Module 1 Simple Interest Bank Discount and Promissiory Notes PDF

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MATH IF (MATHEMATICS OF THE MODERN WORLD)MODULESimple Interest, Bank Discount and Promissory NotesSection 1Simple Interest,Maturity Value,Number of Daysbetween Dates,Section 2Finding Principal,Rate and TimeSection 3Simple interest Notes, Bank Discount Notesand ProceedsSection 3Discounting Promissory...

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MATH IF (MATHEMATICS OF THE MODERN WORLD)

MODULE Simple Interest, Bank Discount and Promissory Notes

Section 1 Simple Interest, Maturity Value, Number of Days b t D t Section 2 Finding Principal, Rate and Time Section 3 Simple interest Notes, Bank Discount Notes and Proceeds

Section 3 Discounting Promissory Notes Before Maturity

COLLEGE OF BUSINESS AND ACCOUNTANCY

COURSE DESCRIPTION The course deals with nature of mathematics, appreciation of its practical, intellectual, and aesthetic dimensions, and appreciation of mathematical tools in daily life. The course then proceeds to survey ways in which mathematics provides a tool for understanding and dealing with various aspects of present day living, such as managing personal finances and making social choices. These aspects will provide opportunities for actually doing mathematics in a broad range of exercises that bring out the various dimensions of mathematics as a way of knowing, and test the students’ understanding and capacity. The course also deals with business mathematics, including mathematics of buying, selling and consumer loans useful for students’ basic understanding of invoices, trade discounts, shipping terms, markups, markdowns, credit terms and loan agreements. This will reinforce and supplement students’ practical knowledge in basic and financial accounting.

COURSE OUTLINE

Week 1-3 Week 4-6

Course Content/Subject Matter A. Simple Interest, Bank Discount and Promissory Notes B. Compound Interest

Week 7-8

C. Annuity Part 1

Week 9

D. Midterm Exam

Week 10-11 Week 12

E. Annuity Part 1 F. Sinking Fund and Amortization

Week 13-14

G. Mathematics of Buying

Week 15-16

H. Mathematics of Selling

Week 17

I. Depreciation

Week 18 One week (or an equivalent of three hours)

J. Final Exam K. Allotted for the Midterm and the Final Exams

RATIONALE

The goal of the course is for students to develop the computational skills they will need to be successful in the world of business along with a better understanding of business concepts and situations that require a mathematical solution. Specifically, the students are expected to understand the concepts on simple interest, simple discount, compound amount, basic concepts on annuities and able to apply this concept in various business transactions in which calculation are required

INSTRUCTIONS TO USERS

Read the main content of the module under developmental activities sections and answer the problems indicated in the closure activities. The learners should have a good background on the following concepts 1. Whole numbers, decimals, fractions, and percent 2. Rules in manipulating equations and formulas. 3. Fluency in calculator use is required.

MODULE 1 LEARNING OBJECTIVES 1. Calculate simple interest and bank discount. 2. Manipulate simple interest and bank discount formula. 3. Apply simple interest and bank discount concepts in discounting promissory notes.

1

Section Simpl

Section Objectives: 1. Solve for simple interest. 2. Calculate maturity value. 3. Use the actual number of days and approximate number of days to find the number of days from one date to another. 4. Find exact and ordinary interest.

Objective 1: Solve for simple interest. Simple interest is interest charged on the entire principal for the entire length of the loan. It is found using the formula shown in the following box. Principal is the loan amount, rate is the interest rate, and time is the length of the loan in years.

Finding Simple Interest

Simple interest = Principal * Rate * Time I=P x R x T

When using the formula I = PRT: 1. Rate (R) must first be changed to a decimal or fraction. 2. Time (T) must first be converted to years. So, for example, a rate of 7.5%, should be changed to .075 and a time of 6 months should be changed to 6/12=0.5 year before using I = PRT.

