LESSON ASSIGNMENT Frames 1-1 through 1-36.
MATERIALS REQUIRED Pencil, paper and eraser.
LESSON OBJECTIVES After completing this lesson, you should be able to:
1-1. Identify by name each number in given addition,
subtraction, multiplication, and division problems.
1-2. Set up and solve given problems involving addition,
subtraction, multiplication, and division of whole numbers.
1-3. Check given problems involving addition,
subtraction, multiplication, and division of whole numbers.
SUGGESTIONS Work the following exercises (numbered
frames) in numerical order. Write the answer in the space
provided in the frame. After you have completed a frame, check
your answer against solution given in the shaded area in the
following frame. The final frame contains review exercises for
Lesson 1. These exercises will help you to achieve the lesson
objectives and prepare for the examination.
NOTE: Do not utilize a calculator. Work the problems out on
paper or in your head.
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FRAME 1-1. PLACE VALUE. Our number system is based upon powers of 10. That is, the value (amount) of a digit (numeral) depends upon its location in the number. Consider a given digit location (its place in the number, not the value of the numeral itself). The digit location to its immediate left is worth ten times as much as the given digit location. The digit place to the immediate right is worth one-tenth as much. This is called place value. For example, in the number 456, the "5" tells how many tens (place value is "10"), the "4" tells how many hundreds (place value is 100, which is 10 x 10) and the 6 tells how many ones (place value is 1, which is 1/10 x 10). This is sometimes called "the base 10 numbering system." The number 456 is equal to 4 x 100 plus 5 x 10 plus 6 x 1 In the number 9724, the digit in the far right tells how many ones (4 x 1). The second digit tells how many tens (2 x 10). The third digit tells how many hundreds (7 x 100). The fourth digit tells how many (9 x ). |
The solution to the exercise in Frame 1-1 is in the shaded area (right side) of Frame 1-2 on the following page. |
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FRAME 1-2. Remember: When dealing with whole numbers (no fractions or decimals), the numeral to the far right tells how many ones (1), the numeral to its left tells how many tens (10 x 1), the next numeral to the left tells how many hundreds (10 x 10), the next numeral tells how many thousands (10 x 100), and so on with the place value increasing by a factor of ten each time. NOTE: In the above statement, number refers to the entire value. Numeral refers to one digit (symbol) within the number. Test your understanding of place values by filling in the blanks below. |
Solution to Frame 1-1.
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FRAME 1-3. PARTS OF A PROBLEM. Just like members of a family can be identified by their relationship to other members (mother, son, aunt, brother, etc.), the different numbers in a math problem can also be identified by their relationships. For example, the number that results when two or more numbers are added together is called the SUM. The numbers that are added together are called ADDENDS. Label the parts of the problem below: 4444 + 333 4777 |
Solution to Frame 1-2.
hundred millions ten thousands |
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FRAME 1-4. The parts in a subtraction problem are the minuend, subtrahend, and the remainder. The remainder is sometimes called the "difference." Example: 978 minuend -243 subtrahend 735 remainder (or difference) The answer in a subtraction problem is called the . The top number is the . The number subtracted from the top number is called the . |
Solution to Frame 1-3.
