GENERAL MATHEMATICS Quarter 2 - Module 6:

Simple and Compound Interests

Department of Education ● Republic of the Philippines

General Mathematics – Grade 11 Alternative Delivery Mode Quarter 2 – Module 6: Simple and Compound Interests First Edition, 2020

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11

General Mathematics

Quarter 2 – Module 6: Simple and Compound Interests

Department of Education • Republic of the Philippines

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encourage teachers and other education stakeholders to email their feedback,

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We value your feedback and recommendations.

LESSON 1: Simple Interest

What I Need to Know …………………………………………………. 1

What I Know …………………………………………………. 1 – 3

What’s In …………………………………………………. 3

What’s New …………………………………………………. 3 – 4

What is It …………………………………………………. 4 – 6

What’s More …………………………………………………. 6

What I Have Learned …………………………………………………. 6 – 7

What I Can Do …………………………………………………. 8

Additional Activities …………………………………………………. 8

Answer Key …………………………………………………. 9

LESSON 2: Compound Interest

What I Need to Know …………………………………………………. 10

What’s In …………………………………………………. 10

What’s New …………………………………………………. 10 - 11

What is It …………………………………………………. 11 – 15

What’s More …………………………………………………. 15

What I Have Learned …………………………………………………. 16 – 17

What I Can Do …………………………………………………. 17

Additional Activities …………………………………………………. 18

Assessment (Post-test) …………………………………………………. 19 – 20

Answer Key …………………………………………………. 21 – 22

References …………………………………………………. 23

TABLE OF CONTENTS

1

In many aspects of modern life, Mathematics plays an important role. In the field of

business, mathematics is essential in analyzing markets, predicting stock market prices,

business decision making, forecasting production, financial analysis, and in business

operation in general. This module will introduce the students to the basic concepts of

business mathematics such as the simple and compound interests.

What I Need to Know

At the end of the lesson, the learners should be able to: 1. define simple interest;

2. compute simple interest, maturity value and present value; and

3. solve problems involving simple interest.

What I Know (Pretest)

Directions: Read each statement carefully. Choose the letter of the correct answer and write it

on a 1 whole sheet of paper.

1.) This refers to the accumulated amount obtained by adding the principal and the

compound interest.

A. Compound amount C. Present value

B. Compound interest D. Simple interest

2.) Date on which the money borrowed or loaned is to be completely repaid.

A. Conversion period C. Maturity date

B. Loan date D. Origin date

3.) What is the formula in computing the simple interest on a given financial transaction?

A. 𝐼𝑠 = 𝑃𝑟𝑡 C. 𝐼𝑠 =𝑃

𝑟𝑡

B. 𝐼𝑠 = 𝑃𝑟2𝑡 D. 𝐼𝑠 =𝑟

𝑃𝑡

4.) This refers to the interest rate per conversion period.

A. Compound interest C. Rate of interest

B. Periodic rate D. Simple interest

Simple Interest Lesson

1

2

5.) This refers to the amount paid or earned for the use of money.

A. Conversion period C. Principal

B. Interest D. Rate

6.) 30 months is equivalent to

A. 2.5 years C. 3 years

B. 2.75 years D. 3.25 years

7.) How much is the simple interest on this financial transaction, P = ₱5,000.00, r = 6%,

and t = 2 years?

A. ₱120.00 C. ₱1,200.00

B. ₱600.00 D. ₱6,000.00

8.) What is the total number of conversion periods when a certain amount is borrowed at

10% compounded monthly for 5 years?

A. 12 C. 24

B. 50 D. 60

9.) How much was the interest if Sophia borrowed ₱45,000.00 and paid a total of

₱55,500.00 at the end of the term?

A. ₱10,500.00 C. ₱11,500.00

B. ₱45,000.00 D. ₱100,500.00

10.) What is the interest rate per conversion period if ₱25,900.00 was invested at 3.5%

compounded annually for 4 years and 6 months?

A. 0.035 C. 0.140

B. 0.350 D. 0.460

11.) John borrowed ₱45,400.00 at 10% simple interest rate. How much should he repay

after 3 years?

A. ₱13,620.00 C. ₱59,020.00

B. ₱46,762.00 D. ₱104,420.00

12.) An interest of ₱760 was earned on an investment for 9 months at 3% interest rate. How

much was invested?

