Civil Engineering - Concrete Technology - Discussion

Yes, P = (Z - X)/(Y - X) * 100.

The given answer is the percentage of fine aggregate to the coarse aggregate but the question is to calculate the percentage of fine aggregate to the total aggregate.

Let p:1 be the proportion of fine to coarse aggregate so the total proportion of combined aggregate is (p+1).

Now p*Y+1*X=(1+p) * Z.
on solving we get p=(Z-X)/(Y-Z).
Now to get the proportion of fine to the total aggregate,
p/(p+1) comes out to be (X-Z)/(X-Y).
And in percentage just multiply it by 100.

How to calculate it? Please explain in detail.

It should be p={(X-Z)/(X-Y)}*100%.

(CALCULATING RATIO OF FINE AGGREGATE TO COARSE AGGREGATE).

Let "x1" is the percentage of fine aggregate to be combined with coarse aggregate. x can be calculated using the formula given below.

x = [(F2-F)/(F-F1)] * 100.

Where,

F1 = Fineness modulus value of the fine aggregate.
F2 = Fineness modulus value of coarse aggregate.
F = Fineness modulus value of the combined aggregate.

After finding the percentage of fine aggregate to be combined with the coarse aggregate by the above formula, their proportion can be calculated.

You are right, thanks @Manohar.

pY+X(1-p)=1 * Z.
It is p = (X-Z) ÷ (X-Y).

Answer is correct.

Because coaser particle will be greater than combined or finer so answer should be coarser-combined/combinef-fine.

Answer is obviously D.

None of the above is correct.

Correct is, (X-Z)/(X-Y).