This site contains a complete, thorough, and concise review of the mathematical information needed to pass the math portion of introductory-level water treatment operation exams. No information contained on this site is useless or unimportant. Anyone who learns all of the information on this site will be able to answer most or all of the mathematical questions on the mathematical portion of his or her water treatment certification exam.
Please read each post carefully. As you finish studying each page, click on older posts at the bottom of the page to reach the posts on the next page. As you study this site, always remember this: Water treatment facility operator certification board members throughout the United States repeatedly say that almost no one fails an introductory-level water treatment facility operator certification exam if he or she performs well on the math portion of the exam.
Sunday, June 7, 2009
Lesson One: Converting Fractions & Percents to Decimals
Fractions are always written with a top number that is separated from a bottom number by a line. This line simply means "divided by." To convert a fraction into a decimal, use your calculator to divide the bottom number into the top number.
PULL OUT YOUR CALCULATOR AND CONVERT THESE FRACTIONS TO DECIMAL FORM.
3/4 (Answer: 3 divided by 4 = .75)
5/8 (Answer: 5 divided by 8 = .625)
1/2 (Answer: 1 divided by 2 = .50)
3/16 (Answer: 3 divided by 16 = .1875)
1/8 (Answer: 1 divided by 8 = .125)
Percentages, when expressed with the % symbol, are converted to decimal form by dropping the % symbol and multiplying the remaining number by .01.
Examples:
5% = 5 x .01 = .05
23% = 23 x .01 = .23
50% = 50 x .01 = .50
70% = 70 x .01 = .70
PULL OUT YOUR CALCULATOR AND CONVERT THESE FRACTIONS TO DECIMAL FORM.
3/4 (Answer: 3 divided by 4 = .75)
5/8 (Answer: 5 divided by 8 = .625)
1/2 (Answer: 1 divided by 2 = .50)
3/16 (Answer: 3 divided by 16 = .1875)
1/8 (Answer: 1 divided by 8 = .125)
Percentages, when expressed with the % symbol, are converted to decimal form by dropping the % symbol and multiplying the remaining number by .01.
Examples:
5% = 5 x .01 = .05
23% = 23 x .01 = .23
50% = 50 x .01 = .50
70% = 70 x .01 = .70
Lesson Two: Calculation of Area
Calculation of area is the most basic and most important concept that must be understood in order to perform introductory-level water treatment math. Area calculations form the single building block upon which four of the six most important introductory-level math formulas are based. Area calulations are needed for such operations as calculating flow through filters, determining filter backwash rates, and calculating detention times in flocculation tanks and sedimentation basins.
On all introductory-level water treatment examinations in the United States, area is expressed in terms of inches and feet. The two key words, which indicate area and which indicate the scope of the area, are square and cubic. When area is expressed in inches, it will be designated as square inches or cubic inches. When area is expressed in feet, it will be designated as square feet or cubic feet.
A square measurement of area is always flat. To remember that the term square means flat, think of ordinary household carpeting, which is flat and which is usually measured in "square feet". A square foot is always 1 foot by 1 foot square. A square inch is always 1 inch by 1 inch square.
A cubic inch or cubic foot is always shaped like a perfect cube with the width, length, and height being exactly equal. A cubic inch is a cube that is 1 inch by 1 inch by 1 inch. A cubic foot is a cube that is 1 foot by 1 foot by 1 foot.
To obtain the square inches or square feet within a flat square or flat rectangular area, multiply the length of one side by the length of the other side. To obtain the cubic inches or cubic feet within a three-dimensional square or three-dimensional rectangular area, multiply the width of the area by the length of the area by the depth of the area. Below are several practice problems to illustrate this procedure.
1. A mixed media filter is 10 feet wide by 15 feet long. How many square feet are in the surface area of the filter? Answer: 10 feet x 15 feet = 150 square feet.
2. A sedimentation basin is 12 feet wide by 60 feet long by 30 feet deep. How many cubic feet of space is in the basin? Answer: 12 feet x 60 feet x 30 feet = 21,600 cubic feet
3. How many square inches are in a rectangle that is 3 feet by 5 inches? Answer: 36 inches x 5 inches = 180 square inches (Don't forget to convert the feet to inches.)
In addition to calculating square and cubic areas in straight-sided figures, introductory-level water treatment students must be able to calculate the area within a circle. Since a circle is always flat, versus a sphere which is three-dimensional, the area within a circle will always be expressed in the units that are used to express flat area - namely, SQUARE inches and SQUARE feet. Introductory-level water treatment students do not need to know the formula for calculating cubic inches and cublic feet within sphere-shaped objects.
