Thea practice test_5 doc

Then add the 1 pound to the 4 pounds:
4 pounds 25 ounces = 4 pounds + 1 pound 9 ounces = 5 pounds 9 ounces
SUBTRACTION WITH MEASUREMENTS
1. Subtract like units if possible.
2. If not, regroup units to allow for subtraction.
3. Write the answer in simplest form.
For example, 6 pounds 2 ounces subtracted from 9 pounds 10 ounces.
9 lb 10 oz Subtract ounces from ounces.
– 6 lb 2 o
z Then subtract pounds from pounds.
3 lb 8 oz
Sometimes, it is necessary to regroup units when subtracting.
Example
Subtract 3 yards 2 feet from 5 yards 1 foot.
Because 2 feet cannot be taken from 1 foot, regroup 1 yard from the 5 yards and convert the 1 yard
to 3 feet. Add 3 feet to 1 foot. Then subtract feet from feet and yards from yards:
5
4
΋
yd 1
4
΋
ft
– 3 y
d 2 ft
1 yd 2ft
5 yards 1 foot – 3 yards 2 feet = 1 yard 2 feet
MULTIPLICATION WITH MEASUREMENTS
1. Multiply like units if units are involved.
2. Simplify the answer.
Example

Multiply 5 feet 7 inches by 3.
5 ft 7 in Multiply 7 inches by 3, then multiply 5 feet by 3. Keep the units separate.
ϫ 3
15 ft 21 in Since 12 inches = 1 foot, simplify 21 inches.
15 ft 21 in = 15 ft + 1 ft 9 in = 16 ft 9 in
– THEA MATH REVIEW–
115
Example
Multiply 9 feet by 4 yards.
First, decide on a common unit: either change the 9 feet to yards, or change the 4 yards to feet. Both
options are explained below:
Option 1:
To change yards to feet, multiply the number of feet in a yard (3) by the number of yards in this
problem (4).
3 feet in a yard ϫ 4 yards = 12 feet
Then multiply: 9 feet ϫ 12 feet = 108 square feet.
(Note: feet ϫ feet = square feet = ft
2
)
Option 2:
To change feet to yards, divide the number of feet given (9), by the number of feet in a yard (3).
9 feet ÷ 3 feet in a yard = 3 yards
Then multiply 3 yards by 4 yards = 12 square yards.
(Note: yards • yards = square yards = yd
2
)
DIVISION WITH MEASUREMENTS
1. Divide into the larger units first.
2. Convert the remainder to the smaller unit.
3. Add the converted remainder to the existing smaller unit if any.

4. Divide into smaller units.
5. Write the answer in simplest form.
Example
Divide 5 quarts 4 ounces by 4.
1. Divide into the larger unit:
1 qt r 1 qt
4ͤ5
ෆ
q
ෆ
t
ෆ
– 4 qt
1 qt
2. Convert the remainder:
1 qt = 32 oz
3. Add remainder to original smaller unit:
32 oz + 4 oz = 36 oz
– THEA MATH REVIEW–
116
4. Divide into smaller units:
36 oz ÷ 4 = 9 oz
5. Write the answer in simplest form:
1 qt 9 oz
Metric Measurements
The metric system is an international system of measurement also called the decimal system. Converting units
in the metric system is much easier than converting units in the customary system of measurement. However, mak-
ing conversions between the two systems is much more difficult. The basic units of the metric system are the meter,
gram, and liter. Here is a general idea of how the two systems compare:
Metric System Customary System

1 meter A meter is a little more than a yard; it is equal to about 39 inches
1 gram A gram is a very small unit of weight; there are about 30 grams
in one ounce.
1 liter A liter is a little more than a quart.
Prefixes are attached to the basic metric units listed above to indicate the amount of each unit. For exam-
ple, the prefix deci means one-tenth (
ᎏ
1
1
0
ᎏ
); therefore, one decigram is one-tenth of a gram, and one decimeter is
one-tenth of a meter. The following six prefixes can be used with every metric unit:
Kilo Hecto Deka Deci Centi Milli
(k) (h) (dk) (d) (c) (m)
1,000 100 10
ᎏ
1
1
0
ᎏ
ᎏ
1
1
00
ᎏ
ᎏ
1,0
1
00

ᎏ
Examples
■
1 hectometer = 1 hm = 100 meters
■
1 millimeter = 1 mm =
ᎏ
1,0
1
00
ᎏ
meter = .001 meter
■
1 dekagram = 1 dkg = 10 grams
■
1 centiliter = 1 cL* =
ᎏ
1
1
00
ᎏ
liter = .01 liter
■
1 kilogram = 1 kg = 1,000 grams
■
1 deciliter = 1 dL* =
ᎏ
1
1
0

