1. Appreciation and Depreciation

2. Appreciation refers to an amount increasing.

3. Appreciation and Depreciation

4. Depreciation refers to an amount decreasing.

5. We often refer to appreciation and depreciation when discussing percentages

6. For example, if a house appreciates in value by 6% each year, then its value increases by 6% each year.

7. On the other hand, if a car depreciates in value by 10% each year, then its value decreases by 10% each year.

8. Example of an appreciation question

9. Jojo purchases a new house for $215000. She expects it to appreciate in value by 8% each year.

10. Assuming she is right, how much will the house be worth

11. a) 1 year from now?

12. Assuming she is right, how much will the house be worth a) 1 year from now? b) 2 years from now?

13. Assuming she is right, how much will the house be worth a) 1 year from now? b) 2 years from now? c) 3 years from now?

14. a) 215000 x 0.08 = 17200

15. Since the value is increasing 8%, we add 17200 to the original value

16. 215000 + 17200 = 232200

17. Therefore, the house is worth $232200 after 1 year.

18. b) 232200 x 0.08 = 18576

19. Since the value is increasing 8%, we add 18576

20. 232200 + 18576 = 250576

21. Therefore, the house is worth $250576 after 2 years.

22. c) 250576 x 0.08 = 20046.08

23. Since the value is increasing 8%, we add 20046.08

24. 250576 + 20046.08 = 270622.08

25. Therefore, the house is worth $270622.08 after 3 years.

26. Example of a Depreciation Question

27. Example: Jojo’s car is worth $10000 today, but it will depreciate by 12% each year.

28. How much will the car be worth:

29. a)1 year from today?

30. How much will the car be worth: a)1 year from today? b) 2 years from today?

31. How much will the car be worth: a)1 year from today? b) 2 years from today? c) 3 years from today?

32. a) 10000 x 0.12 = 1200

33. Since the value is decreasing 12%, we subtract 1200 from the original value

34. 10000 – 1200 = 8800

35. Therefore, the car is worth $8800 after 1 year.

36. b) 8800 x 0.12 = 1056

37. Since the value is decreasing 12%, we subtract 1056

38. 8800 – 1056 = 7744

39. Therefore, the car is worth $7744 after 2 years.

40. c) 7744 x 0.12 = 929.28

41. Since the value is decreasing 12%, we subtract 929.28

42. 7744 – 929.28 = 6814.72

43. Therefore, the car is worth $6814.72 after 3 years.

45. Is there an easier way to do this?

47. Yes there is, Johnny

48. We simply need to convert the percentage that we are appreciating or depreciating to a decimal.

49. Then, if it’s an appreciation question, we add that quantity to 1 to give us our multiplier (i.e., the base of the exponent).

50. On the other hand, if it’s a depreciation question, we subtract from 1 to give us our multiplier (i.e., the base of the exponent).

51. Then, we use a formula similar to the formula that we used for compound interest and for half- lives.

52. For example, go back to our question about Jojo and her house. It is originally worth $215000. This is the “prior” amount, P.

53. We are increasing the value by 8%. So, we convert 8% to a decimal, then add it to 1 (since it’s an appreciation question)

54. We are increasing the value by 8%. So, we convert 8% to a decimal, then add it to 1 1 +.08 = 1.08

55. We are increasing the value by 8%. So, we convert 8% to a decimal, then add it to 1 1 +.08 = 1.08 This is our multiplier

56. 1 After 1 year: A = P (1.08)

57. 1 After 1 year: A = 215000 (1.08)

58. After 1 year: A = 215000 (1.08)

59. After 1 year: A = 232200

60. 2 After 2 years: A = P (1.08)

61. 2 After 2 years: A = 215000 (1.08)

62. After 2 years: A = 215000 (1.1664)

63. After 2 years: A = 250776

64. 3 After 3 years: A = P (1.08)

65. 3 After 3 years: A = 215000 (1.08)

66. After 3 years: A = 215000 (1.259712)

67. After 3 years: A = 270838.08

68. This method allows us to cut right to the answer and not have to figure out EVERY year leading up to our answer

69. For example, when we were discussing the value of Jojo’s car, we calculated that after 3 years, its value would have decreased to $6814.72

70. We can calculate this a lot quicker as follows

71. We know that it is depreciating by 12%

72. We convert 12% to a decimal

73. 12% = 0.12

74. Then, we subtract that amount from 1 to get our multiplier (since it’s a depreciation question)

75. 1 – 0.12 = 0.88

76. 3 After 3 years: A = P (0.88)

77. 3 After 3 years: A = 10000 (0.88)

78. After 3 years: A = 10000 (0.681472)

79. After 3 years: A = 6814.72

80. After 3 years: A = 6814.72 Therefore, after 3 years the car is worth $6814.72

81. Final note about these questions!!!

82. Please note that the question does not have to say “appreciate” or “depreciate”.

83. Other terms such as “increase”, or “decrease”, or “gain value”, or “decline” may be used, as well as others.