ABC/ Mathematics / Chapter 3 / TP 3 - 1 / Rev 1 © 2003 General Physics Corporation OBJECTIVES 1.Without a calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE

ABC/ Mathematics / Chapter 3 / TP 3 - 1 / Rev 1© 2003 General Physics Corporation

OBJECTIVESOBJECTIVES

1. Without a calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE decimal fractions.

2. With an approved calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE decimal fractions.

3. Without a calculator, CALCULATE the average of a series of numbers.

4. With an approved calculator, CALCULATE the average of a series of numbers.

5. Without a calculator, EXPRESS the solution of addition, subtraction, multiplication, and division operations using the appropriate number of significant digits.


DECIMAL PLACE RELATIONSHIPS DECIMAL PLACE RELATIONSHIPS

Fig 3-1


EXAMPLEEXAMPLE

Ex 3-1

The magnitude of 7361.298 is:

Digit Place Value7 1,000 = 7,0003 100 = 3006 10 = 607 1 = 12 1/10 = 2/10 = 200/1,0009 1/100 = 9/100 = 90/1,0008 1/1,000 = 8/1,000 = 8/1,000


EXAMPLEEXAMPLE

Ex 3-1

Sum = 7,361 +

= 7,361

= 7,361.298

000,1

298

000,1

298


EXAMPLEEXAMPLE

Ex 3-2

Find the decimal equivalent of 4

3

434

3

75.04

3

75.000.34

20

8 2

0

20


EXAMPLEEXAMPLE

Ex 3-3

The decimal equivalent of

= 2 3 = 0.6666…..

3

2 is:

3

2

where the ….. indicates successive 6’s.


EXAMPLEEXAMPLE

Ex 3-4

Write the common fraction that is equivalent to the decimal fraction 0.375.

Recall from the previous chapter to reduce the fraction to lowest terms, factor each term into its smallest component.

8

3

125000,1

125375

000,1

375375.0

555222

5553

000,1

375


EXAMPLEEXAMPLE

Ex 3-4

Then cancel out each factor that occurs in each term.

Then multiply the factors in each term together.

8

3

2x2x2

3

Thus simplified, 0.375 in a fractional form is .8

3


EXAMPLEEXAMPLE

Ex 3-5

Find the sum of 39.62, 41.093, and 0.0327.

39.62 41.093+ 0.0327 80.7457


EXAMPLEEXAMPLE

Ex 3-6

Subtraction is done in the same manner as with whole numbers.

32.100– 16.379 15.721


EXAMPLEEXAMPLE

Ex 3-7

Multiply 16.2 and 1.15. Multiply without concern for the decimal.

162 115

810 162 162 18630


EXAMPLEEXAMPLE

Ex 3-8

Multiply 16.2 and 1.15.

10

162

10

2162.16

100

115

100

15115.1

630.18000,1

630,18

100

115

10

162


EXAMPLEEXAMPLE

Ex 3-9

Divide 41.05 by 2.5. Divide without concern for the decimal.

25 160 150 105 100 50 50 0

1642410525


EXAMPLEEXAMPLE

Ex 3-10

Divide 41.05 by 2.5.

100

105,4

100

54105.41

10

25

10

525.2

500,2

050,41

25

10

100

105,4

1025

100105,4

100

642,1 42.16

555522

821552

500,2

050,41


EXAMPLEEXAMPLE

Ex 3-14

Example 3-12Convert 0.25 to a percentage.

0.25 100% = 25%

Example 3-13Convert 2 to a percentage.

2 100% = 200%

Example 3-14Convert 1.25 to a percentage.

1.25 100% = 125%

Ex 3-12Ex 3-13


EXAMPLEEXAMPLE

Ex 3-36

Find the average of the following recorded temperatures: 600° F, 596° F, 597° F, 603° F.

Step 1. After making sure that the individual quantities to be averaged have the same units, add the individual numbers of quantities to be averaged.

