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Practice Tests – Answer Preparing for Technical Training: Essential Skills for Water/Wastewater Operators Pract ice Tests Answer Keys COURSE OUTLINE: Practice Test Module # Name included Module 1: Basic Math Refresher Module 2: Fractions, Decimals and Percents Module 3: Measurement Conversions Module 4: Linear, Area and Volume Calculations Module 5: Solving Equations Chemical Measurements Module 6: Module 7: Hydraulics Wastewater Module 8: Electricity Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 Measurement Conversions Practice Test Answer Key 1 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 – PRACTICE TEST MEASUREMENT CONVERSIONS 1) Convert 100 ml/sec to L/sec. a. 100,000 L/sec b. 10 L/sec c. 0.1 L/sec d. 0.001 L/sec 100 ml x 1 L = 100 ml L = 0.1 L/sec 1 sec 1000 ml 1000 sec ml Write numbers given as a fraction Write the conversion as a fraction (found on the formula sheet) 2) Convert 75 ml/sec to L/sec a. 75,000 L/sec b. 0.075 L/sec c. 0.0075 L/sec d. 750 L/sec 3) Convert 345 m/minute to m/s a. 5.75 m/s b. 20,700 m/s c. 8 m/s d. 4.25 m/s 345 m x 1 minute 1 minute 60 s = 5.75 m/s 2 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 – PRACTICE TEST 4) Convert 1000 ml/s to lpm a. 0.017 lpm b. 16.7 lpm c. 1000 lpm d. 60 lpm 1000 ml x 60 s x 1 litre = 60,000 ml s litre = 60 litre/min 1 s 1 minute 1000 ml 1000 ml s min 5) Convert 9.0032 ml/s to lpm a. 9003.2 lpm b. 540 lpm c. 0.540 lpm d. 0.009 lpm 6) Convert 18 m³/s to lpm a. 18,000 lpm b. 18,000,000 lpm c. 1080 lpm d. 1,080,000 lpm 18 m³ x 1000 L x 60 s = 1,080,000 lpm 1 s 1 min 1 m³ 7) Convert 45.670 m³/s to lpm a. 45,670 lpm b. 45,670,000 lpm c. 2,740,000 lpm d. 2,740 lpm 8) Convert 300 L in 15 seconds to L/min a. 1200 L/min b. 18,000 L/min c. 270,000 L/min d. 2,700,000 L/min 3 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 – PRACTICE TEST 300 L x 60 s = 18,000 L s = 1200 L/min 15 s 1 min 15 s min 9) Convert 76,000,000 L/day to cubic metres per second. a. 0.76 cu. m/s b. 0.88 cu. m/s c. 1267 cu. m/s d. 760 cu. m/s 76,000,000 L x 1 m³ x 1 day = 0.8796 cubic metres per second 1 day 1000 L 86,400 s 10) How many kilograms of water are in a standpipe containing 5678 L? a. 567.8 kilograms b. 5,678 kilograms c. 56,780 kilograms d. 56.78 kilograms 5678 L = ? kg On the formula sheet we know that 1 L of water weighs 1 kg. So 5678 L of water would weigh 5678 kg!! 11) A water meter in a residential home measures that 25 cubic metres of water are used every 30 days. What is the daily water use expressed in cubic metres, and litres? a. 25 m³, 250 L b. 25 m³, 0.00025 L c. 0.83³ m, 830 L d. 0.83³ m, 0.00083 L 25 m³ = 30 days ? m³ 1 day = 0.83333333 m³ /day 0.833333333 m³ = ? L 1 m³ 1000 L = 833.333333 L 4 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 – PRACTICE TEST 12) An empty atmospheric storage tank has a volume of 31.4 m³. How long will it take to fill 90% of the tank volume if a pump is discharging a constant 60 litres per minute into the tank? a. 7 hours 51 minutes b. 8 hours 21 minutes c. 8 hours 23 minutes d. 9 hours 17 minutes 31.4 m³ = ? L 1 m³ 1000 L (90% converted to a decimal is 0.9) = 31,400 L volume of the tank We’re only wanting to fill 90% of it so 31,400 x 0.9 = 28,260 L If the pump discharges 60 L = 1 min 28,260 L ? min = 471 minutes 471 minutes = ? hours 60 min 1 hour = 7 hours and some minutes left over How many minutes in 7 full hours? 