ᎏ
liter = .1 liter
*Notice that liter is abbreviated with a capital letter—L.
– THEA MATH REVIEW–
117
The chart below illustrates some common relationships used in the metric system:
Length Weight Volume
1 km = 1,000 m 1 kg = 1,000 g 1 kL = 1,000 L
1 m = .001 km 1 g = .001 kg 1 L = .001 kL
1 m = 100 cm 1 g = 100 cg 1 L = 100 cL
1 cm = .01 m 1 cg = .01 g 1 cL = .01 L
1 m = 1,000 mm 1 g = 1,000 mg 1 L = 1,000 mL
1 mm = .001 m 1 mg = .001 g 1 mL = .001 L
Conversions within the Metric System
An easy way to do conversions with the metric system is to move the decimal point either to the right or left because
the conversion factor is always ten or a power of ten. Remember, when changing from a large unit to a smaller unit,
multiply. When changing from a small unit to a larger unit, divide.
Making Easy Conversions within the Metric System
When multiplying by a power of ten, move the decimal point to the right, since the number becomes larger. When
dividing by a power of ten, move the decimal point to the left, since the number becomes smaller. (See below.)
To change from a larger unit to a smaller unit, move the decimal point to the right.
→
kilo hecto deka UNIT deci centi milli
←
To change from a smaller unit to a larger unit, move the decimal point to the left.
Example
Change 520 grams to kilograms.
1. Be aware that changing meters to kilometers is going from small units to larger units and, thus,
requires that the decimal point move to the left.
118

Metric Prefixes
An easy way to remember the metric prefixes is to remember the mnemonic: “King Henry Died of
Drinking Chocolate Milk”. The first letter of each word represents a corresponding metric heading from
Kilo down to Milli: ‘King’—Kilo, ‘Henry’—Hecto, ‘Died’—Deka, ‘of’—original unit, ‘Drinking’—Deci,
‘Chocolate’—Centi, and ‘Milk’—Milli.
2. Beginning at the UNIT (for grams), note that the kilo heading is three places away. Therefore, the
decimal point will move three places to the left.
k h dk UNIT d c m
3. Move the decimal point from the end of 520 to the left three places.
520
←
.520
Place the decimal point before the 5: .520
The answer is 520 grams = .520 kilograms.
Example
Ron’s supply truck can hold a total of 20,000 kilograms. If he exceeds that limit, he must buy stabiliz-
ers for the truck that cost $12.80 each. Each stabilizer can hold 100 additional kilograms. If he wants
to pack 22,300,000 grams of supplies, how much money will he have to spend on the stabilizers?
1. First, change 2,300,000 grams to kilograms.
kg hg dkg g dg cg mg
2. Move the decimal point 3 places to the left: 22,300,000 g = 22,300.000 kg = 22,300 kg.
3. Subtract to find the amount over the limit: 22,300 kg – 20,000 kg = 2,300 kg.
4. Because each stabilizer holds 100 kilograms and the supplies exceed the weight limit of the truck by 2,300
kilograms, Ron must purchase 23 stabilizers: 2,300 kg ÷ 100 kg per stabilizer = 23 stabilizers.
5. Each stabilizer costs $12.80, so multiply $12.80 by 23: $12.80 ϫ 23 = $294.40.

Algebra
This section will help in mastering algebraic equations by reviewing variables, cross multiplication, algebraic frac-
tions, reciprocal rules, and exponents. Algebra is arithmetic using letters, called variables, in place of numbers.
By using variables, the general relationships among numbers can be easier to see and understand.

Algebra Terminology
A term of a polynomial is an expression that is composed of variables and their exponents, and coefficients. A vari-
able is a letter that represents an unknown number. Variables are frequently used in equations, formulas, and in
mathematical rules to help illustrate numerical relationships. When a number is placed next to a variable, indi-
cating multiplication, the number is said to be the coefficient of the variable.
– THEA MATH REVIEW–
119
یی ی
یی ی
120
Examples
8c 8 is the coefficient to the variable c.
6ab 6 is the coefficient to both variables, a and b.
THREE KINDS OF
POLYNOMIALS
■
Monomials are single terms that are composed of variables and their exponents and a positive or nega-
tive coefficient. The following are examples of monomials: x,5x,–6y
3
,10x
2
y,7,0.
■
Binomials are two non-like monomial terms separated by + or – signs. The following are examples of
binomials: x + 2, 3x
2
– 5x,–3xy
2
+ 2xy.
■

Trinomials are three non-like monomial terms separated by + or – signs. The following are examples of
trinomials: x
2
+ 2x – 1, 3x
2
– 5x + 4, –3xy
2
+ 2xy – 6x.
■
Monomials, binomials, and trinomials are all examples of polynomials, but we usually reserve the word
polynomial for expressions formed by more three terms.
■
The degree of a polynomial is the largest sum of the terms’ exponents.
Examples
■
The degree of the trinomial x
2
+ 2x – 1 is 2, because the x
2
term has the highest exponent of 2.
■
The degree of the binomial x + 2 is 1, because the x term has the highest exponent of 1.
■
The degree of the binomial –3x
4
y
2
+ 2xy is 6, because the x
4
y

