general mathematics - znnhs · 2021. 1. 22. · php 123,450.00 for 3.25 years at 2 3/4 % simple...

General Mathematics Quarter 2 – Module 2:

Interest, Maturity Value, Future Value, and Present Value in Simple and Compound

Interest

SENIOR HS MODULE DEVELOPMENT TEAM

Author : Bienvenido V. Malit

Co-Author - Content Editor : Ma. Lourdes L. Agkis

Co-Author - Language Reviewer : Efren M. Bacani

Co-Author - Illustrator : Noel S. Sarili

Co-Author - Layout Artist : Cherryl D. Cordova

Team Leaders:

School Head : Omar G. Magcalas

LRMDS Coordinator : Cherryl D. Cordova

DIVISION MANAGEMENT TEAM:

Schools Division Superintendent : Romeo M. Alip, PhD, CESO V

OIC- Asst. Schools Division Superintendent : William Roderick R. Fallorin, CESE

Chief Education Supervisor, CID : Milagros M. Peñaflor, PhD

Education Program Supervisor, LRMDS : Edgar E. Garcia, MITE

Education Program Supervisor, AP/ADM : Romeo M. Layug

Education Program Supervisor, Mathematics : Danilo C. Caysido

Project Development Officer II, LRMDS : Joan T. Briz

Division Librarian II, LRMDS : Rosita P. Serrano

General Mathematics – Grade 11

Alternative Delivery Mode Quarter 2 – Module 2: Interest, Maturity Value, Future Value, and Present Value First Edition, 2020

Republic Act 8293, section 176 states that: No copyright shall subsist in any work

of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties.

Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Published by the Department of Education Secretary: Leonor Magtolis Briones Undersecretary: Diosdado M. San Antonio

Printed in the Philippines by Department of Education – Schools Division of Bataan

Office Address: Provincial Capitol Compound, Balanga City, Bataan Telefax: (047) 237-2102

E-mail Address: [email protected]

mailto:[email protected]

General Mathematics Quarter 2 – Module 2:

Interest, Maturity Value, Future

Value, and Present Value in

Simple and Compound Interest

1

Introductory Message For the facilitator:

Welcome to the General Mathematics – Grade 11 Alternative Delivery Mode (ADM)

Module on Interest, Maturity Value, Future Value, and Present Value in Simple and

Compound Interest!

This module was collaboratively designed, developed, and reviewed by educators

both from public and private institutions to assist you, the teacher, or facilitator in

helping the learners meet the standards set by the K to 12 Curriculum while

overcoming their personal, social, and economic constraints in schooling.

This learning resource hopes to engage the learners in guided and independent

learning activities at their own pace and time. Furthermore, this also aims to help

learners acquire the needed 21st-century skills while taking into consideration

their needs and circumstances.

In addition to the material in the main text, you will also see this box in the body of

the module:

As a facilitator you are expected to orient the learners on how to use this module.

You also need to keep track of the learners' progress while allowing them to

manage their own learning. Furthermore, you are expected to encourage and assist

the learners as they do the tasks included in the module.

Notes to the Teacher

This contains helpful tips or strategies

that will help you in guiding the learners.

2

For the learner:

Welcome to the General Mathematics – Grade 11 Alternative Delivery Mode (ADM)

on Module on Interest, Maturity Value, Future Value, and Present Value in Simple

and Compound Interest!

The hand is one of the most symbolized parts of the human body. It is often used to

depict skill, action, and purpose. Through our hands, we may learn, create, and

accomplish. Hence, the hand in this learning resource signifies that you as a

learner are capable and empowered to successfully achieve the relevant

competencies and skills at your own pace and time. Your academic success lies in

your own hands!

This module was designed to provide you with fun and meaningful opportunities

for guided and independent learning at your own pace and time. You will be

enabled to process the contents of the learning resource while being an active

learner.

This module has the following parts and corresponding icons:

What I Need to Know

This will give you an idea of the skills or

competencies you are expected to learn in

the module.

What I Know

This part includes an activity that aims to

check what you already know about the

lesson to take. If you get all the answers

correct (100%), you may decide to skip this

module.

What’s In

This is a brief drill or review to help you link

the current lesson with the previous one.

What’s New

In this portion, the new lesson will be

introduced to you in various ways such as a

story, a song, a poem, a problem opener, an

activity, or a situation.

