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

2. ABC/ Mathematics / Chapter 3 / TP 3 - 2 / Rev 1 © 2003 General Physics Corporation DECIMAL PLACE RELATIONSHIPS Fig 3-1

3. ABC/ Mathematics / Chapter 3 / TP 3 - 3 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-1 The magnitude of 7361.298 is: DigitPlace Value 7  1,000=7,000 3  100= 300 6  10 = 60 1  1= 1 2  1/10= 2/10 = 200/1,000 9  1/100= 9/100 = 90/1,000 8  1/1,000= 8/1,000 = 8/1,000

4. ABC/ Mathematics / Chapter 3 / TP 3 - 4 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-1 Sum=7,361+ = 7,361 =7,361.298

5. ABC/ Mathematics / Chapter 3 / TP 3 - 5 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-2 Find the decimal equivalent of

6. ABC/ Mathematics / Chapter 3 / TP 3 - 6 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-3 The decimal equivalent of = 2  3 = 0.6666….. is: where the ….. indicates successive 6’s.

7. ABC/ Mathematics / Chapter 3 / TP 3 - 7 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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. ABC/ Mathematics / Chapter 3 / TP 3 - 8 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-4 Then cancel out each factor that occurs in each term. Then multiply the factors in each term together. Thus simplified, 0.375 in a fractional form is

9. ABC/ Mathematics / Chapter 3 / TP 3 - 9 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-5 Find the sum of 39.62, 41.093, and 0.0327. 39.62 41.093 + 0.0327 80.7457

10. ABC/ Mathematics / Chapter 3 / TP 3 - 10 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-6 Subtraction is done in the same manner as with whole numbers. 32.100 – 16.379 15.721

11. ABC/ Mathematics / Chapter 3 / TP 3 - 11 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-7 Multiply 16.2 and 1.15. Multiply without concern for the decimal. 162  115 810 162 18630

12. ABC/ Mathematics / Chapter 3 / TP 3 - 12 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-8 Multiply 16.2 and 1.15.

13. ABC/ Mathematics / Chapter 3 / TP 3 - 13 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-9 Divide 41.05 by 2.5. Divide without concern for the decimal. 25 160 150 105 100 50 0

14. ABC/ Mathematics / Chapter 3 / TP 3 - 14 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-10 Divide 41.05 by 2.5.

15. ABC/ Mathematics / Chapter 3 / TP 3 - 15 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-14 Example 3-12 Convert 0.25 to a percentage. 0.25  100% = 25% Example 3-13 Convert 2 to a percentage. 2  100% = 200% Example 3-14 Convert 1.25 to a percentage. 1.25  100% = 125% Ex 3-12 Ex 3-13

16. ABC/ Mathematics / Chapter 3 / TP 3 - 16 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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

17. ABC/ Mathematics / Chapter 3 / TP 3 - 17 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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.

18. ABC/ Mathematics / Chapter 3 / TP 3 - 18 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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.

19. ABC/ Mathematics / Chapter 3 / TP 3 - 19 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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.

20. ABC/ Mathematics / Chapter 3 / TP 3 - 20 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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.

21. ABC/ Mathematics / Chapter 3 / TP 3 - 21 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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 $137.85 $150.00 $ 1,306.70 4 valves cost $145.25 each 2 valves cost $137.85 each 3 valves cost $150.00 each

22. ABC/ Mathematics / Chapter 3 / TP 3 - 22 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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

23. ABC/ Mathematics / Chapter 3 / TP 3 - 23 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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 in 30 in = 2.5 ft 1.5 ft =18 in 18 in = 1.5 ft

24. ABC/ Mathematics / Chapter 3 / TP 3 - 24 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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 2 4 items Step 3 7.5 ft ÷ 4 = 1.875 ft

25. ABC/ Mathematics / Chapter 3 / TP 3 - 25 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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 2 4 items Step 3 90 in ÷ 4 = 22.5 in

26. ABC/ Mathematics / Chapter 3 / TP 3 - 26 / Rev 1 © 2003 General Physics Corporation THREE TEMPERATURE SCALES Fig 3-2

27. ABC/ Mathematics / Chapter 3 / TP 3 - 27 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-41 Measured Number Most Significant Digits (left) Least Significant Digits Number of Significant Digits 12,345155 123.45155 1,986164 37.806365 201213 300.7374 500.08585 0.00875853

28. ABC/ Mathematics / Chapter 3 / TP 3 - 28 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-41 Measured Number Most Significant Digits (left) Least Significant Digits Number of Significant Digits 900.0309 right ‑ most 0 6 0.0909 right ‑ most 0 2 200221 200.2 right ‑ most 0 3 200.02 right ‑ most 0 4 200.002 right ‑ most 0 5

29. ABC/ Mathematics / Chapter 3 / TP 3 - 29 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-42 Yard Stick Length8.0” Width5.0” Calculated Area40 sq. in. Significant Digits2 Correct Answer40. sq. in.

30. ABC/ Mathematics / Chapter 3 / TP 3 - 30 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-43 Ruler Length8.2” Width4.8” Calculated Area39.36”. Significant Digits2 Correct Answer39 sq. in.

31. ABC/ Mathematics / Chapter 3 / TP 3 - 31 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-44 Micrometer Length8.164” Width4.795” Calculated Area39.14638 sq. in. Significant Digits4 Correct Answer39.15 sq. in.