A. ₱2,052.00 C. ₱20,520.00

B. ₱2,814.81 D. ₱33,777.78

13.) At what simple interest rate was ₱18,350.00 invested if it earned ₱1,025.00 interest

for 1.5 years?

A. 0.0372% C. 3.72% B. 0.1193% D. 11.93%

14.) How much will be the compound interest if ₱30,220.00 is invested at 7% compounded

quarterly for 2 years and 9 months?

A. ₱4,499.21 C. ₱36,574.05

B. ₱6,354.05 D. ₱147,899.31

3

15.) When is ₱78,800.00 due if its present value of ₱61,500.00 is invested at 103

4%

compounded monthly?

A. 2. 32 years C. 10.75 years

B. 6.95 years D. 27.79 years

What’s In

On the previous modules, the basic concepts on functions were introduced. Functions

were used as mathematical models. These are abstract models that use mathematical language

to describe relationships. With the notion of mathematical modeling, mathematics is concerned

not only with the measures of the physical world but it has also expanded its applicability to

sciences, both social and biological, business, and finance. So with this, lessons relating to

business and finance will then be introduced specifically on simple interest.

What’s New

“When you saved money in the bank, you will gained an interest paid by the bank. On

the other hand, when you borrow money, you are charged an interest on the amount you

borrowed. How does gained and charged interests computed?”

A debtor pay the bank an amount which is more than the amount they borrowed. An

investor may withdraw from the bank more than the amount deposited. This additional sum is

called INTEREST.

Definition of terms:

Lender or creditor – person (or institution) who invests the money or makes the funds available.

Borrower or debtor – person (or institution) who owes the money or avails of the funds from

the lender.

Origin or loan date – date on which money is received by the borrower.

Repayment date or maturity date – date on which the money borrowed or loaned is to be

completely repaid.

Time or term (t) – amount of time in years the money is borrowed or invested; length of time

between the origin and maturity dates.

4

Principal or present value (P) – amount of money borrowed or invested on the origin date.

Rate of interest or simply rate (r) – annual rate, usually in percent, charged by the lender, or

rate of increase of the investment.

Interest (I) – amount paid or earned for the use of money.

Maturity Value or Future Value (F) – amount after t years that the lender receives from the

borrower on the maturity date; equal to the sum of principal and the interest earned.

What is It

Simple Interest (Is)

For every financial transaction, whether you borrowed or invested a certain amount P,

a corresponding percentage of the principal called interest is being paid. Simple Interest (Is)

is the interest charged on the principal alone for the entire duration or period t of the loan or

investment, at a particular rate r. After the term of the loan or investment, the maturity value

or future value F is computed by getting the sum of the principal and the interest due.

Formulas:

• 𝑰𝒔 = 𝑷𝒓𝒕

• 𝑭 = 𝑷 + 𝑰𝒔 or 𝑭 = 𝑷 + 𝑷𝒓𝒕 or 𝑭 = 𝑷(𝟏 + 𝒓𝒕)

• 𝑷 =𝑰𝒔

𝒓𝒕 or 𝑷 = 𝑭 − 𝑰𝒔

• 𝒕 =𝑰𝒔

𝑷𝒓

• 𝒓 =𝑰𝒔

𝑷𝒕

***where 𝐼𝑠 − simple interest

𝑃 − principal

𝑟 − rate of interest or simply rate

𝑡 − time (in year)

𝐹 − future value (or maturity value)

Note: If the given time is in months, it can be converted to year(s) by using the formula

𝑡 =𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑛𝑡ℎ𝑠

12 .

5

Example.

Directions: Complete the table below by solving the unknown quantities in each row.

Principal

(P)

Rate

(r)

Time

(t)

Simple Interest

(Is)

Future Value

(F)

1.) ₱500,000.00 12.5% 10 years

2.) 2.5% 4 years ₱1,500.00

3.) ₱36,000.00 1 year and

6 months

₱4,860.00

4.) ₱250,000.00 0.5% ₱1,400.00

5.) ₱10,000.00 4% 5 months

Solution:

