What will be the BOD of the below described sample ?

Question

A municipal sewage sample is diluted to 1 percent by volume prior to BOD5 analysis and initial DO was found to be 8mg/l/ After 5 days the oxygen depletion was determined to be 2mg/l/ What is the BOD5 of the sewage?"

Answer 1

I had to look on the internet to figure out how to solve this problem.

Biochemical oxygen demand is a measure of the amount of oxygen in waste water is used by microbes. An amount of waste water is diluted with purified water to a known volume. The oxygen concentration is measured at the start and then again after 5 days, and the amount of oxygen consumed by the sample is calculated as mg/L of waste water.

Assume that a volume v of sewage sample is diluted 100 fold so that the final volume is 100 x v. As the initial concentration is 8 mg/L, and the final concentration is 2 mg/L, the absolute amount of oxygen used by the sample is given by AbsO2 = (8 - 2) / 1000 x 100 x v. This amount of oxygen was used by v mL of the sample, so the oxygen used by the sample expressed as mg / L of sample is

AbsO2 / v x 1000 or substituting the result for AbsO2, BOD = 6 mg/L /1000 x 100 x v / v x 1000. The v cancels out, so the resulting expression comes out to be 6 mg/L x 100 = 600 mg/L.

If you know the correct answer, please let me know if I am right as this is the first time I did this calculation.