Study Guide 6th grade summer math packet 05/14/2014 Order of Operations with Decimals - A Performing operations with decimals is similar to performing operations with whole numbers. Operations inside parentheses are performed first. Example 1: (7.2 + 3.4) + (2.31 + 5.352) =? (1) 7.2 + 3.4 = 10.6 and 2.31 + 5.352 = 7.662 (2) 10.6 + 7.662 =? (3) 10.6 + 7.662 = 18.262 Step 1: Perform all operations within parentheses. Add 7.2 + 3.4 = 10.6. And add 2.31 + 5.352 = 7.662. Step 2: Rewrite the equation with the new numbers in place of the parentheses. Step 3: Add 10.6 and 7.662. The answer is 18.262 Example 2: (13.295-1.62) - (3.5625 + 5.92) =? (1) 13.295-1.62 = 11.675 and 3.5625 + 5.92 = 9.4825 (2) 11.675-9.4825 =? (3) 11.675-9.4825 = 2.1925 Step 1: Perform operations within parentheses. Subtract 13.295-1.62 = 11.675. And add 3.5625 + 5.92 = 9.4825. Step 2: Rewrite the equation with the new numbers in place of the parentheses. Step 3: Subtract 9.4825 from 11.675. The answer is 2.1925. Compare Decimals - C Comparing decimal numbers involves determining which decimal in a set is either the least or the greatest. Real world problems, also called word or story problems, present decimal problems in text format. Students are required to read passages and determine the questions being asked. They should then identify the elements needed to solve the problem, decide on the correct method to solve each problem, and find a solution. Story problems are often difficult for students to master. It may be beneficial to confirm that the student is comfortable with his or her ability to compare decimals outside of a story problem context. To review this concept with the student, use the example below. Remember, when comparing numbers, line the decimal points up and compare the corresponding columns beginning with the highest place Page 1
value. Comparing Decimals Without a Context: Example 1: Put the decimals below in order from least to greatest: 2.4091, 2.904, 2.09, 2.049 Solution: Step 1: Write the decimals in vertical form, lining up the decimal points. Step 2: Look at the numbers from left to right, stopping when any of the numbers are different. The two is the same in all of the numbers. In the tenths place (to the right of the decimal point), there is 4, 9, 0, and 0. Step 3: The greatest number is 9, so 2.904 is the greatest number in the set. The next highest is the 4, so 2.4091 is next. Step 4: Look at the two numbers with a 0 in the tenths place. In the hundredths place, one number has a 9 and the other number has a 4. Therefore, 2.09 is the third greatest, and 2.049 is the least. The numbers in order from least to greatest are: 2.049, 2.09, 2.4091, and 2.904. Answer: 2.049, 2.09, 2.4091, 2.904. A common mistake that students make is to assume that the numbers after the decimal place follow the rules of counting numbers (1, 2, 3, etc.). Specifically, students assume that the more digits there are in the number, the greater the number is. For example, students know that 215 is greater than 23 because 215 has more digits than 23. However, with decimals, 0.215 is less than 0.23, even though 0.215 contains more digits. Comparing Decimals in a Real World Context: When the student appears to have mastered the skill of comparing decimals, increase the difficulty of the problem by putting the decimals in a real world context. It may be beneficial to create problems that relate to his or her daily activities, such as sports or measurements. Then, help the student determine the correct process to find the solution. Example 2: Kendall is trying to reduce his weight class for wrestling. He wants to determine what time of day he should be weighed so that he will weigh the least. He weighed himself throughout the day and his results are in the table below. At what time did Kendall weigh the least? Page 2
Solution: Step 1: Line up all of the decimal points and move from left to right. The numbers are all identical until the hundredths place. That is, the 70.3 part of the number is common to all of the numbers. Step 2: Look at the hundredths place. There are three numbers with a 0 and one number with a 1 in the hundredths place. Therefore, the 70.31 is the greatest number, and it is eliminated because the question is asking for the smallest weight. Step 3: Look at the thousandths place since all of the other numbers have a 0 in the hundredths place. The thousandths place contains a 6, a 7, and an 8 for the remaining numbers. Regardless of the number that follows it, 6 is the smallest number of the three that are remaining, so 70.3068 kg is the least Kendall weighed during the day. Answer: Kendall weighed the least at 11:30 A.M. Example 3: Naomi's baseball team has won more games than it has lost for four years in a row. Her team has had almost the same record every year as shown in the table below. In what year was the team's record the highest? Solution: Step 1: Line up all of the decimal points and move from left to right. The numbers are all identical until the hundredths place. That is, the 0.6 part of the number is common to all of the numbers. Step 2: Look at the numbers in the hundredths place. There are two numbers with a 3, one with a 2, and one with a 4 in the hundredths place. Therefore, the highest number is 0.64 and the year it occurred was 2000. Answer: The team's record was the highest in 2000. NOTE: When working with real world problems that involve speed, the lowest decimal number is equal to the fastest speed. An activity relating to this skill would be to choose a certain stock from the newspaper or on the Internet (try a search for "stock quotes" and look for one that carries the decimal places out to the hundredths place) and monitor the stock once a day for one week. Determine which day the stock was worth the most and which day it was worth the least. A similar activity would be to have the student track his or her favorite professional sport team's record throughout the season or from year to year. Order Decimals: Story Problem Ordering decimal numbers involves determining which decimals in a set are greater or smaller than the other decimals in the set and ordering them from least to greatest or greatest to least. Real world problems, also called word or story problems, present decimal problems in text format. Story problems require students to read passages and determine the questions being asked. They should then identify the elements needed to solve each problem, decide on the correct method to solve the problem, and find a solution. Page 3