1.) Given: P = ₱500,000.00 ; r = 12.5% or 0.125 ; t = 10 years

𝐼𝑠 = 𝑃𝑟𝑡 𝐹 = 𝑃 + 𝐼𝑠

𝐼𝑠 = ₱500,000.00(0.125)(10) 𝐹 = ₱500,000.00 + ₱625,000.00

𝑰𝒔 = ₱𝟔𝟐𝟓, 𝟎𝟎𝟎. 𝟎𝟎 𝑭 = ₱𝟏, 𝟏𝟐𝟓, 𝟎𝟎𝟎. 𝟎𝟎

2.) Given: r = 2.5% or 0.025 ; t = 4 years ; 𝐼𝑠 = ₱1,500.00

𝑃 =𝐼𝑠

𝑟𝑡 𝐹 = 𝑃 + 𝐼𝑠

𝑃 =₱1,500.00

0.025 (4) 𝐹 = ₱15,000.00 + ₱1,500.00

𝑷 = ₱𝟏𝟓, 𝟎𝟎𝟎. 𝟎𝟎 𝑭 = ₱𝟏𝟔, 𝟓𝟎𝟎. 𝟎𝟎

3.) Given: P = ₱36,000.00 ; t = 1 6

12 years or 1.5 years ; 𝐼𝑠 = ₱4,860.00

𝑟 =𝐼𝑠

𝑃𝑡 𝐹 = 𝑃 + 𝐼𝑠

𝑟 =₱4,860.00

₱36,000.00(1.5 ) 𝐹 = ₱36,000.00 + ₱4,860.00

𝒓 = 𝟎. 𝟎𝟗 or 9% 𝑭 = ₱𝟒𝟎, 𝟖𝟔𝟎. 𝟎𝟎

4.) Given: P = ₱250,000.00 ; r = 0.5% or 0.005 ; 𝐼𝑠 = ₱1,400.00

𝑡 =𝐼𝑠

𝑃𝑟 𝐹 = 𝑃 + 𝐼𝑠

𝑡 =₱1,400.00

₱250,000.00(0.005) 𝐹 = ₱250,000.00 + ₱1,400.00

𝒕 = 𝟏. 𝟏𝟐 years 𝑭 = ₱𝟐𝟓𝟏, 𝟒𝟎𝟎. 𝟎𝟎

6

5.) Given: P = ₱10,000.00 ; r = 4% or 0.04 ; t = 5

12 year

𝐼𝑠 = 𝑃𝑟𝑡 𝐹 = 𝑃 + 𝐼𝑠

𝐼𝑠 = ₱10,000.00 (0.04) (5

12) 𝐹 = ₱10,000.00 + ₱166.67

𝑰𝒔 = ₱𝟏𝟔𝟔. 𝟔𝟕 𝑭 = ₱𝟏𝟎, 𝟏𝟔𝟔. 𝟔𝟕

What’s More

I. Complete the table below by solving the unknown quantities in each row. Write

your complete solutions and answers on a 1 whole sheet of paper.

Principal

(P)

Rate

(r)

Time

(t)

Simple Interest

(Is)

Future Value

(F)

1.) ₱40,000.00 2% 3 years

2.) 10% 5 years ₱2,500.00

3.) ₱100,000.00 1.5 years ₱3,600.00

4.) ₱250,000.00 4.5% ₱15,400.00

5.) ₱12,345.00 8.25% 9 months

II. Solve the future value (refer on test I) using the alternative formulas:

𝐹 = 𝑃 + 𝑃𝑟𝑡 or 𝐹 = 𝑃(1 + 𝑟𝑡)

What I Have Learned

Problems Involving Simple Interest

1. A bank offers 1.5% annual simple interest rate for a particular deposit. How much

interest will be earned if 1 million pesos is deposited in this savings account for 1 year?

Solution:

Given: r = 1.5% or 0.015 ; P = ₱1,000,000.00 ; t = 1 year

𝐼𝑠 = 𝑃𝑟𝑡

𝐼𝑠 = ₱1,000,000.00 (0.015)(1)

𝑰𝒔 = ₱𝟏𝟓, 𝟎𝟎𝟎.00

Therefore, an interest amounting to ₱15,000.00 will be earned if 1 million pesos is

deposited in a savings account for 1 year with 1.5% annual simple interest rate.

7

2. When invested at an annual interest rate of 7%, the amount earned ₱11,200.00 of simple

interest in 2.5 years. How much money was originally invested?