2
term has the highest exponent sum of 6.
LIKE TERMS
If two or more terms have exactly the same variable(s), and these variables are raised to exactly the same expo-
nents, they are said to be like terms. Like terms can be simplified when added and subtracted.
Examples
7x + 3x = 10x
6y
2
– 4y
2
= 2y
2
3cd
2
+ 5c
2
d cannot be simplified. Since the exponent of 2 is on d in 3cd
2
and on c in 5c
2
d, they are not like
terms.
The process of adding and subtracting like terms is called combining like terms. It is important to combine
like terms carefully, making sure that the variables are exactly the same.
Algebraic Expressions
An algebraic expression is a combination of monomials and operations. The difference between algebraic expres-
sions and algebraic equations is that algebraic expressions are evaluated at different given values for variables, while
algebraic equations are solved to determine the value of the variable that makes the equation a true statement.
There is very little difference between expressions and equations, because equations are nothing more than

two expressions set equal to each other. Their usage is subtly different.
– THEA MATH REVIEW–
Example
A mobile phone company charges a $39.99 a month flat fee for the first 600 minutes of calls, with a
charge of $.55 for each minute thereafter.
Write an algebraic expression for the cost of a month’s mobile phone bill:
$39.99 + $.55x,where x represents the number of additional minutes used.
Write an equation for the cost (C) of a month’s mobile phone bill:
C = $39.99 + $.55x,where x represents the number of additional minutes used.
In the above example, you might use the expression $39.99 + $.55x to determine the cost if you are given
the value of x by substituting the value for x. You could also use the equation C = $39.99 + $.55x in the same way,
but you can also use the equation to determine the value of x if you were given the cost.
SIMPLIFYING AND EVALUATING ALGEBRAIC EXPRESSIONS
We can use the mobile phone company example above to illustrate how to simplify algebraic expressions. Alge-
braic expressions are evaluated by a two-step process; substituting the given value(s) into the expression, and then
simplifying the expression by following the order of operations (PEMDAS).
Example
Using the cost expression $39.99 + $.55x, determine the total cost of a month’s mobile phone bill if
the owner made 700 minutes of calls.
Let x represent the number of minutes over 600 used, so in order to find out the difference, subtract
700 – 600; x = 100 minutes over 600 used.
Substitution: Replace x with its value, using parentheses around the value.
$39.99 + $.55x
$39.99 + $.55(100)
Evaluation: PEMDAS tells us to evaluate Parentheses and Exponents first. There is no operation to perform
in the parentheses, and there are no exponents, so the next step is to multiply, and then add.
$39.99 + $.55(100)
$39.99 + $55 = $94.99
The cost of the mobile phone bill for the month is $94.99.
You can evaluate algebraic expressions that contain any number of variables, as long as you are given all of

the values for all of the variables.
– THEA MATH REVIEW–
121
Simple Rules for Working with Linear Equations
A linear equation is an equation whose variables’ highest exponent is 1. It is also called a first-degree equation.
An equation is solved by finding the value of an unknown variable.
1. The equal sign separates an equation into two sides.
2. Whenever an operation is performed on one side, the same operation must be performed on the other
side.
3. The first goal is to get all of the variable terms on one side and all of the numbers (called constants) on
the other side. This is accomplished by undoing the operations that are attaching numbers to the variable,
thereby isolating the variable. The operations are always done in reverse “PEMDAS” order: start by
adding/subtracting, then multiply/divide.
4. The final step often will be to divide each side by the coefficient, the number in front of the variable, leav-
ing the variable alone and equal to a number.
Example
5m + 8 = 48
–8 = –8
ᎏ
5
5
m
ᎏ
=
ᎏ
4
5
0
ᎏ
m = 8

Undo the addition of 8 by subtracting 8 from both sides of the equation. Then undo the multiplication by
5 by dividing by 5 on both sides of the equation. The variable, m, is now isolated on the left side of the equation,
and its value is 8.
Checking Solutions to Equations
To check an equation, substitute the value of the variable into the original equation.
Example
To check the solution of the previous equation, substitute the number 8 for the variable m in
5m + 8 = 48.
5(8) + 8 = 48
40 + 8 = 48
48 = 48
Because this statement is true, the answer m = 8 must be correct.
– THEA MATH REVIEW–
122
ISOLATING VARIABLES USING FRACTIONS
Working with equations that contain fractions is almost exactly the same as working with equations that do not
contain variables, except for the final step. The final step when an equation has no fractions is to divide each side
by the coefficient. When the coefficient of the variable is a fraction, you will instead multiply both sides by the recip-
rocal of the coefficient. Technically, you could still divide both sides by the coefficient, but that involves division
of fractions which can be trickier.
Example
ᎏ
2
3
ᎏ
m +
ᎏ
1
2
ᎏ

=12
–
ᎏ
1
2
ᎏ
=–
ᎏ
1
2
ᎏ
ᎏ
2
3
ᎏ
m =11
ᎏ
1
2
ᎏ
ᎏ
3
2
ᎏ
•
ᎏ
2
3
ᎏ
m =11

ᎏ
1
2
ᎏ
•
ᎏ
3
2
ᎏ
ᎏ
3
2
ᎏ
•
ᎏ
2
3
ᎏ
m =
ᎏ
2
2
3
ᎏ
•
ᎏ
3
2
ᎏ
m =

ᎏ
6
4
9
ᎏ
Undo the addition of
ᎏ
1
2
ᎏ
by subtracting
ᎏ
1
2
ᎏ
from both sides of the equation. Multiply both sides by the recip-
rocal of the coefficient. Convert the 11
ᎏ
1
2
ᎏ
to an improper fraction to facilitate multiplication. The variable m is now
isolated on the left side of the equation, and its value is
ᎏ
6
4
9
ᎏ
.
Equations with More than One Variable

