Download Aptitude Crack
img not found img not found

Waste Water Engineering - Engineering

Q1:

Afive day B.O.D. at 15°C of the sewage of a town is 100 kg/day. If the 5 day B.O.D. per head at 15°C for standard sewage is 0.1 kg/day, the population equivalent is

A 100

B 1000

C 5000

D 10000

ANS:D - 10000

Bod of 1 day= 0.1 kg but waste per day is 100 so bod = 100*0.1
Waste per day is 100 so 5 days waste = 5*100
And 5 days bod = 5 * 0.1( 0.1= 1 days bod).
Total population = (bod per waste * waste per 5 day)/ 5 days bod.
= (100*0.1 * 5 *100)/( 0.1* 5)
= 10,000.