Solution:

Given: r = 7% or 0.07 ; 𝐼𝑠 = ₱11,200.00 ; t = 2.5 years

𝑃 =𝐼𝑠

𝑟𝑡

𝑃 =₱11,200.00

0.07(2.5)

𝑷 = ₱64,000.00

Therefore, the amount of money originally invested was ₱64,000.00.

3. Ricky borrowed ₱25,000.00 and paid ₱1,250.00 interest for 6 months. What was the

rate of interest?

Solution:

Given: P = ₱25,000.00 ; 𝐼𝑠 = ₱1,250.00 ; t = 6

12 year or 0.5 year

𝑟 =𝐼𝑠

𝑃𝑡

𝑟 =₱1,250.00

₱25,000.00(0.5)

𝒓 = 0.1 or 10%

Therefore, the rate of interest was 0.1 or 10%.

4. How long in years will it take for ₱17,300.00 to amount to ₱20,000.00 at 11.25%

simple interest?

Solution:

Given: P = ₱17,300.00 ; F = ₱20,000.00 ; r = 11.25% or 0.1125

𝐼𝑠 = 𝐹 − 𝑃

𝐼𝑠 = ₱20,000.00 − ₱17,300.00

𝐼𝑠 = ₱2,700

𝑡 =𝐼𝑠

𝑃𝑟

𝑡 =₱2,700

₱17,300.00(0.1125)

𝒕 = 𝟏. 𝟑𝟗 years

Therefore, it will take 1.39 years for ₱17,300.00 to amount to ₱20,000.00.

8

What I Can Do

Answer the following problems involving simple interest. Write your complete

solutions and answers on a 1 whole sheet of paper.

1. Find the simple interest on a loan of ₱65,000.00 if the loan is given at a rate of 2% and

is due in 5 years and 3 months?

2. How much money will you have after 4 years if you deposited ₱10,000.00 in a bank

that pays 6% simple interest?

Additional Activities

Formulate own three problems that involve simple interest. Write the problems and its

complete solutions and answers on 1 whole sheet of paper.

9

Answer Key

What I Know (Pretest)

1. A

2. C

3. A

4. B

5. B

6. A

7. B

8. D

9. A

10. A

11. C

12. D

13. C

14. B

15. A

What’s More

I.

1. Is = ₱2,400.00 F = ₱42,400.00

2. P = ₱5,000.00 F = ₱7,500.00

3. r = 0.024 or 2.4% F = ₱103,600.00

4. t = 1.37 years F = ₱265,400.00

5. Is = ₱763.85 F = ₱13,108.85

II.

*The same answers on test I (values of F).

What I Can Do

1. Is = ₱6,825.00

2. F = ₱12,400.00

Additional Activities

*Answers may vary

10

Compound Interest

What I Need to Know

At the end of the lesson, the learners should be able to: 1. define compound interest;

2. compute compound interest, maturity value and present value;

3. differentiate simple from compound interest; and

4. solve problems involving compound interest.

What’s In

Problems involving simple interest were discussed on the previous lesson. Simple

interest is the interest charged on the principal alone for the entire length of the loan or

investment. Several formulas were introduced to solve problems involving simple interest. The

second type of interest that will be discussed on this lesson is the compound interest. For many

long-term financial transactions, compound interest is used instead of simple interest.

What’s New

“Suppose you won ₱10,000.00 and you plan to invest if for 5 years. A cooperative

group offers 2% simple interest rate per year. A bank offers 2% compounded annually. Which

will you choose and why?”

Definition of terms:

Compound amount (F) – also called maturity value, it is an accumulated amount obtained by

adding the principal and the compound interest.

Conversion period (m) – the number of times in a year the interest will be compounded.

The following are the common conversion periods in a year:

annually : m = 1

semi-annually : m = 2

quarterly : m = 4

monthly : m = 12

Lesson

2

11

Number of conversion periods (n) – the total number of times interest is calculated for the

entire term of the investment or loan.

Annual interest rate or nominal rate (r) – the stated rate of interest per year.

Periodic rate (i) – the interest rate per conversion period.

Present value of F (P) – this is the principal P, that will accumulate to F if there is an interest

at periodic rate i for n conversion periods.

What is It

Compound Interest (Ic)

Compound interest (Ic) is usually used by banks in calculating interest for long-term

investments and loans such as savings account and time deposits. In this type of interest, the

interest due at stipulated interval is added to the principal and earns interest thereafter. It implies

that the principal increases over a period of time, resulting to an increase in interest earned at

every compounding period. Thus, compound interest is an interest resulting from the periodic

addition of simple interest to the principal amount or simply the difference between the

compound amount and the original principal.

