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 [...]...– THEA MATH REVIEW – Reciprocal Rules There are special rules for the sum and difference of reciprocals The reciprocal of 3 is 1 of x is ᎏxᎏ 1 1 y 1 1 y 1 ᎏᎏ 3 and the reciprocal x ■ y+x If x and y are not... 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 = 15 125 – THEA MATH REVIEW – 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... 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? 126 – THEA MATH REVIEW – 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... percentage of the employees will receive raises? 1 Translate 15 is % of 75 15 = x ϫ 75 2 Solve 15 75x ᎏᎏ = ᎏ ᎏ 75 75 20 = x 20% = x 15 is 20% of 75 Therefore, 20% of the employees will receive raises 127 – THEA MATH REVIEW – 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 (ᎏxᎏ)... 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: employees ᎏᎏ hours 300 800 → ᎏᎏ = ᎏxᎏ 58.5 128 – THEA MATH REVIEW – Cross-multiply to solve: (800)(58.5)= 300x 46,800 ᎏᎏ 300 300x = ᎏ0ᎏ 3 0 156 = x Therefore, it would take 156 hours to train 800 employees ■ Two quantities are said to vary inversely... unit, movement, and work-output Rate is defined as a comparison of two quantities with different units of measure x units ᎏ Rate = ᎏnits yu Examples dollars cost miles ᎏᎏ, ᎏᎏ, ᎏᎏ hour pound hour 129 – THEA MATH REVIEW – C OST P ER U NIT Some problems will require the calculation of unit cost Example If 100 square feet cost $1,000, how much does 1 square foot cost? Total cost ᎏᎏ # of square feet $1,000... 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 130 – THEA MATH REVIEW – W ORK -O UTPUT 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... 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 17 131 – THEA MATH REVIEW – Example 3 What is the next number in the sequence ᎏ4ᎏ, 3, 12, 48? Each number in the sequence can be multiplied by the number 4 to get the next number in the 3 sequence: ᎏ4ᎏ ϫ 4 = 3,... 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 132 – THEA MATH REVIEW – For example, the following four representations are equivalent to the same function: Equation Word Problem Javier has one more than two times the number of books Susanna has y = 2x +... 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 133 – THEA MATH REVIEW – x y x y x y 5 2 –2 5 –2 3 3 –1 –1 6 –1 3 2 0 0 7 0 3 6 1 1 8 1 3 5 4 2 9 2 3 In this table, every x-value {–2, –1, 0, 1, 2, 3} has one corresponding y-value, even though that value is . 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. 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 54 32 5 –1 –2 –3–4 x 5 5 In this table, the x-value of 5 has two corresponding y-values, 2. 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