Story problems are often difficult for students to master. It may be beneficial to confirm that the student is comfortable with ordering decimals outside of a real world context. Remember, when comparing numbers, line the decimal points up and compare the corresponding columns beginning with the highest place value. Ordering Decimals Without a Context: Example 1: Put the decimals below in order from least to greatest: 2.4091, 2.904, 2.09, 2.049 Solution: Step 1: Write the decimals in vertical form, lining up the decimal points. Step 2: Look at the numbers from left to right, stopping when any of the numbers are different. The two is the same in all of the numbers. In the tenths place (to the right of the decimal point), there is 4, 9, 0, and 0. Step 3: The greatest number is 9, so 2.904 is the greatest number in the set. The next highest is the 4, so 2.4091 is next. Step 4: Look at the two numbers with a 0 in the tenths place. In the hundredths place, one number has a 9 and the other number has a 4. Therefore, 2.09 is the third greatest, and 2.049 is the least. The numbers in order from least to greatest are: 2.049, 2.09, 2.4091, and 2.904. Answer: 2.049, 2.09, 2.4091, 2.904. A common mistake students make is to assume that the numbers after the decimal place follow the rules of counting numbers (1, 2, 3, etc.). Specifically, students assume that the more digits there are in the number, the greater the number is. For example, students know that 215 is greater than 23 because 215 has more digits than 23. However, with decimals, 0.215 is less than 0.23, even though 0.215 contains more digits. Ordering Decimals in a Real World Context: When the student appears to have mastered the skill of comparing decimals, increase the difficulty of the problem by putting the decimals in a real world context. It may be beneficial to create problems that relate to his or her daily activities, such as sports or measurements. Then, help the student determine the correct process to find the solution. Example 2: Paul is trying to determine which breakfast cereal costs the least per ounce. By his calculations, he determines that Crispy O's cost $0.2073 per ounce, Rice Puffs cost $0.273 per ounce, Sugar Beads cost $0.0273 per ounce, and Frosty Orbs cost $0.07723 per ounce. Order the cereals from most expensive to least expensive. Solution: Step 1: Write the decimals in vertical form, lining up the decimal points. Page 4
Step 2: Look at the numbers from left to right, stopping when any of the numbers are different. The numbers are all the same until the tenths place. In the tenths place, we have two numbers with a 2 and two with a 0. Step 3: Look at the two numbers with the 2 in the tenths place because these are the two largest numbers. In the hundredths place for these two numbers, one has a 0 and the other has a 7. Therefore, 0.273 is the greatest, followed by 0.2073. Step 4: Look at the two numbers with a 0 in the tenths place. In the hundredths place, one has a 2 and the other a 7. Therefore, the third largest number is 0.07723 and the smallest is 0.0273. The numbers in order from most expensive to least expensive are: 0.273, 0.2073, 0.07723, 0.0273. Answer: Rice Puffs, Crispy O's, Frosty Orbs, Sugar Beads NOTE: Although the student is used to seeing dollar amounts written to the hundredths place, when dealing with unit rates (per ounce, per pound, per unit, etc.) decimal values are often extended further. Example 3: The manager at Fizz Cola tests the machine once each hour to make sure it is performing as it should. He has measured the amount of cola the two-liter bottles contained and found the following results. 8:00 A.M. - the machine filled the container with 2.0023 liters of cola 9:00 A.M. - the machine filled the container with 2.0203 liters of cola 10:00 A.M. - the machine filled the container with 1.9708 liters of cola 11:00 A.M. - the machine filled the container with 1.981 liters of cola 12:00 P.M. -the machine filled the container with 2.023 liters of cola Put the times in order so that the time when the bottle was filled with the most cola comes first and the time the bottle was filled with the least cola comes last. Solution: Step 1: Write the decimals in vertical form, lining up the decimal points. Step 2: Look at the numbers from left to right, stopping when any of the numbers are different. Begin by comparing the numbers that have a 2 in the ones place because they are greater than those that have a 1 in the ones place. Step 3: For the three numbers that have a 2 in the ones place, compare the tenths place. All three numbers have a 0 in the tenths place, so look at the hundredths place. The numbers 2.0203 and 2.023 have a 2 in the hundredths place, so they are greater than 2.0023. Since 2.023 has a 3 in the thousandths place, it is greater than 2.0203 which has a 0 in the thousandths place. The order of the greatest numbers is: 2.023, 2.0203, 2.0023. Step 4: Now compare 1.9708 and 1.981. The two numbers are the same until the hundredths place. Since 1.981 has an 8 in the hundredths place, it is greater than 1.9708 which has a 7 in the hundredths place. This means that the order of the amounts is: 2.023, 2.0203, 2.0023, 1.981, 1.9708. Page 5
Answer: 12:00 P.M., 9:00 A.M., 8:00 A.M., 11:00 A.M., 10:00 A.M. NOTE: When working with real world problems that involve speed, the lowest decimal number is equal to the fastest speed. An activity relating to this skill would be to choose a certain stock from the newspaper or on the Internet (try a search for "stock quotes" and look for one that carries the decimal places out to the hundredths place) and monitor the stock once a day for one week. Order the stock prices from least to greatest or greatest to least. A similar activity would be to have the student track his or her favorite professional sport team's record throughout the season or from year to year. Divide Fractions: Story Problems - B Story problems, also called word problems, relate division of fractions and mixed numbers to real world situations. Story problems require students to read passages and determine the question being asked. Students should then identify the elements needed to solve the problem, decide on the correct method to solve each problem, and find a solution. Story problems are often difficult for students to master. It may be beneficial to confirm that the student is comfortable with division of fractions and mixed numbers outside of a real world context. To review this concept with the student, use the examples below. Dividing Fractions: Example 1: Divide and reduce your answer to lowest terms. Step 1: Rewrite the problem as a multiplication problem. Dividing by a fraction is the same