The problem below is an example of compound interest.

Example:

₱50,000.00 was loaned for a period of 3 years with 5% interest compounded annually.

What amount of money will be needed to repay the loan?

Principal at

the start of

the year

Interest Amount at the end of the

year

First Year ₱50,000.00 ₱50,000 × 0.05 × 1 = ₱2,500.00 ₱50,000 + 2 500 =

₱52,500.00

Second Year ₱52,500.00 ₱52,500 × 0.05 × 1 = ₱2,625.00 ₱52,500 + 2625 =

₱55,125.00

Third Year ₱55,125.00 ₱55,125 × 0.05 × 1 = ₱2,756.25 ₱55,125 + 2 756.25 =

₱57,881.25

The required answer to the problem is ₱57,881.25.

12

As shown in the table, the amount at the end of the year is equal to the sum of the

principal and the interest for that year.

Thus,

Amount for First Year : A = 50000 + (50000 × 0.05)

= 50000 (1 + 0.05)

Amount for Second Year: A = 50000 (1 + 0.05) + (50000 (1 + 0.05)(0.05)) = 50000 (1 + 0.05) (1 + 0.05)

= 50000 ( 1 + 0.05)2

Amount for Third Year: A = 50000 (1 + 0.05)2 + (50000 ( 1 + 0.05)2(0.05))

= 50000 (1 + 0.05)2 (1 + 0.05)

= 50000 (1 + 0.05)3

Generally, when interest is compounded annually for n years, the amount A = P( 1 + i) n.

Computation of the compound amount by the method shown above is tedious and time-

consuming. The formulas below will greatly ease computations.

Formulas:

• 𝑭 = 𝑷(𝟏 + 𝒊)𝒏 or 𝑭 = 𝑷 (𝟏 +𝒓

𝒎)

𝒎𝒕

• 𝑷 = 𝑭(𝟏 + 𝒊)−𝒏 or 𝑷 = 𝑭 (𝟏 +𝒓

𝒎)

−𝒎𝒕

𝑰𝒄 = 𝑭 − 𝑷 or

𝑰𝒄 = 𝑷[(𝟏 + 𝒊)𝒏 − 𝟏]

• 𝒓 = 𝒎 [(𝑭

𝑷)

𝟏

𝒏− 𝟏]

• 𝒕 =𝐥𝐨𝐠(

𝑭

𝑷)

𝒎[𝐥𝐨𝐠(𝟏+𝒊)]

***where

𝐼𝑐 − compound interest

𝑃 − present value of F

𝑟 − annual interest rate

𝑡 − time (per year)

𝐹 − compound amount or maturity value

𝑚 − conversion period

annually : m = 1

semi-annually : m = 2

quarterly : m = 4

monthly : m = 12

𝑛 − total number of conversion periods (𝑛 = 𝑚𝑡)

𝑖 − periodic rate (𝑖 =𝑟

𝑚)

13

Examples:

1.) Find the compound amount and interest earned on ₱15,000.00 for 1 year at

(a) 7% compounded semi-annually and

(b) 7% compounded quarterly.

Solution:

(a) Given: P = ₱15,000.00 t = 1 year

r = 7% or 0.07 m = 2

𝑖 =𝑟

𝑚=

0.07

2= 0.035 𝑛 = 𝑚𝑡 = 2(1) = 2

𝐹 = 𝑃(1 + 𝑖)𝑛 𝐼𝑐 = 𝐹 − 𝑃

𝐹 = ₱15,000.00(1 + 0.035 )2 𝐼𝑐 = ₱16,068.38 − ₱15,000.00

𝑭 = ₱𝟏𝟔, 𝟎𝟔𝟖. 𝟑𝟖 𝑰𝒄 = ₱𝟏, 𝟎𝟔𝟖. 𝟑𝟖

Therefore, the compound amount and the interest are ₱16,068.38 and ₱1,068.38,

respectively.

