1. MAT 110 Workshop
Created by Michael Brown, Haden McDonald & Myra Bentley
for use by the Center for Academic Support

2. Unit 4: Finance
Introduction

3. Definitions
Interest: The amount paid to a lender for use of their money, or the amount your account grows due to an investment.
Principal: Usually the initial amount borrowed from a lender or the amount deposited to the bank in a saving account.
Interest Rate: The amount of interest charged on the principal. This rate can be simple, or compound, and can be annual (yearly) or many other variants, monthly, quarterly, semi annual, etc., and it must be expressed in decimal form when performing calculations.
Future Value: The amount that will be present in an account or owed on a loan in the future.
Annuity: A type of compound interest, where payments are made at regular periods rather than in one sum.
Finance Charge: Another term for the interest charged on a loan.
Amortization: The process of paying off a loan in a series of regular payments.
Present Value: In the case of amortization, the amount initially borrowed or loaned.
APR: Annual percentage rate, a measure of the true interest being charged on a transaction.

4. Simple Interest
Interest is the money that one person (a borrower) pays to another (a lender) to use the lender’s money.
The amount you deposit in a bank account is called the principal.
The bank specifies an interest rate for that account as a percentage of your deposit.

5. Simple Interest

6. Simple Interest
Example: $500 is deposited in a bank account paying 6% simple interest, how much interest will the deposit earn in 4 years?
Solution:
P = $500 (principal)
r = 0.06 (rate)
t = 4 (time)

7. Simple Interest
To find the amount that will be in your account at some time in the future, called the future value (or sometimes called the future amount) we add the principal and the interest earned. We will represent future value by A.

8. Simple Interest

9. Simple Interest
Example: $1,000 is deposited in a bank account paying 3% annual interest for 6 years. Compute the future value of this account.
Solution:
We see that P = 1,000, r = 0.03, and t = 6.

10. Simple Interest
Example: $2,500 is needed in 2 years for a purchase. A bank offers a certificate of deposit (CD) that pays 4% annual interest computed using simple interest. How much must be put in this CD now to have the necessary money in 2 years?
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11. Simple Interest
Solution:
We see that A = 2,500, r = 0.04, and t = 2.

12. Simple Interest Problems
A) Find the amount due if $200 is borrowed for 14 months at 9% simple interest.
B) If $700 is borrowed and the interest after 18 months is $136.5, what is the annual interest rate for a simple interest loan?
C) An investment pays simple interest, and triples in 14 years. What is the interest rate?

13. Simple Interest Problems
A) 200(1+.09* )= $221.00
B) ( )=700(1+r ) so the rate is .13 or 13%
C) 3=1(1+r*14) the rate is or %

14. Compound Interest
Interest that is paid on principal plus previously earned interest is called compound interest.
If the interest is added yearly, we say that the interest is compounded annually.
If the interest is added every three months, we say the interest is compounded quarterly.

15. Compound Interest
Example: $2,000 is deposited for 3 years in a bank account that pays 10% annual interest, compounded annually. How much will be in the account at the end of 3 years?
Solution:
Calculations, one year at a time, are in the table.

16. Compound Interest
For annual compounding:

17. Solving for Unknowns in the Compound Interest Formula

18. Solving for Unknowns in the Compound Interest Formula
Example: An item is $3,500 with no payments due for 6 months. Although no payments are made, the dealer is not loaning the money for 6 months for nothing. $3,500 was borrowed and, in 6 months, the payments will be based upon that fact. If the interest rate is 12%, compounded monthly, what interest will accumulate on the purchase over the next 6 months?
(continued on next slide)

19. Solving for Unknowns in the Compound Interest Formula
Solution:
We see that P = 3,500, r = 0.12, m = 12, and n = 6.

20. Solving for Unknowns in the Compound Interest Formula
Example: Upon a child’s birth, a deposit is made into an account. Assume that the account has an annual interest rate of 4.8% and that the compounding is done quarterly. How much must be deposited now so that the child will have $60,000 at age 18?
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21. Solving for Unknowns in the Compound Interest Formula
Solution:
We see that A = 60,000, r = 0.048, m = 72, and n = 4.

22. Solving for Unknowns in the Compound Interest Formula
To solve the equation for n, you will need to use a property of logarithmic functions. Many calculators have a key labeled either “log” or “log x,” which stands for the common logarithmic function.

23. Solving for Unknowns in the Compound Interest Formula
Example: $2,800 is needed for a purchase. The buyer has $2,500 and will invest the money at 9% annual interest, compounded monthly. In how many months will the $2,500 become $2,800?
(continued on next slide)

24. Solving for Unknowns in the Compound Interest Formula
Solution:
We see that A = 2,800, P = 2,500, r = 0.09, and m = 12. We wish to solve for n.

25. Solving for Unknowns in the Compound Interest Formula
Example: $1.4 million is to be invested now to be paid as $1.68 million in 2 years. What rate of investment will yield $1.68 million in 2 years? Assume an annual interest rate that is compounded monthly.
(continued on next slide)

26. Solving for Unknowns in the Compound Interest Formula
Solution:
We see that A = 1.68, P = 1.4, m = 12, and n = 24. We wish to solve for r.
r =

27. Calculating the Time for an Investment to Double
The Rule of 70 – To estimate the doubling time of a quantity, divide 70 by the growth rate.