7 hours x 60 minutes = 420 minutes So 471 minutes – 420 minutes = 51 minutes left over. Answer: 7 hours and 51 minutes 13) A chemical solution contains 2.5 lbs per Imperial gallon. What is this in g/L? a. 250 g/L b. 2.5 g/L c. 300 g/L d. 25 g/L grams on the formula sheet is abbreviated “gr” 2.5 lbs x 453.6 g x 1 gal = 1134 lbs g gal = 249.77973 g/L 1 gal 1 lb 4.54 L 4.54 gal lbs L 5 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 – PRACTICE TEST 14) A small pump can discharge 17 litres of water every two days. Calculate that in litres per minute. a. 1.7 lpm b. 0.012 lpm c. 0.024 lpm d. 0.00035 lpm 17 L x 1 day = 0.0059 litres per minute 2 days 1440 min 15) 890,000 litres of water flows through a pipe every hour. How many litres will flow through that pipe every second? a. 10.3 L/s b. 247 L/s c. 323 L/s d. 476 L/s 16) How many kilograms of a chemical applied at the rate of 3 mg/L are required to dose 200,000 litres? a. 0.2 kg b. 0.6 kg c. 4 kg d. 8 kg 3 mg = 1L ? 200,000 L = 600,000 mg Convert to kg: 600,000 mg = ? kg = 0.6 kg 1,000,000 mg 1 kg 6 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 – PRACTICE TEST 17) 25o Centigrade is equal to how many degrees Fahrenheit? a. 101o b. 91o c. 11o d. 77o Fahrenheit = (25 x 9/5 ) + 32 (Change the fraction into a decimal 9 ÷ 5 = 1.8) = (25 x 1.8) + 32 = 45 + 32 = 77 ⁰F 18) Convert 60.5 degrees Fahrenheit to degrees Celsius. a. 15.8 degrees Celsius b. 20.6 degrees Celsius c. 72.0 degrees Celsius d. 101.2 degrees Celsius Celsius = (60.5 – 32) x 5/9 (Change the fraction into a decimal 5 ÷ 9 = 0.55555556) = (28.5) x 0.55555556 = 15.8 ⁰C 19) Convert 88 degrees Celsius to Fahrenheit. a. 190 ⁰F b. 120 ⁰F c. 31 ⁰F d. 36 ⁰F 20) Convert 0 ⁰C to ⁰F a. 0 ⁰F b. 32 ⁰F c. 34 ⁰F d. 58 ⁰F 21) Convert 90 ⁰F to ⁰C a. 58 ⁰C b. 32 ⁰C c. 104 ⁰C d. 30 ⁰C 7 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 3 – PRACTICE TEST 8 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 4 Linear, Area and Volume Practice Test Answer Key 1 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 4 – PRACTICE TEST LINEAR, AREA AND VOLUME 1) If a clarifier has a diameter of 20.7 m, and a height of 26.2 m, what is the surface area of the water within the clarifier? a. 1707.18 m2 b. 336.36 m2 c. 1349.60 m2 d. 19.88 m2 Area of circle = π D² 4 = 3.14 (20.7)² 4 = 3.14 (428.49) 4 = 1345.4586 4 Area = 336.36465 m² 2) What is the volume of water that is in a tank that is 38.1 m long and 21 m wide, and has a depth of 5.2 meters? a. 4,160,520 litres b. 4,629,002 litres c. 1,741,445 litres d. 555,036 litres Volume = L x W x H From formula sheet 1000 L = 1 m³ = 38.1 m x 21 m x 5.2 m So 1000 L = 1 m³ ? 4160.52 m³ = 4160.52 m³ Cross multiply, then divide by the third # Volume = 4,160,520 L 2 Preparing for Technical Training: Essential Skills for Water/Wastewater Operators MODULE 4 – PRACTICE TEST 3) If a storage tank is 23 m long, 11 m wide, and 4.25 m deep how many litres of water would it take to overflow the tank? a. 1,075,254 litres b. 889,534 litres c. 1,160 litres d. 18,598 litres Convert to Litres Volume = L x W x H 1000 L = 1 m³ ? 1075.25 m³ = 23 m x 11 m x 4.25 m Volume = 1,075,250 L = 1075.25 m³ *any more then this volume of water would overflow the storage tank. 4) An empty atmospheric storage tank is 2 m in diameter and 10 m high. How long will it take to fill 90% of the tank volume if a pump is discharging a constant 60 litres per minute into the tank? a. 7 hours 51 minutes b. 8 hours 21 minutes c. 8 hours 23 minutes d. 9 hours 17 minutes Volume = π D² x H Calculate 90% of this volume 4 31,400 L x 0.9 = 28,260 L = 3.14 (2)² x 10 m 4 If the pump discharges 60 L in 1 min, then how long would it take to fill 28,260 L? = 31.4 m³ 60 L = 1 min.
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