+ addend sum |
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FRAME 1-5. A multiplication problems consist of a multiplicand, a multiplier, and a product. The multiplicand (top number) is the number to be multiplied. The multiplier (second number) is the number doing the multiplying. The answer is called the product. The numbers being multiplied together (the multiplicand and the multiplier) are sometimes referred to as "factors." Label the parts of the following multiplication problem. 45 x 4 180 In this problem, the "45" and "4" can also be called . |
Solution to Frame 1-4
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FRAME 1-6. Division is used to determine the number of times one number is contained in another number. If you were to divide 18 by 6, you might ask yourself, "How many groups of 6's are there in 18?" Your answer would be 3. The answer is called the quotient. The number that is being divided is called the dividend. The divisor is the number that is divided into the dividend. In the problem 18 ) 6 = 3 (18 divided by 6 equals 3), "6" is the , "18" is the , and "3" is the . |
Solution to Frame 1-5
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FRAME 1-7. If the divisor does not go into the dividend evenly, an amount is left over. The quantity left is called the remainder. For example, when the number 19 is divided by 6, the quotient is 3 with a remainder of 1. (There are three groups of sixes and another group consisting of the one.) The remainder is usually expressed as a fraction (remainder over divisor). Label the parts of the problem 442 ¸ 15 below. 29 442 15/ 442 30 15 142 135 29 7 7 |
Solution to Frame 1-6
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FRAME 1-8. The answer to the problem 19 ) 6 can be written as 3r1 (three with a remainder of 1) or as 3 1/6 (three and one-sixth). The second method (whole number and a fraction) is the preferred method of stating the answer. Another way of expressing the answer (especially if you are using a calculator) is as a decimal (see Frame 1-24). The answer to the problem 442 ) 15 can be written as or as . |
Solution to Frame 1-7442 - dividend |
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FRAME 1-9. ADDITION. Any problem in addition, subtraction, multiplication, or division must be set up correctly in order to solve it. In addition, you must put like units under like units (ones under ones, tens under tens, hundreds under hundreds, etc.). Set up and work this addition problem: 3 + 212 + 21 = ? |
Solution to Frame 1-8 29r7 (29 with a remainder of 7) or 29 7/15 (twenty-nine and seven-fifteenths.) |
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FRAME 1-10. In the previous addition problem, none of the columns added up to more than 9. If the sum of a column is more than 9, write down the last (right) digit under the column and add the remaining digit(s) to the next column (the column to the left). REMEMBER: When adding, begin with the column on the far right and then go to the left. 1 3 2 6 In the first (right) column, 6 + 6 = 12. + 1 4 6 The "2" is written below the column and the "1" is carried to the next column where it is added with the "2" and the "4." (The "1," "2," and "4" are all tens.)4 7 2 A more complicated addition problem is shown below. 1 1 2 3 6 7 7 + 6 + 6 + 8 = 27 Write "7" carry "2" 1, 4 1 6 2 + 6 + 1 + 7 + 0 = 16 Write "6" carry "1" 6, 5 7 6 1 + 3 + 4 + 5 + 1 = 14 Write "4" carry "1" 2, 1 0 8 1 + 0 + 1 + 6 + 2 = 10 Write "10" (There 10, 4 6 7 is no column to the left.) NOTE: You do not have to write the commas when adding, but may help to keep the columns straight. Now you add these numbers: "1459," "38," and "327." |
Solution to Frame 1-9 3 003 |
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FRAME 1-11 Another way of thinking about the problem 367 + 1416 + 6576 + 2108 is 7 + 6 + 6 + 8 = 27 60 + 10 + 70 + 00 = 140 300 + 400 + 500 + 100 = 1300 0000 + 1000 + 6000 + 2000 = 9000 10467 |
Solution to Frame 1-10
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FRAME 1-12. SUBTRACTION. Now, let's subtract whole numbers. Just as in addition, like units must go under like units (ones under ones, tens under tens, etc.). Set up and work this subtraction problem in the space to the right: 3697 - 375 |
Solution to Frame 1-11
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FRAME 1-13. If the top numeral in the column is smaller than the numeral beneath it, then you must "borrow 10" from the top numeral in the column to the left. (Remember that each numeral in the left column has a place value that is ten times greater. Therefore, to "borrow 10," decrease the numeral in the left column by "1" and add "10" to the top numeral in the column with which you are working. In the problem "43 minus 17," four tens become three tens and three ones becomes 13 ones.) This problem is worked below. Another (longer) procedure for performing the same operation is shown to the right. 3 13 -17 -17 - (10 + 7) - (10 + 7) -10 - 7 26 20 + 6 = 26 Work this problem using the "borrowing" method: 542 minus 264. NOTE: You will have to borrow more than one time. |