The very clever formula for calculating area within a circle was devised over two thousand years ago by one of the greatest mathematicians in the ancient world. The simplified version of this formula, which all water treatment schools in the United States teach, is "D2 x .785". D2 is the diameter of the circle squared. To square a number, just multiply the number times itself.
The diameter of a circle might best be described as the full width of the circle. The radius of a circle is one-half of the full diameter, much like the spoke on a bicycle wheel is one-half the full width of the circular rim. Sometimes, water treatment math problems only give the length of a circle's radius. To obtain the circle's diameter, just multiply the radius by two.
There are three primary applications on introductory-level water treatment exams that require the use of the formula to calculate area within a circle. The first application involves the calculation of area within a circular basin, such as a solids-contact basin. The second application involves the calculation of area within a structure that has a round opening, such as a tube, hose, or pipe. The third application involves the calculation of area within a circular bulk storage tank or a circular day tank.
Below are two practice problems involving the area within a circle. Study these examples carefully.
1. A large circular basin is 32 feet across and 50 feet deep. How many cubic feet of water does the basin hold?
ANSWER: The first step is to find the area in square feet for the flat circle that forms the top surface of the basin. To do this, square the 32-foot diameter and multiply the answer by .785. (32 x 32 = 1024 x .785 = 803.84 square feet.) Next, multiply this answer by the depth of the basin to obtain the three-dimensional cubic area within the basin. 803.84 x 50 = 40,192 cubic feet.
2. How many cubic inches of water does a 100-foot length of 8-inch pipe hold?
ANSWER: First, find the flat area that forms the surface opening of the 8-inch pipe. (8 x 8 = 64 x .785 = 50.24 square inches.) Next, because inches can only be multipled by inches, convert the 100-foot length to inches. (100 x 12 inches = 1200 inches.) Now, multiply the flat surface area by the three-dimensional length. (50.24 square inches x 1200 inches of depth = 60,288 cubic inches.) The 100-foot segment of 8-inch pipe in the question will hold 60,288 cubic inches of water.
On all introductory-level water treatment examinations in the United States, area is expressed in terms of inches and feet. The two key words, which indicate area and which indicate the scope of the area, are square and cubic. When area is expressed in inches, it will be designated as square inches or cubic inches. When area is expressed in feet, it will be designated as square feet or cubic feet.
A square measurement of area is always flat. To remember that the term square means flat, think of ordinary household carpeting, which is flat and which is usually measured in "square feet". A square foot is always 1 foot by 1 foot square. A square inch is always 1 inch by 1 inch square.
A cubic inch or cubic foot is always shaped like a perfect cube with the width, length, and height being exactly equal. A cubic inch is a cube that is 1 inch by 1 inch by 1 inch. A cubic foot is a cube that is 1 foot by 1 foot by 1 foot.
To obtain the square inches or square feet within a flat square or flat rectangular area, multiply the length of one side by the length of the other side. To obtain the cubic inches or cubic feet within a three-dimensional square or three-dimensional rectangular area, multiply the width of the area by the length of the area by the depth of the area. Below are several practice problems to illustrate this procedure.
1. A mixed media filter is 10 feet wide by 15 feet long. How many square feet are in the surface area of the filter? Answer: 10 feet x 15 feet = 150 square feet.
2. A sedimentation basin is 12 feet wide by 60 feet long by 30 feet deep. How many cubic feet of space is in the basin? Answer: 12 feet x 60 feet x 30 feet = 21,600 cubic feet
3. How many square inches are in a rectangle that is 3 feet by 5 inches? Answer: 36 inches x 5 inches = 180 square inches (Don't forget to convert the feet to inches.)
In addition to calculating square and cubic areas in straight-sided figures, introductory-level water treatment students must be able to calculate the area within a circle. Since a circle is always flat, versus a sphere which is three-dimensional, the area within a circle will always be expressed in the units that are used to express flat area - namely, SQUARE inches and SQUARE feet. Introductory-level water treatment students do not need to know the formula for calculating cubic inches and cublic feet within sphere-shaped objects.
The very clever formula for calculating area within a circle was devised over two thousand years ago by one of the greatest mathematicians in the ancient world. The simplified version of this formula, which all water treatment schools in the United States teach, is "D2 x .785". D2 is the diameter of the circle squared. To square a number, just multiply the number times itself.
