Time for Math: 1. The effluent of a wastewater treatment plant is 35,000 gpd. How would you show the 35,000 gal as a decimal fraction of 1 mil gal? a. 0.35 mil gal. b. 0.035 mil gal. c. 3.50 mil gal. d. 0.0035 mil gal. e. 0.00035 mil gal. 2. An operator of a small plant performs four BOD tests during a certain month. obtaining the following results: 29.0, 22.5, 27.0, and 26.3. What is the average BOD for the month? a. 25.5 b. 27.8 c. 30.5 d. 21.9 e. 26.2 3. An operator of a small treatment works takes two BOD tests every week. At the end of the month he finds he has the following results: 19.3, 21.4, 17.0, 29.9, 24.0, and 21.0. What is the average BOD for the month? a. 17.0 b. 19.8 c. 22.1 d. 23.9 e. 24.2 4. A tank holding 2,340 gal fills in 12 min. What is the rate of flow in gallons per minute? a. 195 gpm. b. 234 gpm. c. 284 gpm. d. 362 gpm. e. 412 gpm. 5. A sludge drying bed 60 x 90 ft is filled to a depth of 12 in. When the sludge is ready for removal, it is found that its volume has been reduced by 70 percent. How many cubic feet of dried sludge are there on the bed? a. 540 cu ft. b. 1,010 cu ft. c. 1,620 cu ft. d. 2,420 cu ft. e. 3,780 cu ft. Answers: 1. B (35,000 gpd) / (1,000,000) = or just shuffle 6 places to the left and move decimal point 2. E 29.0 + 22.5 + 27.0 + 26.3 = (104.8) / (4) = 3. C 19.3 + 21.4 + 17.0 + 29.9 + 24.0 + 21.0 = (132.6) / 6 = 4. A Known – 2,340 gallons, 12 min Unknown – gallons per minute or gal/min = creates its own formula (2,340 gal) / (12 min) = 5. C Convert 12” to ft = (12”) / (12 in/ft) = 1 ft X 60 ft X 90 ft = 5400 cuft Question asks how much sludge is left behind and tells us that it has been reduced 70%. That means there is 30% left behind – convert 30% to a decimal so it can be used (30 %) / 100 = 0.30 or shuffle decimal two places to the left = 0.30 Now complete the problem = 5,400 cuft X 0.30 =
Posted on: Mon, 17 Nov 2014 01:44:06 +0000
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