600 + 596 + 597 + 603 = 2,396

Step 2. Count the number of numbers or quantities to be averaged.

The number of items is 4.

Step 3. Divide the sum found in Step 1 by the number counted in Step 2.

2,396 ÷ 4 = 599° F


EXAMPLEEXAMPLE

Ex 3-37

The level in a tank is recorded once a day. The recorded level in the tank since the last addition over the last several days has been 500 gals, 490 gals, 487 gals, 485 gals, and 480 gals. Calculate the average tank level.

Step 1. All levels have the same units (gals) so the individual numbers can be averaged.ADD the individual numbers.

500 + 490 + 487 + 485 + 480 = 2,442

Step 2. Count the quantities to be averaged.The number of items is 5.


EXAMPLEEXAMPLE

Ex 3-37

Step 3. Divide the sum found in Step 1 by the quantities in Step 2.

2,442 ÷ 5 = 488.4

The average tank level for the recorded days is 488.4 gals.


EXAMPLEEXAMPLE

Ex 3-38

Given the following price list of replacement pumps, find the average cost.$10,200; $11,300; $9,900; $12,000; $18,000; $7,600

Step 1. After making sure that the individual quantities to be averaged have the same units, add the individual numbers or quantities to be averaged.10,200 + 11,300 + 9,900 + 12,000 + 18,000 + 7,600 =

69,000

Step 2. Count the numbers or quantities to be averaged.

Total number of prices is 6.


EXAMPLEEXAMPLE

Ex 3-38

Step 3:. Divide the sum found in Step 1 by the number counted in Step 2.

69,000 ÷ 6 = 11,500

Thus, the average price of the replacement pump is $11,500.


EXAMPLEEXAMPLE

Ex 3-39

The Purchasing department ordered replacement valves for the upcoming outage

Step 1. The valves have the same units ($). Add the quantities to be averaged.

$ 145.25$ 145.25$ 145.25$ 145.25$ 137.85$ 137.85$ 150.00$ 150.00$ 150.00$ 1,306.70

• 4 valves cost $145.25 each• 2 valves cost $137.85 each• 3 valves cost $150.00 each


EXAMPLEEXAMPLE

Ex 3-39

Step 2. Count the quantities to be averaged.4 + 2 + 3 = 9

Step 3. Divide the sum found in Step 1 ($1,306.70) by the number counted in Step 2 (9).

$1,306.70 ÷ 9 = $145.19

The average price for the valves ordered is $145.19.

Alternately, Step 1 could have been performed as(4)($145.25) + (2)($137.85) + (3)($150.00)= ($581.00) +($275.70) + ($450.00) = $1,306.70


EXAMPLEEXAMPLE

Ex 3-40

Calculate the average of the following lengths.2 ft, 30 in, 1.5 ft, 18 in

Step 1. The items to be averaged have different units (ft and in). Determine which unit would be preferred. (In this case it doesn’t matter.)

2 ft = 24 in30 in = 2.5 ft

1.5 ft =18 in18 in = 1.5 ft


EXAMPLEEXAMPLE

Ex 3-40

You could convert all to feet, and sum and average.

Step 1. 2 ft + 2.5 ft + 1.5 ft + 1.5 ft = 7.5 ft

Step 24 items

Step 37.5 ft ÷ 4 = 1.875 ft


EXAMPLEEXAMPLE

Ex 3-40

Alternately, you could convert to inches, sum and average.

Step 1.24 in + 30 in + 18 in + 18 in = 90 in.