as multiplying by its reciprocal. To find the reciprocal, simply switch the numerator and the denominator. In this case, the second fraction, 8/9, becomes 9/8 and the symbol becomes a symbol. Step 2: Multiply the numerators (2 9 = 18) and the denominators (5 8 = 40). Step 3: Reduce the fraction to lowest terms. A fraction is in lowest terms when the numerator and denominator do not have a common factor greater than 1. The numbers 18 and 40 can both be divided by 2, so complete this division to reduce the fraction. *An alternate way to reduce the fraction is to reduce before multiplication is completed. This can be done by finding a numerator and a denominator that are both evenly divisible by the same number, such as the numerator 2 and the denominator 8. Both the 2 and the 8 can be divided by 2. Complete the divisions: 2 2 = 1 and 8 2 = 4, so the new problem would be (1 9)/(5 4) and the answer would be 9/20. Dividing Mixed Numbers: Example 2: Divide and reduce your answer to lowest terms. Page 6
Step 1: Rewrite the mixed numbers as improper fractions. For example, to change 6 3/5 into an improper fraction, multiply the denominator by the whole number (5 6 = 30). Then add that product and the numerator (30 + 3 = 33). The denominator remains the same and 6 3/5 can be written as 33/5. Similarly, 3 2/3 becomes 11/3. Step 2: Rewrite the problem as a multiplication problem. In this case, the second fraction, 11/3, becomes 3/11 and the symbol becomes a symbol. Then, multiply the numerators and denominators. Step 3: Reduce the fraction to lowest terms. The numbers 99 and 55 can both be divided by 11, so complete this division to reduce the fraction. Step 4: Rewrite the fraction as a mixed number. Divide the numerator by the denominator. Write the quotient as the whole number and leave the remainder in the numerator position. For example, 9 divided by 5 is 1 with a remainder of 4. The denominator remains 5. Remember, when the division involves whole numbers, use 1 as the denominator for the whole number. When given a problem that involves dividing a fraction by a mixed number or a mixed number by a fraction, the process is very similar to the above process of dividing two mixed numbers. The only difference is that in Step 1, only one mixed number needs to be converted into an improper fraction because the other number is already in the correct form. Dividing Fractions and Mixed Numbers in a Real World Context: Example 3: Solution: Step 1: Determine the problem that you are trying to solve. For this problem we want to break 27 3/5 feet into pieces that are 1 2/7 feet long. Therefore we need to divide 27 3/5 by 1 2/7. Step 2: Rewrite the mixed numbers as improper fractions. To change 27 3/5 into an improper fraction, multiply the denominator by the whole number (5 27 = 135). Then add that product to the numerator (135 + 3 = 138). The denominator remains the same and 27 3/5 can be written as 138/5. Similarly, 1 2/7 becomes 9/7. Step 3: Rewrite the problem as a multiplication problem. In this case, the second fraction, 9/7, becomes 7/9 and the symbol becomes a symbol. Then, multiply the numerators and denominators. Step 4: Reduce the fraction to lowest terms. The numbers 966 and 45 can both be divided by 3, so complete this division to reduce the fraction. Page 7
Step 5: Rewrite the fraction as a mixed number. Divide the numerator by the denominator. Write the quotient as the whole number and leave the remainder in the numerator position. In this example, 322 divided by 15 is 21 with a remainder of 7. Example 4: Solution: Step 1: Determine the problem that you are trying to solve. For this problem we want to break 10 3/4 hours into 5/12 of an hour pieces. Therefore we need to divide 10 3/4 by 5/12. Step 2: Rewrite the mixed numbers as improper fractions. To change 10 3/4 into an improper fraction, multiply the denominator by the whole number (4 10 = 40). Then, add that product to the numerator (40 + 3 = 43). The denominator remains the same and 10 3/4 can be written as 43/4. Step 3: Rewrite the problem as a multiplication problem. In this case, the second fraction, 5/12, becomes 12/5 and the symbol becomes a symbol. Then, multiply the numerators and denominators. Step 4: Reduce the fraction to lowest terms. The numbers 516 and 20 can both be divided by 4, so complete this division to reduce the fraction. Step 5: Rewrite the fraction as a mixed number. Divide the numerator by the denominator. Write the quotient as the whole number and leave the remainder in the numerator position. In this example, 129 divided by 5 is 25 with a remainder of 4. An that activity will give the student practice with this skill is to have the student start with a standard 8 1/2 11 inch piece of paper. For the short dimension, compute how many 1/4-inch pieces will fit across and then measure with a ruler to check if he or she is correct. Repeat the process using different increments such as 1 2/3 inches, 4/5 of an inch, and 3 5/6 inches. Compute the result first and measure with a ruler to check. Multiply Whole and Mixed Numbers A mixed number is a number written as a whole number followed by a fraction. Students must be comfortable rewriting mixed numbers as improper fractions and vice versa. Improper fractions are fractions in which the numerator is larger than the denominator. This form allows students to easily multiply two fractions, two mixed numbers, two whole numbers, or any combination of the three. Remember: In order to write a mixed number as an improper fraction, multiply the denominator by the whole number and then add the numerator. The result then becomes the numerator and the original denominator remains the same. Page 8
Step 1: Multiply the denominator (9) by the whole number (8) and then add the numerator (5). Step 2: Rewrite the fraction using the result from step one (77) as the numerator and 9 (the original denominator) as the denominator. In order to write an improper fraction as a mixed number, divide the numerator by the denominator. The new quotient becomes the whole number and the remainder becomes the numerator of the mixed number. The denominator remains the same. Step 1: Divide the numerator (77) by the denominator (9). Step 2: Write the mixed number using the whole number quotient, 8, as the whole number and the remainder, 5, as the numerator. The denominator remains the same. Once the student is comfortable with converting mixed fractions to improper fractions and improper fractions to mixed numbers, he or she is ready to move on to multiplying a mixed number by a whole number. Example 1: Multiply. Reduce your answer to lowest terms. Step 1: Rewrite the problem. Step 2: Rewrite 4 7/9 as an improper fraction and 5 as a fraction. Remember, to change a whole number into a fraction, write the whole number as the numerator with a denominator of 1. Step 3: Rewrite the problem using the new forms of 5/1 and 43/9. Step 4: Multiply the numerators and the denominators. Step 5: In order to reduce the answer to lowest terms, the student will need to change the answer of 215/9 back into a mixed number. The first step to doing this is to divide the numerator by the denominator. Step 6: Since 215 divided by 9 is 23 r8, 23 becomes the whole number and 8 becomes the numerator. The denominator will remain 9. The answer 23 and 8/9 cannot be reduced any further since 8 and 9 do not share any common factors. Example 2: Multiply. Reduce your answer to lowest terms. Step 1: Rewrite the problem. Page 9