28. Calculating the Time for an Investment to Double
Example: Let’s say that you have $1,500 to invest. We want to find the time in years that it takes the investment to double with a growth rate of 3.5% and annual compounding, first by using the compounding formula then by using the Rule of 70.
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29. Calculating the Time for an Investment to Double
Solution:
P = $1, Initial investment
A = $3,000 - Double original amount of $1,500
R = Annual interest rate
M = 1- Compounding, once a year
Using the future value formula:
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30. Calculating the Time for an Investment to Double
Solution:
Using the Rule of 70: 70/3.5 = 20 yrs

31. Calculating the Time for an Investment to Double
Example: Collectors often buy classic DVDs hoping to get a good return on their investment. DeDe has invested $8,500 in old Stars Wars DVDs, which are currently increasing in value at 6.5% per year. If this increase continues at the same rate, in how many years will DeDe’s investment double?
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32. Calculating the Time for an Investment to Double
Solution: We are looking for a future value of 2 x $8,500 or $17,000. We can use the future value formula given by:
A = 17,000 - Double original amount of $8,500
P = 8, Initial investment
R = Annual interest rate
M = 1- Compounding, once a year
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33. Calculating the Time for an Investment to Double
Solution:
Now we will solve the equation for n
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34. Calculating the Time for an Investment to Double
Solution: solve the equation for n
DeDe’s investment will double in about 11 years.
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35. Calculating the Time for an Investment to Double
Solution: Using the Rule of 70:
70/6.5 =
Again we see DeDe’s investment will double in about 11 years.

36. Compound Interest
If $800 is invested at 6% compounded quarterly, what is the interest earned after:
A) 5 years
B) 4 years

37. Compound Interest A) 800 (1+ 0.06 4 ) 4∗5 = $1077.48
= $ the interest is $277.48
B) 800 ( ) 4∗4 = $
= $215.19

38. Compounded Interest
Sam invests $ 7500 into an account earning 9% interest compounded quarterly. How long will it take to double his money?

39. Compounded Interest Use the formula 15000=7500 (1+ 0.09 4 ) 4𝑡
2= ( ) 4𝑡
Ln(2)=4t ln( )
Ln(2)/ln( )=4t
=4t
T= /4
T= years

40. Compound Continuous
A special type of compound interest - compounded continuously - uses the above formula, where e is just Euler’s constant (~2.71).
Algebraically continuous interest problems can be worked exactly the same as compound m-thly interest problems.
Note that for compound continuous problems, using the natural log when solving for r or t simplifies the process by the identity: ln(e) = 1.

41. Compounded Continuously
If Hannah places $ 7500 into an account earning 7% interest compounded continuously, how much will she have after 7 years?

42. Compounded Continuously
Use formula A=Pe^(rt)
A=7500e^(.07*7)
A=7500e^(.49)
A=$

43. Compounded Continuously
Kyle invests in an account earning 7.5% interest compounded continuously. How long will it take to double his investment?

44. Compounded Continuously
Use formula A=Pe^(rt)
2=1e(.075t)
2=e(.075t)
ln(2) = .075t ln(e)
ln(2)=.075t
t= ln(2)/.075
t= years

45. Annuities
An annuity is an interest-bearing account into which we make a series of payments of the same size.
If one payment is made at the end of every compounding period, the annuity is called an ordinary annuity.
The future value of an annuity is the amount in the account, including interest, after making all payments.

46. Annuities
Suppose that in January you begin making payments of $100 at the end of each month into an account paying 12% yearly interest compounded monthly.
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47. Annuities We may express the value of this annuity as
If we compute how much each deposit contributes to the account and sum these amounts, we will have the value of the annuity on July 1.
We may express the value of this annuity as
Factoring out 100, we can write this in the form

48. Annuities
Example:
Show that
Solution: Multiply by
This shows that

49. Annuities From the previous example, we have
Returning to our earlier example, we have
and multiplying by $100, we obtain

50. Annuities

51. Annuities
Example: A payment of $50 is made at the end of each month into an account paying a 6% annual interest rate, compounded monthly. How much will be in that account after 3 years?
Solution:
We see that R = 50, = and n =
(continued on next slide)

52. Annuities
Using the formula for finding the future value of an ordinary annuity, we get

53. Sinking Funds
You may want to save regularly to have a fixed amount available in the future. The account that you establish for your deposits is called a sinking fund.
Because a sinking fund is a special type of annuity, it is not necessary to find a new formula. We can use the formula for calculating the future value of
an ordinary annuity that we have stated earlier. In this case, we will know the value of A and we will want to find R.