Equations can have more than one variable. Each variable represents a different value, although it is possible that
the variables have the same value.
Remember that like terms have the same variable and exponent. All of the rules for working with variables
apply in equations that contain more than one variable, but you must remember not to combine terms that are
not alike.
Equations with more than one variable cannot be “solved,”because if there is more than one variable in an equa-
tion there is usually an infinite number of values for the variables that would make the equation true. Instead, we are
often required to “solve for a variable,” which instead means to isolate that variable on one side of the equation.
Example
Solve for y in the equation 2x + 3y = 5.
There are an infinite number of values for x and y that that satisfy the equation. Instead, we are
asked to isolate y on one side of the equation.
2x + 3y =5
– 2x =– 2x
ᎏ
3
3
y
ᎏ
=
ᎏ
–2x
3
+5
ᎏ
y =
ᎏ
–2x
3
+5

ᎏ
– THEA MATH REVIEW–
123
Cross Multiplying
Since algebra uses percents and proportions, it is necessary to learn how to cross multiply. You can solve an equa-
tion that sets one fraction equal to another by cross multiplication. Cross multiplication involves setting the cross
products of opposite pairs of terms equal to each other.
Example
ᎏ
1
x
0
ᎏ
=
ᎏ
1
7
0
0
0
ᎏ
100x = 700
ᎏ
1
1
0
0
0
0
x

ᎏ
=
ᎏ
7
1
0
0
0
0
ᎏ
x =7
Algebraic Fractions
Working with algebraic fractions is very similar to working with fractions in arithmetic. The difference is that alge-
braic fractions contain algebraic expressions in the numerator and/or denominator.
Example
A hotel currently has only one-fifth of their rooms available. If x represents the total number of
rooms in the hotel, find an expression for the number of rooms that will be available if another
tenth of the total rooms are reserved.
Since x represents the total number of rooms,
ᎏ
5
x
ᎏ
(or
ᎏ
1
5
ᎏ
x) represents the number of available rooms.
One tenth of the total rooms in the hotel would be represented by the fraction

ᎏ
1
x
0
ᎏ
. To find the new
number of available rooms, find the difference:
ᎏ
5
x
ᎏ
–
ᎏ
1
x
0
ᎏ
.
Write
ᎏ
5
x
ᎏ
–
ᎏ
1
x
0
ᎏ
as a single fraction.

Just like in arithmetic, the first step is to find the LCD of 5 and 10, which is 10. Then change each
fraction into an equivalent fraction that has 10 as a denominator.
ᎏ
5
x
ᎏ
–
ᎏ
1
x
0
ᎏ
=
ᎏ
5
x(
(
2
2
)
)
ᎏ
–
ᎏ
1
x
0
ᎏ
=
ᎏ

1
2
0
x
ᎏ
–
ᎏ
1
x
0
ᎏ
=
ᎏ
1
x
0
ᎏ
Therefore,
ᎏ
1
x
0
ᎏ
rooms will be available after another tenth of the rooms are reserved.
– THEA MATH REVIEW–
124
Reciprocal Rules
There are special rules for the sum and difference of reciprocals. The reciprocal of 3 is
ᎏ
1

3
ᎏ
and the reciprocal
of x is
ᎏ
1
x
ᎏ
.
■
If x and y are not 0, then
ᎏ
1
x
ᎏ
+
ᎏ
1
y
ᎏ
=
ᎏ
x
y
y
ᎏ
+
ᎏ
x
x

y
ᎏ
=
ᎏ
y
x
+
y
x
ᎏ
.
■
If x and y are not 0, then
ᎏ
1
x
ᎏ
–
ᎏ
1
y
ᎏ
=
ᎏ
x
y
y
ᎏ
–
ᎏ

x
x
y
ᎏ
=
ᎏ
y
x
–
y
x
ᎏ
.
Translating Words into Numbers
The most important skill needed for word problems is being able to translate words into mathematical operations.
The following will be helpful in achieving this goal by providing common examples of English phrases and their
mathematical equivalents.
Phrases meaning addition: increased by; sum of; more than; exceeds by.
Examples
A number increased by five: x + 5.
The sum of two numbers: x + y.
Ten more than a number: x + 10.
Phrases meaning subtraction: decreased by; difference of; less than; diminished by.
Examples
10 less than a number: x – 10.
The difference of two numbers: x – y.
Phrases meaning multiplication: times; times the sum/difference; product; of.
Examples
Three times a number: 3x.
Twenty percent of 50: 20% ϫ 50.