Step 2: Rewrite 3 4/9 as an improper fraction and 27 as a fraction. Step 3: Rewrite the problem using the new forms of 27/1 and 31/9. Step 4: Before multiplying the numerators and the denominators, use cross cancellation to make the numbers in the problem more manageable. Since the numerator (27) and denominator (9) are both divisible by 9, divide both numbers by 9 and then perform the multiplication. 3 31 = 93 and 1 1 = 1. Step 5: Since 93/1 is a whole number written as a fraction, remove the denominator to write 93 in its whole number form. Answer: 93 ***Cross cancellation is the process of reducing the numbers within a multiplication of fractions problem before multiplying the fractions. If a numerator and a denominator of any of the fractions to be multiplied can be divided by the same number, this division can be performed before the fractions are multiplied. An activity to reinforce the concept of multiplying mixed numbers by a whole number is to create two stacks of flash cards. Write only whole numbers on the cards in one stack and only mixed numbers on the cards in the other stack. Place each stack separately into two brown paper bags. Have the student randomly pull one whole number and one fraction from each bag and calculate the product of the two numbers. Multiply Fractions: Story Problems - B In this skill, story problems, also called word problems, relate multiplication of fractions to actual situations. Operational symbols, such as the multiplication symbol,, are replaced with text. The problems in this skill set require the students to reduce. Many students find story problems challenging. It may be useful to first confirm that students are comfortable with multiplying fractions. Then, they can move on to problems that are presented in story problem context. The following is an example of the multiplication of two fractions with no reducing required. Example 1: Solution: Multiply the numerators (the numbers on the top of the fraction), 7 2 = 14, and the denominators (the numbers on the bottom of the fraction), 9 5 = 45. The solution is 14/45. Answer: The following is an example of the multiplication of two fractions with reducing required. Example 2: Page 10
Step 1: Use cross cancellation to reduce the fractions before multiplying. Look for any numerator that has a common factor with any denominator. In this example, 3 and 9 share the factor 3, so it can be divided into both numbers to reduce them. Perform the divisions. 3 3 = 1 and 9 3 = 3. Also, 5 and 10 share the factor 5, so it can be divided into both numbers to reduce them. Perform the divisions. 5 5 = 1 and 10 5 = 2. Step 2: Multiply the numerators, 1 1 = 1, and the denominators, 2 3 = 6. The solution is 1/6. Answer: An alternative method when multiplying fractions that can be reduced is to multiply through as usual and look for common factors in the resulting fraction. Multiplication of fractions, especially in the context of a story problem, can be confusing for students. It is important for them to remember that when multiplying two fractions between 0 and 1, the product is always a smaller fraction. This result is the opposite of what students are used to with whole number multiplication. The following is an example of the multiplication of two fractions in the context of a story problem with no reducing required. Example 3: Step 1: Determine that multiplication is required to solve the problem. To relate the bottom snowball to the top snowball, it is necessary to relate the bottom snowball to the middle snowball (1/3) and then the middle snowball to the top snowball (2/5). Multiply these two fractions to find the solution. Step 2: Multiply the numerators, 1 2 = 2, and the denominators, 3 5 = 15. The solution is 2/15. The following is an example of multiplying two fractions in the context of a story problem that requires reducing. Example 4: Page 11
Step 1: Determine that multiplication is required to solve the problem. To relate Carrie's whole day to her time spent at lunch and in between classes, it is necessary to relate her whole day to the school day (1/4) and then the school day to the time spent at lunch and in between classes (2/7). The solution needed is 2/7 of 1/4 of her school day. Multiply 2/7 and 1/4 to find the solution. Step 2: Simplify the fractions using cross cancellation. The numerator 2 and the denominator 4 both have a factor of 2 in common. Perform the divisions. 2 2 = 1 and 4 2 = 2. Step 3: Multiply the numerators, 1 1 = 1, and the denominators, 2 7 = 14. The solution is 1/14. An activity to help reinforce this skill is to create scenarios with the student. These can be scenarios such as determining the fraction of the front lawn that has flower beds. Then, determining the fraction of the flower beds with a specific type of flower (such as daffodils). Finally, have the student determine the fraction of the front yard that has the specific type of flower (daffodils). For example, if 1/8 of the front yard has flower beds and 2/3 of the flower beds are filled with daffodils, then 2/24 or 1/12 of the entire front yard is a flower bed with daffodils. Number Machines A function is a rule that turns a number of one set into a number of another set. At this grade level, functions are often referred to as number machines. The idea is that an input value (an initial number) enters the number machine. One or more operations are then performed on the number, and a new number is sent out. The end result is called the output value. This study guide will focus on determining the output values of single operation number machines. Normally, the student will see a number machine represented as a table of input and output values or represented using arrows. The student should first ask himself or herself what operation is being performed on the numbers in column A (or the numbers in front of the arrow) in order for the result to be the numbers in column B (or the numbers after the arrow). The operation could be addition, subtraction, multiplication or division. Once the student has an idea, he or she must test that idea on each number. When the operation is correct for all given values, the student should perform that operation on the number in question. Example 1: A number machine changes the numbers in column A into the numbers in column B. What will the number 17 be changed into? Page 12