Example 1: To start her business, Jessica Hernandez needs to borrow P 85,000 for 9 months. Her bank would not lend her the money since she has no experience or assets. She found an individual who would lend her the money at 18.5%, However, her uncle agreed to cosign on a loan for her, meaning that he would have to pay the loan if Jessica failed to do so. On this basis, the bank would lend the money at 10%, simple interest. Find the interest at ( a) 18.5%, and b) 10%. (c) Then find the amount saved using the lower interest rate. Solution: (a)

(b)

(c)

I =PRT I = P 85,000 * .185 * 9/12

Convert 18.5% to .185

I = P 11,793.75

simple interest

I =PRT I = P 85,000 * .10 * 9/12

Convert 10% to .10 (or .1)

I = P 6375

simple interest

Difference = P 11,793.75 – P 6375 = P 5418.75

Quick Check 1: Find the interest on a loan of P 14,680 for 6 months at 9%.

OBJECTIVE 2: Calculate maturity value. The amount that must be repaid when the loan is due is the maturity value of the loan. Find this value by adding principal and interest. Finding Maturity Maturity value = Principal + Interest M=P+I Alternatively, M=P+I = P + PRT M = P( 1+ RT)

Example 2: Tom Swift needs to borrow P 28,300 to remodel his bookstore so that he can serve coffee to customers as they browse or sit and read. He borrows the funds for 10 months at an interest rate of 9.25%. Find the interest due on the loan and the maturity value at the end of 10 months. Solution: Interest due is found using I = PRT, where T must be in years 10 months = 10/12 year and R=9.25% =0.0925 Interest = PRT I = P 28,300 * 0.0925 *10/12 = P 2181.46 (rounded) Maturity value = P + I M = P 28,300 + P 2181.46 = P 30,481.46 Alternatively, Maturity value = P (1+RT) M= P 28,300 (1+.0925*10/12) =P 30,481.46 Quick Check 2 Find the maturity value of a loan of P 48,600 at 9, for 8 months.

OBJECTIVE 3: Use the actual number of days and approximate number of days to find the number of days from one date to another. Up to this point, the period of the loan was given in months, but it can also be given in days. Or a loan may be due at a fixed date, such as April 17, and we may have to figure out the number of days until the loan must be paid off. Actual Number of Days. To compute the actual number of days, count every day up to the repayment date. Approximate Number of Days. To compute the approximate number of days, assume every month has 30 days.

Example 3: Find the approximate and actual number of days of the following. (a) March 24 to July 22 (b) April 4 to October 10 (c) March 15 to Dec 30

Solution: a. March 24 to July 22

Month March April May June July Total

Actual Number of Days 7 30 31 30 22 120

Approximate Number of Days 6 30 30 30 22 118

Note: For March, under actual number of days, there are 7 days between March 24 to March 31. While under approximate number of days, there are only 6 days since we assumed that there are only 30 days in a month.

b. April 4 to October 10

Month April May June July August Septembe r October Total

Actual Number of Days 26 30 30 31 31

Approximate Number of Days 26 30 30 30 30

30 10 188

30 10 186

c) March 15 to Dec 20

Month March April May June July August Septembe r October November December Total

Actual Number of Days 16 30 31 30 31 31

Approximate Number of Days 15 30 30 30 30 30

30 31 30 20 280

30 30 30 20 275

OBJECTIVE 4: Find exact and ordinary interest A simple interest rate is given as an annual rate, such as 7, per year. Since the rate is per year, time must also be given in years or fraction of a year when using I = PRT. If time is given in number of days, first change it to a fraction of a year. Finding Time in Fraction of a Year

T=

Number of days ∈theloan period Number of days∈a year

Exact interest calculations require the use of the exact number of days in the year, 365 or 366 if a leap year. Ordinary interest, or banker’s interest, calculations require the use of 360 days. Banks commonly used 360 days in a year for interest calculations before calculators and computers became widely available. Today, many institutions, the government, and the Federal Reserve Banks and Central Banks use the exact number of days in a year in interest calculations. However, some banks and financial institutions still use 360 days. You need to be able to use both.

