Simple And Compound Interest Module

Introduction Money! Who doesn’t want it? Money become essential thing in everyone’s life. You can buy what you want if you have lot of it. People used to shop, pay for their debts, bills, fand the like. If you have plenty of money, it seems that life is happier and enjoyable. Having enough knowledge how to handle money will bring you success in the financial aspect of life. Becoming familiar with money matter might help you to decide the best option how to spend, save or invest it. Money is not free to borrow. People can always find a use for money, so it costs to borrow money. If you borrow money from private persons, banks or any financial institutions or lending companies, you are expected to pay back more than you borrowed. How much you more you pay back depends on your agreement as to how much interest rate agreed upon. In the same manner, the banks will pay you interest with the saving accounts or money you deposited with them. As you go on with the module, you will learn how invested or borrowed money getting higher for a particular period of time. The given enhancement exercises will test yourself to enhance your skills in applying the concepts you have learned after each lesson. Are you now excited to take your race in interest? Read and analyze carefully each lesson and you will be amazed on how these concepts excite the cells of your brain.

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Learning Competencies: The students expected to:

are

 illustrate simple and compound interests  distinguish between simple and compound interests  compute interest, maturity value, and present value in simple interest environment and compound interest environment  solve problems involving simple and compound interests

Terms to Ponder On!  Interest  Maturity (Future) Value  Present Value  Simple Interest  Compound Interest  Nominal rate  Effective rate  Equivalent rate

Before dealing with the lessons of this module, let us try to measure your prior knowledge on Interest.

PRE-TEST Direction: Write the letter of the correct answer on the space provided before each number. _____ 1. Determine the simple interest to be paid by Pia Catriona if she borrowed Php 2,345.00 1 at 3 5 % for 10 months. A. Php 26.35

B. Php 67.25

C. Php 62.53

D. 65.32

_____ 2. How much was borrowed by Cardo Dalisay if the interest he paid after 9 months at 12.25% simple interest is Php 2,468? A. Php 28, 266.59

B. Php 26, 268.95

C. Php 22, 686.95

D. Php 26, 862.59

_____ 3. At what rate of interest did Ahmad invest the amount of Php 24, 500.00 for it to earn a simple interest of Php 4, 545.00 for 3.5 years? A. 3.5%

B. 4.5%

C. 5.3%

D. 5.5%

_____ 4. How long should Maria Linda invest the amount of Php 85, 430.00 to earn an interest 1 of Php 5, 555.00 at 3 4 % simple interest? A. 1.5 years

B. 2 years

C. 2.5 years

D. 3 years

_____ 5. How much is the total amount would Mrs. Soriano receive if she invest Php 123, 3 450.00 for 3.25 years at 2 4 % simple interest? A. Php 134, 483.34

B. Php 143, 384.34

C. Php 144, 834.43

D. Php 148, 343.43

_____ 6. Which of the following are NOT true? I. Principal is the money borrowed or invested on the origin data. II. Origin date is a date on which money is paid by the borrower. III. Interest is an amount paid or earned for the use of money. IV. Simple interest is an interest that is computed on the principal and then subtracted to it. A. I and II

B. II and III

C. III and IV

D. II and IV

_____ 7. Analyze the two statements below.

Statement 1: Equivalent rates refer to two annual rates with different conversion periods that will earn the same maturity value for the same time/term.

Statement 2: Effective rate is a rate when compounded annually will give different compound each year with the nominal rate. 2

A. Both statements are true B. Both statements are false C. Statement 2 is true but statement 1 is false D. Statement 1 is true but statement 2 is false For numbers 8 and 9, consider the statement below.

Gazini Ganados is thinking of investing an amount of Php 36, 960.00 for 5.75 years at 3 an interest rate of 8 5 % compounded annually. _____ 8. Find the maturity value (F). A. Php 53, 995.49

B. Php 55, 399.94

C. Php 59, 395.49

D. Php 59, 953.94

C. Php 22, 435.49

D. Php 25, 234.94

____ 9. Find the compound interest (Ic). A. Php 24, 243.94

B. Php 23, 542.49

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_____10. Mr. Duterte invested Php 234, 500.00 at 3 4 % compounded annually. He plans to get this amount after 45 months for his daughter’s debut party. How much will he get? A. Php 263, 184.20

B. Php 264, 381.20

C. Php 268, 143.20

D. Php 281, 346.20

_____ 11. Find the interest if Php 987, 650.00 is deposited in a bank at 4.15% compounded monthly for 39 months. A. Php 140, 349.50

B. Php 142, 347.50

C. Php 811, 479.50

D. Php 991, 542.50

_____ 12. Patricia Magtanong borrows Php 481, 215.00 and promises to pay the principal and 1 interest at 9 10 % compounded quarterly. How much must she pay after 7 years and 9 months? A. Php 966,480.93

B. Php 948,660.39

C. Php 986,406.93

D. Php 906,468.39

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_____ 13. If money is worth 6 5 % compounded semi-annually, find the present value of Php 500, 500.00 due at the end of 5 years and 3 months. A. Php 391, 53.20

B. Php 352, 320.91

C. Php 325, 193.10

D. Php 330, 252.19

_____ 14. How long will it take Php 7, 350.00 to amount to Php 18, 500.00 if invested in a bank at 8% compounded monthly? A. 8.51 years

B. 15.18 years

C. 11.58 years

D. 18.15 years

_____ 15. At what interest rate will Php 2, 050.00 amount Php 3, 875.00 in 4 years and 6 months, if interest is compounded semi-annually? A. 1.46%

B. 1.66%

C. 16.46% 3

D. 14.66%

_____ 16. If interest is compounded quarterly, find the nominal rate which will yield an 1 effective rate of 4 4 %. A. 1.48%

B. 1.84%

C. 4.18%

D. 4.81%

_____ 17. When interest is compounded monthly, find the effective rate corresponding to the nominal rate 3%. A. 4.03%

B. 3.04%

C. 3.4%

D. 4.3%

_____ 18. Maine Mendoza is investing Php 500, 000.00 at 15% converted semi-annually for 5 years. At what rate compounded monthly could she just as well invest her money? A. 14.55%

B. 15.54%

C. 1.45%

D. 5.54%

_____ 19. If you are to invest Php 130, 130.00 at 13% converted monthly for 6 years and 6 months, what simple interest rate could you just as well invest your principal for the same period of time? A. 18.72%

B. 20.27%

C. 22.70%

D. 27.22%

_____ 20. Vice Ganda plans to invest Php 100, 000.00 at 12.5% simple interest for 36 months. At what interest compounded quarterly could she just as well invest for the same period of time? A. 6.17%

B. 7.01%

C. 10.76%

D. 16.07%

=== END OF THE TEST ===

To check your answers, refer to page 62 of this module. Have you done a good job? If you are not satisfied with your score, do your best in dealing with the succeeding parts of this module. The lessons were prepared for you to enhance your knowledge on interest. Hoping that you will enjoy your race!

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Lesson 1 ILLUSTRATING SIMPLE AND COMPOUND INTEREST

Basic Concepts Almost everyday money is borrowed and loaned in thousands of transactions amounting, in total, to hundreds of millions of pesos. A bank is a financial institution which is involved in borrowing and lending money. Banks take customer deposits in return for paying customers an annual interest payment. The bank then uses the majority of these deposits to lend to other customers for a variety of loans. Borrowing and lending are two sides of the same transaction. To the lender loan represents an investment in a debt obligation. The interest charged provides income form the investment. The rate of return on the investment is equal to the rate of interest charged to the customer. In view of the fact that borrowing and lending are so central to our daily lives and in the economic system, it is essential to familiarize with the definition of terms and learn the computation of interest in borrowing/lending transactions.

Definition of Terms  Debtor or Maker – an individual or institution that borrows money for any purpose  Lender –an individual or financial institution which loans the money  Interest (I) – the payment for the use of borrowed money or the amount earned on invested money  Principal (P) – the capital or sum of money borrowed or invested  Rate of Interest (r) – this is a fractional part of the principal that is paid on the loan or investment which is usually expressed as percent  Time (t) – the number of years for which money is borrowed or invested  Maturity Value or Future Value (F) – the sum of the principal and the interest accumulated over a certain period

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To illustrate the difference between simple and compound interest, and better understand the concept of compound interest, consider the following example. Lovely and Lorainne have Php 1, 000.00 each. Lovely lent her money to a friend who agreed to pay her back after 5 years with 5% simple interest. Lorraine, on the other hand, deposited her money to a savings account and left it there for 5 years to earn 5% compounded annually. Guide Questions: 1. Who do you think will earn higher after 5 years? ___________________________________________________________________________________ 2. How much is their money’s worth after 5 years? ___________________________________________________________________________________ ___________________________________________________________________________________

The chart below shows how the simple and compound interests are computed. Lovely’s investment at 5% simple interest Time (t) 0 1

2

3

4

5

Maturity Value =Php 1,000.00 =1,000 + 1,000 (0.05) (1) =1,000 + 50 =1,050.00 =1,000 + 1,000 (0.05) (2) =1,000 + 1,000 (0.1) =1,000 + 100 =1,100.00 =1,000 + 1,000 (0.05) (3) =1,000 + 1,000 (0.15) =1,000 + 150 =1,150.00 =1,000 + 1,000 (0.05) (4) =1,000 + 1,000(0.2) =1,000 + 200 =1,200.00 =1,000 + 1,000 (.005) (5) =1,000 + 1,000 (0.25) =1,000 + 250 =1,250.00