Step 24 items

Step 390 in ÷ 4 = 22.5 in


THREE TEMPERATURE SCALES THREE TEMPERATURE SCALES

Fig 3-2

110

100

90

80

70

60

Outdoor Thermometer

102.0

101.0

100.0

99.0

98.0

97.0

Fever Thermometer

600

500

400

300

200

100

Oven Thermometer


EXAMPLEEXAMPLE

Ex 3-41

Measured Number

MostSignificant

Digits(left)

LeastSignificant

Digits

Numberof

SignificantDigits

12,345 1 5 5

123.45 1 5 5

1,986 1 6 4

37.806 3 6 5

201 2 1 3

300.7 3 7 4

500.08 5 8 5

0.00875 8 5 3


EXAMPLEEXAMPLE

Ex 3-41

Measured Number

MostSignificant

Digits(left)

LeastSignificant

Digits

Numberof

SignificantDigits

900.030 9 right‑most 0 6

0.090 9 right‑most 0 2

200 2 2 1

200. 2 right‑most 0 3

200.0 2 right‑most 0 4

200.00 2 right‑most 0 5


EXAMPLEEXAMPLE

Ex 3-42

Yard Stick

Length 8.0”

Width 5.0”

Calculated Area 40 sq. in.

Significant Digits 2

Correct Answer 40. sq. in.


EXAMPLEEXAMPLE

Ex 3-43

Ruler

Length 8.2”

Width 4.8”

Calculated Area 39.36”.


Correct Answer 39 sq. in.


EXAMPLEEXAMPLE

Ex 3-44

Micrometer

Length 8.164”

Width 4.795”

Calculated Area 39.14638 sq. in.


Correct Answer 39.15 sq. in.


EXAMPLEEXAMPLE

Ex 3-45

Round to two figures0.193 0.19

Round to five figures157,632 157,630

Round to three figures7,591 7,590

Round to four figures0.98764 0.9876


EXAMPLEEXAMPLE

Ex 3-46

Round to one figure0.193 0.2

Round to three figures157,632 158,000

Round to two figures7,591 7,600

Round to four figures0.98764 0.9877


EXAMPLE EXAMPLE

Ex 3-47


Round to four figures195,753 195,800

Round to one figure7,591 8,000

Round to three figures 0.98751 0.988

Round to two figures18,501 19,000

Round to four figures 19,555,005 19,560,000


EXAMPLEEXAMPLE

Ex 3-48


Round to four figures195,750 195,800

Round to one figure7,500 8,000

Round to one figure 6,500 6,000

Round to three figures0.98450 0.984


EXAMPLEEXAMPLE

Ex 3-49

Add 0.0146, 0.0950, and 0.43

Step 1Set up to do the math as normal.

0.01460.0950

+ 0.043 Step 2Identify the least significant digits in each term to be added or subtracted. 0.0146 the six is the least significant digit 0.0950 the zero is the least significant digit+ 0.043 the three is the least significant digit


EXAMPLEEXAMPLE

Ex 3-49

Step 3Draw a vertical line to the right of the term with the least accuracy (the least significant digit that is farthest to the left). 0.0146 0.0950 + 0.043

Step 4Do the math as normal. 0.0146 0.0950 + 0.043 0.1526


EXAMPLEEXAMPLE

Ex 3-49

Step 5Round the digit to the left of the line following the rounding rules.

0.1526 rounds to 0.153 (Rounding Rule 1)


EXAMPLEEXAMPLE

Ex 3-50EX 3-51

Example 3-50Add 850 and 5.90

850 + 5.90 855.90

855.90 rounds to 860 (Rounding Rule 3)

Example 3-51Subtract 5.6 from 875

875 – 5.6 869.4

869.4 rounds to 869 (Rounding Rule 1)


EXAMPLEEXAMPLE

Ex 3-52EX 3-53

Example 3-52Subtract 85 from 1,000,000

1,000,000– 85

999,915999,915 rounds to 1,000,000 (Rounding Rule 2)

Example 3-53Subtract 0.375 from 0.5

0.5– 0.375 0.125

0.125 rounds to 0.1 (Rounding Rule 1)


EXAMPLEEXAMPLE

Ex 3-54

The altimeter in an airplane reads to the nearest 100 ft. A cargo plane is cruising at 35,500 ft. Inside the cargo plane are crates. The crates are each 4 ft tall. These crates are stacked five high. On top of the highest crate is a ball bearing which measures 0.350 inches in diameter. How far is the top of the ball bearing from the ground?