What is It

This section provides a brief discussion of

the lesson. This aims to help you discover

and understand new concepts and skills.

What’s More

This comprises activities for independent

practice to solidify your understanding and

skills of the topic. You may check the

answers to the exercises using the Answer

Key at the end of the module.

What I Have Learned

This includes questions or blank

sentence/paragraph to be filled in to process

what you learned from the lesson.

What I Can Do

This section provides an activity which will

help you transfer your new knowledge or

3

skill into real life situations or concerns.

Assessment

This is a task which aims to evaluate your

level of mastery in achieving the learning

competency.

Additional Activities

In this portion, another activity will be given

to you to enrich your knowledge or skill of

the lesson learned. This also tends retention

of learned concepts.

Answer Key

This contains answers to all activities in the

module.

At the end of this module you will also find:

The following are some reminders in using this module:

1. Use the module with care. Do not put unnecessary mark/s on any part of

the module. Use a separate sheet of paper in answering the exercises.

2. Don’t forget to answer What I Know before moving on to the other activities

included in the module.

3. Read the instruction carefully before doing each task.

4. Observe honesty and integrity in doing the tasks and checking your

answers.

5. Finish the task at hand before proceeding to the next.

6. Return this module to your teacher/facilitator once you are through with it.

If you encounter any difficulty in answering the tasks in this module, do not

hesitate to consult your teacher or facilitator. Always bear in mind that you are

not alone.

We hope that through this material, you will experience meaningful learning

and gain a deep understanding of the relevant competencies. You can do it!

References This is a list of all sources used in

developing this module.

4

What I Need to Know

This module is made for the learners to be able to demonstrate an understanding of

the key concepts of simple and compound interest, and simple and general

annuities.

At the end of this module, the learners will be able to:

1. Investigate problems involving simple and compound interests;

2. Computes interest, maturity value, and present value in simple interest and

compound interest environment (M11GM-IIa-b-1); and

3. Solve problems involving simple and compound interests (M11GM-IIb-2).

What I Know

Before dealing with the lessons of this module, try to measure your prior knowledge

about Interest. Direction: Write the letter of the correct answer on the space provided before each

number.

1. Determine the simple interest that Pia Catriona needed to pay if she borrowed Php2, 345 at 3 1/5 % for10 months.

A. Php26.35 B. Php67.25 C. Php62.53 D. 65.32

2. How much Cardo Dalisay borrowed 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. What is the interest rate did Ahmad receive, after investing the amount of Php 24, 500.00 which earned a simple interest of Php 4, 545.00 after 3.5 years?

A. 3.5% B. 4.5% C. 5.3% D. 5.5% 4. How long should it takes if Maria Linda invests the amount of Php 85, 430.00 to

earn an interest of Php 5, 555.00 at 3 1/4 % simple interest?

A. 1.5 years B. 2 years C. 2.5 years D. 3 years 5. How much is the total amount should Mrs. Soriano receives, if she will invest

Php 123,450.00 for 3.25 years at 2 3/4 % simple interest? A. Php 134, 483.34 B. Php 143, 384.34

C. Php 144, 834.43 D. Php 148, 343.43

5

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.

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 wants to invest money amounting Php 36, 960.00 for 5.75 years at the interest rate of 8 3/5 % 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

9. Find the compound interest (Ic) A. Php 24, 243.94 B. Php 23, 542.49

C. Php 22, 435.49 D. Php 25, 234.94 10. Mr. Duterte invested Php 234, 500.00 at 3 1/4 % compounded annually. He

decides to get this amount for his daughter’s debut party. How much will he get after 45 months?

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 she compromises to pay the

principal and interest at 9 1/10 % compounded quarterly. How much she will

pay after 7 years and 9months? A. Php 966,480.93 B. Php 948,660.39

C. Php 986,406.93 D. Php 906,468.39 13. If a certain amount of money is worth 6 4/5% compounded semi-annually, find

the present value of Php500, 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 takes an amount of Php 7, 350.00 to Php 18, 500.00 if it is

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 an amount of Php 2, 050.00 to Php 3, 875.00 in 4 years and 6months, if the given interest is compounded semi-annually?

A. 1.46% B. 1.66% C. 16.46% D. 14.66%

6

Lesson

1&2 Illustrate and Distinguish Simple and Compound

Interests 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 that is involved in borrowing and lending money. Bank stake 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, the

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.