Simple Interest (Is) 0 Php 50.00

Lorainne’s investment at 5% compounded annually Time (t) 0 1

2 Php 100.00

3 Php 150.00

4 Php 200.00

5 Php 250.00

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Maturity Value =Php 1,000.00 =1,000 + 1,000 (0.05) (1) =1,000 + 50 =1,050.00

Compound Interest (Ic) 0 Php 50.00

=1,050 + 1,050 (0.05) (1) =1,050 + 52.50 =1,102.50

Php 102.50

=1,102.50 + 1,102.50 (0.05) (1) =1,102.50 + 55.13 =1,157.63

Php 157.63

=1,157.63 + 1,157.63 (0.05) (1) =1,157.63 + 57.88 =1,215.51

Php 215.51

=1,215.51 + 1,215.51 (0.05) (1) =1,215.51 + 60.78 =1,276.29

Php 276.29

Questions: 1. How much money will Lovely have after 5 years? How much interest she gained? ___________________________________________________________________________________ 2. How much money will Lorainne have after 5 years? How much interest she gained? ___________________________________________________________________________________ 3. What have you noticed with the computation? ___________________________________________________________________________________ ___________________________________________________________________________________ ___________________________________________________________________________________

Think of It With simple interest, interest stays the same as time passes. With compound interest, interest grows as time passes because previous interest is compounded or added to the principal to earn interest. The principal continuously changes during the time period.

Bear in Mind Compound interest is higher than simple interest when principal, rate and time are the same. However, when the interest is only compounded for one period, say 1 year, the compound interest and the simple interest are the same. We have seen the difference between simple and compound interest using computation. This time, let us study the comparison of facts underlying between simple and compound interests.

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8

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How do you find the comparison? Can you distinguish now between simple and compound interests? ___________________________________________________________________________________ ___________________________________________________________________________________ ___________________________________________________________________________________

Enhancement Exercise 1.1 A. Illustrate the given situation below: 1. Gazini is planning to invest her separation fee from her working company after she resigned as Payroll Supervisor. She received Php 230,000.00 and plan to invest it for 5 years. Which investment is better? A bank offers 7.5% simple interest a year or a lending company offers 7.5% compounded annually? 2. Bea Patricia wants her Php 35,700.00 to gain more when she used to invest it for 3 years. She is looking for a bank with best offers and services. Will she go for ABC Bank offering 4.25% simple interest rate per year or XYZ Bank offering 4.35% compounded annually? Justify your answer.

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B. Write SI if the statement below refers to simple interest and CI if compound interest. _____ 1. The product of principal, rate and time. _____ 2. It is useful for investing since it allows the fund to grow at a faster rate. _____ 3. Most of the car loans calculated based on it. _____ 4. Computation is very easy and easy to understand. _____ 5. Interest is charged on the principal and the interest amount. _____ 6. Principal and interest growth is rapid and increases at a fast pace. _____ 7. The principal keeps on changing due to the addition of accrued interest in the entire period. _____ 8. The principal is constant. _____ 9. It offers a comparatively high return to the lenders. _____ 10. It is the interest which is a percentage of both principal and accrued interest.

The Philippines isn’t the only country that uses peso as a currency. In 16th century, the Spanish peso had a value of eight pieces of real, the Spanish Empire’s currency. Because of this, most of the colonies of the Spanish empire adapted the currency, despite some countries discarding it. As of the moment, Argentina, Chile, Colombia, Cuba, Dominican Republic, Mexico, the Philippines, and Uruguay still use the peso.

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Lesson 2 SIMPLE INTEREST

Basic Concepts Interest is the fee for use of money that is either borrowed, lent or loaned, invested or saved over a period of time. Lenders ask for interest to make up for the inconvenience of not being able to use the money during the time the money is still with the borrower or on loan, to help make the risk of lending worth taking in case the borrower fails to pay back on time and to help make up for the inflation as it reduces the purchasing power of the amount borrowed. Simple Interest (Is) is the amount paid for the borrowed money or the amount earned for the deposited or invested money. It is calculated as a percentage of the original amount borrowed or deposited over a period of time. Calculating Simple Interest

The equation to compute for simple interest is – Is = Prt where Is = interest P = principal or the original amount borrowed or deposited r = percentage rate of interest t = term or period or length of borrowing time until payment Let us work on the following examples to illustrate the application of the above. Example 1: Mary Anthonette borrows Php 10, 000.00 at 10.25% interest rate a year. a. What is the principal (P)? b. What is the interest rate (r)? c. What is the time or borrowing period (t)? d. What is the interest after a year? 12

Solution: a. P = Php10, 000.00 b. r = 10.25% or 0.1025 c. t = 1 year d. Substituting the values in the given formula Is = Prt = (10, 000)(0.1025)(1) = Php 1, 025.00 1

Example 2. Kara Mia deposits Php 111, 110.00 in an account at her bank at 5 5 % interest rate. How much interest will she earn in 5.25 years? Given: P = 111,110 1 r = 5 % = 5.2% or 0.052 5 t = 5.25 years Find: Is

Solution: Is = Prt = (111, 110)(0.052)(5.25) = (111, 110)(0.273) = Php 30, 333.03 Example 3: Find the interest paid by Kathryn Nadine on Php 150, 150.00 that 3 she borrowed for three years and 9 months at 1 % simple interest. 4 Given: Note: P = 150, 150 3 When term is r = 1 % or 1.75% or 0.0175 4

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t = 3 years and 9 months = 3 = 3.75 years 12 Find: Is

Solution: Is = Prt = (150,150)(0.0175)(3.75) = (150,150)(0.065625) = Php 9,853.59

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expressed in month (m), convert in 𝑚 year(s) by t=12

Example 4: If Sahaya borrowed Php 125, 500.00 at an annual simple interest 3 rate if 7 % for 42 months, how much interest should she pay? 4 Given: P = 125, 500 3 r=7 % 7.75% or 0.0775 4

42

t = 42 months = 3.5 years 12 Find: Is Solution: Is = Prt = (125,500)(0.0775)(3.5) = (125,500)(0.27125) = Php 34,041.88

Enhancement Exercise 2.1 Answer each of the following problems by identifying the given information, required unknown, formula(s) to be used. Compute for the unknown then label your final answer properly. 1. Mrs. Batungbakal gave a Php 100, 000.00 to Mrs. Macaspac at 18%. How much was the interest after 45 days? 2. How much interest is paid after 8 months if Mr. Tom McKey borrowed P515, 515.00 at 3.25%? 3. Cory Kong deposited Php 202, 020.00 at a rural bank in Tayug paying 1.5% per annum simple interest. How much interest will she earn after 2 years and 6 months?

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Finding Principal, Rate, and Time

From the equation of the simple interest Is = Prt, you can also solve for the other variables; P, r and t. Equations for principal, rate, and time can be formulated using the Golden Rule of Equations –

“Equality is preserved if both sides of the equation are changed the same way.” To solve for P, 𝐼𝑠 𝑟𝑡

𝑃=

=

𝑃𝑟𝑡 𝑟𝑡



𝐼𝑠 𝑟𝑡

Equations for rate (r) and time (t) can be in a similar manner. To easily remember things, use the drawn circle below. Cover the letter to get its formula. Covering P, you will have

Is P r t

Covering r, you will have Covering t, you will have

𝐼𝑠

𝑰

𝐼𝑠

𝑰

𝐼𝑠

𝑰𝒔

𝒔 . Hence, P = 𝑟𝑡 𝒓𝒕 𝒔 . Hence, r = 𝑃𝑡 𝑷𝒕

. Hence, t = 𝑃𝑟

𝑷𝒓

Let us work on the following examples to illustrate the application of the above. Example 1: To earn Php 1, 500.00 in five years, how much money should you deposit in a bank at 3.5% simple interest? Given: Is = 1, 500 r = 3.5% or 0.035 t = 5 years Find: P 15

Solution: 𝐼𝑠

1,500

1,500

= (0.035)(5) = 0.175 = Php 8, 571.43 𝑟𝑡

P=

Example 2: How much was borrowed by Manisan if the interest she paid after 3 months at 12.5% simple interest is Php 3, 030.00? Given: Is = 3, 030 r = 12.5% or 0.125 3 t = 3 months = 0.25 year 12 Find: P Solution: P=

𝐼𝑠

3,030

3.030

= (0.125)(0.25) = 0.03125 = Php 96, 960.00 𝑟𝑡

Example 3: If a principal of Php 500, 005.00 earns interest of Php 50, 005.00 in 6 years and 6 months, what interest rate is in effect? Note:

Given: Is = 50, 005 P = 500, 005 t = 6 years and 6 months Find: r Solution: r=

𝐼𝑠

50,005

6

6 12

= 6.5 years

Rate is usually expressed in percentage form. Multiply the decimal number by 100 and attach % symbol.