(A) 31 (B) 51 (C) 3 (D) 47 (A) is not correct. Although 12 + 24 = 36 in the first row, the rule "add 24" does not hold true with the remaining numbers. (B) is correct. Each number in column A produces the number in column B when multiplied by the number 3. Therefore 17 3 = 51. (C) is not correct. The number 3 is the factor that should be multiplied by the numbers in column A; it is not the product. (D) is not correct. Although 15 + 30 = 45 in the third row, the rule "add 30" does not hold true with the remaining numbers. Example 2: The following number machine changes 70 into 56, 25 into 11, and 39 into 25. What number will 18 be changed into? (A) 4 (B) 14 (C) 24 (D) 32 (A) is correct. Each number before the arrow is turned into the number after the arrow by subtracting the number 14. Therefore 18-14 = 4. (B) is not correct. The number 14 should be subtracted from each input number; it is not the output value. (C) is not correct. Since it has been concluded that the rule for the number machine is - 14 and 18-14 = 4, 24 is not the correct number. (D) is not correct. Although 14, which is the correct value, is being used, the operation of addition is incorrect. A fun way to help the student become familiar with number machines is to have him or her create one. Have the student think of a number and an operation. Provide the student with a random input value (a whole number is suggested, but not necessary). Have the student calculate the output value according to the number machine. To make the activity more difficult, have the number machine use more than one operation on a single input value. The rules become harder to figure out when two or more operations are used. Calculating With Exponents In this study guide, students will learn how to perform basic calculations with exponents. An exponent is a number that represents repeated multiplication. It tells the student how many times the base is used Page 13
as a factor. A factor is a number that is multiplied by another number. For example, the base number 2 with an exponent of 3 is equal to 2 2 2. It is usually written in the following format: Before calculating exponents within an expression, find the equivalent whole number forms of these exponential numbers: The most common error among students learning about exponents is to multiply the base number by the exponent. That is, many students will calculate 8 to the 3rd power as 8 3 = 24, instead of 8 8 8 = 512. Calculating Exponents Within An Expression When working with exponents within an expression, the student must remember the rules for the order of operations. The order of operations can be remembered with the phrase "Please Excuse My Dear Aunt Sally." The student should always perform operations in the following order: P - Parentheses E - Exponents M/D - Multiplication/Division in order from left to right A/S - Addition/Subtraction in order from left to right Example 1: Subtract. 5,000-8 4 = (1) 8 4 = 8 8 8 8 = 4,096 (2) 5,000-4,096 = 904 Step 1: According to the rules for order of operations, calculate the exponent first. Step 2: Subtract 4,096 from 5,000. Answer: 5,000-8 4 = 904 Example 2: Add. 9 2 + 3 4 = (1) 9 9 = 81 (2) 3 3 3 3 = 81 (3) 81 + 81 = 162 Step 1: Since both terms in the expression contain exponents, the student should work from left to right. Calculate 9 2 first. Page 14
Step 2: Next, calculate 3 4. Step 3: Finally, add 81 + 81. Answer: 9 2 + 3 4 =162 To help the student practice calculating exponents try the following activity. Begin with several blank index cards. First, write the numbers 1-10 on ten different cards. Next, write exponent of 1, exponent of 2...up to exponent of 10 on ten different cards. Then, write ten whole numbers of your choice on ten different cards. Finally, write the addition and subtraction symbols on two different cards. (For a challenge you may want to include multiplication and division.) Sort the cards into 4 separate piles (single-digit whole numbers, exponents, whole numbers, and operational symbols). Turn all the cards face down. Have the student randomly select one card from each pile and then combine the numbers and symbols to create an expression. Finally, have the student calculate the various exponential expressions. Add Decimals: Story Problems - C Story problems, also called word problems, relate addition of decimal numbers to actual situations. Operational symbols, such as the addition (+) symbol, are replaced with text. Word problems in this skill also deal with money. Story problems are often very difficult for students to master. It may be beneficial for you to create problems that students can easily relate to, and help the student determine the correct formulas. Example: Fred ran 8.971 miles on Saturday and 5.363 miles on Sunday. How many miles did Fred run in all? Step 1: Identify the equation and remember to line up the decimal points. Step 2: Add the numbers in the thousandths position (1 + 3 = 4). Write the 4 in the thousandths position below the line. Step 3: Add the numbers in the hundredths position (7 + 6 = 13). Write the 3 in the hundredths position below the line. Carry the 1 to the next column (tenths). Step 4: Add the numbers in the tenths column, including the number carried over from the previous column (9 + 3 + 1 = 13). Write the 3 in the tenths position (below the line). Carry the 1 to the next column (ones). Bring the decimal point down. Step 5: Add the numbers in the ones position, including the number carried over from the previous column (8 + 5 + 1 = 14). Write the 14 to the left of the decimal point (below the line). Answer: Fred ran 14.334 miles. Subtract Fractions: Mixed Numbers - C Subtracting mixed fractions requires a solid understanding of adding fractions and the multiplication table. If the numerator of a fraction is less than the denominator, the fraction is called a proper fraction. If the numerator is equal to or greater than the denominator, the fraction is called an improper fraction. An improper fraction Page 15