Since 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.

Activity 1: RECALL AND MOVE FORWARD

DIRECTION: Write SI if the statement below refers to simple interest and CI if compound interest.

_____ 1. It is 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 are 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.

What’s In

7

What’s New

Activity 2: CONCEPTUAL DRILL/ SOLVE AND DECODE A WORD

DIRECTION: A. Write TRUE if the statement is correct otherwise, write FALSE.

________1. If the interest is compounded monthly for 3 years, then the number of interest periods is 36.

________2. The periodic rate of an annual interest rate of 6% compounded quarterly is still 6%.

________3. The interest rate per period of an annual interest rate of 6%

compounded quarterly is 1.5%. ________4. If Php2000 is deposited for 2 years with an annual simple interest rate of

1%, then it will earn a total of Php20 in the given time. ________5. If Php2000 is deposited for 2 years with an annual simple interest rate of

1%, then it will earn a total of Php40 in the given time.

DIRECTION:

Solve the worded problem analytically. Match the appropriate letter that corresponds to your answer in each item. Write it on the space provided

inside the box.

Hint: It is the percentage of a certain amount called the principal amount.

i n t e r e s t

1 2 3 4 5 6 7 8

COLUMN A COLUMN B

1. Danica invested Php50, 000, and was offered an interest rate

of 3% a year. How much will she earn in a year?

a. Php9, 800

2. Jeff was convinced by his friend to invest in his company. His investment in his friend’s company will earn a simple interest

rate of 5% per year. If Jeff decided to invest Php250, 000 for 5 years, how much interest will he get?

b. Php12, 500

3. Debbie invested Php100, 000, and earned an interest rate of 1% per annum. How much will she earn in 2 years?

c. Php2, 500

4. Ronda invested Php50, 000 with an interest rate of 5% a year.

How much will he earn in a year?

d. Php62, 000

5. Annaliza wants to put her savings in a trust fund with a simple interest rate of 2% a year. If she decides to invest

Php500, 000 for 3 years, how much interest will she earn?

e. Php1, 500

6.Francisca invested Php150, 000 with an interest rate of 10% a

year. How much will she earn in a year?

f. Php62, 500

g. Php5, 000

h. Php30, 000

i. Php15, 000

j. Php2, 000

8

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.

3.1 Calculating Simple Interest The Equation to compute for simple interest is

Where

I = 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

You may now 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?

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 I = Prt= (10, 000)(0.1025)(1)

= Php 1, 025.00

Example 2: Kara Mia deposits Php111, 110.00 in her bank

account at 5 1/5% interest rate. How much

interest will she earn in 5.25 years? Given: P = 111,110

r = 5 1/5% = 5.2% or 0.052 t = 5.25 years

Find:I

Solution: I = Prt= (111, 110)(0.052)(5.25)

= (111, 110)(0.273)

= Php 30, 333.03

What is It

Lesson

3 Interest, Maturity Value, Future Value and Present Value in Simple

Interest

9

Example 3:

Find the interest paid by Kathryn Nadine on Php150, 150.00 that she borrowed for

three years and 9 months at 1 3/4 % simple interest.

Given:

P = 150, 150

r = 1 3/4% or 1.75% or 0.0175 t = 3 years and 9 months = 3 9/12= 3.75

years Find: I

Solution:

I = Prt = (150,150) (0.0175)(3.75)

= (150,150) (0.065625) = Php9, 853.59

Example 4: If Sahaya borrowed Php125, 500.00 at an

annual simple interest rate of 7 3/4 % for

42 months, how much interest should she pay?

Given: P = 125, 500

r = 7 3/4 % 7.75% or 0.0775 t = 42 months

= 3.5 years Find: I

Solution: Prt= (125,500)(0.0775)(3.5)

= (125,500)(0.27125)

= Php 34,041.88

N o t e :

When the term is expressed in a month (m), convert in year(s) by t= m/12

ENHANCEMENT EXERCISE 1: Answer each of the following worded problems. Compute for the

unknown then label your final answer properly. 1. Mrs. Batungbakal gave a Php100, 000.00 to Mrs. Macaspac at 18%. How much

was the interest after 45 days?