50,005

= (500,005)(6.5)= 3,250,032.5 = 0.0154 or 1.54% 𝑃𝑡

Example 4: Ian Gabriel is charging a simple interest amounting to Php 3, 500.00 for Php 35, 000.00 which John Kerby will borrow and promise to pay after 5 years. What interest rate is used? Given: Is = 3, 500 P = 35, 000 t = 5 years Find: r 16

Solution: r=

𝐼𝑠

3,500

3,500

= (35,000)(5)= 175,000 = 0.02 or 2% 𝑃𝑡

Example 5: How long will it take Php 30, 000.00 invested at 9% simple interest to earn Php 5, 500.00? Given: P = 30, 000 r = 9% or 0.09 Is = 5, 500 Find: t Solution: t=

𝐼𝑠

5,500

5,500

= (30,000)(0.09) = 2,700 = 2.04 years 𝑃𝑟 1

Example 6: How long will a principal earn an interest equal to half of it at 5 4 % simple interest? Given: P=P 1 r=54% 5.25% = 0.0525 1

Is = P or 0.5P 2 Find: t Solution: t=

𝐼𝑠

0.5𝑃

0.5

= 𝑃(0.0525) = 0.0525 = 9.52 years 𝑃𝑟

Enhancement Exercises 2.2 Solve each of the following completely. 1. Find the missing value.

Simple Interest (Is)

Principal (P)

rate (r)

time (t)

Php 385.00

(a)

9.5%

2 years

Php 185.00

Php 2, 500.00

(b)

3 years and 3 months

Php 4, 640.62

Php 15, 000.00 17

1

11 % 4

3 4

(c)

2. Four months after borrowing money, Rodrigo Robredo pays an interest of 1 Php 2, 750.00. How much did he borrow if the simple interest rate is 12 %? 8

3. How much was borrowed by Leni Duterte if the interest at 15

1 5

% after 3

years and 4 months is Php 3, 040.50? 4. Cynthia Poe borrowed Php 45, 000.00 for 2 years and 8 months and paid Php 3, 750.00 interest. What was the simple interest rate? 5. At what simple interest rate should Grace Villar invest her Php 25, 500 so that it earns Php 5,500.00 in 5 years and 5 months? 5

6. Sisi Rondina loans Php 18, 500.00 at 9 % simple interest, how long will it 6

take her to get Php 6,800.00 interest? 7. How long will it take Php 369, 000.00 to earn Php 96, 300.00 if Eya Laure 1 invested it at 8 %? 8

Maturity (Future) Value with Simple Interest

Basic Concepts When the time period of a loan reaches its maturity date, the loan is said to mature. In that period, the borrower repays the principal and the interest. The total repayment is known as the maturity value (or future value). To find the maturity (future) value, use the formula F = P + Is = P + Prt = P (1+rt) < by factoring > where P = principal r = interest rate (convert to decimal) t = time (in years) 18

Note: Although maturity value and future value mean the same thing, maturity value is used for loans while future value is for invested or saved money. Although principal and present value mean the same thing, principal is usually used for loans while present value is for invested or saved money.

Let us take the following examples for application of the formula. Example 1: Find the maturity (future) value: Principal (P) Php 12, 345.00 Php 67, 890.00

rate (r) 2.15% 0.1%

time (t) 1.25 years 2 years and 9 months

Maturity Value (F) (a) (b)

Solution: a.) F = P (1 + rt) = 12, 345 [(1 + (0.0215)(1.25)] = 12, 345 (1 + 0.026875) = 12, 345 (1.026875) = Php 12, 676.77

2 yrs. and 9 mos. 3

= 2 4 years = 2.75

b.) F = P (1 + rt) = 67, 890 [(1 + (0.001)(2.75)] = 67, 890 (1 + 0.00275) = 67, 890 (1.00275) = Php 68, 076.70 Example 2: How much should Lyzza Mae pay for a Php 10, 000.00 loan charged 1 at 16 % at the end of 1 year and 6 months? 2

Given: P = 10, 000 1 r = 16 % 2

16.5% = 0.165

t = 1 year and 6 months Find: F

1

19

6 12

= 1.5 years

Solution: F = P (1 + rt) = (10, 000) [1+ (0.165)(1.5)] = (10, 000)(1 + 0.2475) = (10, 000)(1.2475) = Php 12, 475.00 Example 3: Six years and nine months ago, Bianca borrowed Php 25, 000.00 from Bernadeth with the promise that Bianca pay Bernadeth the principal plus 5 accumulated interest at 9 % simple interest now. What amount is due? 8 Given: P = 25, 000 5 r =9 % 9.625% = 0.09625 8

t = 6 years and 9 months Find: F

6

9 12

= 6.75 years

Solution: F = P (1 + rt) = (25, 000) [1+ (0.09625)(6.75)] = (25, 000)(1 + 0.6496875) = (25, 000)(1.6496875) = Php 41, 242.19

Enhancement Exercise 2.3 Solve the following problems with complete solution. 1. Find the maturity (future) value (F). P Php 123, 450.00 Php 456, 780.00

r .25% 2.5%

Php 789, 100.00

2 %

2 5

t 15 months 2 years and 3 months

F (a) (b)

3 years and 6 months

(c)

2. Jamaica Rimorin invested Php 100,000.00 in a bank that pay 15 annum simple interest. How much will she have after 90 days? 20

1 2

% per

3. Reuben’s mother borrows Php 35,000.00 to fix their car. The bank charges 5% interest for two years. What is the total amount that his mother will owe the bank?

Finding the Present Value, Rate and Time in Simple Interest From the formula F = P (1 + rt), we can also be derived other equations. Using the Golden Rule of Equations, we will arrive with the following formulas: For Present value (P), we have P=

𝐹 1+𝑟𝑡

In finding the rate (r), the formula is r=

𝐹−𝑃 𝑃𝑡

In computing number of years (t), use t=

𝐹−𝑃 𝑃𝑟

Let us take the following examples for you to understand the application of the above formulas. Example 1: Complete the table below by solving for the unknown. F Php 12, 345.00 Php 246, 810.00 Php3, 691, 215.00

P (a) Php 151,515.00 Php 1,234,560.00

r 1.05% (b) 3 12 % 20

t 51 months 9 years and 9 months (c)

Solution: a.) P =

𝐹

12,345

12,345

12,345

= [1+(0.0105)(4.25)] = (1+0.044625) = 1.044625 1+𝑟𝑡 = Php 11, 817.64 21

b.) r =

c.) t =

𝐹−𝑃 𝑃𝑡

𝐹−𝑃 𝑃𝑟

=

=

246,810−151,515 (151,515)(9.75)

95,295

= 1,477,271.25 = 0.0645 or 6.45%

3,691,215−1,234,560 (1,234,560)(0.1215)

2,456,715

= 149,999.04 = 16.38 years

Example 2: Marvin and Alyssa Marie want to invest now in preparation for their daughter’s debut 10 years hence. They want to have Php 150, 000.00 when that time comes. How much should they invest at an investment paying 15% for 5 years? Given: F = 150, 000 r = 15% = 0.15 t = 5 years Find: P Solution: P=

𝐹

150,000

150,000

= [1+(0.15)(5)] = (1+0.75) = 1+𝑟𝑡

150,000 1.75

= Php 85, 714.29 Example 3: At what rate will Php 20,000.00 investment yield Php 85, 000.00 in 15 years at simple interest? Given: F = 85, 000 P = 20, 000 t = 15 years Find: r Solution: r=

𝐹−𝑃 𝑃𝑡

=

85,000−20,000 (20,000)(15)

65,000

= 300,000

= 0.22 or 22%

Example 4: How long will an investment of Php 543, 200.00 yield Php 876, 500.00 at simple interest rate of 5% annually? 22

Given: P = 543, 200 F = 876, 500 r = 5% = 0.05 Find: t Solution: t=

𝐹−𝑃 𝑃𝑟

=

876, 500−543,200 (543,200)(0.05)

=

333,300 27,160

= 12. 27 years

Example 5: At what simple interest rate will an amount of money double itself in 10 years? Given: F = 2P P=P t = 10 years Find: r Solution: r=

𝐹−𝑃 𝑃𝑡

2𝑃−𝑃

𝑃

1

= (𝑃)(10) = 10𝑃 = 10 = 0.1 or 10%

Enrichment Exercise 2.4 Answer the following with complete solution. 1. Complete the table below. F Php 1, 010.10 Php 202, 020.20 Php5, 050, 500.50

P Php 864.20 Php 100, 100.00 (c)

r 5.15% (b) 3 4 % 5

t (a) 6 years and 6 months 8.5 years

2. How much must Coco Dalisay invest today in order to have Php 22, 500.00 available in 3 years if money is worth 13% simple interest? 3. Kawhi Leonard issued a check for Php 16, 800.00 to settle a loan of Php 12, 750.00 he got two years ago. How much simple interest rate was he charged?

23

4. When will Php 9, 750.00 double itself if Kyla Ashley invested it at the rate of 7% simple interest? 5. In order to have Php 460, 000.00 five years and eight months from now, how 3 much must a businessman invest today if the simple interest is 16 %? 5

6. How long will it take Php 80, 000.00 to accumulate to Php 96, 000.00 at 8% simple interest rate? 7. At what interest rate will Php 104, 000.00 increase to Php 128, 310.00 in 2 years and 9 months?