Alternative solution for solving the compound interest:

𝐼𝑐 = 𝑃[(1 + 𝑖)𝑛 − 1]

𝐼𝑐 = ₱15,000.00[(1 + 0.035)2 − 1]

𝑰𝒄 = ₱𝟏, 𝟎𝟔𝟖. 𝟑𝟖

(b) Given: P = ₱15,000.00 t = 1 year

r = 7% or 0.07 m = 4

𝑖 =𝑟

𝑚=

0.07

4= 0.0175 𝑛 = 𝑚𝑡 = 4(1) = 4

𝐹 = 𝑃(1 + 𝑖)𝑛 𝐼𝑐 = 𝐹 − 𝑃

𝐹 = ₱15,000.00(1 + 0.0175 )4 𝐼𝑐 = ₱16,077.89 − ₱15,000.00

𝑭 = ₱𝟏𝟔, 𝟎𝟕𝟕. 𝟖𝟗 𝑰𝒄 = ₱𝟏, 𝟎𝟕𝟕. 𝟖𝟗

Therefore, the compound amount and the interest are ₱16,077.89 and ₱1,077.89,

respectively.

Alternative solution for solving the compound interest:

𝐼𝑐 = 𝑃[(1 + 𝑖)𝑛 − 1]

𝐼𝑐 = ₱15,000.00[(1 + 0.0175 )4 − 1]

𝑰𝒄 = ₱𝟏, 𝟎𝟕𝟕. 𝟖𝟗

14

2.) Find the present value of ₱12,850.00 due in 3 years if the interest rate is 6%

compounded monthly.

Solution:

(a) Given: F = ₱12,850.00 t = 3 years

r = 6% or 0.06 m = 12

𝑖 =𝑟

𝑚=

0.06

12= 0.005 𝑛 = 𝑚𝑡 = 12(3) = 36

𝑃 = 𝐹(1 + 𝑖)−𝑛

𝑃 = ₱12,850.00(1 + 0.005)−36

𝑷 = ₱𝟏𝟎, 𝟕𝟑𝟖. 𝟎𝟒

Therefore, the present value is ₱10,738.04.

3.) At what rate of interest compounded semi-annually will ₱14,300.00 accumulate to

₱17,000.00 in 2 years and 6 months?

Solution:

Given: P = ₱14,300.00 m = 2

F = ₱17,000.00 t = 26

12 years or 2.5 years

𝑛 = 𝑚𝑡 = 2(2.5) = 5

𝑟 = 𝑚 [(𝐹

𝑃)

1𝑛

− 1]

𝑟 = 2 [(₱17,000.00

₱14,300.00)

15

− 1]

𝒓 = 𝟎. 𝟎𝟕𝟎𝟒 or 7. 𝟎𝟒%

Therefore, the rate of interest will be 0.0704 or 7.04%.

15

4.) How many years will it take for ₱13,000.00 to become ₱20,000.00 at 12.5%

compounded annually?

Solution:

Given: P = ₱13,000.00 m = 1

F = ₱20,000.00 r = 12.5% or 0.125

𝑖 =𝑟

𝑚=

0.125

1= 0.125

𝑡 =log (

𝐹𝑃)

𝑚[log(1 + 𝑖)]

𝑡 =log (

₱20,000.00₱13,000.00)

1[log(1 + 0.125)]

𝒕 =3.66 years

Therefore, ₱13,000.00 will become ₱20,000.00 in 3.66 years.

What’s More

Solve what is asked in each item. Write your complete solutions and answers on a 1

whole sheet of paper.

1.) Find the final or compound amount of ₱15,900.00 at 5.5% interest compounded

annually for 18 months.

2.) Find the interest on ₱25,750.00 for 3 years at 8% compounded quarterly.

3.) Find the present value of ₱150,000.00 at 15% interest compounded monthly for 6 years.

4.) At what rate of interest compounded semi-annually will ₱21,590.00 accumulate to

₱26,900.00 in 2 years?

5.) How long will it take for ₱3,700.00 accumulate to ₱6,900.00 at 4.5% compounded

quarterly?

16

What I Have Learned

Problems Involving Compound Interest

1. Joseph borrows ₱50,000.00 and promise to pay the principal and interest at 12%

compounded monthly. How much must he repay after 6 years?