2. How much interest interest will be paid by Mr. Tom McKey, if he borrowed Php515, 515 at 3.25% after 8 months?

3. Cory Kong deposited Php202, 020 at a rural bank in Tayug and she is paying 1.5%per annum simple interest. How much interest will she earn after 2 years

and 6 months?

3.2 Maturity (Future) Value with Simple Interest

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 + I

= P + Prt = P (1+rt) < by factoring >

10

Where P = principal

r = interest rate (convert to decimal) t = time (in years)

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.

While principal and present value mean the same thing also, principal is usually used for loans while present value is for invested or saved money.

Take the following examples for application of the formula.

Example 1: Find the maturity (future) value:

Principal (P) Rate (r) Time (t) Maturity Value (F)

Php12, 345 2.15% 1.25 years (a)

Php67, 890 0.1% 2 years and 9 mos. (b)

Solution:

a. F = P (1 + rt )

= 12, 345 [(1 + (0.0215) (1.25)]

= 12, 345 (1 + 0.026875) = 12, 345 (1.026875)

= Php12, 676.77

b. F = P (1 + rt) = 67, 890 [(1 + (0.001) (2.75)] 2 years and 9 mos. = 2 ¾ years

= 67, 890 (1 + 0.00275) = 2. 75 = 67, 890 (1.00275)

= Php68, 076.70

Example 2:

How much should Lizzy Mae pay for a Php10, 000.00 loan, if charged at 16.5% at the end of 1 year and 6 months?

Given:

P = 10, 000 r = 16.5% = 0.165

t = 1 year and 6 months

= 1 6/12 = 1.5 years

Find: F

Example 3:

Six years and nine months ago, Bianca borrowed Php25, 000 from Bernadette with

the promise that Bianca will pay Bernadette the principal amount plus accumulated interest at 9 5/8% simple interest now. What amount is due?

Solution: F = P (1 + rt )

= (10, 000) [1+ (0.165)(1.5)] = (10, 000)(1 + 0.2475)

= (10, 000)(1.2475)

= Php12, 475.00

11

Given: P = 25, 000

r = 9 5/8% =9.625% = 0.09625 t = 6 years and 9 months = 6 9/12 = 6.75 years

Find: F

ENHANCEMENT EXERCISE 2: Solve the following problems with complete solution.

1. Find the maturity (future) value (F).

Principal (P) Rate (r) Time (t) Maturity Value (F)

Php123, 450 .25% 15 months (a)

Php456, 780 2.5% 2 years and 3 mos. (b)

Php789, 100 2 2/5% 3 years and 6 mos. (c)

2. Jamaica Rimorin invested the amount of Php100, 000 in a bank that offers 15

1/2% per annum simple interest. How much will she earned after 90 days? 3. Reuben’s mother borrowed Php35, 000 to fix their car. The bank charges 5%

interest for two years. What is the total amount that his mother will owe the

bank?

3.3 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

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 P r t

Php12, 345 (a) 1.05% 51 months

Php246, 810 Php151,515 (b) 9 years and 9 months

Php3, 691, 215 Php1,234,560 12 3/20% (c)

Solution: a. P= F / (1+rt) = 12, 345 / [1 + .0105(4.25)]= 12, 345 / 1.004625

=Php11, 817.64

Solution:

F = P (1 + rt) = (25, 000) [1+ (0.09625)(6.75)]

= (25, 000)(1 + 0.6496875)

= (25, 000)(1.6496875) = Php41, 242.19

12

b. r= (F-P) / Pt = (264,810 – 151, 515) / (151,515)(9.75) =95,295 / 1, 477, 271.25

=0.0645 or 6.45% c. t= (F-P) / Pr =(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. Hence, they want to have Php150, 000.00 after 10 years. How much should

they invest at an investment paying 15% for 5years? Given: F = 150, 000 r = 15% = 0.15 t = 5 years Find: P

Example 3: At what rate will the amount of Php20, 000 investment yields Php85, 000 in15

years at simple interest?

Given:

F = 85, 000 P = 20, 000

t = 15 years Find: r

Example 4:

How long will an investment of Php543, 200 yields Php876, 500 at simple interest rate of 5% annually?

Given:

P= 543, 200 F= 876, 500

r= 5% = 0.05 Find: t=?

ENHANCEMENT EXERCISE 3: Complete the table below.

