Civil Engineering - Waste Water Engineering - Discussion

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

[A]. 25%
[B]. 50%
[C]. 10%
[D]. 5%.

Answer: Option B

Explanation:

No answer description available for this question.

Priyanka said: (Jun 4, 2015)  
Will you get me the explanation?

Raihan said: (Jul 26, 2015)  
V2/V1 = (100-p1) / (100-p2) = 1/2.

Vishal Shah said: (Jul 22, 2017)  
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%.

Kartik said: (Jun 22, 2019)  
Thank you @Vishal Shah.

Abinash Das said: (Dec 13, 2019)  
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?

Rabin Das said: (Feb 1, 2020)  
Thanks all.

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