Solution:

Given: P = ₱50,000.00 m = 12

r = 12% or 0.12 t = 6 years

𝑖 =𝑟

𝑚=

0.12

12= 0.01 𝑛 = 𝑚𝑡 = 12(6) = 72

𝐹 = 𝑃(1 + 𝑖)𝑛

𝐹 = ₱50,000.00(1 + 0.01 )72

𝑭 = ₱𝟏𝟎𝟐, 𝟑𝟓𝟒. 𝟗𝟕

Therefore, Joseph must repay ₱102,354.97 after 6 years.

2. A loan ₱125,000.00 at 8% compounded quarterly was paid back with an amount of

₱176,000.00 at the end of the period. For how long was the money borrowed?

Solution:

Given: P = ₱125,000.00 r = 8% or 0.08

F = ₱176,000.00 m = 4

𝑖 =𝑟

𝑚=

0.08

4= 0.02

𝑡 =log (

𝐹𝑃

)

𝑚[log(1 + 𝑖)]

𝑡 =log (

₱176,000.00₱125,000.00)

4[log(1 + 0.02)]

𝒕 =4.32 years

Therefore, the money was borrowed for 4.32 years.

17

3. How much must be invested today in a savings account in order to have ₱50,800.00

in 6 years and 9 months if money earns 5.4% compounded semi-annually?

Solution:

Given: F = ₱50,800.00 t = 69

12 years or 6.75 years

r = 5.4% or 0.054 m = 2

𝑖 =𝑟

𝑚=

0.054

2= 0.027 𝑛 = 𝑚𝑡 = 2(6.75) = 13.5

𝑃 = 𝐹(1 + 𝑖)−𝑛

𝑃 = ₱50,800.00(1 + 0.027)−13.5

𝑷 = ₱𝟑𝟓, 𝟒𝟓𝟑. 𝟕𝟗

Therefore, an amount of ₱35,453.79 must be invested today.

What I Can Do

Answer the following problems involving compound interest. Write your complete

solutions and answers on a 1 whole sheet of paper.

1. In a certain bank, Justine invested ₱100,000.00 in a time deposit that pays 0.5%

compounded annually. How much will be his money after 5 years? How much interest

will he gain?

2. Recca borrows ₱35,400.00 and agrees to pay ₱47,500.00 after 2 years. At what rate,

compounded monthly, is the interest computed?

18

Additional Activities

Directions: Answer the following activities and write answers on a 1 whole sheet of paper.

I. Arrange the jumbled letters to form a word/s related to business mathematics.

1.) T M T U R I Y A A E D T

2.) N E E T T R I S E R T A

3.) I I P P C N L R A

4.) O O U D C M N P N U T M A O

5.) I P S E M L T E N T S R E I

6.) E V M N T T N E I S

7.) T E U Y A Q R L R

8.) C M C U E T U L A A

9.) R W O R B O R E

10.) M O C U D O N P T N T R E E S I

II. Suppose you are working in a bank that encourages customers to save money for

their future. Your bank manager requested you to form a team and formulate a

slogan to be used for this savings campaign. Write a short phrase or slogan to

persuade people to open a savings account in the bank where you work.

III. Complete the Simple Interest Vs Compound Interest Comparison Chart

below.

Simple Interest Vs Compound Interest

Basis for Comparison Simple Interest Compound Interest

1. Meaning

2. Formula

3. Principal

4. Amount of Interests

19

Assessment (Post-test)

Directions: Read each statement carefully. Choose the letter of the correct answer and write it

on a 1 whole sheet of paper.

1.) Date on which money is received by the borrower.

A. Conversion period C. Maturity date

B. Loan date D. Repayment date

2.) 31

5% is equivalent to

A. 0.0032 C. 0.32

B. 0.032 D. 3.2

3.) This refers to the interest charged on the principal alone for the entire duration or period

of the loan or investment.

A. Compound interest C. Interest rate

B. Future value D. Simple interest

4.) This refers to the number of years for which the money is borrowed or invested.

A. Conversion period C. Principal

B. Interest rate D. Time

5.) An interest resulting from the periodic addition of simple interest to the principal

amount.

A. Compound amount C. Interest rate

B. Compound interest D. Simple interest

6.) What is the formula in computing the present value of F in a financial transaction

involving compound interest?

A. 𝑃 = 𝐹(1 + 𝑖)−𝑛 C. 𝑃 = 𝐹(1 − 𝑖)−𝑛

B. 𝑃 = 𝐹(1 + 𝑖)𝑛 D. 𝑃 = 𝐹(1 − 𝑖)𝑛

7.) How much was the interest if Althea invested ₱30,400.00 and received a total of

₱40,300.00 at the end of the term?