F P r t

Php1, 010.10 Php864.20 5.15% (a)

Php202, 020.20 Php100, 100 (b) 6 years &6 mos

Php5, 050, 500.50 (c) 4 3/5% 8.5 years

Solution:

P= F / (1+rt) =150, 000 / [1+(0.15)(5)]

= 150, 000/ (1+0.75) = 150, 000 / 1.75

= Php85, 714.29

Solution:

r = (F-P) / Pt = (85,000-20, 000) / (20, 000)(15)

= 65, 000 / 300, 000 = 0.22 or 22%

Solution:

t = (F-P) / Pr = (867, 500 – 543, 200) / (543, 200)(0.05)

= 333, 300 / 27, 160

= 12.27 years

13

Lesson

4 Interest, Maturity Value, Future Value, and Present Value in

Compound Interest C o m p o u n d I n t e r e s t 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

Example 1: Find the maturity value and compound interest on the principal Php30, 000 if

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

Example 2:

How much money will you have in a bank after 5 years, when your mother deposited Php15, 000 at an annual interest rate of 1 ¼ % compounded yearly?

Given: P = 15, 000

r = 1 ¼ % = 1.25% = 0.0125 t = 5 years

Find: F

Solution:

F = P (1 + r)t = 30, 000 (1 + 0.045)3

= 30, 000 (1.045)3= 30, 000 (1.141166125)

Ic = Php 34, 234.98– 30, 000= Php 4, 234.98

Solution:

F = P (1 + r)t

= 15, 000 (1 + 0.0125)3

= 15, 000 (1.0125)3

= 15, 000 (1.037970703)

= Php15, 569.56

14

Example 3. Bryan Bagunas invested Php 123, 450.00 in a bank at 2 ¾ %compounded annually

for 69 months. Find:

a. maturity (future) value b. compound interest

Given:

P = 123, 450

r = 2 ¾ % = 2.75 % = 0.0275 t = 69 months = 69/12 = 5.75 years

Find: F and Ic

ENHANCEMENT EXERCISE 4: 1. Complete the table by finding the unknown values.

P r t F Ic

Php5, 790 1.3% 75 mos. (a1) (a2)

Php35, 790 .13% 7.75 mos. (b1) (b2)

Php135, 790 1 3/5 % 5 years and 6 mos.

(c1) (c2)

2. To what sum of money of Php71, 200 will accumulate in 2 years and 9 months

at 7% compounded annually? How much is the compound interest?

6. How much will be paid on the maturity date December 27, 2019, for a loan of

Php45, 000 that is made on June 27, 2015 with interest rate at 12 ½ % converted annually?

4.1 Finding the Present Value at Compound Interest From the formula, F = P (1 + r) t, you can derive the formula in finding the present

value.

Take F = P (1 + r) t then

F /(1+rt) t = P(1+rt) t / (1+r) t <divide both sides by (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 Php34, 500.00 due in 3 years and 3months, if money is

worth 3.45% compounded annually?

Given: F = 34, 500

t = 3 years and 3 months 3 = 3 3/12 = 3.25 years r = 3.45% or 0.0345

Find: P

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

= Php20, 840.13

Solution:

P = F / (1+r) t = 34, 500 / (1 + 0.0345) 3.25

= 34, 500 / (1.0345) 3.25

= 34, 500 / 1.116539544

= Php30, 899.04

15

Example 2: How much Manilyn should put in a time deposit in a bank that pays 2.15% compounded

annually so that she will have Php876, 500 after 81months?

Given: r = 2.15% or 0.0215

F = 876, 500 t = 81 months = 81/12 = 6.75 years

Find: P

ENHANCEMENT EXERCISE 5

Solve the following problems with complete solution. 1. How much money must be invested to obtain the amount of Php757, 575 in 7

years and 6 months if money earns 5.25% compounded annually?

2. Keifer Ravena wants his investment grows to Php696, 969 in 6 years and 9 months. How much should he invest in his account that pays 6.9%compounded

annually?

3. How much Ramil de Jesus must be deposited in a bank that offers 6 3/5 % compounded annually so that after 69 months, he will have Php909, 090?