Solution to Frame 1-12 |
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FRAME 1-14. Check your understanding of the borrowing method by working the same problem using the longer method shown in Frame 1-13. |
Solution to Frame 1-13 4 13 12 |
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FRAME 1-15. What happens if you need to borrow from the column on the left, but there is a zero in that column? You must go one column more to the left, "borrow 10" in order to change the zero into "10," and then borrow from the "10." For example, subtract "7" from "403." 3 9 13 - 7 - 7 - 7 - 000 - 00 - 7 3 9 6 300 + 90 + 6 = 396 Now you solve this one: 5003 - 1009 |
Solution to Frame 1-14 500 + 40 + 2 |
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FRAME 1-16. MULTIPLICATION. Multiplication is actually a shortened form of addition. For example: 9 x 4 = 9 + 9 + 9 + 9 = 36. Also note that 9 x 4 = 4 x 9 = 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 = 36. NOTE: When a multiplication problem is set up, the number with the most digits is usually chosen to be the multiplicand (the number on top). The problem 367 x 97 can also be solved by writing 367 down times and adding the numbers together or by writing 97 down times and adding the numbers together. |
Solution to Frame 1-154 9 9 13
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FRAME 1-17. Like addition, you need to carry when the product is more than nine. Remember to add the number you carried after you multiply. For example: 24 x 3. 1 24 4 x 3 = 12. Write the "2" and carry the 1 (really 10), then add x3 2 x 3 = 6 the number carried (1) to the product (6) 72 [really six 10's]. 6 + 1 = 7. Write down the "7". Another way of stating the problem is: 24 = 20 + 4 x3 x3 x3 60 + 12 = 72 Now you work this problem: 3,415 x 4 |
Solution to Frame 1-16 97 367 (You can see that multiplication takes less time and results in fewer math errors.) |
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FRAME 1-18. When the multiplier has more than one digit, the multiplicand is multiplied by each digit in the multiplier beginning with the right digit of the multiplier. The products are added together to obtain the final answer. Note that the right digit of the product is in the same column as the digit of the multiplier being used. You may wish to fill in the empty digit places in the product with zeros to help keep the columns straight. For example: 621 x 27. 1 1 621 621 621 621 621 x 27 or x 27 x27 = x 7 + x 20 4347 4347 4347 + 12420 = 16,767 1242 12420 16767 16767 You work this problem: 367 x 97 |
Solution to Frame 1-171 2 1 3 6 6 0 5 x 4 = 20 |
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FRAME 1-19. Normally, each group of three digits (beginning at the decimal point and going to the left) is set off by a comma in order to make reading easier (separate thousands from hundreds, millions from thousands, billions from millions, etc.). If the product has only four digits (less than 10,000), the comma may be present or absent (4,500 or 4500). If the number has five or more digits, include the comma(s) in the product. Numbers of three or less digits (less than 1,000) have no commas. NOTE: If the number is a decimal number (discussed in Lesson 3), the above comments apply only to digits to the left of the decimal point. Commas are not placed between digits located to the right of the decimal point. Remember to keep your columns straight. This is especially true if the multiplier contains a zero. Also remember that the product of any number multiplied by zero is zero. NOTE: You may wish to remove the commas from the factors when multiplying if that will help you to keep the columns straighter and keep you from becoming confused. Work this problem: 23,042 x 1,020 |
Solution to Frame 1-18
6 6 3 5,5 9 9 |
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FRAME 1-20. DIVISION. The division problem "250 ) 25 = 10" (which is read "250 divided by 25 equals 10") means that 250 contains 10 sets (groups) of 25. 10 25 / 250 25 00 00 Notice that the number to the right of the division sign () ) goes outside the division block. When "250 ) 25" is changed to 25/250, it is read as "25 divided into 250." Remember, when dividing, you divide into the dividend going from left to right (go from large place value to smaller place value), a change from addition, subtraction, and multiplication. Set up this problem: You have 156 eggs. You want to put them into egg cartons (12 eggs per carton). How many cartons do you need? |
Solution to Frame 1-1923042
23042 00 000 |