The very first currency where the name “Filipinas” appeared in the coins produced from 1861 to 1868. Considered as the first series of Philippine currency, these coins issued by the Spanish colonizers had denominations in 1, 2, and 4 pesos. In 1897, “Filipinas” was changed to “Islas Filipinas” to reflect the archipelagic nature of the country. This was the last currency issued by the Spanish empire in the Philippines until their departure in 1898.

24

Lesson 3 COMPOUND INTEREST Basic Concepts Compound Interest is the interest resulting from the periodic addition of simple interest to the principal. The formula for compound interest (Ic) is Ic = F – P where Ic = compound interest F = maturity (future) value P = principal The time interval between succeeding interest calculations is called the conversion period or compounding period. The interest earned during a period is “converted” to principal at the end of the period because the principal and the interest are combined and treated as the new principal for the succeeding period. The effect of converting interest to principal is that the interest earned in a period will also earned interest in all succeeding periods. The resulting value is called maturity (future) value and is designated by F. The formula for maturity (future) value is – F = P (1 + r)t where F = maturity (future) value P = principal or original amount of money r = interest rate t = term or number of years Let us take the following examples to apply the above formulas. Example 1: Find the maturity value and compound interest on the principal Php 30, 000.00 borrowed at 4.5% compounded annually for 3 years. Given: P = 30,000 r = 4.5% = 0.045 t = 3 years Find: F and Ic 25

Solution: F = P (1 + r)t = 30, 000 (1 + 0.045)3 = 30, 000 (1.045)3 = 30, 000 (1.141166125) = Php 34, 234.98 Ic = F – P = 34, 234.98 – 30, 000 = Php 4, 234.98

Example 2. How much will you have in your bank after 5 years when your 1 mother deposited Php 15, 000.00 at an annual interest rate of 1 % 4 compounded yearly? Given: P = 15, 000 1 r = 1 % = 1.25% = 0.0125 4 t = 5 years Find: F Solution: F = P (1 + r)t = 15, 000 (1 + 0.0125)3 = 15, 000 (1.0125)3 = 15, 000 (1.037970703) = Php 15, 569.56 Example 3. Bryan Bagunas invested Php 123, 450.00 in a bank at 2 compounded annually for 69 months. Find: a. maturity (future) value b. compound interest Given: P = 123, 450 3 r = 2 % = 2.75 % = 0.0275 4

t = 69 months Find: F and Ic

69 12

= 5.75 years 26

3 4

%

Solution: F = P (1 + r)t = 123, 450 (1 + 0.0275)5.75 = 123, 450 (1.0275)5.75 = 123, 450 (1.168814325) = Php 144, 290.13 Ic = F – P = 144, 290.13 – 123, 450 = Php 20, 840.13

Enhancement Exercise 3.1 Solve for the following problems. 1. Complete the table by finding the unknown values. P Php 5, 790.00 Php 35, 790.00 Php 135, 790.00

r 1.3% .13% 3 1 % 5

t 75 months 7.75 years 5 years and 6 months

F a1 b1 c1

Ic a2 b2 c2

2. To what sum of money will Php 71, 200.00 accumulate in 2 years and 9 months at 7% compounded annually? How much is the compound interest? 3. How much must be paid on the maturity date, December 27, 2019, on a loan 1 of Php 45, 000.00 made on June 27, 2015 with interest rate at 12 % 2 converted annually?

27

Finding the Present Value at Compound Interest From the formula, F = P (1 + r)t, we can derive the formula in finding the present value. Take F = P (1 + r)t then 𝐹 (1+𝑟)𝑡

=

𝑃 (1+𝑟)𝑡 (1+𝑟)𝑡



so you will get, P=

𝐹 (1+𝑟)𝑡

To fully understand the application, take the following solved examples. Example 1: Find the present value of Php 34, 500.00 due in 3 years and 3 months if money is worth 3.45% compounded annually? Given: F = 34, 500 3 t = 3 years and 3 months 3 = 3.25 years 12 r = 3.45% or 0.0345 Find: P Solution: P= =

= =

𝐹 (1+𝑟)𝑡 34,500 (1+0.0345)3.25 34,500 (1.0345)3.25 34,500

1.116539544

= Php 30, 899.04 Example 2: How much should Manisan place in a time deposit in a bank that pay 2.15% compounded annually so that she will have Php 876, 500.00 after 81 months? 28

Given: r = 2.15% or 0.0215 F = 876, 500 t = 81 months Find: P Solution: P= =

= =

81 12

= 6.75 years

𝐹 (1+𝑟)𝑡 876,500 (1+0.0215)6.75 876,500 (1.0215)6.75 876,500

1.15440714

= Php 759, 264.19

Enhancement Exercise 3.2 Solve the following problems with complete solution. 1. How much money must be invested to obtain an amount of Php 757, 575.00 in 7 years and 6 months if money earns 5.25% compounded annually? 2. Keifer Ravena aims to have his investment grow to Php 696, 969.00 in 6 years and 9 months. How much should he invest in an account that pays 6.9% compounded annually? 3

3. What amount must be deposited by Ramil de Jesus in a bank that pays 6 % 5 compounded annually so that after 69 months he will have Php 909, 090.00?

29

Lesson 4 COMPOUNDING MORE THAN ONCE A YEAR

Basic Concepts The compound frequency (or conversion frequency) is the number of compounding that take place a year. The number of conversion periods for one year is denoted by m, while the total number of conversion periods for the whole investment term is denoted by n.

Compounding or conversion frequency

Number of compounding(s) or conversion(s) per year

Annually Semi-annually Quarterly Monthly Weekly Daily

1 2 4 12 52 365

The formula for the total number of conversion periods for the whole term is – n = mt where n = total number of conversion periods for the whole term m = number of conversion periods per year t = time period (term) of the loan or investment The interest rate (im) is usually expressed as an annual or yearly rate, and must be changed to the interest rate per conversion period or periodic rate (j). The formula to be used is – j=

𝒊(𝒎) 𝒎

where j = rate of interest for each conversion period i(m) = annual rate of interest m = frequency of conversion 30

Finding the Maturity (Future) Value in Compound Interest The maturity (future) value, compounding m times a year can be computed using the formula – F = P (1 + j)n F = P (1 +

𝑖 (𝑚) 𝑚

mt

)

where F = maturity (future) value P = principal (m) i = nominal rate of interest (annual rate) m = frequency of conversion t = term/time in years Let us take the following examples for the application of the above formula. Example 1: Find the maturity value of Php 565, 565.00 invested for 2 years and 6 months at 5.25% compounded quarterly. Given: P = 565, 565 6 t = 2 years and 6 months 2 = 2.5 years 12 m=4 i(m) = i(4) = 5.25% or 0.0525 Find: F Solution: Compute for the interest rate in a conversion period j=

𝑖 (𝑚) 𝑚

=

𝑖 (4) 4

=

0.0525 4

= 0.013125

Compute for the total number of conversion periods n = mt = (4)(2.5) = 10 31

Compute for the maturity (future) value F = P (1 + j)n = 565, 565 (1 + 0.013125)10 = 565, 565 (1.013125)10 = 565, 565 (1.130279603) = Php 644, 336.67 Example 2: On January 23, 2013, Julius Balagot borrowed Php 15, 000.00 and promised to pay the principal and interest at 9% compounded semi-annually on October 23, 2017. How much will he pay? Given: P = 15, 000 m=2 (m) i = i(2) = 9% = 0.09 t = from January 23, 2013 up to October 23, 2017 9 = 4 years and 9 months 4 = 4.75 years 12 Find: F Solution: Compute first j and n before F j=

𝑖 (𝑚) 𝑚

=

𝑖 (2) 2

=

0.09 2

= 0.045

n = mt = (2)(4.75) = 9.5 F = P (1 + j)n = 15, 000 (1 + 0.045)9.5 = 15, 000(1.045)9.5 = 15, 000 (1.519164346) = Php 22, 787.47 Example 3. Find the compound amount and interest on Php 175, 250.00 for 15 3 years and 6 months at 6 % compounded quarterly. 4 Given: P = 175, 250 m=4 3 i(m) = i(4) = 6 % = 6.75 or 0.0675 4

t = 15 years and 6 months Find: F 32

15

6 12

= 15.5 years

Solution: Compute first j and n before F and Ic j=

𝑖 (𝑚) 𝑚

=

𝑖 (4) 4

=

0.0675 4

= 0.016875

n = mt = (4)(15.5) = 62 F = P (1 + j)n = 175, 250 (1 + 0.016875)62 = 175, 250(1.016875)62 = 175, 250 (2.822210202) = Php 494, 592. 34 Ic = F – P = 494, 592. 34 – 175, 250 = 319, 342.34 Example 4: Vice Ganda borrows Php 246, 810.00 and agrees to repay the loan 1 with interest in 15 months. If money is worth 3 % per annum compounded 5 quarterly, find: a. total amount due in 15 months b. compound interest for 15 months Given: P = 246, 810 t = 15 months

15 12

1

= 1.25 years

i(m) = i(4) = 3 % = 3.2% or 0.032 5 m=4 Find: F and Ic Solution: Compute j and n before F and then Ic. j=