4.2 COMPOUNDING MORE THAN ONCE A YEAR

The compound frequency (or conversion frequency) is the number of compounding that takes 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 1

Semi-Annually 2

Quarterly 4

Monthly 12

Weekly 52

Daily 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= im / m

Where:

Solution:

P = F / (1+r) t = 876, 500 / (1+0.0215)6.75

= 876, 500 / (1.0215)6.75 = 876, 500 / 1.15440714

= Php759, 264.19

16

j = rate of interest for each conversion period im = annual rate of interest

m = frequency of conversion

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 (+j) n

F= P ( 1 + im/m) mt

Let us take the following examples for the application of the above formula.

Example 1: Find the maturity value of Php565, 565 invested for 2 years and 6 months

at 5.25% compounded quarterly.

Given: P = 565, 565

t = 2 years and 6 months= 2 6/12 = 2.5 years m = 4

im = i4 = 5.25% or 0.0525

Find: F

Solution: Compute for the interest rate in a conversion period

j= im / m = i4 / 4 = 0.0525/4 = 0.013125

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

Compute for the maturity (future) value

F= P (+j) n = 15, 000 (1 + 0.045)9.5 = 15, 000(1.045)9.5 = 15, 000 (1.519164346) = Php22, 787.47

Example 3.

Find the compound amount and interest on Php175, 250 for 15years and 6 months

at 6 ¾ % compounded quarterly.

Given: P = 175, 250

m = 4 im = i4 = 6 ¾ % = 6.75 or 0.0675

t = 15 years and 6 months = 15 6/12 = 15.5 years

Find: F

Solution:

Where: F = maturity (future) value

P = principal im= nominal rate of interest (annual

rate)

m = frequency of conversion t = term/time in years

17

Compute first j and n before F and Ic

j= im / m = i4 /4= 0.0675 / 4 = 0.016875

n = mt = (4) (15.5) = 62

F= P (+j) n = 175, 250 (1 + 0.016875)62 = 175, 250(1.016875)62= 175, 250 (2.822210202)

= Php494, 592. 34

Ic = F– P= 494, 592. 34 –175, 250 = 319, 342.34

ENHANCEMENT EXERCISE 6

Solve each of the following: 1. Find the compound amount and interest on Php 175, 250.00 for 15 years and6 months at

6 ¾ % compounded quarterly.

2. On January 23, 2013, Julius Paclibar borrowed Php15, 000 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 Php565, 565 invested for 2 years and 6 months at 5.25% compounded quarterly.

4.3 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 <divide both sides by (1+j)n>

So you will get,

P= F / (1+j)n

Where P = present value F = maturity (future) value

j = im / m = rate of interest per conversion period

n = mt = total number of conversion period Apply the formula with the following examples:

Example 1: Cardo and Aryana aim to accumulate Php3, 691, 215 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

18

t = 15 years

Solution: a. When m =12 and im = i12 = 5% 0r 0.05

j = im / m = i12 / 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 = Php1, 735, 922.27

b. When m = 4 and im = i4 = 5% or 0.05

j = im / m = i4 / 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 = Php1, 751, 731.05

c. When m = 2 and im = i2 = 5% or 0.05

j = im / m = i2 / 2 = 0.05/2 = 0.025

n = mt = (2) (15) = 30 P= F / (1+j)n = 3, 691, 215 / (1+0.025)30

= 3, 691, 215 / (1.025)30

= 3, 691, 215 / 2.097567579 = Php1, 759, 759.75 Example 2:

A financial obligation of Php25, 500 is due on January 6, 2019. What is the value of this obligation on July 6, 2013 at 9% compounded semi-annually?

Given: F = 25, 500

im = i2 = 9% or 0.09 m = 2

t = July 6, 2013 to January 6, 2019 (5 years and 6 months)= 5 6/12 = 5.5 years Find: P

Solution:

When m = 2 and i2 = i2 = 9% or 0.09

j = im / m = i2 / 2 = 0.09/2 = 0.045

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

19

P= F / (1+j)n = 25, 500 / (1+0.045)11

= 25, 500 / (1.045)11

=25, 500 / 1.622853046 = Php15, 713.07

ENHANCEMENT EXERCISE 7 Solve each of the following: 1. Find the present value of the following if the compounded amount is Php690,

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.

2. How much should Jeuz Batalla needs to invest today to buy his only son a brand new Toyota car worth Php1, 234, 567 as a graduation gift 8.5 years from

now if money is worth 9 ¾ % compounded quarterly?