The diameter of a circle might best be described as the full width of the circle. The radius of a circle is one-half of the full diameter, much like the spoke on a bicycle wheel is one-half the full width of the circular rim. Sometimes, water treatment math problems only give the length of a circle's radius. To obtain the circle's diameter, just multiply the radius by two.
There are three primary applications on introductory-level water treatment exams that require the use of the formula to calculate area within a circle. The first application involves the calculation of area within a circular basin, such as a solids-contact basin. The second application involves the calculation of area within a structure that has a round opening, such as a tube, hose, or pipe. The third application involves the calculation of area within a circular bulk storage tank or a circular day tank.
Below are two practice problems involving the area within a circle. Study these examples carefully.
1. A large circular basin is 32 feet across and 50 feet deep. How many cubic feet of water does the basin hold?
ANSWER: The first step is to find the area in square feet for the flat circle that forms the top surface of the basin. To do this, square the 32-foot diameter and multiply the answer by .785. (32 x 32 = 1024 x .785 = 803.84 square feet.) Next, multiply this answer by the depth of the basin to obtain the three-dimensional cubic area within the basin. 803.84 x 50 = 40,192 cubic feet.
2. How many cubic inches of water does a 100-foot length of 8-inch pipe hold?
ANSWER: First, find the flat area that forms the surface opening of the 8-inch pipe. (8 x 8 = 64 x .785 = 50.24 square inches.) Next, because inches can only be multipled by inches, convert the 100-foot length to inches. (100 x 12 inches = 1200 inches.) Now, multiply the flat surface area by the three-dimensional length. (50.24 square inches x 1200 inches of depth = 60,288 cubic inches.) The 100-foot segment of 8-inch pipe in the question will hold 60,288 cubic inches of water.
Lesson Three: Learn These Two Constants Now
Learn these two constants now. Repeat them in your mind several times a day. Learn to recite them with the same ease in which you recite your telephone number.
1. There are 8.34 pounds of water in one gallon of water.
2. There are 7.48 gallons of water in one cubic foot of water.
If you have trouble remembering numbers, think of them as time. Think of 8.34 as 8:34 pm or 7.48 as 7:48 am.
1. There are 8.34 pounds of water in one gallon of water.
2. There are 7.48 gallons of water in one cubic foot of water.
If you have trouble remembering numbers, think of them as time. Think of 8.34 as 8:34 pm or 7.48 as 7:48 am.
Lesson Four: Million-Gallon Notation
On introductory-level water treatment exams, flow is expressed as "number of gallons within a given frame of time." The four primary expressions of flow on introductory-level exams are gallons per minute (gpm), gallons per hour (gph), gallons per day (gpd), and million-gallons per day (MGD) .
When flow is expressed as gpm, gph, or gpd, the number before the gpm, gph, or gpd is written without any type of abbreviated notation. As examples, one thousand gallons per minute is written as "1,000 gpm", one hundred thousand gallons per hour is written as "100,000 gph", and ten million gallons per day is written as "10,000,000 gpd".
When flow is expressed as MGD, the full six-digit million number is never written before the MDG. A flow of ten million gallons per day, correctly written as 10,000,000 gpd, is not correctly written as 10,000,000 MDG. Ten million gallons per day, written in the MGD form, is "10 MDG".
Before beginning the practice portion of this post, it is important to understand the difference between MGD and MG. Both expressions are written with the same type of abbreviated notation, but MGD contains a timeframe (per day), which denotes flow. The term "MG", or million-gallons, has no timeframe attached to it. The term "MG" can express flow, under the right conditions, but it can also be used to denote capacity or quantity, such as a 2.5 MG clearwell or a 1.25 MG elevated tank.
For introductory-level water treatment operators, writing one million gallons per day as 1 MGD is easy. Writing ten and one-half million gallons per day as 10.5 MGD is also easy. But many introductory-level water treatment operators start to experience a problem when they have to write less than one million gallons per day in the MGD form.
To convert gallons into million-gallon notation, multiply the gallons by .000001 and add the abbreviation MGD or MG, depending on the context in which the million-gallon notation is used. The figure .000001 is a decimal point that is followed by five zeros and a 1. Below are several examples to illustrate how easily gallons can be converted into the million-gallon form.
