Thea practice test_5 doc

Bothoptions 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 thisproblem 4.. Note: feet feet = square feet = ft2

Then add the 1 pound to the 4 pounds:

4 pounds 25 ounces = 4 pounds + 1 pound 9 ounces = 5 pounds 9 ounces

S UBTRACTION WITH M EASUREMENTS

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 oz 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:

54

 yd 1 ft4

– 3 yd 2 ft

1 yd 2ft

5 yards 1 foot – 3 yards 2 feet = 1 yard 2 feet

M ULTIPLICATION WITH M EASUREMENTS

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

 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

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 Bothoptions 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 thisproblem (4)

3 feet in a yard  4 yards = 12 feet

Then multiply: 9 feet  12 feet = 108 square feet

(Note: feet  feet = square feet = ft2)

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 = yd2)

D IVISION WITH M EASUREMENTS

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 qt45qt

– 4 qt

1 qt

2 Convert the remainder:

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, 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:

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 (110); therefore, one decigram is one-tenth of a gram, and one decimeter isone-tenth of a meter The following six prefixes can be used with every metric unit:

1

1 0

■ 1 deciliter = 1 dL* = 110liter = 1 liter

*Notice that liter is abbreviated with a capital letter—L.

The chart below illustrates some common relationships used in the metric system:

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 becausethe 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 Whendividing 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

An easy way to remember the metric prefixes is to remember the mnemonic: “King Henry Died ofDrinking Chocolate Milk” The first letter of each word represents a corresponding metric heading fromKilo 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, thedecimal point will move three places to the left.

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 ers for the truck that cost $12.80 each Each stabilizer can hold 100 additional kilograms If he wants

stabiliz-to pack 22,300,000 grams of supplies, how much money will he have stabiliz-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  23 = $294.40

 A l g e b r a

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 able is a letter that represents an unknown number Variables are frequently used in equations, formulas, and in

vari-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.

  

  

8c 8 is the coefficient to the variable c.

6ab 6 is the coefficient to both variables, a and b.

T HREE K INDS OF P OLYNOMIALS

■ 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, –6y3, 10x2y, 7, 0.

■ Binomials are two non-like monomial terms separated by + or – signs The following are examples of

binomials: x + 2, 3x2– 5x, –3xy2+ 2xy.

■ Trinomials are three non-like monomial terms separated by + or – signs The following are examples of

trinomials: x2+ 2x – 1, 3x2– 5x + 4, –3xy2+ 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 x2+ 2x – 1 is 2, because the x2term 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 –3x4y2+ 2xy is 6, because the x4y2term has the highest exponent sum of 6

L IKE T ERMS

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.

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 sions and algebraic equations is that algebraic expressions are evaluated at different given values for variables, while

A mobile phone company charges a $39.99 a month flat fee for the first 600 minutes of calls, with acharge 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.

S IMPLIFYING AND E VALUATING A LGEBRAIC E XPRESSIONS

We can use the mobile phone company example above to illustrate how to simplify algebraic expressions braic expressions are evaluated by a two-step process; substituting the given value(s) into the expression, and thensimplifying the expression by following the order of operations (PEMDAS)

Alge-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 ofthe values for all of the variables

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 byadding/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



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

I SOLATING V ARIABLES U SING F RACTIONS

Working with equations that contain fractions is almost exactly the same as working with equations that do notcontain 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 rocal of the coefficient Technically, you could still divide both sides by the coefficient, but that involves division

recip-of fractions which can be trickier

recip-isolated on the left side of the equation, and its value is 649

Equations with More than One Variable

Equations can have more than one variable Each variable represents a different value, although it is possible thatthe variables have the same value

Remember that like terms have the same variable and exponent All of the rules for working with variablesapply in equations that contain more than one variable, but you must remember not to combine terms that arenot alike

Equations with more than one variable cannot be “solved,”because if there is more than one variable in an tion there is usually an infinite number of values for the variables that would make the equation true Instead, we areoften required to “solve for a variable,” which instead means to isolate that variable on one side of the equation

equa-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

3

3y=– 2x3

+ 5



y =–2x3+ 5

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 fracequa-tion equal to another by cross multiplicaequa-tion Cross multiplicaequa-tion involves setting the cross

products of opposite pairs of terms equal to each other

Example

 1

x

0

= 1

7 0

0 0

x

=71

0 0

0 0

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,5x(or 1

5 x) represents the number of available rooms.

One tenth of the total rooms in the hotel would be represented by the fraction 1x0 To find the new

number of available rooms, find the difference:5x– 

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

x(

(

2 2

) )

–  1

x

0



=120x–  1

Reciprocal Rules

There are special rules for the sum and difference of reciprocals The reciprocal of 3 is 13and the reciprocal

of x is 1x

■ If x and y are not 0, then 1x+ 1y= x y y+ x xy= y x +y x

■ If x and y are not 0, then 1x– 1y= x y y– x xy= y x–yx

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 theirmathematical 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%  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.

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 andunknowns 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

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 thefirst stage How many houses are in the first stage?

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?

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 willreceive raises?

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 (x y ) 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:



x = 4

Therefore, there are (7)(4) = 28 screws and (8)(4) = 32 nails

Check: 28 + 32 = 60 screws,2382= 78

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 equaldirection 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:

→300 = 800

employees



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 oppositedirections This means that as one quantity increases, the other decreases, or as one decreases, the otherincreases

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  6 days = 6 people  x days

2  6 = 6x

16 2

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.

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