can be rewritten as a mixed fraction. For example, 5/3 is an improper fraction. It can be rewritten as 1 2/3, which is a mixed fraction. The following is a step-by-step example of subtracting two mixed fractions. Example 1: Step 1: Write the problem vertically. Step 2: Separate the problem into subtraction of whole numbers and subtraction of fractions. Step 3: Find a common denominator (a common multiple of the denominators of two or more fractions) for the fractions. For this problem, the common denominator is 45. Multiply 4/5 by 9/9 to get 36/45. Multiply 2/9 by 5/5 to get 10/45. Step 4: Subtract the whole numbers (7-3 = 4). Subtract the numerators (36-10 = 26). The denominator remains the same (45). The following is a step-by-step example of subtracting two mixed fractions when the second fraction is larger than the first. Example 2: Step 1: Write the problem vertically. Step 2: Separate the problem into subtraction of whole numbers and subtraction of fractions. Step 3: Find a common denominator for the fractions. For this problem, the lowest common denominator is 12. Multiply 2/3 by 4/4 to get 8/12. Multiply 3/4 by 3/3 to get 9/12. Step 4: Rewrite the problem with the fractions having common denominators. Step 5: Since the top fraction (8/12) is smaller than the bottom fraction (9/12), trade (or borrow) one whole (12/12) from the 5 and add 12/12 to the 8/12 to get 20/12. Step 6: Subtract the whole numbers (4-2 = 2). Subtract the numerators of the fractions (20-9 = 11). The denominator (12) remains the same. Lowest Terms - B A fraction is comprised of two parts: a numerator (the top number) and a denominator (the bottom number). For example, in the fraction 2/3 the "2" is the numerator and the "3" is the denominator. A fraction is in lowest terms when the numerator and the denominator do not have a common factor greater than one. Page 16
Example 1: Reduce the fraction to lowest terms. Step 1: Divide the numerator and the denominator by a common factor (2). Step 2: The fraction (15/36) is still not in lowest terms. Divide the numerator and the denominator by a common factor (3). An improper fraction is a fraction in which the numerator is greater than or equal to the denominator. All improper fractions can be rewritten as mixed numbers or as whole numbers. The following is an example of how to write an improper fraction as a mixed number. Example 2: Reduce all fractions to lowest terms. Solution: 18 will divide into 55 three times (18 x 3 = 54) with one left over. Write the 3 as a whole number and then make the fraction 1/18. Compare Whole Number Equations - A Students are asked to place an ordering symbol (<, >, or =) in a number sentence to make the sentence true. It may be helpful to review the ordering symbols with the student. Example 1: 20,253 + 1,000? 20,042 + 1,211 (1) 20,253 + 1,000 = 21,253 (2) 20,042 + 1,211 = 21,253 (3) 21,253? 21,253 Step 1: Simplify the expression on the left. Step 2: Simplify the expression on the right. Step 3: Rewrite the mathematical sentence with the new numbers and determine which symbol to place between the two numbers. The answer is: 20,253 + 1,000 = 20,042 + 1,211. Example 2: 8,640 + 936? 7,851 + 1,460 Page 17
(1) 8,640 + 936 = 9,576 (2) 7,851 + 1,460 = 9,311 (3) 9,576? 9,311 Step 1: Simplify the expression on the left. Step 2: Simplify the expression on the right. Step 3: Rewrite the mathematical sentence with the new numbers and determine which symbol to place between the two numbers. The answer is: 8,640 + 936 > 7,851 + 1,460. Multiple Operations: Whole Numbers - A Students are assessed on the ability to add and subtract multiple-digit numbers (237 + 56) where regrouping (carrying, borrowing, or renaming) is required. Regrouping is necessary when the sum of the numbers in a specific column position is equal to or greater than ten. When performing multiple operations on expressions, it is important to remember that operations inside parentheses are completed first. If there are two sets of parentheses, then the set that comes first when reading from left to right is completed first. Once the operations in all sets of parentheses are completed, addition or subtraction may be completed in order from left to right. Example 1: (38 + 15) - (42 + 3) =? (1) 38 + 15 = 53 (2) 42 + 3 = 45 (3) 53-45 = 8 Step 1: Since there are two sets of parentheses in this problem, we will complete the operations in the set on the left first. 38 + 15 = 53. Step 2: Complete the operations in the second set of parentheses. 42 + 3 = 45. Step 3: The final step is to subtract the answer to the second set of parentheses from the answer to the first set of parentheses. 53-45 = 8. Answer: 8 Example 2: 21,368 + (56,435-24,789) =? (1) 56,435-24,789 = 31,646 (2) 21,368 + 31,646 = 53,014 Step 1: Subtract the numbers inside the parentheses. Step 2: Add the difference of the numbers inside the parentheses to 21,368. Answer: 53,014 Subtract Decimals: Ten Thousandths Subtracting decimal numbers with more than one decimal position (columns of numbers) is very similar to Page 18
subtracting whole numbers. Subtracting decimal numbers requires the ability to regroup (carry, borrow, or rename) when the number being subtracted is greater than the other number. Example 1: Solve 7.3639-0.4312 =? Step 1: Rewrite the problem vertically. Always line up the decimal points. Step 2: Begin by subtracting the ten thousandths column (9-2 = 7). Put the 7 in the ten thousandths column.step 3: Subtract the thousandths column (3-1 = 2). Put the 2 in the thousandths column. Step 4: Subtract the hundredths column (6-3 = 3). Put the 3 in the hundredths column. Step 5: Subtract the tenths column. Regrouping must occur because you cannot subtract 4 from 3. Borrow 1 from the ones column, changing the 7 to a 6. Give the one to the tenths column, creating 13. Subtract the tenths column (13-4 = 9). Put the 9 in the tenths column. Bring down the decimal point. Step 6: Subtract the ones column (6-0 = 6). Put the 6 in the ones column. Answer: 7.3639-0.4312 = 6.9327 Order of Operations with Decimals - B Performing operations with decimals is similar to performing operations with whole numbers. Operations inside parentheses are performed first. Example 1: (2.3 x 5.4) (1.3 x 1.1) =? (1) 2.3 x 5.4 = 12.42 and 1.3 x 1.1 = 1.43 (2) 12.42 1.43 =? (3) 12.42 1.43 = 8.685314685 Step 1: Perform all operations in parentheses. Multiply 2.3 x 5.4. And multiply 1.3 x 1.1. Step 2: Rewrite the equation with the new numbers in place of the parentheses. Step 3: Divide 12.42 by 1.43. The answer is 8.685314685. Example 2: 34.25 (1.23 x 2.12) =? (1) 1.23 x 2.12 = 2.6076 (2) 34.25 2.6076 =? (3) 34.25 2.6076 = 13.13468323 Step 1: Perform all operations in parentheses. Multiply 1.23 x 2.12. Step 2: Rewrite the equation with the new number in place of the parentheses. Step 3: Divide 34.25 by 2.6076. The answer is 8.685314685. Page 19