𝑖 (𝑚) 𝑚

=

𝑖 (4) 4

=

0.032 4

= 0.008

n = mt = (4)(1.25) = 5

33

a.) F = P (1 + j)n = 246, 810 (1 + 0.008)5 = 246, 810(1.008)5 = 246, 810 (1.040645141) = Php 256, 841. 63 b.) Ic = F – P = 256, 841. 63 – 246, 810 = 10,031 .63 1

Example 5: Joenel Fernando deposited Php 1, 234, 560.00 at 4 % per annum 4 compounded monthly. Find the future value for a. 21 months b. 4.25 years c. 6 years and 9 months Given: P = 1, 234, 560 1 i(m) = i(12) = 4 % = 4.25% or 0.0425 4 m = 12 Solution: a.) for 21 months j=

𝑖 (𝑚) 𝑚

=

𝑖 (12) 12

21 12

=

= 1.75 years

0.0425 12

= 0.0035

n = mt = (12)(1.75) = 21 F = P (1 + j)n = 1, 234, 560 (1 + 0.0035)21 = 1, 234, 560 (1.0035)21 = 1, 234, 560 (1.076130433) = Php 1, 328, 547. 59

34

b.) for 4.25 years j=

𝑖 (𝑚) 𝑚

=

𝑖 (12) 12

=

0.0425 12

= 0.0035s

n = mt = (12)(4.25) = 51 F = P (1 + j)n = 1, 234, 560 (1 + 0.0035)51 = 1, 234, 560 (1.0035)51 = 1, 234, 560 (1.19505039) = Php 1, 475, 361. 41 c.) for 6 years and 9 months j=

𝑖 (𝑚) 𝑚

=

𝑖 (12) 12

=

0.0425 12

6

9 12

= 6.75 years

= 0.0035

n = mt = (12)(6.75) = 81 F = P (1 + j)n = 1, 234, 560 (1 + 0.0035)81 = 1, 234, 560 (1.0035)81 = 1, 234, 560 (1.327111837) = Php 1, 638, 399. 19

Example 6: Christian Kim Refugia placed a principal of Php 8, 765.00 in an 1 interest-bearing account. How much is in the account after 7 years if interest 2 is earned at – a. 3.5% compounded monthly b. 4.15% compounded quarterly? 3 c. 5 % compounded semi-annually? 4

Given: P = 8, 765 1 t = 7 years = 7.5 years 2

35

Solution: a.) When m = 12, i(m) = i(12) = 3.5% = 0.035 j=

𝑖 (𝑚) 𝑚

=

𝑖 (12) 12

=

0.035 12

= 0.0029

n = mt = (12)(7.5) = 90 F = P (1 + j)n = 8, 765 (1 + 0.0029)90 = 8, 765 (1.0029)90 = 8, 765 (1.297737392) = Php 11, 374. 67 b.) When m = 4, j=

𝑖 (𝑚) 𝑚

=

𝑖 (4) 4

=

0.0575 4

= 0.010375

n = mt = (4)(7.5) = 30 F = P (1 + j)n = 8, 765 (1 + 0.010375)30 = 8, 765 (1.010375)30 = 8, 765 (1.362943191) = Php 11, 946. 20 c.) When m = 2, j=

𝑖 (𝑚) 𝑚

=

𝑖 (2) 2

=

0.0575 2

= 0.02875

n = mt = (2)(7.5) = 15 F = P (1 + j)n = 8, 765 (1 + 0.02875)15 = 8, 765 (1.02875)15 = 8, 765 (1.529846028) = Php 13, 409. 10 36

Enhancement Exercise 4.1 Solve each of the following: 1. Find the compound amount and interest on Php 175, 250.00 for 15 years and 3 6 months at 6 % compounded quarterly. 4

2. On January 23, 2013, Julius Balagot borrowed Php 15, 000.00 and promised to pay the principal and interest at 9% compounded semi-annually on October 23, 2017. How much will he pay? 3. Find the maturity value of Php 565, 565.00 invested for 2 years and 6 months at 5.25% compounded quarterly.

Finding the Present Value in Compound Interest From the formula, F = P (1 + j)n, we can derive the formula in finding the present value. Take F = P (1 + j)n then 𝐹 (1+𝑗)𝑛

=

𝑃 (1+𝑗)𝑛



(1+𝑗)𝑛

so you will get, P=

𝐹 (1+𝑗)𝑛

where P = present value F = maturity (future) value 𝑖 (𝑚)

j= = rate of interest per conversion period 𝑚 n = mt = total number of conversion period 37

Let us apply the formula with the following examples: Example 1: Cardo and Aryana aim to accumulate Php 3, 691, 215.00 in 15 years. Which investment will require the largest present value? a. 5% compounded monthly b. 5% compounded quarterly c. 5% compounded semi-annually Given: P = 3, 691, 215 t = 15 years Solution: a. When m = 12 and i(m) = i(12) = 5% or 0.05 j=

𝑖 (𝑚) 𝑚

=

𝑖 (12) 12

=

0.05 12

= 0.0042

n = mt = (12)(15) = 180 P=

𝐹 (1+𝑗)𝑛

=

3,691,215 (1+0.0042)180

=

3,691,215 (1.0042)

180 =

3,691,215 2.126371129

= Php 1, 735, 922.27 b. When m = 4 and i(m) = i(4) = 5% or 0.05 j=

𝑖 (𝑚) 𝑚

=

𝑖 (4) 4

=

0.05 4

= 0.0125

n = mt = (4)(15) = 60 P=

𝐹 (1+𝑗)𝑛

=

3,691,215 (1+0.0125)60

=

3,691,215 (1.0125)60

=

3,691,215 2.107181347

= Php 1, 751, 731. 05 c. When m = 2 and i(m) = i(2) = 5% or 0.05 j=

𝑖 (𝑚) 𝑚

=

𝑖 (2) 2

=

0.05 2

n = mt = (2)(15) = 30 38

= 0.025

P=

𝐹 (1+𝑗)𝑛

=

3,691,215 (1+0.025)30

=

3,691,215 (1.025)30

=

3,691,215 2.097567579

Largest present value

= Php 1, 759, 759.75 Example 2: A financial obligation of Php 25, 500.00 is due on January 6, 2019. What is the value of this obligation on July 6, 2013 at 9% compounded semiannually? Given: F = 25, 500 i(m) = i(2) = 9% or 0.09 m=2 t = July 6, 2013 to January 6, 2019 (5 years and 6 months) 6 = 5 = 5.5 years 12 Find: P Solution: When m = 2 and i(2) = i(2) = 9% or 0.09 j=

𝑖 (𝑚) 𝑚

=

𝑖 (2) 2

=

0.09 2

= 0.045

n = mt = (2)(5.5) = 11 P=

𝐹 (1+𝑗)𝑛

=

25,500 (1+0.045)11

=

25,500 (1.045)11

=

25,500 1.622853046

= Php 15, 713. 07

Enhancement Exercises 4.2 Solve each of the following: 1. Find the present value of the following if the compounded amount is Php 690, 690.00 at 6.09%; a. compounded monthly for 6 years and 9 months. b. compounded quarterly for 9 years and 6 months. c. compounded semi-annually for 6 years and 6 months.

39

2. How much should Jeuz Batalla invest today to provide his only son a brand new Toyota car worth Php 1, 234, 567.00 as a graduation gift 8.5 years from 3 now if money is worth 9 % compounded quarterly? 4

Finding the Time in Compound Interest We will start with this section by deriving the formula of time period t in the compound interest formula. Use the formula, F = P (1 + j)n

then

F = P (1 + j)mt



𝐹 𝑃

log log

𝐹 𝑃 𝐹 𝑃

= (1 +j)mt



= log (1 +j)mt



= mt log (1 +j)



log

𝐹 𝑃

𝑚 log(1+𝑗)

t=

=t

log

𝐹 𝑃

𝑚 log(1+𝑗)





We can now apply the formula in computing the time period of a compound interest problem in the following examples.