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FRAME 1-21. Set up and work this problem: 406 ) 15 |
Solution to Frame 1-2012/ 156 |
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FRAME 1-22. Let's review the problem given in Frame 1-21 in case you had any problems. Remember, begin at the first (left) digit of the dividend and work to the right. 15/ 406 How many 15's in 4? Answer: 0 Write "0" directly above the "4." 0 15/ 406 Multiply your answer by the divisor and -0 subtract (0x15=0; 4-0=4). 4 02 15/ 406 Bring down the next digit (0) from the dividend and put it after 0 the remainder (4). The "working dividend" is now 40. 40 How many 15's in 40? Answer: 2. Write the "2" in the 30 quotient immediately above the number previously What is your next step?10 brought down (the "0"). Multiply your answer by the |
Solution to
Frame 1-21 27r1 15/ 406 27 with a remainder of one OR 27 1/15 (twenty-seven and one-fifteenth) |
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FRAME 1-23. 027 After you bring down the "6" from the dividend and enter 15/406 it after the remainder, you have a new "working dividend" 0 of 106. 40 How many 15's are in 106? Answer: 7. 30 Write the 7 over the 6. 106 Then multiply the divisor by the last answer (15x7 = 105) 105 and subtract (106 - 105 = 1). NOTE: Often the initial zero (or zeros) in the quotient is not written. This example shows the zero in the quotient to help make the process clearer. Also, when the numbers contain commas, you may prefer to delete the commas when setting up the division problem.1 There are no more unused digits in the dividend. If there The answer to how many 15's are in 406 is 27, with 1 left over. The remainder may be written as 27r1. This means "27 with a remainder of l." More often, the remainder is written as a fraction. The top number of the fraction is the remainder and the bottom number is the divisor (see Frame 1-7). Therefore, the answer to 406 ) 15 can be written as . |
Solution to
Frame 1-22 Bring down the next digit from the dividend (the "6"). |
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FRAME 1-24. If you wanted to express the answer in decimal form rather than using a fraction, put a decimal point after the last digit in the dividend (after the "6"), add zeros to the right of the decimal point (they do not change the value of the dividend), and keep dividing. NOTE: Don't forget to put a decimal point in the quotient directly above the decimal point in the dividend. Solve the problem so that the quotient is in decimal form instead of "remainder" or "fraction" form. The decimals and the first zero have already been placed. This procedure is discussed in more detail in Lesson 3 (Decimals). 27. 15/ 406.0 30 106 105 10 |
Solution to Frame 1-2327 1/15 |
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FRAME 1-25. Set up and solve this problem: 2,461 ) 23. |
Solution to
Frame 1-24 You haven't finished the problem yet! The answer is 27.0666666666666 with the 6's continuing forever. Knowing how to stop (rounding) is covered in Lesson 3. |
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FRAME 1-26. CHECKING. Once you have worked a problem, how do you know if you arrived at the correct answer? One way is to do the problem over again from the beginning (and hope that you don't make the same mistake twice). Another way is to check (verify) your answer by rearranging the problem and solving the new problem. For example, when you subtract, you can check your answer (the difference) by adding the difference to the subtrahend. If your subtraction was correct, the resulting sum will be the minuend. Look at this problem: 300 (minuend) Check: 135 (difference) -165 (subtrahend) + 165 (subtrahend) 135 (difference) 300 (minuend) Solve and check this problem: 455 Check: - 50 |
Solution to Frame 1-25
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FRAME 1-27. To check an addition problem, simply reverse the order of the addends and add again. For example: 473 361 + 361 is checked in this manner: + 473 834 834 NOTE: When more than two numbers are added together, there is more than one way to rearrange the addends. The easiest may be to begin with the bottom and add upward (with the top number being added last). The term "adding up" comes from an older method in which the sum was written at the top instead of the bottom. NOTE: Filling in spaces with zeros can help to keep columns straight. Solve and check: 34 + 121 + 87. |
455 Check: 405 |
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FRAME 1-28. Just as subtraction problems can be checked using addition, so addition problems can be checked using subtraction. Subtracting one addend from the sum will yield the remaining addend. (This method of checking is used less than the reverse adding method.) Check: 473 + 361 = 834 |
Solution to Frame 1-27
1 1 242 |