Five times the sum of a number and three: 5(x + 3).
Phrases meaning “equals”: is; result is.
Examples
15 is 14 plus 1: 15 = 14 + 1.
10 more than 2 times a number is 15: 2x + 10 = 15.
– THEA MATH REVIEW–
125
Assigning Variables in Word Problems
It may be necessary to create and assign variables in a word problem. To do this, first identify any knowns and
unknowns. The known may not be a specific numerical value, but the problem should indicate something about
its value. Then let x represent the unknown you know the least about.
Examples
Max has worked for three more years than Ricky.
Unknown: Ricky’s work experience = x
Known: Max’s experience is three more years = x + 3
Heidi made twice as many sales as Rebecca.
Unknown: number of sales Rebecca made = x
Known: number of sales Heidi made is twice Rebecca’s amount = 2x
There are six less than four times the number of pens than pencils.
Unknown: the number of pencils = x
Known: the number of pens = 4x – 6
Todd has assembled five more than three times the number of cabinets that Andrew has.
Unknown: the number of cabinets Andrew has assembled = x
Known: the number of cabinets Todd has assembled is five more than 3 times the number Andrew
has assembled = 3x + 5
Percentage Problems
To solve percentage problems, determine what information has been given in the problem and fill this informa-
tion into the following template:
____ is ____% of ____
Then translate this information into a one-step equation and solve. In translating, remember that is trans-

lates to “=” and of translates to “ϫ”. Use a variable to represent the unknown quantity.
Examples
A) Finding a percentage of a given number:
In a new housing development there will be 50 houses; 40% of the houses must be completed in the
first stage. How many houses are in the first stage?
– THEA MATH REVIEW–
126
1. Translate.
____ is 40% of 50.
x is .40 ϫ 50.
2. Solve.
x = .40 ϫ 50
x = 20
20 is 40% of 50. There are 20 houses in the first stage.
B) Finding a number when a percentage is given:
40% of the cars on the lot have been sold. If 24 were sold, how many total cars are there on the lot?
1. Translate.
24 is 40% of ____.
24 = .40 ϫ x.
2. Solve.
ᎏ
.
2
4
4
0
ᎏ
=
ᎏ
.

.
4
4
0
0
x
ᎏ
60 = x
24 is 40% of 60. There were 60 total cars on the lot.
C) Finding what percentage one number is of another:
Matt has 75 employees. He is going to give 15 of them raises. What percentage of the employees will
receive raises?
1. Translate.
15 is ____% of 75.
15 = x ϫ 75.
2. Solve.
ᎏ
1
7
5
5
ᎏ
=
ᎏ
7
7
5
5
x
ᎏ

.20 = x
20% = x
15 is 20% of 75. Therefore, 20% of the employees will receive raises.
– THEA MATH REVIEW–
127
Problems Involving Ratio
A ratio is a comparison of two quantities measured in the same units. It is symbolized by the use of a colon—x:y.
Ratios can also be expressed as fractions (
ᎏ
x
y
ᎏ
) or using words (x to y).
Ratio problems are solved using the concept of multiples.
Example
A bag contains 60 screws and nails. The ratio of the number of screws to nails is 7:8. How many of
each kind are there in the bag?
From the problem, it is known that 7 and 8 share a multiple and that the sum of their product is 60.
Whenever you see the word ratio in a problem, place an “x” next to each of the numbers in the ratio,
and those are your unknowns.
Let 7x = the number of screws.
Let 8x = the number of nails.
Write and solve the following equation:
7x + 8x =60
ᎏ
1
1
5
5
x

ᎏ
=
ᎏ
6
1
0
5
ᎏ
x =4
Therefore, there are (7)(4) = 28 screws and (8)(4) = 32 nails.
Check: 28 + 32 = 60 screws,
ᎏ
2
3
8
2
ᎏ
=
ᎏ
7
8
ᎏ
.
Problems Involving Variation
Variation is a term referring to a constant ratio in the change of a quantity.
■
Two quantities are said to vary directly if their ratios are constant. Both variables change in an equal
direction. In other words, two quantities vary directly if an increase in one causes an increase in the other.
This is also true if a decrease in one causes a decrease in the other.
Example

If it takes 300 new employees a total of 58.5 hours to train, how many hours of training will it take
for 800 employees?
Since each employee needs about the same amount of training, you know that they vary directly.
Therefore, you can set the problem up the following way:
→
ᎏ
5
3
8
0
.
0
5
ᎏ
=
ᎏ
80
x
0
ᎏ
employees
ᎏᎏ
hours
– THEA MATH REVIEW–
128
Cross-multiply to solve:
(800)(58.5)= 300x
ᎏ
46
3

,
0
8
0
00
ᎏ
=
ᎏ
3
3
0
0
0
0
x
ᎏ
156 = x
Therefore, it would take 156 hours to train 800 employees.
■
Two quantities are said to vary inversely if their products are constant. The variables change in opposite
directions. This means that as one quantity increases, the other decreases, or as one decreases, the other
increases.
Example
If two people plant a field in six days, how many days will it take six people to plant the same field?
(Assume each person is working at the same rate.)
As the number of people planting increases, the days needed to plant decreases. Therefore, the
relationship between the number of people and days varies inversely. Because the field remains
constant, the two products can be set equal to each other.
2 people ϫ 6 days = 6 people ϫ x days
2 ϫ 6=6x