Add 35,500 ft + (5 × 4 ft) + 0.350 in.

If we convert 0.350 in. to ft the problem becomes:

35,500 ft 20 ft+ 0.029166 ft 35,520.029166 ft


EXAMPLEEXAMPLE

Ex 3-54

The altimeter is the least accurate measurement, and it controls the accuracy of the answer. Since the plane’s altitude is only accurate to within 100 ft, this controls the accuracy of the addition problem. By rule 2 above, the sum has the same accuracy as the least accurate measurement. The altimeter cannot distinguish between 35,500 ft and 35,520.029 ft.

Therefore the answer is 35,500 ft, not 35,520.029 ft.


EXAMPLEEXAMPLE

Ex 3-55

A scale used to weigh trucks is marked in tons (T). In an hour, the following weights are recorded:

18T, 22T, 17T, 19T, 25T, 30T, 11T, 8T.

a. Calculate the total weight of the trucks weighed.

The total is 150 T.

b. Calculate the average weight of all the trucks weighed.

The average is 18.75 T = 19 T

Since the scale can only measure to the nearest ton, the average cannot be more accurate than the scale.


EXAMPLEEXAMPLE

Ex 3-56

Multiply 3.3 and 0.025.

Step 13.3 has two significant digits (rule 3).0.025 has three significant digits (rule 3).3.3 0.025 = 0.0825

Step 3The answer contains two significant digits.

Step 4The most significant digit is 8.

Step 50.0825 rounds to 0.082 (rounding rule 3b)


EXAMPLEEXAMPLE

Ex 3-57

Multiply 1,780.0 and 0.050.

Step 11,780.0 has five significant digits (rule 3).0.050 has three significant digits (rule 3).

Step 2(1,780)(0.050) = 89

Step 3Answer has three significant digits.

Step 4The most significant digit is the 8 followed by two other significant digits. 89.0

Step 589 rounds to 89.0 (rounding rule 1) (Note decimal and zero)


EXAMPLEEXAMPLE

Ex 3-58

Divide 23.5 into 180,000

Step 123.5 has three significant digits (rule 3)180,000 has two significant digits (rule 2)

Step 2180,000 ÷ 23.5 = 7,659.5745

Step 3Answer will contain two significant digits.

Step 4The most significant digit is the first 7 followed by one other significant digit.

Step 57,659.5745 rounds to 7,700 (rounding rule 3a).

(Note no decimal.)


EXAMPLE EXAMPLE

Ex 3-59

Divide 888 by 464.

Step 1888 has three significant digits (rule 2).464 has three significant digits (rule 2).

Step 2888 ÷ 464 = 1.9137931

Step 3Answer will have three significant digits.

Step 4The most significant digit is one and is followed by two more significant digits.

Step 51.9137931 rounds to 1.91 (rounding rule 2).


EXAMPLEEXAMPLE

Ex 3-60

Calculate the area of a rectangle measuring 4.473 in by 6.238 in.

4.473 in 6.238 in= 27.902574 in2

= 27.90 in2

Both measurements have the same accuracy, four significant digits.

.


EXAMPLEEXAMPLE

Ex 3-61

Calculate the area of a rectangle measuring 9.825 in by 3.0 in.

9.825 in 3.0 in = 29.475 in2

= 29 in2

The measurement with the least accuracy has two significant digits. Therefore the answer must have two significant digits.

Documents

ABC/ Mathematics / Chapter 3 / TP 3 - 1 / Rev 1 © 2003 General Physics Corporation OBJECTIVES 1.Without a calculator; ADD, SUBTRACT, MULTIPLY, and DIVIDE