LESSON ASSIGNMENT
 
LESSON 1 Review of Whole Numbers.


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.


thousands 
1000

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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.


4,
444,444,444
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
+ 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


remainder (or difference)
minuend
subtrahend

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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


multiplicand
x multiplier
product
factors

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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


6 - divisor
18 - dividend
3 - quotient

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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 1-7

442 - dividend
15 - divisor
29 - quotient
7 - remainder

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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
4 7 2 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.)
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
212 or 212
21 021
236 236
3 ones + 2 ones +
1 one = 6 ones.
0 tens +1 ten +
2 tens = 3 tens.
0 hundreds +
2 hundreds +
0 hundreds =
3 hundreds.

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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


1459
38
327
1824

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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


No problem
was given.

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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
  43 43 = 40 + 3 = 30 + 13 = 30 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
3697 3697
- 375 OR -0375
3322 3322
OR
0007 - 0005 = 0002
0090 - 0070 = 0020
0600 - 0300 = 0300
3000 - 0000 = 3000
3697 - 0375 = 3322

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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
 
5 4 2
- 2 6 4
  2 7 8

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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
4 0 3 = 400 +00 +3 = 300 + 100 + 3 = 300 + 90 + 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
-(200 + 60 + 4)
500 + 30 + 12
-(200 + 60 + 4)
400 + 130 + 12
-200 - 60 - 4
200 + 070 + 08 =278

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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 1-15

4 9 9 13
 
5 0 0 3
-1 0 0 9
 
3 9 9 4


4000+900+90+13
-1000 -000 -00 -9
3000+900+90 + 4

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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 1-17

1 2
 
3 4 1 5
       x 4
1 3 6 6 0

5 x 4 = 20
10 x 4 = 40
400 x 4 = 1,600
3000 x 4 = 12,000
13,660

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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
4 4

3 6 7
x 9 7
2 5 6 9 (367x7)
3 3 0 3 0 (367x90)

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 1-19

23042
x 1020
  46084

23042
23502840
OR
 23 042
x 1 020

00 000
460 840
0 000 000
23 042 000
23 502 840
(23,502,840)

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FRAME 1-21.
Set up and work this
problem: 406 ) 15

 Solution to Frame 1-20
12/ 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
10 brought down (the "0"). Multiply your answer by the
divisor and subtract (2x15=30; 40-30=10).
What is your next step?
 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).
1 There are no more unused digits in the dividend. If there
were, you would continue the above procedures until all of
the digits of the dividend have been used.
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.
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 1-23
27 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


107
23/ 2461
23
16
0
161
161
0

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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.

 
Solution to
Frame 1-26

455 Check: 405
- 50 + 50
405 455

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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
34 Check: 034
121 121
+ 87 087
242 1 1

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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.

 
Solution to
Frame 1-28

7 13 7 13
 
8 3 4 or  8 3 4
- 4 7 3   - 3 6 1
3 6 1       4 7 3

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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 1-29
374r2 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
107
161
23
2461
+0
2461

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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 1-31

232 176
176 232
1392 352


16240 5280
23200 35200
40832 40832

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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 1-32

25 CHECK
12 12
50 25/ 300
25 25
300 50
50
0
OR
25
12/ 300

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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) = __________

 
Solution to
Frame 1-33

a. 31 (21 + 10)
b. 3 (4 – 1)

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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 1-34

a. 52 (26 x 2)
b. 5 (20 ) 4)

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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. 455 x 33 =
 
b. 3,690 - 2,460 =
 
c. 44 + 275 + 9 =
 
d. 400 ) 50 =
 
2. Solve and check each of the problems below.

a. 3/406 Check:
 
 
b. 389 Check:
     27
+ 122
 
c. 47 Check:
  x 22
 
 
d. 996 Check:
    - 57
 
3.

a. 16 + 4 ) 2 = _________
b. 2 (4 + 3) = __________
c. a (c + d) = __________ 

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.

 
Solution to Frame 1-35

408
ac + ad + bc + bd =
(10)(30) + (10)(4) +
(2)(30) + (2)(4) =
300 + 40 + 60 + 8 = 408
12
x 34
48
360
408