For Exact interest: Use 365 days (or 366 days if a leap year). I=

P∗ R∗Actual Number of Days 365

I=

P∗R∗Approximate Number of Days 365

Using the concept in counting of number of dates between dates and two ways to compute the interest (Exact and Ordinary method). We can calculate the Simple Interest in four ways.

For Ordinary interest: Use 365 days (or 366 days if a leap year).

This is known as the Banker’s Rule

Example 5: Radio station KOMA borrowed P 148,500 on May 12 with interest due on August 27. If the interest rate is 10%, find the interest on the loan using (a) exact interest and (b) ordinary interest using actual number of days and approximate number of days. Solution: Number of days between May 12 to August 27.

Month May June July August Total

Actual Number of Days 19 30 31 27 107

Approximate Number of Days 18 30 30 27 105

a. Exact Interest The exact interest is found from I = PRT with P = P 148,500, R = 0.10, and T =107/365.

I= I=

P∗ R∗Actual Number of Days 365

(P 148,500)∗(0.10 )∗107 365 I = P 4353.29 (rounded)

I=

P∗R∗Approximate Number of Days 365

I=

(P 148,500)∗(0.10 )∗105 365 I = P 4,271.92 (rounded)

b. Ordinary Interest Find ordinary interest with the same formula and values, except T = 107/360.

I= I=

P∗ R∗Actual Number of Days 360

(P 148,500)∗(0.10 )∗107 360 I =P 4,413.75 (rounded)

I=

P∗R∗Approximate Number of Days 365

I=

(P 148,500)∗(0.10 )∗105 365 I = P 4,331.25 (rounded)

Note: Use banker’s rule throughout the remainder of the book unless stated otherwise.

Quick Check 5 Find the exact and ordinary interest for a 200-day loan of P 19,500 at 9, to the nearest cent. Then find the difference between the two interest amounts using actual and approximate number of days.

Section 1 Exercise. Provide a short solution as shown by the solved problem in item 1. Find simple interest and maturity value to the nearest cent. Interest

Maturity Value

2. P 10,200 at 9.5%, for 10 months

_____________

______________

3. P 5500 at 8%, for 1 year

_____________

______________

4. P 18,500 at 7.5%, for 1 ¼ years

_____________

______________

Find the exact and approximate number of days from the first date to the second 1.

February 15 to April 24

_____________

______________

2.

May 22 to August 30

_____________

______________

3.

December 1 to March 10 of the following year

_____________

______________

4.

October 12 to February 22 of the following year

_____________

______________

Find (a) the exact interest and (b) the ordinary interest for each of the following to the nearest cent. Then find (c) the amount by which the ordinary interest is larger.

1.

P 185,000 at 7.5% for 180 days

a. ___________________ b. ___________________ c. ___________________

2.

P 29,500 for 11.25% for 120 days

a. ___________________ b. ___________________ c. ___________________

3.

P 52,610 at 8 ½ %, for 82 days

a. ___________________ b. ___________________

c. ___________________

4.

P 52,000 at 8 ¾ % for 200 days

a. ___________________ b. ___________________ c. ___________________

Find the date due, the amount of interest (use banker’s rule and rounded to the nearest cent if necessary), and the maturity value Date Loan Face Term of Maturity Was Made Value Value 1. Mar. 12 P 4800 220 days 9%

Date Loan Rate

Loan

Is Due

__________

__________

2. Jan. 3

P 12,000

100 days

9.8%

__________

__________

3. Nov. 10

P 6300

180 days

9¼%

__________

__________

4. July 14

P 20,40090 days

11 ¾ %

__________

__________

Solve the following application problems. Round dollar amounts to the nearest cent. 1. Wells Fargo Bank borrows $ 25,000,000 at 4% for 90 days from a bank in Chicago Find (a) the interest and (b) the maturity value . .