ᎏ
1
6
2
ᎏ
=
ᎏ
6
6
x
ᎏ
2=x
Thus, it would take 6 people 2 days to plant the same field.
Rate Problems
In general, there are three different types of rate problems likely to be encountered in the workplace: cost per unit,
movement, and work-output. Rate is defined as a comparison of two quantities with different units of measure.
Rate =
ᎏ
x
y u
u
n
n
i
i
t
t
s
s
ᎏ

Examples
ᎏ
d
h
o
o
ll
u
a
r
rs
ᎏ
,
ᎏ
po
co
u
s
n
t
d
ᎏ
,
ᎏ
m
ho
il
u
e
r

s
ᎏ
– THEA MATH REVIEW–
129
COST
PER UNIT
Some problems will require the calculation of unit cost.
Example
If 100 square feet cost $1,000, how much does 1 square foot cost?
=
ᎏ
$
1
1
0
,
0
00
ft
0
2
ᎏ
= $10 per square foot
MOVEMENT
In working with movement problems, it is important to use the following formula:
(Rate)(Time) = Distance
Example
A courier traveling at 15 mph traveled from his base to a company in
ᎏ
1

4
ᎏ
of an hour less than it took
when the courier traveled 12 mph. How far away was his drop off?
First, write what is known and unknown.
Unknown: time for courier traveling 12 mph = x.
Known: time for courier traveling 15 mph = x –
ᎏ
1
4
ᎏ
.
Then, use the formula (Rate)(Time) = Distance to find expressions for the distance traveled at each
rate:
12 mph for x hours = a distance of 12x miles.
15 miles per hour for x –
ᎏ
1
4
ᎏ
hours = a distance of 15x –
ᎏ
1
4
5
ᎏ
miles.
The distance traveled is the same, therefore, make the two expressions equal to each other:
12x =15x – 3.75
–15x = –15x

ᎏ
–
–
3
3
x
ᎏ
=
ᎏ
–3
–
.
3
75
ᎏ
x = 1.25
Be careful, 1.25 is not the distance; it is the time. Now you must plug the time into the formula
(Rate)(Time) = Distance. Either rate can be used.
12x = distance
12(1.25) = distance
15 miles = distance
Total cost
ᎏᎏ
# of square feet
– THEA MATH REVIEW–
130
WORK-OUTPUT
Work-output problems are word problems that deal with the rate of work. The following formula can be used on
these problems:
(Rate of Work)(Time Worked) = Job or Part of Job Completed

Example
Danette can wash and wax 2 cars in 6 hours, and Judy can wash and wax the same two cars in 4
hours. If Danette and Judy work together, how long will it take to wash and wax one car?
Since Danette can wash and wax 2 cars in 6 hours, her rate of work is
ᎏ
6
2
h
c
o
a
u
rs
rs
ᎏ
, or one car every three
hours. Judy’s rate of work is therefore,
ᎏ
4
2
h
c
o
a
u
rs
rs
ᎏ
, or one car every two hours. In this problem, making a
chart will help:

Rate Time = Part of job completed
Danette
ᎏ
1
3
ᎏ
x =
ᎏ
1
3
ᎏ
x
Judy
ᎏ
1
2
ᎏ
x =
ᎏ
1
2
ᎏ
x
Since they are both working on only one car, you can set the equation equal to one:
Danette’s part + Judy’s part = 1 car:
ᎏ
1
3
ᎏ
x +

ᎏ
1
2
ᎏ
x = 1
Solve by using 6 as the LCD for 3 and 2 and clear the fractions by multiplying by the LCD:
6(
ᎏ
1
3
ᎏ
x) + 6(
ᎏ
1
2
ᎏ
x) = 6(1)
2x + 3x =6
ᎏ
5
5
x
ᎏ
=
ᎏ
6
5
ᎏ
x =1
ᎏ

1
5
ᎏ
Thus, it will take Judy and Danette 1
ᎏ
1
5
ᎏ
hours to wash and wax one car.
Patterns and Functions
The ability to detect patterns in numbers is a very important mathematical skill. Patterns exist everywhere in nature,
business, and finance.
When you are asked to find a pattern in a series of numbers, look to see if there is some common number
you can add, subtract, multiply, or divide each number in the pattern by to give you the next number in the series.
For example, in the sequence 5, 8, 11, 14 . . . you can add three to each number in the sequence to get the
next number in the sequence. The next number in the sequence is 17.
– THEA MATH REVIEW–
131
Example
What is the next number in the sequence
ᎏ
3
4
ᎏ
,3,12,48?
Each number in the sequence can be multiplied by the number 4 to get the next number in the
sequence:
ᎏ
3
4