32. ABC/ Mathematics / Chapter 3 / TP 3 - 32 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-45 Round to two figures 0.1930.19 Round to five figures 157,632157,630 Round to three figures 7,5917,590 Round to four figures 0.987640.9876

33. ABC/ Mathematics / Chapter 3 / TP 3 - 33 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-46 Round to one figure 0.1930.2 Round to three figures 157,632158,000 Round to two figures 7,5917,600 Round to four figures 0.987640.9877

34. ABC/ Mathematics / Chapter 3 / TP 3 - 34 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-47 Round to two figures 0.18530.19 Round to four figures 195,753195,800 Round to one figure 7,5918,000 Round to three figures 0.98751 0.988 Round to two figures 18,50119,000 Round to four figures 19,555,005 19,560,000

35. ABC/ Mathematics / Chapter 3 / TP 3 - 35 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-48 Round to two figures 0.185000.18 Round to four figures 195,750195,800 Round to one figure 7,5008,000 Round to one figure 6,500 6,000 Round to three figures 0.984500.984

36. ABC/ Mathematics / Chapter 3 / TP 3 - 36 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-49 Add 0.0146, 0.0950, and 0.43 Step 1 Set up to do the math as normal. 0.0146 0.0950 + 0.043 Step 2 Identify the least significant digits in each term to be added or subtracted. 0.0146the six is the least significant digit 0.0950the zero is the least significant digit + 0.043 the three is the least significant digit

37. ABC/ Mathematics / Chapter 3 / TP 3 - 37 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-49 Step 3 Draw 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 4 Do the math as normal. 0.0146 0.0950 + 0.043 0.1526

38. ABC/ Mathematics / Chapter 3 / TP 3 - 38 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-49 Step 5 Round the digit to the left of the line following the rounding rules. 0.1526 rounds to 0.153 (Rounding Rule 1)

39. ABC/ Mathematics / Chapter 3 / TP 3 - 39 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-50 EX 3-51 Example 3-50 Add 850 and 5.90 850 + 5.90 855.90 855.90 rounds to 860 (Rounding Rule 3) Example 3-51 Subtract 5.6 from 875 875 – 5.6 869.4 869.4 rounds to 869 (Rounding Rule 1)

40. ABC/ Mathematics / Chapter 3 / TP 3 - 40 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-52 EX 3-53 Example 3-52 Subtract 85 from 1,000,000 1,000,000 – 85 999,915 999,915 rounds to 1,000,000 (Rounding Rule 2) Example 3-53 Subtract 0.375 from 0.5 0.5 – 0.375 0.125 0.125 rounds to 0.1 (Rounding Rule 1)

41. ABC/ Mathematics / Chapter 3 / TP 3 - 41 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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

42. ABC/ Mathematics / Chapter 3 / TP 3 - 42 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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.

43. ABC/ Mathematics / Chapter 3 / TP 3 - 43 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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.

44. ABC/ Mathematics / Chapter 3 / TP 3 - 44 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-56 Multiply 3.3 and 0.025. Step 1 3.3 has two significant digits (rule 3). 0.025 has three significant digits (rule 3). 3.3  0.025 = 0.0825 Step 3 The answer contains two significant digits. Step 4 The most significant digit is 8. Step 5 0.0825 rounds to 0.082 (rounding rule 3b)

45. ABC/ Mathematics / Chapter 3 / TP 3 - 45 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-57 Multiply 1,780.0 and 0.050. Step 1 1,780.0 has five significant digits (rule 3). 0.050 has three significant digits (rule 3). Step 2 (1,780)(0.050) = 89 Step 3 Answer has three significant digits. Step 4 The most significant digit is the 8 followed by two other significant digits. 89.0 Step 5 89 rounds to 89.0 (rounding rule 1) (Note decimal and zero)

46. ABC/ Mathematics / Chapter 3 / TP 3 - 46 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-58 Divide 23.5 into 180,000 Step 1 23.5 has three significant digits (rule 3) 180,000 has two significant digits (rule 2) Step 2 180,000 ÷ 23.5 = 7,659.5745 Step 3 Answer will contain two significant digits. Step 4 The most significant digit is the first 7 followed by one other significant digit. Step 5 7,659.5745 rounds to 7,700 (rounding rule 3a). (Note no decimal.)

47. ABC/ Mathematics / Chapter 3 / TP 3 - 47 / Rev 1 © 2003 General Physics Corporation EXAMPLE Ex 3-59 Divide 888 by 464. Step 1 888 has three significant digits (rule 2). 464 has three significant digits (rule 2). Step 2 888 ÷ 464 = 1.9137931 Step 3 Answer will have three significant digits. Step 4 The most significant digit is one and is followed by two more significant digits. Step 5 1.9137931 rounds to 1.91 (rounding rule 2).

48. ABC/ Mathematics / Chapter 3 / TP 3 - 48 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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 in 2 = 27.90 in 2 Both measurements have the same accuracy, four significant digits..

49. ABC/ Mathematics / Chapter 3 / TP 3 - 49 / Rev 1 © 2003 General Physics Corporation EXAMPLE 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 in 2 = 29 in 2 The measurement with the least accuracy has two significant digits. Therefore the answer must have two significant digits.