PULL OUT YOUR CALCULATOR AND TRY EACH OF THESE PROBLEMS FOR YOURSELF. YOU MUST KNOW THIS INFORMATION IN ORDER TO ANSWER SEVERAL QUESTIONS ON YOUR EXAM. THESE EXAMPLES ARE PROBABLY MORE COMPLICATED THAN THE QUESTIONS ON THE EXAM, BUT IF YOU CAN PERFORM THESE PROBLEMS, YOU'LL HAVE NO DIFFICULTY AT ALL WITH ANY OF THE MILLION-GALLON CONVERSIONS ON YOUR EXAM.
3,250,000 gallons x .000001 = 3.25 MG
45,000 gallons x .000001 = .045 MG
999,999 gallons x .000001 = .999999 MG
(This is actually one gallon away from being 1 MG.)
1 gallon x .000001 = .000001 MG
(This is 999,999 gallons away from being 1 MG.)
655,000 gallons x .000001 = .655 MG
351,200 gallons x .000001 = .3512 MG
17,600 gallons x .000001 = .0176 MG
18,379,000 gallons x .000001 = 18.379 MG
To check and verify that your calculation is correct, multiply your MG figure by 1,000,000. You should get the original figure with which you started. You will need to use this procedure also if you are asked to convert MG or MGD into gallons.
3.25 MG x 1,000,000 = 3,250,000 gallons
.045 MG x 1,000,000 = 45,000 gallons
.999999 MG x 1,000,000 = 999,999 gallons
.000001 MG x 1,000,000 = 1 gallon
.655 MG x 1,000,000 = 655,000 gallons
.3512 MG x 1,000,000 = 351,200 gallons
.0176 MG x 1,000,000 = 17,600 gallons
18.379 MG x 1,000,000 = 18,379,000 gallons
VERY IMPORTANT NOTE: Whenever a test question or math problem requires an answer that involves a large number of gallons, be sure to ascertain if the answer should be in gallons or in million-gallons. Frequently, introductory-level math questions will ask "how many million gallons" are involved in a certain situation. But test questions can also ask "how many gallons" are involved in a situation. Always assess if questions involving large numbers of gallons are asking for the answer to be in gallons or in million gallons.
When flow is expressed as gpm, gph, or gpd, the number before the gpm, gph, or gpd is written without any type of abbreviated notation. As examples, one thousand gallons per minute is written as "1,000 gpm", one hundred thousand gallons per hour is written as "100,000 gph", and ten million gallons per day is written as "10,000,000 gpd".
When flow is expressed as MGD, the full six-digit million number is never written before the MDG. A flow of ten million gallons per day, correctly written as 10,000,000 gpd, is not correctly written as 10,000,000 MDG. Ten million gallons per day, written in the MGD form, is "10 MDG".
Before beginning the practice portion of this post, it is important to understand the difference between MGD and MG. Both expressions are written with the same type of abbreviated notation, but MGD contains a timeframe (per day), which denotes flow. The term "MG", or million-gallons, has no timeframe attached to it. The term "MG" can express flow, under the right conditions, but it can also be used to denote capacity or quantity, such as a 2.5 MG clearwell or a 1.25 MG elevated tank.
For introductory-level water treatment operators, writing one million gallons per day as 1 MGD is easy. Writing ten and one-half million gallons per day as 10.5 MGD is also easy. But many introductory-level water treatment operators start to experience a problem when they have to write less than one million gallons per day in the MGD form.
To convert gallons into million-gallon notation, multiply the gallons by .000001 and add the abbreviation MGD or MG, depending on the context in which the million-gallon notation is used. The figure .000001 is a decimal point that is followed by five zeros and a 1. Below are several examples to illustrate how easily gallons can be converted into the million-gallon form.
PULL OUT YOUR CALCULATOR AND TRY EACH OF THESE PROBLEMS FOR YOURSELF. YOU MUST KNOW THIS INFORMATION IN ORDER TO ANSWER SEVERAL QUESTIONS ON YOUR EXAM. THESE EXAMPLES ARE PROBABLY MORE COMPLICATED THAN THE QUESTIONS ON THE EXAM, BUT IF YOU CAN PERFORM THESE PROBLEMS, YOU'LL HAVE NO DIFFICULTY AT ALL WITH ANY OF THE MILLION-GALLON CONVERSIONS ON YOUR EXAM.
3,250,000 gallons x .000001 = 3.25 MG
45,000 gallons x .000001 = .045 MG
999,999 gallons x .000001 = .999999 MG
(This is actually one gallon away from being 1 MG.)