54. Sinking Funds
Example: Assume that you wish to save $1,800 in a sinking fund in 2 years. The account pays 6% compounded quarterly and you will also make payments quarterly. What should be your
monthly payment?
Solution: Recall the formula for finding the future value of an ordinary annuity:
(continued on next slide)

55. Sinking Funds
We see that A = 1,800, = and n =

56. Sinking Funds
Example: Suppose you have decided to retire as soon as you have saved $1,000,000. Your plan is to put $200 each month into an ordinary annuity that pays an annual interest rate of 8%. In how many years will you be able to retire?
Solution:
We see that A = 1,000,000, = and R = 200.
(continued on next slide)

57. Sinking Funds
We solve this equation for n.

58. Ordinary Annuity Example
Assume that you make monthly payments of $ 725 into an ordinary annuity paying 6% compounded monthly. How much will be in the account after 8 years?

59. Ordinary Annuity Example
Use the Formula
A=725 ( ) 12∗8 −
A=$

60. Amortization
The process of paying off a loan (plus interest) by making a series of regular, equal payments is called
amortization, and such a loan is called an amortized loan.

61. Amortization
Example: An amortized loan of $10,000 is made to pay off a car in 4 years. If the yearly interest rate is 18%, what is your monthly payment?
Solution: We know the following values.
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62. Amortization
We must solve for R in the equation

63. Amortization Schedule
Example: $5,000 is borrowed at a 12% annual interest rate, and will be paid back in three equal monthly installments of $1, Construct an amortization schedule for this loan.
(continued on next slide)

64. Amortization Schedule
Solution:
First month’s interest:
Money applied to principal:
$1, – $50 = $1,
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65. Amortization Schedule
Second month’s balance:
$5,000 – $1, = $3,
The table shows the rest of the computations for this problem.

66. Amortization Schedule
Example: You wish to borrow $120,000 to buy a house. A bank offers a 30-year mortgage at an annual rate of 7%. The monthly payment is $ Construct an amortization schedule for the first three payments on this loan.
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67. Amortization Schedule
Solution:

68. Finding the Present Value of an Annuity

69. Finding the Present Value of an Annuity
Example: You can afford to spend $200 each month on car payments. A bank offers you a 4-year car loan with an annual rate of 12%. what is the present value of this annuity?
Solution: We can use the formula for finding payments on an amortized loan:
(continued on next slide)

70. Finding the Present Value of an Annuity
We know and

71. Finding the Unpaid Balance of a Loan
Example: Suppose you have a 30-year mortgage for $100,000 at an annual interest rate of 9%. After 10 years, you refinance. How much remains to be paid on your mortgage?
The remaining 20 years is financed at an annual interest rate of 7.2%. What are the monthly payments?
How much will you save in interest in 20 years by paying the lower rate?
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72. Finding the Unpaid Balance of a Loan
Solution: We can find that the monthly payment is $ based on the fact that your loan was a 30-year loan.
The unpaid balance U on the loan is
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73. Finding the Unpaid Balance of a Loan
We know
and
Therefore, you still owe $89, on this mortgage.
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74. Finding the Unpaid Balance of a Loan
In essence, you are taking out a new loan with
and
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75. Finding the Unpaid Balance of a Loan
Payment reduction: $ – $ = $ per month.
Total amount paid over 20 years at the old interest rate: 240 × $ = $193,
Total amount paid over 20 years at the new interest rate: 240 × $ = $168,
Amount saved: $193, – $168, = $24,

76. Simple/Compound Interest Review/Solutions
$800 is currently in a bank account that has been earning 5% simple interest for 6 years. What principal was originally deposited into this account?
In 5 years a principal of $1000, earning interest compounded continuously, has doubled in value. What is the interest rate on this account?
P = $615.38
For the first problem:
A = 800, r = .05, t = 6, P = A / (1 + rt)
For the second problem:
A = 2P, P = 1000, t = 5, r = ln(2P/P) / t
r =.1386 or 13.86%

77. Annuities/Amortization Review/Solutions
A payment of $125 is made at the end of each month into an account paying a 7% annual interest rate, compounded monthly. How much will be in that account after 20 years?
A = $
An amortized loan of $45,000 is made to pay off a large debt in 10 years. If the yearly interest rate is 15%, what is your monthly payment?
First:
R = 125, r = .07, m = 12, t = 20
Second:
P = 45000, r = .15, m = 12, t = 10
R = $726.01

78. Schedules/Unpaid Balance Review
$4,000 is borrowed at a 8% annual interest rate, and will be paid back in 6 years with monthly installments. What is the monthly payment? What is the unpaid balance after the third payment?
Construct the first 3 lines of the amortization schedule for this loan to verify your result.

79. Unpaid Balance Solutions
P = $4,000, r = 8%, t = 6 years, m = 12
a) monthly payment? b) unpaid balance after the third payment?
R = $70.13
U = $
What is the payment?
P = 4000, r = .08, m = 12, t = 6.
What is the balance after the third payment?
P = 4000, r = .08, m = 12, n = 3, R = 70.13

80. Amortization Schedule Solutions
Payment #
Amount of Payment
Interest Payment
Applied to Principal
Balance $ 1
$70.13
$26.67
$43.46 $ 2
$26.38
$43.75 $ 3
$26.09
$44.04
Amount of Payment = R
Interest Payment = Balance * r * t
Applied to Principal = Amount of Payment - Interest Payment
Balance = Balance - Applied to Principal