40

Example 1: How long will it take Php 7, 350.00 to amount to Php 18, 500.00, if invested at 8% compounded monthly? Given: P = 7, 350 F = 18, 500 i(12) = 8% or 0.08 m = 12 Find: t Solution: Compute j before t, j=

t=

𝑖 (𝑚)

=

𝑚

log

=

=

=

12

=

0.08 12

= 0.0067

𝐹 𝑃

𝑚 log(1+𝑗) log

=

𝑖 (12)

18,500 7,350

12 log(1+0,0067) 0.4001 12 log(1.0067) 0.4001 12 (0.0029) 0.4001 0.0348

= 11.5 years Example 2: When is Php 15, 250.00 due if its present value is Php 12, 650.00 1 when money is worth 5 % compounded semi-annually? 2 Given: P = 12, 650 F = 15, 250 1 i(2) = 5 % = 5.5% or 0.055 2 m=2 41

Find: t Solution: Compute j before t, j=

t=

𝑖 (𝑚) 𝑚

=

log

=

= =

2

=

0.055 2

= 0.0275

𝐹 𝑃

𝑚 log(1+𝑗) log

=

𝑖 (2)

15,250 12,650

2 log(1+0,0275) 0.0812

2 log(1.0275) 0.0812 2 (0.0118) 0.0812 0.0236

= 3.44 years Example 3: Kyla Ashley borrowed Php 9, 250.00 from Alyssa Grace with the agreement that interest is charged at 8% compounded quarterly. If the maturity value of her loan is Php 11, 500.00 when is it due? Given: P = 9, 250 F = 11, 500 i(4) = 8% or 0.08 m=4 Find: t Solution: Compute j before t, j=

𝑖 (𝑚) 𝑚

=

𝑖 (4) 4

=

0.08 4

= 0.02 42

t=

log

𝑚 log(1+𝑗) log

=

=

= =

𝐹 𝑃

11,500 9,250

4 log(1+0,02) 0.0946 4 log(1.02) 0.0946 4 (0.0086) 0.0946

0.0344

= 2.75 years

Enhancement Exercise 4.3 Solve the following interest problems with complete solution. 1

1. Olivia Gutoc receives a loan of Php 18, 790.00 with interest at 7 % 4 compounded monthly. She promised to pay her creditor in full on the day when Php 26, 275.00 will be due. How long does it take for Olivia pay the debt? 2. Mark Joshua deposits Php 100, 000.00 in a savings account that pays 13.5% interest converted semi-annually. If he decides to withdraw his money when it grows to Php 150, 000.00, when should he withdraw his money? 3. Misuari Refugia received a loan of Php 290, 290.00 from Elmer Bondoc with interest at 10% converted quarterly. He promised to pay Elmer in full on the day when Php 360, 360.00 will be due. When should Misuari pay?

43

Finding Interest Rate (r) In this section, we will derive the formula of nominal rate (i) in the compound interest formula. F = P (1 + j)n F = P (1 +

<maturity (future) value formula>

𝑖 (𝑚) 𝑚

)

mt


𝑖 (𝑚) 𝑚

and n by mt>

mt 𝐹 𝑃

= (1 +

𝑚𝑡

𝑖 (𝑚) 𝑚

𝐹

𝑖 (𝑚)

𝑃

𝑚

𝐹

𝑖 (𝑚)

𝑃

𝑚

√ = (1 +

𝑚𝑡

√ –1= 𝑚𝑡



) )

𝐹

m ( √ – 1) = i(m) 𝑃

𝑚𝑡

𝐹

i(m) = m ( √ – 1) 𝑃

<extract the mt root of both sides>

<subtract both sides by 1>

<multiply both sides by m>



Now we can use the formula in finding the compound interest rate in the preceding examples. Example 1: At what interest rate will Php 2, 050.00 amount Php 3, 875.00 in 4 years and 6 months, if interest is computed semi-annually? Given: P = 2, 050 F = 3, 875 t = 4 years and 6 months = 4.5 years m=2 Find: i(2) 44

Solution: 𝑚𝑡

𝐹

i(m) = m ( √ – 1) 𝑃

i(2) = 2 (

(2)(4.5)

3,875

√2,050 – 1)

3,875

9

=2( √ – 1) 2,050 = 2 (1.073307601 - 1) = 2 (0.073307601) = 0.1466 or 14.66% Example 2: Joshua Dioniso, who invested Php 15, 500.00, had Php 19, 900.00 returned to him 4 years and 3 months later. At what rate of interest converted monthly did his money earn? Given: P = 15, 500 F = 19, 900 t = 4 years and 3 months = 4.25 years m = 12 Find: i(12) Solution: 𝑚𝑡

𝐹

i(m) = m ( √ – 1) 𝑃

i(12) = 12 (

(12)(4.25)

51

19,900

√15,500 – 1)

19,900

= 12 ( √ – 1) 15,500 = 12 (1.004911625 - 1) = 12 (0. 004911625) = 0.0589 or 5.89% Example 3: Roldan Andaya placed Php 34, 400.00 in an investment firm. If this amounted to Php 42, 200.00 after 4 years and 9 months, find the rate used if it is converted quarterly.

45

Given: P = 34, 400 F = 42, 200 t = 4 years and 9 months = 4.75 years m=4 Find: i(4) Solution: 𝑚𝑡

𝐹

i(m) = m ( √ – 1) 𝑃

i(4) = 4 (

(4)(4.75)

42,200

√34,400 – 1)

19

42,200

=4( √ – 1) 34,400 = 4 (1.010814035– 1) = 4 (0.010814035) = 0.0433 or 4.33%

Enhancement Exercise 4.4 Solve each problem with complete solution. 1. If Php 18, 850.00 amounts to Php 20, 500.00 in 3 years and 3 months, what is the rate of interest compounded quarterly? 2. For a sum of Php 9, 500.00 to triple itself in 6 years and 6 months, what must be the rate of interest compounded semi-annually? 3. On April 21, 2014 Trixie Riogelon invested Php 250, 000. If she expects to receive Php 680, 000.00 on July 21, 2019, what interest rate compounded monthly is used?

46

Nominal Rate (i) Aside from periodic rate (j), the other kind of rate associated with compound interest problems is the nominal rate. Nominal rate is the rate quoted when interest is compounded more than once a year. It is denoted by (i) and the number of times per year this rate is compounded or the frequency of conversion is represented by (m). In any investment problem, if the rate is not specified as nominal or effective rate, it is assumed that the given rate is nominal. To determine the nominal rate (i), if the effective rate (e) is given, we apply the formula: 𝑚

i = m [ √(1 + 𝑒) – 1] where i = nominal rate m = conversion period per year e = effective rate Let us try to illustrate using the given examples. Example 1: What nominal rate compounded quarterly, will yield the effective 1 rate 3 %? 5 Given: m=4 1 e = 3 % = 3.2% or 0.032 5

Find: i Solution: 𝑚 i = m [ √(1 + 𝑒) – 1] 4 = 4 [√(1 + 0.032) – 1] 4 = 4 [√(1.032) – 1] = 4 (1.0079– 1) = 4 (0.0079) = 0.0316 or 3.16% 47

Example 2: If interest is compounded monthly find the nominal rate if the effective rate is 9.75% Given: m = 12 e = 9.75% or 0.0975 Find: i Solution: 𝑚 i = m [ √(1 + 𝑒) – 1] 12 = 12 [ √(1 + 0.0975) – 1] 12 = 12 [ √(1.0975) – 1] = 12 (1.0078– 1) = 12 (0.0078) = 0.0936 or 9.36%

Enhancement Exercise 4.5 Solve the following completely. 1. What nominal rate compounded semi-annually, will yield an effective rate 1 of 4 %? 2

2. If interest is compounded quarterly, find the nominal rate which will yield an 3 effective rate of 2 %? 4

3. When interest is compounded monthly, find the nominal rate if the effective rate is 5.25%?

.

48

Effective Rate (e) When the conversion period is one year, the quoted annual rate of interest is known as the effective rate. To determine the effective rate (e) when nominal rate (i) is given, we apply the formula: 𝑖 e = [(1 + )𝑚 – 1)] 𝑚 where e = effective rate i = nominal rate m = conversion period per year

Note: A. When interest is compounded annually, the effective rate is equal to the nominal rate, but when it is compounded more than once a year, the effective rate is higher than the nominal rate. B. When comparing two different rates compounded at different conversion periods per year, we compute their respective effective rates. Let us try to illustrate the application of the above in the following examples. Example 1: Find the effective rate corresponding to the rate 4% compounded quarterly. Given: m=4 i = 4% or 0.04 Find: e Solution: e = [(1 +

𝑖

)𝑚 – 1] 𝑚 0.04 4 + ) – 1] 4 4

= [(1 = [(1 + 0.01 ) – 1] = [(1.01)4 - 1] = 1.04060401 – 1 = 0.0406 or 4.06%

49

Example 2: When interest is compounded monthly, find the effective rate corresponding to the nominal rate 3. Given: m = 12 i = 3% or 0.03 Find: e Solution: e = [(1 +

𝑖

)𝑚 – 1] 𝑚 0.03 12 + ) – 1] 12 12

= [(1 = [(1 + 0.0025 ) – 1] = [(1.0025)12 - 1] = 1.0304 – 1 = 0.0304 or 3.04%

Enhancement Exercise 4.6 Solve the following problems: 1. What effective rate can give you the same final amount as when you place your money at 15% compounded monthly? 2. If interest is converted quarterly, find the effective rate when the nominal rate is 4.5%. 3. If you are to invest, which would you prefer: depositing your money at 10% 1 compounded monthly or at 10 % compounded semi-annually? 2

50

Equivalent Rates (eq) If two rates produce equal interests on the same principal in the same period of time, they are said to be equivalent rates (eq).

Case 1: To determine the compound interest rate equivalent to another compound interest rate, we apply the formula 𝑐

eq = c1 [(1 +

𝐺 𝑐2 )1 𝑐2

– 1]

where eq = the missing or unknown rate c1 = conversion period of the missing rate G = given compound interest rate c2 = conversion period of the given rate

Example 1: What rate compounded semi-annually is equivalent to 6% compounded monthly? Given: c1 = 2 G = 6% or 0.06 c2 = 12 Find: eq Solution: eq = c1 [(1

𝑐

𝐺 2 + ) 𝑐1 – 𝑐2 0.06 12

1]

= 2 [(1 + ) 2 – 1] 12 = 2 [(1 + 0.005)6 – 1] = 2 [(1.005)6 – 1] = 2 (1.030377509– 1) = 2 (0.030377509) = 0.0608 or 6.08%

51

Example 2: Complete the table below by computing for the rates equivalent to the following nominal rate.