1 gallon x .000001 = .000001 MG
(This is 999,999 gallons away from being 1 MG.)
655,000 gallons x .000001 = .655 MG
351,200 gallons x .000001 = .3512 MG
17,600 gallons x .000001 = .0176 MG
18,379,000 gallons x .000001 = 18.379 MG
To check and verify that your calculation is correct, multiply your MG figure by 1,000,000. You should get the original figure with which you started. You will need to use this procedure also if you are asked to convert MG or MGD into gallons.
3.25 MG x 1,000,000 = 3,250,000 gallons
.045 MG x 1,000,000 = 45,000 gallons
.999999 MG x 1,000,000 = 999,999 gallons
.000001 MG x 1,000,000 = 1 gallon
.655 MG x 1,000,000 = 655,000 gallons
.3512 MG x 1,000,000 = 351,200 gallons
.0176 MG x 1,000,000 = 17,600 gallons
18.379 MG x 1,000,000 = 18,379,000 gallons
VERY IMPORTANT NOTE: Whenever a test question or math problem requires an answer that involves a large number of gallons, be sure to ascertain if the answer should be in gallons or in million-gallons. Frequently, introductory-level math questions will ask "how many million gallons" are involved in a certain situation. But test questions can also ask "how many gallons" are involved in a situation. Always assess if questions involving large numbers of gallons are asking for the answer to be in gallons or in million gallons.
Lesson Five: Rounding Off Numbers
Very often, the multiple-choice answers on the math portion of water treatment facility operator certification exams are supplied in a "rounded off" form. For this reason, the answers on certification tests may be very close, but not identical, to the answers that appear on your calculator screen. When selecting a multiple-choice answer on your certification exam, choose the multiple-choice answer that is closest to the figure on your calculator screen.
EXAMPLES OF ROUNDED-OFF NUMBERS:
310,020 gallons might be written as 310,000 gallons.
1206 cubic feet might be written as 1200 cubic feet.
65.34 mg/l might be written as 65 mg/l.
8.247 MGD might be written as 8.2 or 8.25 MGD.
Note: Math questions and answers on introductory-level water treatment operator certification exams are not intended to be particularly tricky. There will generally be enough difference in the multiple-choice answers on these exams, so there will be no doubt as to which answer is correct.
EXAMPLES OF ROUNDED-OFF NUMBERS:
310,020 gallons might be written as 310,000 gallons.
1206 cubic feet might be written as 1200 cubic feet.
65.34 mg/l might be written as 65 mg/l.
8.247 MGD might be written as 8.2 or 8.25 MGD.
Note: Math questions and answers on introductory-level water treatment operator certification exams are not intended to be particularly tricky. There will generally be enough difference in the multiple-choice answers on these exams, so there will be no doubt as to which answer is correct.
Lesson Six: Averaging a Set of Numbers
To average a set of numbers, add the numbers together and divide the total by the number of numbers that were added together. Below are a couple of practical examples to illustrate this procedure.
1. A water treatment facility feeds an alum dosage of 45 mg/l on Sunday, Monday, and Tuesday and feeds an alum dosage of 75 mg/l on Wednesday, Thursday, Friday, and Saturday. What was the water treatment facility's average daily alum dosage for the week?
Answer: 45 + 45 + 45 + 75 + 75 + 75 + 75 = 435 divided by 7 = 62 mg/l (rounded off)
2. A water treatment plant has six filters. At the end of the day, the filter hours on these filters are 14.5, 59.0, 71.5, 2.0, 88.5, and 33.0 hours. What is the filter-hour average for these filters?
Answer: 14.5 + 59.0 + 71.5 + 2.0 + 88.5 + 33.0 = 268.5 divided by 6 = 44.75 hours
1. A water treatment facility feeds an alum dosage of 45 mg/l on Sunday, Monday, and Tuesday and feeds an alum dosage of 75 mg/l on Wednesday, Thursday, Friday, and Saturday. What was the water treatment facility's average daily alum dosage for the week?
Answer: 45 + 45 + 45 + 75 + 75 + 75 + 75 = 435 divided by 7 = 62 mg/l (rounded off)
2. A water treatment plant has six filters. At the end of the day, the filter hours on these filters are 14.5, 59.0, 71.5, 2.0, 88.5, and 33.0 hours. What is the filter-hour average for these filters?
Answer: 14.5 + 59.0 + 71.5 + 2.0 + 88.5 + 33.0 = 268.5 divided by 6 = 44.75 hours
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