 100 = 40
(6/20)* 100 = 30
(4/20)*100 = 20. And so on, if u have to find cumulative % retained u have to add the presently available value with all previous values hence first value is 10.
Second value is 10 + 40 = 50.
Third value is 10+ 40 +30 = 80.
Fourth value is 10+40+30+20 = 100 and so on.
(2/20) * 100 = 10
(8/20) * 100 = 40
(6/20)* 100 = 30
(4/20)*100 = 20. And so on, if u have to find cumulative % retained u have to add the presently available value with all previous values hence first value is 10.
Second value is 10 + 40 = 50.
Third value is 10+ 40 +30 = 80.
Fourth value is 10+40+30+20 = 100 and so on.
Turab said:
1 decade ago
Please explain how the last column how it comes to b 10 50 80 and 100?
KISHORE said:
1 decade ago
% RETAINED CUMULATIVE % RETAINED
80 MM 0 0 0
40 2 10 10
20 8 40 50
10 6 30 80
4.75 4 20 100
2.36 0 0 100
1.18 0 0 100
600 0 0 100
300 0 0 100
150 0 0 100
TOTAL 740.
FINENESS MODULUS = 740/100 = 7.4.
80 MM 0 0 0
40 2 10 10
20 8 40 50
10 6 30 80
4.75 4 20 100
2.36 0 0 100
1.18 0 0 100
600 0 0 100
300 0 0 100
150 0 0 100
TOTAL 740.
FINENESS MODULUS = 740/100 = 7.4.
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