ᎏ
ϫ 4 = 3, 3 ϫ 4 = 12, 12 ϫ 4 = 48, so the next number in the sequence is 48 ϫ 4 = 192.
Sometimes it is not that simple. You may need to look for a combination of multiplying and adding, divid-
ing and subtracting, or some combination of other operations.
Example
What is the next number in the sequence 0, 1, 2, 5, 26?
Keep trying various operations until you find one that works. In this case, the correct procedure is
to square the term and add 1: 0
2
+ 1 = 1, 1
2
+ 1 = 2, 2
2
+ 1 = 5, 5
2
+ 1 = 26, so the next number in
the sequence is 26
2
+ 1 = 677.
P
ROPERTIES OF FUNCTIONS
A function is a relationship between two variables x and y where for each value of x, there is one and only one
value of y. Functions can be represented in four ways:
■
a table or chart
■
an equation
■
a word problem
■

a graph
– THEA MATH REVIEW–
132
For example, the following four representations are equivalent to the same function:
Helpful hints for determining if a relation is a function:
■
If you can isolate y in terms of x using only one equation, it is a function.
■
If the equation contains y
2
, it will not be a function.
■
If you can draw a vertical line anywhere on a graph such that it touches the graph in more than one place,
it is not a function.
■
If there is a value for x that has more than one y-value assigned to it, it is not a function.
Word Problem
Table
Graph
Equation
Javier has one more than two times the
number of books Susanna has.
y = 2x + 1
x
y
–3
–5
–2
–3
–1

–1
0
1
1
3
2
5
1
4
3
2
–5
–1
–2
–3
–4
1
5
4
32
–5
–1
–2
–3–4
5
y
x
– THEA MATH REVIEW–
133
In this graph, there

is at least one vertical
line that can be drawn
(the dotted line) that intersects
the graph in more than one
place. This is not a function.
In this graph, there is no
vertical line that can be drawn
that intersects the graph
in more than one place.
This is a function.
y
1
4
3
2
–5
–1
–2
–3
–4
1
5
4
32
–5
–1–2–3–4
x
y
1
4

3
2
–5
–1
–2
–3
–4
1
5432
–5
–1
–2
–3–4
x
5
5
In this table, the
x-value of 5 has two
corresponding y-values,
2 and 4. Therefore,
it is not a function.
x
y
5
2
3
–1
2
0
6

1
5
4
In this table, every
x-value
{–2, –1, 0, 1, 2, 3}
has one corresponding
y-value. This is
a function.
In this table, every
x-value
{–2, –1, 0, 1, 2, 3}
has one corresponding
y-value, even though
that value is 3
in every case.
This is a function.
x
y
–2
5
–1
6
0
7
1
8
2
9
x

y
–2
3
–1
3
0
3
1
3
2
3
– THEA MATH REVIEW–
134
Examples
x = 5 Contains no variable y, so you cannot isolate y. This is not a function.
2x + 3y = 5 Isolate y:
2x + 3y =5
–2x –2x
ᎏ
3
3
y
ᎏ
=
ᎏ
–2x
3
+5
ᎏ
y =–

ᎏ
3
2
x
ᎏ
+
ᎏ
5
3
ᎏ
This is a linear function, of the form y = mx + b.
x
2
+ y
2
= 36 Contains y
2
, so it is not a function.
|y| = 5 There is no way to isolate y with a single equation, therefore it is not a function.
FUNCTION
NOTATION
Instead of using the variable y, often you will see the variable f(x). This is shorthand for “function of x” to auto-
matically indicate that an equation is a function. This can be confusing; f(x) does not indicate two variables f and
x multiplied together, it is a notation that means the single variable y.
Although it may seem that f(x) is not an efficient shorthand (it has more characters than y), it is very elo-
quent way to indicate that you are being given expressions to evaluate. For example, if you are given the equation
f(x) = 5x – 2, and you are being asked to determine the value of the equation at x = 2, you need to write “evalu-
ate the equation f(x) = 5x – 2 when x = 2.” This is very wordy. With function notation, you only need to write
“determine f(2).” The x in f(x) is replaced with a 2, indicating that the value of x is 2. This means that f(2) = 5(2)
– 2 = 10 – 2 = 8.

All you need to do when given an equation f(x) and told to evaluate f(value), replace the value for every occur-
rence of x in the equation.
Example
Given the equation f(x) = 2x
2
+ 3x + 1, determine f(0) and f(–1).
f(0) means replace the value 0 for every occurrence of x in the equation and evaluate.
f(0) = 2(0)
2
+ 3(0) + 1
= 0 + 0 + 1
=1
f(–1) means replace the value –1 for every occurrence of x in the equation and evaluate.
f(0) = 2(–1)
2
+ 3(–1) + 1
= 2(1) + –3 + 1
= 2 – 3 + 1
=0
– THEA MATH REVIEW–
135
FAMILIES OF
FUNCTIONS
There are a number of different types, or families, of functions. Each function family has a certain equation and
its graph takes on a certain appearance. You can tell what type of function an equation is by just looking at the
equation or its graph.
These are the shapes that various functions have. They can appear thinner or wider, higher or lower, or upside
down.
Systems of Equations
A system of equations is a set of two or more equations with the same solution. Two methods for solving a sys-

tem of equations are substitution and elimination.
SUBSTITUTION
Substitution involves solving for one variable in terms of another and then substituting that expression into the
second equation.
Linear Function
f(x) = mx + b
y = mx + b
Constant Function
f(x) = c
y = c
The equation contains no
variable x.
Quadratic Function
f(x) = ax
2
+ bx + c
y = ax
2
+ bx + c
This is the function name
for a parabola.
Square Root Function
The equation has to
contain a square root
symbol.
Cubic Function
f(x) = ax
3
+ bx
2

