# 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

Discussion:

23 comments Page 1 of 3.
SHASHANK said:
2 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%.

P2=6.8.

P1= 2.8.

P=5.4.

X= 6.8-5.4/5.4 - 2.8 * 100.

= 50%.

(3)

Pankaj Nagar said:
3 years ago

Here, The percentage of the aggregate of F.M. 2.6 to be combined.

i.e If the total amount of F.M 2.6 is 100% now then how much % of Aggregate will be required to obtain an F.M 5.4 with 100% amount of F. M. 6.8 aggregates.

i.e If the total amount of F.M 2.6 is 100% now then how much % of Aggregate will be required to obtain an F.M 5.4 with 100% amount of F. M. 6.8 aggregates.

Khaled Mahmoud said:
3 years ago

Can anyone give me feedback about the previous percent 50%, Is this percent is relative to the total mix volume or related to the total volume of aggregates (coarse and fine)? Please explain me.

Dawood khan said:
3 years ago

5O is the correct answer.

FM= x-z/z-y x100.

X= coarse agg which is 6.8.

Y = fine agg which is 2.6.

Z= combined or obtained agg which is= 5.4.

Now put the values in the above formula,

= 6.8-5.4/5.4-2.6x100.

=1.4/2.8x100,

= 0.5x100,

= 50.

FM= x-z/z-y x100.

X= coarse agg which is 6.8.

Y = fine agg which is 2.6.

Z= combined or obtained agg which is= 5.4.

Now put the values in the above formula,

= 6.8-5.4/5.4-2.6x100.

=1.4/2.8x100,

= 0.5x100,

= 50.

(4)

Rahul said:
4 years ago

@Mrityunjoy mete.

First, find the percentage of fine aggregate by FMc-FMcom/FMcm-FMf.

Then the percentage of coarse aggregate will be 100- the percentage of fine aggregate.

First, find the percentage of fine aggregate by FMc-FMcom/FMcm-FMf.

Then the percentage of coarse aggregate will be 100- the percentage of fine aggregate.

Khongthaw said:
4 years ago

Let Fine aggregates be P1.

Course aggregates be P2.

Combined aggregates be P.

Given: P1=2.6, P2=6.8 and P=5.4.

Fineness modulus = (P2 - P/P-P1) * 100.

= 50%.

Course aggregates be P2.

Combined aggregates be P.

Given: P1=2.6, P2=6.8 and P=5.4.

Fineness modulus = (P2 - P/P-P1) * 100.

= 50%.

(1)

Anand Ratan said:
5 years ago

X = course aggregates.

Y = fine aggregates.

Z = combined aggregates.

Formula ,

Fine modules ( % )= ( x - z / z - y ) x 100

=( 6.8 - 5.4 / 5.2 - 2.6 ) x 100

= 50%

Correct answer is C .

Y = fine aggregates.

Z = combined aggregates.

Formula ,

Fine modules ( % )= ( x - z / z - y ) x 100

=( 6.8 - 5.4 / 5.2 - 2.6 ) x 100

= 50%

Correct answer is C .

Anoms said:
5 years ago

(coarse -Combine)/(Combine-Fine) * 100.

So, (6.8-5.4)/(5.4-2.6)*100 = 50%.

So, (6.8-5.4)/(5.4-2.6)*100 = 50%.

Chintu said:
5 years ago

Let the proportion of FA is =p.

Then, proportion of CA will be =1-p , as sum of proportion is always 1.

So, 2.6p+6.8(1-p)=5.4 , by solving it we get, p=1/3=0.33, i.e in percentage it is 33.33% .

But in option appropriate answer will be 30%.

Then, proportion of CA will be =1-p , as sum of proportion is always 1.

So, 2.6p+6.8(1-p)=5.4 , by solving it we get, p=1/3=0.33, i.e in percentage it is 33.33% .

But in option appropriate answer will be 30%.

Chintu said:
5 years ago

Agree @Naveen Kallan.

The Answer is 33.33%.

The Answer is 33.33%.

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