Given Interest Rate 12% compounded monthly 8% compounded semi-annually

Equivalent Interest Rate (a) ________ semi-annually (b) ________ compounded quarterly

(a). Given: c1 = 2 G = 12% or 0.12 c2 = 12 Find: eq Solution: 𝑐

eq = c1 [(1 +

𝐺 𝑐2 )1– 𝑐2 0.12 12

1]

= 2 [(1 + ) 2 – 1] 12 = 2 [(1 + 0.01)6 – 1] = 2 [(1.01)6 – 1] = 2 (1.061520151– 1) = 2 (0. 061520151) = 0.123 or 12.3% (b). Given: c1 = 4 G = 8% or 0.08 c2 = 2 Find: eq Solution: 𝑐

eq = c1 [(1

𝐺 2 + ) 𝑐1 𝑐2 0.08 2

= 4 [(1 +

2

– 1]

)4 – 1] 1 2

= 4 [(1 + 0.04) – 1] 1

= 4 [(1.04)2 – 1] = 4 (1.019803903– 1) = 4 (0. 019803903) = 0.0792 or 7.92% 52

Enhancement Exercise 4.7 Solve each of the following problem with complete solution. 1. Complete the table below.

Given Interest Rate 6.09% compounded monthly 9.06% compounded quarterly

Equivalent Interest Rate (a) ________ compounded quarterly (b) ________ compounded semi-annually 3

2. What rate compounded quarterly is equivalent to 5 % compound annually? 4

Case 2: To determine the simple interest rate equivalent to compound rate, we apply the formula: rs =

[(1+𝑗)𝑛 −1] 𝑡

where rs = missing or unknown simple interest rate 𝑖 j = = periodic rate 𝑚 n = total number of conversion periods for the whole term t = term or time of the investment

Example 1: Find the simple interest rate equivalent to 9.6%, m=6 for 9 years and 6 months. Given: i = 9.6% or 0.096 m=6 6 t = 9 years and 6 months = 9 years = 9.5 years 12 Find: rs 53

Solution: Computer first for j and n before rs. j=

𝑖 𝑚

=

0.096 6

= 0.016

n = mt = (6)(9.5) = 57 rs =

= = = =

[(1+𝑗)𝑛 −1] 𝑡 [(1+0.016)57 −1] 9.5 [(1.016)57 −1] 9.5 2.471390387−1] 9.5 1.471390387 9.5

= 0.1549 or 15.49% 1

Example 2: What simple interest rate is equivalent to 5 % compounded semi2

1

annually , if money is invested for 5 years? 2

Given: 1 i = 5 % = 5.5% or 0.055 2 m=2 1 t = 5 years = 5.5 years 2 Find: rs Solution: Computer first for j and n before rs. j=

𝑖 𝑚

=

0.055 2

= 0.0275

n = mt = (2)(5.5) = 11

54

rs = = = = = =

[(1+𝑗)𝑛 −1] 𝑡

[(1+0.0275)11 −1] 5.5 [(1.0275)11 −1] 5.5 [(1.0275)11 −1] 5.5 1.347721436 − 1 5.5 0.347721436 5.5

= 0.0632 or 6.32%

Enhancement Exercise 4.8 Solve the following interest problem with complete solution. 1. Find the simple interest rate equivalent to 3.45%, m=2 for 4 years and 6 months. 1

2. If money deposited in a bank is 2 % compounded monthly for 3 2 what simple interest rate is equivalent to it? 3. What simple interest rate is equivalent to 2 1

money is invested for 5 years? 4

55

1 2

1 4

years,

% compounded quarterly, if

Case 3: To determine the compound interest rate equivalent to a given simple interest rate, apply the formula rc = m[ 𝑛√(1 + 𝑟𝑠 𝑡) – 1] where rc = missing or unknown compound interest rate m = conversion period of missing rate n = mt = number of conversion periods for the whole term rs =given simple interest rate t = term or time of the investment Let us take the following examples for the application of the formula. Example 1: Find the rate compounded monthly equivalent to simple interest rate 6.5 % for 4 years. Given: rs = 6.5% or 0.065 t = 4 years m = 12 Find: rc Solution: Compute first for n before rc. n = mt = (12)(4) = 48 rc = m[ 𝑛√(1 + 𝑟𝑠 𝑡) – 1] = 12[ 48√(1 + (0.065)(4) – 1] = 12[ 48√(1 + 0.26) – 1] 48 = 12( √1.26 – 1) = 12(1.004826437 - 1) = 12(0.004826437) = 0.0579 or 5.79% Example 2: What rate compounded semi-annually is equivalent to simple 1 interest rate 3 % for 42 months ? 4 Given: 1 rs = 3 % = 3.25% or 0.0325 4

42

t = 42 months = = 3.5 years 12 m=2 56

Find: rc Solution: Compute first for n before rc. n = mt = (2)(3.5) = 7 rc = m[ 𝑛√(1 + 𝑟𝑠 𝑡) – 1] 7 = 2[√(1 + (0.0325)(3. 5) – 1] 7 = 2[√(1 + 0.11375) – 1] 7 = 2(√1.11375 – 1) = 2(1.015509428 - 1) = 2(0.01550942759) = 0.031 or 3.1% 1

Example 3: Alwean Serios plans to invest Php 100,000.00 at 12 % simple 2 interest for three years. At what compound interest m = 4 could he just as well invest for the same period of time? Given: 1 rs = 12 % = 12.5% or 0.125 2 t = 3 years m=4 Find: rc Solution: Compute first for n before rc. n = mt = (4)(3) = 12 rc = m[ 𝑛√(1 + 𝑟𝑠 𝑡) – 1] = 4[ 12√(1 + (0.125)(3) – 1] = 4[ 12√(1 + 0.375) – 1] 12 = 4( √1.375 – 1) = 4(1.026893074 - 1) = 4(0.026893074) = 0.1076 or 10.76% 57

Enhancement Exercise 4.9 Solve each of the following: 1. Find the rate compounded monthly equivalent to simple interest rate 3.75% for 5.25 years. 1

2. What rate compounded quarterly is equivalent to simple interest rate 5 % 2 for 66 months? 1

3. Joshua Dionisio plans to invest Php 100, 000.00 in a bank at 12 % simple 2 interest for three years and 6 months. At what compound interest m = 2 could he just as well invest for the same period of time?

Now that you have reached the finish line of your race, I think you are now ready to deal with the post-test of this module.

Do your best and Good luck!

58

POST-TEST Direction: Write the letter of the correct answer on the space provided before each number. _____ 1. Determine the simple interest to be paid by Stephen Karl if he borrowed Php 8, 765.00 2 at 7 5 % for 8 months. A. Php 234.14

B. Php 324.41

C. Php 432.41

D. Php 441.23

_____ 2. How much was borrowed by Ms. Gray if the interest she paid after 15 months at 10.5% simple interest is Php 6, 420.00? A. Php 41, 948.29

B. Php 44, 198.92

C. Php 49, 149.92

D. Php 48, 914.29

_____ 3. At what rate of interest did Alari invest the amount of Php 56, 780.00 for it to earn a simple interest of Php 5, 555.00 for 4.75 years? A. 2.65%

B. 6.25%

C. 2.06%

D. 6.02%

_____ 4. How long should Sisi Rondina invest the amount of Php 65, 120.00 to earn an 3 interest of Php 5, 225.00 at 4 5 % simple interest? A. 1.47 years

B. 1.74 years

C. 4.17 years

D. 7.41 years

_____ 5. How much is the total amount would Mrs. Supremido receive if she invest Php 12, 1 890.00 for 1.75 years at 3 2 % simple interest? A. Php 13, 679.51

B. Php 16, 379.15

C. Php 17, 936.15

D. Php 19, 763.51

_____ 6. Which of the following are NOT true? I. Principal is the money given or paid invested on the origin data. II. Origin date is a date on which money is paid by the borrower. III. Interest is an amount paid or earned for the use of money. IV. Simple interest is an interest that is computed on the principal and then added to it. A. I and II

B. II and III

C. III and IV

D. II and IV

_____ 7. Analyze the two statements below.

Statement 1: Equivalent rates refer to two annual rates with different conversion periods that will earn the same maturity value for the same time/term.

Statement 2: Effective rate is a rate when compounded annually will give the same compound each year with the nominal rate. 59

A. Both statements are true B. Both statements are false C. Statement 1 is true but statement 2 is false D. Statement 2 is true but statement 1 is false For numbers 8 and 9, consider the statement below.