Lowest Common Denominator - B The lowest common denominator (LCD) of two or more fractions is the least common multiple of the fractions' denominators. Find the least common multiple of two denominators by finding the smallest non-zero number that is a multiple of each of the denominators. To find the LCD of two or more fractions, make a list of each of the denominators' multiples. Then determine the smallest number that is a multiple of all of the denominators. Example 1: Multiples of the denominators: Multiples of 3: 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36,... Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36,... Multiples of 9: 9, 18, 27, 36,... The smallest common multiple is 36. This multiple (36) is the lowest common denominator for the three fractions. Adding Integers Integers are the set of positive and negative whole numbers, including zero. To add integers, students must understand how integers appear on a number line. Numbers to the right of 0 on a number line are positive and numbers to the left of 0 are negative. The number -3 is a negative integer and the number 3 is a positive integer. The number zero is neither positive nor negative, it is neutral. It may be beneficial to verify that the student understands integers by having him or her create a number line. Label points to the left of 0 "negative," and points to the right of the 0 "positive." The following is an example of a number line: Confirm that the student understands that -4 is less than 4. Once he or she is comfortable with the concept of integers, introduce adding and subtracting. For example, -4 + 2. Start at -4 and move 2 places to the right (because we are adding). The answer is -2. When adding two integers with the same sign, add their absolute values. Then give the sum (answer) the sign of the integers. Answer: -5-3 + -2 =? -3 + -2 =? 3 +2 = 5, then make the result negative. Page 20
When adding integers with different signs, first find their absolute values. Then subtract the lesser absolute value from the greater absolute value, and give the result the sign of the integer with the greater absolute value. -7 + 3 =? -7 = 7 and 3 = 3 (find the absolute values) 7-3 =? (subtract the lesser from the greater) 7-3 = 4-7 + 3 = -4 (The result is given the sign of the greater integer.) Understanding Integers Integers are the set of positive and negative whole numbers, including zero. Numbers to the right of 0 on a number line are positive and numbers to the left of 0 are negative. The number -3 is a negative integer and the number 3 is a positive integer. The number zero is neither positive nor negative, it is neutral. Each point on the number line below represents an integer. Confirm that the student understands that -4 is less than 4, 0 is greater than -5, and -5 is less than -3. The following list provides the definitions for the commonly used ordering symbols. Percent of a Number Percent means "per one hundred." For example, if 7 out of 100 students ate pizza for lunch, then 7% (7 percent) of the students ate pizza for lunch. The student should understand how to determine the percent of a number. Example: Find 25% of 48. Step 1: Change the percent amount to a fraction (remember percent means "per one hundred"). Step 2: Multiply 25/100 by 48. Change 48 into a fraction by making its denominator 1. Multiply numerator by numerator (25 x 48 = 1,200). Multiply denominator by denominator (100 x 1 = 100). Step 3: Reduce the product to its lowest terms (1,200 100 = 12). Answer: 12 Page 21
Story Problems Percents - A Percent means "per one hundred." For example, if 7 out of 100 students ate pizza for lunch, then 7% (7 percent) of the students ate pizza for lunch. Students must be able to find percentages when given a story problem. A discount is the amount the price is reduced when an item goes on sale. A store might reduce its original price by a percentage. There are two ways to determine the sale price of an item. We can find the amount of the discount and subtract that from the original price to find the sale price, or we can multiply the regular price by the percentage that will be paid to find the sale price. Example 1: Jordan works at a bookstore. He gets a 25% discount on any book he buys. He wants to buy a book in which the regular price is $35.50. How much will the book cost Jordan? Method 1 (1) 25% x $35.50 = 0.25 x $35.50 = 8.875 ~ $8.88 (2) $35.50 - $8.88 = $26.62 Step 1: Find the amount of discount by multiplying the percent of Jordan's discount by the price of the book. You must convert the percent into a decimal number before multiplying it by the price of the book. This can be accomplished by moving the decimal point of the percentage to the left two places. Round your answer to the nearest cent. Step 2: Subtract the discount from the original amount. Method 2 (1) 100% - 25% = 75% (2) $35.50 x 75% = $35.50 x 0.75 = 26.625 (3) 26.625 approximately equals $26.63 Step 1: Determine the percentage Jordan will pay for the book by subtracting his discount from 100%. Jordan will pay 75% of the price. Step 2: Multiply the original price by the percentage Jordan will pay. You must convert the percent into a decimal number before multiplying it by the price of the book. This can be accomplished by moving the decimal point of the percent to the left two places. Step 3: Round 26.625 to the nearest hundredth to determine the price of the book. It does not matter which method you use, Jordan will pay $26.63 for the book. NOTE: The answers for method 1 and method 2 do not match because they were rounded in different places. Bar Graphs - C A bar graph is a drawing used to show and compare information. A bar graph has rectangular bars at different heights to show/compare data. An interesting method for increasing the student's understanding of graphs is to help him or her develop a graph for a school project or event, such as a magazine sale. The following is an example of a bar graph. Three athletes are selling cookbooks to raise money for Page 22
