# Civil Engineering - Waste Water Engineering - Discussion

Discussion Forum : Waste Water Engineering - Section 3 (Q.No. 2)
2.
The moisture content of a sludge is reduced from 90% to 80% in a sludge digestion tank. The percentage decrease in the volume of sludge, is
25%
50%
10%
5%.
Explanation:
No answer description is available. Let's discuss.
Discussion:
7 comments Page 1 of 1.

Prashanta Das said:   1 year ago

Rabin Das said:   4 years ago
Thanks all.
(1)

Abinash Das said:   4 years ago
The answer will be the same if the moisture content is reduced from 95% to 90%?? As per your explanation.

V1/V2 = (100-Initial) / (100- Final)
= 100-95 / 100-90
= 5/10
= 1/2.

Is it right?
(1)

Kartik said:   5 years ago
Thank you @Vishal Shah.

Vishal Shah said:   7 years ago
Initial = 90%.
Final = 80%.

V2/V1 = (100 - Initial) / (100- Final).
= 100 - 90 / 100 - 80,
= 10 / 20,
= 1/2.

For % multiply by 100.
Ans is 50%.
(1)

Raihan said:   9 years ago
V2/V1 = (100-p1) / (100-p2) = 1/2.

Priyanka said:   9 years ago
Will you get me the explanation?