Ion Perez is thinking of investing an amount of Php 36, 912.00 for 5.5 years at an 1 interest rate of 8 2 % compounded annually. _____ 8. Find the maturity value (F). A. Php 51, 378.57

B. Php 53, 718.75

C. Php 57, 813.75

D. Php 58, 731.57

C. Php 20, 901.75

D. Php 29, 001.57

____ 9. Find the compound interest (Ic). A. Php 27, 090.15

B. Php 21, 009.57

3

_____10. Oliver Almario invested Php 102, 030.00 at 3 4 % compounded annually. He plans to get this amount after 60 months for his son’s 7th birthday. How much will he get? A. Php 120, 625.42

B. Php 122, 650.24

C. Php 165, 022.43

D. Php 164, 502.21

_____11. Find the interest if Php 87, 650.00 is deposited in a bank at 4.5% compounded monthly for 36 months. A. Php 16, 234.23

B. Php 12, 643.32

C. Php 14, 263.23

D. Php 13, 246.32

_____ 12. Kung Fu Reyes borrows Php 101, 010.00 and promises to pay the principal and 1 interest at 9 5 % compounded quarterly. How much must she pay after 6 years and 3 months? A. Php 178, 342.65

B. Php 176, 824.85

C. Php 174, 628.58

D. Php 172, 468.85

2

_____ 13. If money is worth 4 5 % compounded semi-annually, find the present value of Php 300, 300.00 due at the end of 3 years and 6 months. A. Php 267, 886.69

B. Php 257, 868.96

C. Php 288, 567.96

D. Php 286, 857.69

_____14. How long will it take Php 3, 690.00 to amount to Php 8, 910.00 if invested in a bank at 5 % compounded monthly? A. 17.67 years

B. 16.77 years

C. 15.62 years

D. 14.76 years

_____ 15. At what interest rate will Php 1, 010.00 amount Php 2, 020.00 in 3 years and 3 months, if interest is compounded semi-annually? A. 25.12%

B. 24.41%

C. 23.31% 60

D. 22.51%

_____ 16. If interest is compounded quarterly, find the nominal rate which will yield an 1 effective rate of 3 2 %. A. 1.54%

B. 2.35%

C. 3.45%

D. 4.53%

_____ 17. When interest is compounded monthly, find the effective rate corresponding to the nominal rate 4 %. A. 3.02%

B. 4.07%

C. 5.05%

D. 6.09%

_____ 18. Alden Richards is investing Php 100, 000.00 at 10 % converted semi-annually for 10 years. At what rate compounded monthly could she just as well invest her money? A. 9.8%

B. 8.7%

C. 7.6%

D. 6.5%

_____ 19. If you are to invest Php 110, 110.00 at 15% converted monthly for 9 years and 9 months, what simple interest rate could you just as well invest your principal for the same period of time? A. 2.36%

B. 3.36%

C. 4.63%

D. 5.63%

_____ 20. June Mar Fajardo plans to invest Php 500, 000.00 at 5% simple interest for 30 months. At what interest compounded quarterly could she just as well invest for the same period of time? A. 2.47%

B. 3.74%

C. 4.74%

D. 5.47%

=== END OF THE TEST ===

How did you find the post-test? See page 65 of this module for key to correction. I hope you have done well on this part of module. I hope this module was able to help you in enhancing your knowledge on interest.

61

Pre-Test: 1. C 2. D 3. C 4. B 5. A

6. D 7. D 8. C 9. C 10. B

11. B 12. A 13. B 14. C 15. D

16. C 17. B 18. A 19. B 20. C

Enhancement Exercise 1.1 1. Gazini’s investment at 7.5% simple interest Time (t) 0 1

2

3

4

5

Maturity Value =Php 230,000.00 =230,000 + 230,000 (0.075) (1) =230,000 + 17, 250 =247, 250 =230,000 + 230,000 (0.075) (2) =230,000 + 230,000 (0.15) =230,000 + 34, 500 =264, 500.00 =230,000 + 230,000 (0.075) (3) =230,000 + 230,000 (0.225) =230,000 + 51, 750 =281, 750.00 =230,000 + 230,000 (0.075) (4) =230,000 + 230,000 (0.3) =230,000 + 69, 000 =299,000.00 =230,000 + 230,000 (.075) (5) =230,000 + 230,000 (0.375) =230,000 + 86, 250 =316, 250.00

Simple Interest (Is) 0 Php 17, 250.00

Php 34, 500.00

Php 51, 750.00

Php 69, 000.00

Php 86, 250.00

Gazini’s investment at 7.5% compounded annually Time (t) 0 1

Maturity Value =Php 230,000.00 =230,000 + 230,000 (0.075) (1) =230,000 + 17,250 =247, 250.00

Compound Interest (Ic) 0 Php 17, 250.00

2

=247, 250 + 247, 250 (0.075) (1) =247, 250 + 18,543.75 =265, 793.75

Php 35, 793.75

3

=265, 793.75 + 265, 793.75 (0.075) (1) =265, 793.75 + 19, 934.53 =285, 728.28

Php 55, 728.28

4

=285, 728.28 + 285, 728.28 (0.075) (1) =285, 728.28 + 21, 429.62 =307, 157.90

Php 77, 157.90

5

=307, 157.90 + 307, 157.90 (0.075) (1) =307, 157.90 + 23, 036.84 =330, 194. 74

Php 100, 194.74

It is better to invest in a lending company offering 7.5% compounded annually as shown in the above chart. 62

2. Bea Patricia’s investment at 4.25% simple interest Time (t) 0 1

2

3

Maturity Value =Php 35,700.00 =35,700 + 35,700 (0.0425) (1) =35,700 + 1, 517.25 =1,050.00 =35,700 + 35,700 (0.0425) (2) =35,700 + 35,700 (0.085) =35,700 + 3, 034.50 =38, 734.50 =35,700 + 35,700 (0.0425) (3) =35,700 + 35,700 (0.1275) =35, 700 + 4551.75 =40, 251.75

Simple Interest (Is) 0

Bea Patricia’s investment at 4.25% compounded annually Time (t) 0

Php 1, 517. 25

Php 3, 034. 50

Php 4, 551.75

=Php 35,700.00

Compound Interest (Ic) 0

1

=35,700 + 35,700 (0.0425) (1) =35,700 + 1, 517.25 =37, 217.25

Php 1, 517.25

2

=37, 217.25 + 37, 217.25 (0.0425) (1) =37, 217.25 + 1, 581.73 =40, 380.71

Php 4, 680.71

3

=40, 380.71 + 40, 380.71 (0.0425) (1) =40, 380.71 + 1, 716.18 =42, 096.89

Php 6, 396.89

Maturity Value

Bea Patricia should go for XYZ Bank offering 4.35% compounded annually because she will get Php 6, 396.89 at the end of three years compared to ABC Bank which will give her Php 4, 551.75 only at the end of 3 years. B. 1. SI 2. CI 3. SI 4. SI 5. CI

6. CI 7. CI 8. SI 9. CI 10. CI

Enhancement Exercise 2.1 1. Php 2, 250.00 2. Php 11, 169.49 3. Php 7, 575.75

Enhancement Exercise 2.2 1. (a) Php 1, 473.68 2. Php 68, 041.24 3. Php 6, 001.00 4. 3.125% 5. 3.98%

(b) 2.28%

63

(c) 2.75 years

6. 3.74 years 7. 3.21 years

Enhancement Exercise 2.3 1. (a) Php 123, 835.78 (b) Php 482, 473.88 (c) Php 855, 384.40 2. Php 103, 875.00 3. Php 38, 500.00

Enhancement Exercise 2.4 1. (a) 3.28 years 2. Php 16, 187.05 3. 15.88% 4. 14.29 years 5. Php 296, 137.34 6. 2.5 years 7. 8.5%

(b) 15.66%

(c) Php 3, 630, 841.48

Enhancement Exercise 3.1 1. Maturity /Future Value (F) a1. Php 6, 276.79 b1. Php 36, 152.17 c1. Php 148, 177.81 2. F = Php 85, 760.12; Ic = Php 14, 560.12 3. Php 76, 453.76 Enhancement Exercise 3.2 1. Php 516, 130.89 2. Php 444, 236.21 3. Php 629, 510.46

Enhancement Exercise 4.1 1. F = Php 494, 592.34; Ic = 319, 342.34 2. Php 22, 787.47 3. Php 644, 336.67 64

Compound Interest (Ic) a2. Php 486.79 b2. Php 362.17 c2. Php 12, 387. 81

Enhancement Exercise 4.2 1. (a) Php 458, 360.36 (b) Php 388, 969.57 (c) Php 467, 663.16 2. Php 544, 389.39

Enhancement Exercise 4.3 1. 4.6 years

2. 3.1 years

3. 2.18 years

2. 17.64%

3. 19.21%

2. 2.72%

3. 5.13%

2. 4.58%

3. 10.47%

Enhancement Exercise 4.4 1. 2.59%

Enhancement Exercise 4.5 1. 4.45%

Enhancement Exercise 4.6 1. 16.08%

Enhancement Exercise 4.7 1. (a) 6.12% 2. 5.63%

(b) 9.16%

Enhancement Exercise 4.8 1. 3.74%

2. 2.42%

3. 2.66%

2. 4.83%

3. 10.64%

Enhancement Exercise 4.9 1. 3.43%

Post-Test: 1. C 2. D 3. C 4. B 5. A

6. A 7. A 8. C 9. C 10. B

11. B 12. A 13. B 14. A 15. D

65

16. C 17. B 18. A 19. B 20. C