new uniforms. The left side of the graph represents the number of cookbooks sold. The bottom of the graph represents the athletes. Example 1: According to the above graph, who sold the most cookbooks? Answer: Zoe sold the most cookbooks because the bar above Zoe's name is the tallest (12). Example 2: According to the above graph, how many cookbooks have Zoe and Ann sold? Zoe sold 12 cookbooks. Ann sold 4 cookbooks. 12 + 4 = 16 Answer: Zoe and Ann have sold 16 cookbooks. Expanded Notation - E Expanded notation is the format for writing numbers so that each digit shows a place value. For example, 753 = 700 + 50 + 3. Students with an understanding of expanded notation will have a strong foundation for learning exponential notation. It may be helpful to discuss the concept of expanded notation with the student. Once he or she understands the theory, develop a series of numbers and help the student create expanded notation for each number. The following samples may be helpful: Number Expanded Notation 5,672 5,000 + 600 + 70 + 2 113 100 + 10 + 3 289 200 + 80 + 9 19 10 + 9 3 3 Example: Find the expanded form: 36,971. A. 30,000 + 6,000 + 900 + 71 B. 30,000 + 600 + 900 + 70 + 1 C. 3,000 + 6,000 + 900 + 70 + 1 D. 30,000 + 6,000 + 900 + 70 + 1 Solution: Line up the numbers and add to get 36,971. Answer: D Exponential Notation - B Exponents communicate the number of times a base number is used as a factor. The base number 5 to the 3rd power (an exponent of 3) translates to 5 x 5 x 5. (5 to the 3rd power is not 5 x 3.) The result of 5 to the 3rd Page 23
power is 125. To perform operations with exponents, exponential properties and scientific notation must be understood. Have the student find the equivalent whole number forms of these exponential numbers: Scientific notation is based upon exponential properties and is used to communicate very large or very small numbers. To write a large number using scientific notation, count the digits (from right to left) to be represented by a power of 10. 123,000,000 can be written in scientific notation as 1.23 x 10 to the 8th power. To write a small number, count the digits from left to right. The very small number, 0.0000008 can be written in scientific notation as 8 x 10 to the negative 7th power. The most common error among students learning about exponents is multiplying the base number by the exponent. That is, many students will calculate 8 to the 3rd power as 8 x 3 = 24. The correct answer is 8 x 8 x 8 = 512. Rounding and Estimation - D Rounding and estimation are used to express numbers to the nearest tenth, hundredth, thousandth, and so forth. In many real world applications for mathematics, rounding and estimation are used to make numbers more manageable and understandable. For instance, television producers often round large numbers so that they can be stated by reporters in a simple manner. If the United States produced 134,995,659 ounces of gold, a reporter might state, "The United States' gold production this year was 135,000,000 ounces." An interesting method for improving the student's rounding and estimation skills is to create a list of numbers. Help the student round each number. Remember, numbers less than 5 are rounded down, while numbers 5 or greater are rounded up (in both cases, you are looking one place to the right of the place value you wish to round). EXAMPLES: 34 rounded to the nearest ten is 30. 37 rounded to the nearest ten is 40. To help the student round decimal numbers, review decimal places with him or her. For example, in the number 6,879.342 6 = thousands 8 = hundreds 7 = tens 9 = ones 3 = tenths 4 = hundredths 2 = thousandths Page 24
When the student understands decimal places, ask him or her to round 6,879.342 to the nearest tenth. Look at the first digit to the right of the tenths place. Since the hundredths place is less than 5, the tenths place will remain unchanged. 6,879.342 rounded to the tenth place is 6,879.3 Comparison Comparisons at this grade level involve whole numbers, fractions, decimals, percents, and integers. Students must compare the value of given equations. It may be necessary to review integers, decimals, fractions, and percents with the student. Help the student understand that integers include whole numbers, their opposites, and zero. Practice integers using a number line by plotting points such as -5, -1, 0, and 2. As the student practices comparisons, remind him or her that fractions, percentages, and decimals represent portions or parts and that for every fraction, percentage, or decimal, there is a corresponding portion. The fraction 1/2 communicates a specific portion of something, but this specific portion can also be communicated by the percentage 50%. Decimals are portions communicated in columns (place values) which represent an underwritten denominator of 10 or a power of 10. 3.23 expresses 3 wholes and 23 hundredths of a whole. To compare fractions, percentages, and decimals, they must be converted to the same form. The fraction 3/4 can be compared to the percentage 60% by converting the percentage to fraction form. Percentage means "per one hundred," so 60% = 60/100. Find a common denominator between 60/100 and 3/4. 100 can be divided by 4, so the fractions become: 60/100 compared to 75/100. From this comparison we can see that 75/100 or 3/4 represents a greater portion than 60/100 or 60%. To compare the decimal 0.74 to the fraction 3/4, convert the decimal to fraction form. 0.74 becomes 74 hundredths or 74/100. We already know that 3/4 = 75/100. Therefore, 3/4 represents a greater portion than 0.74 or 74/100. Ratio/Proportion - B A ratio is a comparison of two numbers expressed as a quotient. They can be written in three ways: a fraction (3/5), a ratio (3:5), or a phrase (3 to 5). Like fractions, ratios refer to a specific comparison. The ratios 3/5, 3:5, and 3 to 5 (as in "the ratio of cellos to violins was 3 to 5") all express the same ratio or comparison. A proportion reflects the equivalency of two ratios. The ratio 3/5 expresses the same proportion as the ratio 15/25. To understand how ratios operate, students need to understand equivalent fractions. Fractions represent portions or parts. For every fraction, there is a corresponding portion. The fraction 1/2 communicates a specific portion of something, but this specific portion can also be communicated by the fractions 2/4, 3/6, 8/16, 10/20, etc. All of these fractions are equal to 1/2 because the relationship between the numerator and denominator in 1/2 is the same relationship between the numerators and denominators in 2/4, 3/6, 8/16, and 10/20. Ratios and proportions operate in a similar manner. The ratio 2:5 communicates a specific portion. The ratio 4:10 communicates the same portion. Example 1: Sandra has 15 lollipops and 25 jellybeans. What is the ratio of lollipops to jellybeans? There are 15 lollipops to 25 jellybeans, so the ratio of lollipops to jellybeans is 15:25. Answer: 15:25 Page 25