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FRAME 1-29. Division can be checked by multiplying the quotient (without the remainder) by the divisor and adding the remainder (if any). The resulting number will be the dividend. For example: 5 5 quotient 3/ 17 Check: x3 divisor 15 15 2 +2 remainder 17 dividend Solve and check this problem: 4,864 ) 13. |
7 13 7 13 |
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FRAME 1-30. You can also check a division problem by multiplying the divisor by the quotient (without the remainder) and adding the remainder. Check the division problem in Frame 1-25 using this method. NOTE: If the remainder is expressed as a fraction, multiply the entire quotient by the divisor to obtain the dividend. Multiplying fractions is discussed later. |
Solution to Frame 1-29374r2 CHECK 13/ 4864 374 39 x13 96 1122 91 3740 54 4862 52 +2 2 4864 |
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FRAME 1-31. Multiplication can be checked by switching the factors (for example, the product of 232 x 176 should be the same as the product of 176 x 232). Work both problems and see for yourself. |
Solution to Frame 1-30
23 |
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FRAME 1-32. Multiplication can also be checked by dividing the product by either the multiplicand or the multiplier. If you divide the product by the multiplicand, the quotient will be the multiplier. If you divide the product by the multiplier, the quotient will be the multiplicand. For example: 2 x 3 = 6 can be checked by dividing: 3 2 2/ 6 OR 3/ 6 Remember that you may use either one of the factors as the divisor, but the quotient must be the other factor. Solve and check this problem: 25 x 12 |
Solution to Frame 1-31232 176
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FRAME 1-33. COMBINED OPERATIONS. Sometimes a problem requires you to do two or more different operations. For example: 3 x 5 + 1 requires multiplication and addition. Is the answer 16 (15 + 1) or 18 (3 x 6)? The rule is that, unless the problem indicates otherwise, you should multiply and divide first. After these operations are completed, then you add and subtract. Practice by solving these problems. a. 21 + 5 x 2 = __________ b. 20 ) 5 – 1 = __________ |
Solution to Frame 1-3225 CHECK |
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FRAME 1-34. The previous frame tells you to multiply/divide, then add/subtract "unless the problem indicates otherwise." What does this mean? Sometimes, you need to add or subtract first, then multiply or divide. This is usually indicated by inclosing the operation to be done first in parentheses ( ). For example, 3 x 5 + 1 = 16 because you follow the basic rule of multiply and divide first, but 3 x (5 +1) indicates that you are to perform the addition function first. In this instance, the answer is 18 (3 x 6 = 18). Practice by solving these problems. a. (21 + 5) x 2 = __________ b. 20 ) (5 – 1) = __________ |
a. 31 (21 + 10) |
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FRAME 1-35. Sometimes parentheses are used to indicate multiplication. For example, 6 x 7 can also be written as (6)(7). Parentheses are often used in algebra in which letters are used to represent numbers. This allows a general formula to be developed. For example, the quantity a+b multiplied by the quantity c+d can be represented by: (a + b) (c + d) = ac + ad + bc + bd NOTE: "ac" means the quantity "a" multiplied by the quantity "c;" "ad" means the quantity "a" multiplied by the quantity "d;" "bc" means the quantity "b" multiplied by the quantity "c; and "cd" means the quantity "c" multiplied by the quantity "d." Similarly, "a" times the quantity "b+c" can be written as "a(b+c)." Test the general formula "(a + b) (c + d) = ac + ad + bc + bd" by letting a = 10, b = 2, c = 30 and d = 4; then check by multiplying 12 x 34. |
Solution to Frame 1-34a. 52 (26 x 2) |
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FRAME 1-36. SELF TEST. You have completed the section on adding, subtracting, multiplying, and dividing whole numbers and checking your answers. If you feel that you need more review on solving and/or checking problems, look over the appropriate frames again. Then work the following self-test exercises shown below. The solutions are found on the following page. NOTE: When there is more than one method of checking an answer, only one or two methods may be shown. 1. Set up and solve the problems below.
a. 16 + 4 ) 2 = _________ This lesson may have appeared too simple for you (and perhaps it was), but it serves as a foundation for the lessons that follow. If you have learned other methods of solving these types of problems, you may use them on tests; however, be sure that they work. If you missed any problem(s), review the appropriate lesson frames and rework the problem(s) before going to the next lesson. |
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