What’s More

Activity 3

A. Determine the simple interest for these loans.

1. Php450 at 7% for 2 years 2. Php5, 200 at 4% for 3 years

3. Php1, 300 at 5% for 6 years 4. Php5, 400 at 3.5% for 6 months

5. Php600 at 4% for 9 months.

B. Solve for the following: 1. Php500 invested at 4% compounded annually for 10 years.

2. Php600 invested at 6% compounded annually for 6 years. 3. Php750 invested at 3% compounded annually for 8 years.

4. Php1500 invested at 4% compounded semiannually for 7 years. 5. Php900 invested at 6% compounded semiannually for 5 years.

20

What I Have Learned

Interest, Maturity Value, Future Value and Present Value in Simple

and Compound Interest!

Topic Basic Concepts

Calculating Simple Interest

Simple Interest (Is) is calculated as a

percentage of the original

amount borrowed or deposited over a period of time.

Maturity (Future) Value with Simple

Interest

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).

Finding the Present Value, Rate and

Time in Simple Interest

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

Compound Interest The formula for compound interest (Ic) is Ic = F– P

Where Ic = compound interest

F = maturity (future) value P = principal

Finding the Present Value at

Compound Interest

The formula in finding the present value.

P = F / (1+r) t

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 (+j) n

F= P ( 1 + im/m) mt

Finding the Present Value in Compound

Interest

Formula in finding the present value.

P= F / (1+j)n

21

Solve the following:

1. Ben deposited Php6, 500 in a simple interest account that pays 2.8% interest

annually. If Ben leaves the money in the account for 12 years, how much

interest will he earn?

2. Steph took out a simple interest loan that charges 8.5% interest annually. If her

original loan amount was for Php30, 000 and she is paying off the loan over 20 years, how much will she pay altogether?

3. Derek deposited Php3, 000 in a simple interest account 3 years ago. He has

earned Php360 in interest so far. What is the interest rate for his account?

4. Clint earned Php1, 155 in interest from his simple interest account over the last

5 years. The account pays 1.25% interest annually. How much did Clint originally deposit in his account?

5. Hector took out a small loan of Php900 for 18 months. The simple interest rate

on the loan was 10.5%. How much will he pay in interest on the loan?

6. If you deposit Php4000 into an account paying 6% annual interest

compounded quarterly, how much money will be in the account after 5 years?


compounded monthly, how much money will be in the account after 7 years?

8. How much money would you need to deposit today at 9% annual interest

compounded monthly to have Php12000 in the account after 6 years?


compounded monthly, how long until there is Php8000 in the account?

10. If you deposit Php8000 into an account paying 7% annual interest compounded quarterly, how long until there is Php12400 in the account?

Assessment

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

Php8,765 at 7 2/5% for 8months.

A. Php234.14 B. Php324.41 C. Php432.41 D. Php441.23 2. How much was borrowed by Ms. Gray if the interest she paid after 15 months at

10.5% simple interest is Php6, 420? A. Php41, 948.29 B. Php44, 198.92 C. Php49, 149.92 D. Php48, 914.29

What I Can Do

22

3. At what rate of interest did Alari invest the amount of Php56, 780.00 for it to earn a simple interest of Php5, 555 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 Php65, 120 to earn an

interest of Php5, 225.00 at 4 3/5 % 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 invests Php12, 890 for 1.75 years at 3 ½ % simple interest?

A. Php13, 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 in 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: An effective rate is a rate when compounded annually will give the

same compound each year with the nominal rate. 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 wants to invest an amount of Php36, 912 for 5.5 years at an interest rate of 8 1/2 % compounded annually.

8. Find the maturity value (F).

A. Php51, 378.57 B. Php53, 718.75

C. Php57, 813.75 D. Php58, 731.57 9. Find the compound interest (Ic).

A. Php27, 090.15 B. Php21, 009.57 C. Php20, 901.75 D. Php29, 001.57

10. Oliver Almario invested Php102, 030 at 3 ¾% 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. Php164, 502.21

11. Find the interest if Php87, 650 is deposited in a bank at 4.5%compounded monthly for 36 months.


12. Kung Fu Reyes borrows Php 101, 010.00, and promises to pay the principal and interest at 9 1/5 % compounded quarterly. How much must will he pay after 6

years and 3 months?


