Civil Engineering - Concrete Technology - Discussion

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

[A]. 7.30
[B]. 7.35
[C]. 7.40
[D]. 7.45
[E]. none of these.

Answer: Option C

Explanation:

No answer description available for this question.

Kishore said: (Jul 25, 2013)  
% 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.

Turab said: (Nov 15, 2013)  
Please explain how the last column how it comes to b 10 50 80 and 100?

Maha said: (Nov 17, 2013)  
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.

Rhythm said: (Jan 18, 2015)  
But the answer is in 7.35 and your answer is 7.4 which is correct?

Rohan said: (Jan 22, 2015)  
But the answer is nearly to correct so answer is true.

Tanu said: (Jan 22, 2015)  
But here both 7.35 and 7.40 options are given. Finally what will we do?

Nitin said: (Jan 28, 2015)  
Yes the answer should be 7.40 means option C is correct.

Nikita said: (Jan 31, 2015)  
Please tell me how you get the sum 740?

Snehal Nikam said: (Feb 1, 2015)  
How it is possible?

Hardik said: (Feb 2, 2015)  
Approximately maximum near by value is taken as A answer.

Nirav Bhatt said: (May 22, 2015)  
Pass Wt Cum Wt. %Cum %cum.

Retained % passing.

0 0 0 100.
2 2 10 90.
8 10 40 50.
6 16 30 20.
4 20 20 0.
--- --- --- ---.
20 48 100 260.

Sieves used are 10.

Hence FM = (10*100-260)/100 = 7.4.

In method suggested by @Kishor, the total is done individually for each sieve where as here it is done by using the nos. sieve.

Both the methods are correct. So the Answer is 7.4.

Jagadeesh said: (Nov 3, 2015)  
Sieve weight % weight cumulative.

80 0 0 0.

40 2 10 10.

20 8 40 50.

10 6 30 80.

7.5 4 20 100.

4.75 0 0 100.

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 of given course aggregate is 740/100 = 7.4.

Sanu Sonkamble said: (Jul 27, 2016)  
Thanks Kishorn & Jagadeesh.

Vinod said: (Aug 7, 2016)  
Fineness modulus of given course aggregate is 740/100 =7.4.

Bapureddy said: (Aug 8, 2016)  
I didn't get it. Please, Can anyone explain it clearly?

Naman said: (Oct 1, 2016)  
I didn't get this, Please explain briefly.

Sanjana said: (Dec 2, 2016)  
Can anyone explain last column?

Sudip said: (Dec 3, 2016)  
Sieve.size wt. retained(kg) %wt.retained(%) cumulative%

80 0 0 0.

40 2 10 10.

20 8 40 50.

10 6 30 80.

7.5 4 20 100.

4.75 0 0 100.

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 of given course aggregate is 740/100 = 7.4.
For wt. retained in %, for 40mm sieve size, (2/20)*100% where 20 is the total wt.

Abi said: (Dec 24, 2016)  
Can't understand. Please explain this.

Umang said: (Dec 27, 2016)  
Thanks for the explanation @Maha.

Moreri said: (Jan 16, 2017)  
Why do we divide 740 by 100?

Dangerous said: (Feb 8, 2017)  
Sieve Retained Reained perce Cumaltive size
80 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
740
Fineness modulus =740/100,
= 7.4.

Harsha G said: (Feb 23, 2017)  
SIEVE WEIGHT RET CUMULATIVE WT CUM.%
SIZE
80 0 0 0
40 2 2 10
20 8 10 50
10 6 16 80
4.75 4 20 100
2.36 0 20 100
1.18 0 20 100
600M 0 20 100
300M 0 20 100
150M 0 20 100


ADD CUM % RET AND DIVIDE BY 100 740/100 = 7.4.

Fineness modulus is an empirical factor obtained by adding the cumulative percentages of aggregate retained on each of the standard sieves ranging from 80 mm to 150 micron and dividing this sum by 100.

CUM% = (CUM WT / TOTAL WT )* 100.

Virendra Kumar said: (Jun 23, 2017)  
How to total 740 and what is 740? please explain.

Kavita said: (Jul 27, 2017)  
@ALL.

Please, Refer Is code 2386.

Pranit Vitekar said: (Aug 7, 2017)  
Seive. Weight % weight. Cumulative % wt
size. Retained retained. Retained
80. 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 cumulative % weight retained= 740.
Fineness modulus = total cumulative % weight retained/ 100.
FM=740/100.
7.40 is the correct answer.

Ritesh said: (Sep 29, 2017)  
Please explain this. I can't understand,

How get 740?

Kavita said: (Dec 20, 2017)  
Please explian clearly.

Priyanka said: (Mar 5, 2018)  
How to get 7.40?

Rahul said: (Aug 2, 2018)  
The fm of coarse aggregate is C+500/100.
The fm of fine aggregate is f/100.

So, what we try on sieve 4.75 lower all are fine particles as per code.

Shailendra said: (Aug 3, 2018)  
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.

Ganga said: (Aug 5, 2018)  
Well explained @Shailendra.

Munir said: (Sep 21, 2018)  
Thanks for the answers.

Santosh said: (Nov 21, 2018)  
Can't understand it, please explain it clearly.

Bhavin said: (Feb 4, 2019)  
As per my knowledge the answer is 7.35.

Ravi said: (Mar 23, 2019)  
Thanks @Shailendra.

Ajay said: (Jun 8, 2019)  
Thank you all.

Naveen said: (Aug 24, 2019)  
Thanks @Shailendra.

Muhammad Saqib said: (Sep 20, 2019)  
Answer Will be 7.4.

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