+ cx + d
y = ax
3
+ bx
2
+ cx + d
Absolute Value Function
The equation has to have
an absolute value symbol
in it.
– THEA MATH REVIEW–
136
Example
2p + q = 11 and p + 2q = 13
■
First, choose an equation and rewrite it, isolating one variable in terms of the other. It does not matter
which variable you choose.
2p + q = 11 becomes q = 11 – 2p.
■
Second, substitute 11 – 2p for q in the other equation and solve:
p + 2(11 – 2p) = 13
p + 22 – 4p = 13
22 – 3p = 13
22 = 13 + 3p
9 = 3p
p = 3
■
Now substitute this answer into either original equation for p to find q.
2p + q = 11
2(3) + q = 11

6 + q = 11
q = 5
■
Thus, p = 3 and q = 5.
ELIMINATION
The elimination method involves writing one equation over another and then adding or subtracting the like terms
on the same sides of the equal sign so that one letter is eliminated.
Example
x – 9 = 2y and x – 3 = 5y
■
Rewrite each equation in the same form.
x – 9 = 2y becomes x – 2y = 9 and x – 3 = 5y becomes x – 5y = 3.
■
If you subtract the two equations, the x terms will be eliminated, leaving only one variable:
Subtract:
x – 2y = 9
–(x – 5y = 3)
ᎏ
3
3
y
ᎏ
=
ᎏ
6
3
ᎏ
y = 2 is the answer.
– THEA MATH REVIEW–
137

■
Substitute 2 for y in one of the original equations and solve for x.
x – 9 = 2y
x – 9 = 2(2)
x – 9 = 4
x – 9 + 9 = 4 + 9
x = 13
■
The answer to the system of equations is y = 2 and x = 13.
If the variables do not have the same or opposite coefficients as in the example above, adding or subtract-
ing will not eliminate a variable. In this situation, it is first necessary to multiply one or both of the equations by
some constant or constants so that the coefficients of one of the variables are the same or opposite. There are many
different ways you can choose to do this.
Example
3x + y = 13
x + 6y = –7
We need to multiply one or both of the equations by some constant that will give equal or opposite coeffi-
cients of one of the variable. One way to do this is to multiply every term in the second equation by –3.
3x + y = 13
–3(x + 6y = –7) → –3x – 18y = 21
Now if you add the two equations, the “x” terms will be eliminated, leaving only one variable. Continue as
in the example above.
3x + y =13
–3x – 18y =21
ᎏ
–
–
1
1
7

7
y
ᎏ
=
ᎏ
–
3
1
4
7
ᎏ
y = –2 is the answer.
■
Substitute –2 for y in one of the original equations and solve for x.
3x + y =13
3x + (–2) = 13
3x + (–2) + –2 = 13 + –2
3x =11
x =
ᎏ
1
3
1
ᎏ
■
The answer to the system of equations is y = –2 and x =
ᎏ
1
3
1

ᎏ
.
– THEA MATH REVIEW–
138
Inequalities
Linear inequalities are solved in much the same way as simple equations. The most important difference is that
when an inequality is multiplied or divided by a negative number, the inequality symbol changes direction.
Example
10 > 5 so (10)(–3) < (5)(–3)
–30 < –15
SOLVING
LINEAR INEQUALITIES
To solve a linear inequality, isolate the letter and solve the same way as you would in a linear equation. Remem-
ber to reverse the direction of the inequality sign if you divide or multiply both sides of the equation by a nega-
tive number.
Example
If 7 – 2x > 21, find x.
■
Isolate the variable.
7 – 2x > 21
–7 –7
–2x > 14
■
Because you are dividing by a negative number, the direction of the inequality symbol changes direction.
ᎏ
–
–
2
2
x

ᎏ
>
ᎏ
–
14
2
ᎏ
x < –7
■
The answer consists of all real numbers less than –7.
SOLVING COMPOUND INEQUALITIES
To solve an inequality that has the form c < ax + b < d, isolate the letter by performing the same operation on each
part of the equation.
Example
If –10 < –5y – 5 < 15, find y.
■
Add five to each member of the inequality.
–10 + 5 < –5y – 5 + 5 < 15 + 5 – 5 < –5y < 20
■
Divide each term by –5, changing the direction of both inequality symbols:
ᎏ
–
–
5
5
ᎏ
< –
ᎏ
–
5

5
y
ᎏ
<
ᎏ
–
20
5
ᎏ
= 1 > y > –4
The solution consists of all real numbers less than 1 and greater than –4.
– THEA MATH REVIEW–
139