13. If money is worth 4 2/5 % compounded semi-annually, find the present value of Php300, 300 due at the end of 3 years, and 6 months.


23

14. How long will it takes the amount of Php3, 690 to Php8, 910, if it is 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 Php1, 010 grow to an amount of Php2, 020 in 3 years and 3months, if interest is compounded semi-annually?

A. 25.12% B. 24.41% C. 23.31% D. 22.51% 16. If interest is compounded quarterly, find the nominal rate which will yield an

effective rate of 3 ½ %.

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 of 4 %. A. 3.02% B. 4.07% C. 5.05% D. 6.09%

18. Alden Richards is investing Php100, 000 at 10 % converted semi-annually for 10 years. At what rate compounded monthly he could have in investing his

money?

A. 9.8% B. 8.7% C. 7.6% D. 6.5% 19. If you are to invest Php110, 110.00 at 15% converted monthly for 9 years and 9

months, what simple interest rate could have, as well as your invested 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 Php500, 000.00 at 5% simple interest for 30

months. At what interest compounded quarterly he could have, as well as his investment for the same period of time?

A. 2.47% B. 3.74% C. 4.74% D. 5.47%

Solve the problem analytically.

1. Find the interest on a used car loan of Php5000 at a rate of 16% for a period of

8 months.

2. Find the amount owed on an investment of Php10, 000 into a money market

account that pays a simple interest rate of 1.75% over a 39 week period. 3. How much should you deposit initially in an account paying 10% compounded

semi-annually in order to have Php1, 000, 000 in 30 years? b) compounded monthly? c) compounded daily? d) continuous compounding

Additional Activities

24

Answer Key

What’s New:

Activity 2 -A

1.True

2.True

3.False

4.False

5.True

B

1. I

2. N

3. T

4. E

5. R

6. E

7. S

8. T

What I Know:

1.C 11. B

2.D 12. A

3.C 13. B

4.B 14. C

5.A 15. D

6.D

7.D

8.C

9.C

10.B

What’s In:

ACTIVITY 1

1.SI

2.CI

3.SI

4.SI

5.CI

6.CI

7.CI

8.SI

9.CI

10.CI

Additional Activities:

- depende sa sagot

ng bata

Assessment:

1.C 11. B

2.D 12. A

3.C 13. B

4.B 14. A

5.A 15. D

6.A 16. C

7.A 17. B

8.C 18. A

9.C 19. B

10.B 20. C

What I Can Do

1.Php2,184

2.Php81, 000

3.4%

4.Php18, 480

5.Php141.75

6.Php5, 387.42

7.Php11,

358.24

8.Php7, 007.08

9.7.9

10.6.3

What is It: Enhancement Exercise 1

1.Php2, 250

2.Php11, 169.49

3.Php7, 575.75

Enhancement Exercise 2

1.

A.Php123, 835.78

B.Php482, 473.88

C.Php155, 384.40

2.Php103, 875

3.Php38, 500


a.3.28 years

b.15.66%

c.Php3, 630, 841.48


A1. Php6, 276.79

A2. Php486.79

B1. Php36, 152.17

B2. Php362.17

C1. Php148, 177.81

C2. Php12, 387.81


1.Php516, 130.89

2.Php444, 236.21

3.Php629, 510.46


1.F=Php494, 592.34

Ic=Php319, 342.34

2.Php22, 787.47

3.Php644, 336.67


1.

a. Php458, 360.36

b. Php388, 969.57

c. Php467, 633.16

2. Php544, 389.39

What’s More:

Activity 3

A.

1. Php63

2. Php624

3. Php390

4. Php94.50

5. Php18

B.

1. Php740.12

2 .Php851.11

3. Php950.08

4. Php1, 979.22

5. Php1, 209.52

25

References

Phoenix (2016), Next Century Mathematics General Mathematics.

Vibal (2016), General Mathematics

Rex (2015), Math of Investment, Fifth Edition

For inquiries or feedback, please write or call: Department of Education – Region III, Schools Division of Bataan - Curriculum Implementation Division Learning Resources Management and Development Section (LRMDS)

Provincial Capitol Compound, Balanga City, Bataan

Telefax: (047) 237-2102

Email Address: [email protected]

mailto:[email protected]

general mathematics - znnhs · 2021. 1. 22. · php 123,450.00 for 3.25 years at 2 3/4 % simple...

Documents