A. ₱9,900.00 C. ₱40,300.00

B. ₱30,400.00 D. ₱70,700.00

8.) How much is the future value on this financial transaction, P = ₱10,000.00, r = 5%,

and t = 3 years?

A. ₱1,500.00 C. ₱21,500.00

B. ₱11,500.00 D. ₱25,000.00

9.) What is the total number of conversion periods when a certain amount is borrowed at

5.5% compounded quarterly for 4 years?

A. 4 C. 16

B. 12 D. 22

20

10.) What is the interest rate per conversion period if ₱29,500.00 was invested at 2.5%

compounded semi-annually for 5 years and 4 months?

A. 0.0025 C. 0.025

B. 0.0125 D. 2.5

11.) Edgardo invested ₱15,600.00 at 10.25% interest rate. How long will take for his

investment to earn an interest of ₱5,055.00?

A. 0.32 years C. 6.29 years

B. 3.16 years D. 30.11 years

12.) Find the simple interest on a loan of ₱65,000.00 if the loan is given at a rate of 8% and

is due in 6 years and 3 months.

A. ₱3,250.00 C. ₱32,500.00

B. ₱31,200.00 D. ₱46,800.00

13.) Jamaico made a loan of ₱20,450.00 from a bank that charges 3% simple interest. How

much must he pay the bank after 2 years?

A. ₱1,227.00 C. ₱32,720.00

B. ₱21,677.00 D. ₱42,127.00

14.) At what interest rate compounded semi-annually will ₱15,000.00 accummulate to

₱25,000.00 in 10 years?

A. 2.05% C. 4.05%

B. 2.59% D. 5.17%

15.) ABC University anticipates additional expenses of ₱367,800.00 for a new equipment

needed for offering a new course 5 years from now. How much should be invested in

an account that earns 12% compounded monthly?

A. ₱62,427.83 C. ₱202,455.37

B. ₱165,344.63 D. ₱668,181.05

21

Answer Key

What’s More 1. F = ₱17,229.62

2. 𝐼𝑐 = ₱6,907.23

3. P = ₱61,326.61

4. r = 0.1130 or 11.30%

5. t = 13.93 years

What I Can Do

1. F = ₱102,525.13 𝐼𝑐 = ₱2,525.13

2. r = 0.1479 or 14.79%

Additional Activities

I.

1. MATURITY DATE

2. INTEREST RATE

3. PRINCIPAL

4. COMPOUND AMOUNT

5. SIMPLE INTEREST

6. INVESTMENT

7. QUARTERLY

8. ACCUMULATE

9. BORROWER

10. COMPOUND INTEREST

II.

*Answers may vary

III.

1. Definitions of Simple Interest and Compound Interest

2. Formulas of Simple Interest and Compound Interest

3. Simple Interest – Constant;

Compound Interest – Changing during the entire term of loan or investment

4. Simple Interest – uniform; Compound Interest – Increasing rapidly

22

Assessment (Post-test)

1. B

2. B

3. D

4. D

5. B

6. A

7. A

8. B

9. C

10. B

11. B

12. C

13. B

14. D

15. C

23

References:

[1] Teaching Guide for Senior High School – General Mathematics

by Commission on Higher Education (2016)

pages 156-185

[2] General Mathematics for Senior High School

by Aoanan, Grace O.; Plarizan, Ma. Lourdes P.; Regidor, Beverly T.;

Simbulas, Lolly Jean C.

C & E Publishing, Inc. (2016)

839 EDSA, South Triangle, Quezon City

pages 225-229; 233-235; 243-257

[3] General Mathematics for Senior High School

by Bagano, Dominic C.; Bansa, Hazam E.; Cabrera, Milna K.;

Ganir, Florence D.; Partible, Fe G,; Zorilla, Roland S.

MUTYA Publishing House, Inc. (2016)

105 Engineering Road, Araneta University Village, Potrero, Malabon City

pages 93-97; 102-109; 114-128; 131-136; 140-145

[4] General Mathematics

by Oronce, Orlando A

Rex Book Store, Inc. (2016)

856 Nicanor Reyes Sr. St., Sampaloc, Manila

pages 196 - 199

1

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