2. On October 15, IBM borrows $ 45,000,000 at 8% from a bank in San Francisco and agrees to repay the loan in 120 days using ordinary interest. Find (a) the due date and (b) the maturity value.

3. Joe Simpson’s property tax is $ 3416.05 and is due on April 15. He does not pay until July 23. The county adds a penalty of 9.3%simple interest on his unpaid tax. Find the penalty using exact interest.

2

Section Findin

pal, Rate and Time

.

Section Objectives 1. Find the principal. 2. Find the rate. 3. Find the time.

Principal (P), rate (R), and time (T) were given for all problems in Section 9.1, and we calculated interest. In this section, interest is given, and we solve for principal, rate, or time.

OBJECTIVE 1 Find the principal. The principal (P) is found by dividing both sides of the simple interest equation I = PRT by RT.

The various forms of the simple interest equation can be remembered using the circle sketch shown above. In the sketch, I (interest) is in the top half of the circle, with P (principal), R (rate), and T (time) in the bottom half of the circle. Find the formula for any

one variable by covering the letter in the circle and then reading the remaining letters, noticing their position. For example, cover P and you are left with

P=

I RT

.

I Interest ∨P= RT Rate x Time (¿ years )

Example 1: Gilbert Construction Company borrows funds at 10%, for 50 days to finish building a home. Find the principal that results in interest of P 50,000.

Solution Write the rate as 0.10, the time as 50/360, and then use the formula for principal P= P=

I RT P 50,000 =P 3,600,000 50 ) (0.10)( 360

The principal is P 3,600,000. Check the answer using I = PRT. The principal is P P 3,600,000, the rate is 10%, and the time is 54/360 year. The interest should be, and is, P 50,000.

I =PRT=P

3,600,000∗0.10∗50 =P50,000 360

Quick Check 1 A 90-day loan with a rate of 12, results in interest of P 285. Find the principal.

Example 2: Frank Thomas took out a loan to pay his college tuition on February 2. The loan is due to be repaid on April 15. The interest on the loan is P 151.20 at a rate of 10.5%. Find the principal. Solution First find the number of days.

Month February March April Total

Actual Number of Days 26 31 15 72

Days remaining in February

Days from February 2 to April 15 Next find the principal.

P= P=

I RT P 151.20 =P 7200 72 (0.105)( ) 360

The principal is P 7200. Check the answer using the formula for simple interest. I =PRT=P

7200∗0.105∗72 =P151.20 360

Quick Check 2 A loan made on May 12 must be repaid on December 18. Find the principal given that the rate is 9%, and the interest at maturity is P 1551.

Principal (P) can also be determined if Maturity Value is given instead of I. From the formula, M = P( 1+ RT) The formula for P is, M P= (1+ RT )

Example 3: How a father must invest today at 15% simple interest in order to have P 245, 000 for the education of his son five years later? Solution: In the problem, R=15%, T=5 years, and M= P 245,000 and P is unknown. P=

M ( 1+RT )

P=

P 245,000 (1+( 0.15 )(5 years ))

¿ P140,000 The father needs to invest P 140,000 at present to gain a total amount of P 245,000 pesos five years after.

OBJECTIVE 2: Find the rate. Solve the formula I = PRT for rate (R) by dividing both sides of the equation by PT. The rate found in this manner will be the annual interest rate.

R=

I Interest ∨R= Principal x Time (¿ years ) PT

Example 4: An exchange student from the United States living in Brazil deposits P 5000 in U.S. currency in a Brazilian bank for 45 days. Find the rate if the interest is P 75 in U.S. currency. Solution R= R=

I PT P 75 45 (P 5000)( ) 360

=0.12

Convert 0.12 to a percent to get 12,. Check the answer using the simple interest formula.

Quick Check 3 A 120-day loan for P 15,000 has interest of P 412.50. Find the rate.

Example 5: Blaine Plumbing kept extra cash of P 86,500 in an account from June 1 to August 16. Find the rate if the company earned P 365.22 in interest during this period of time. Round to the nearest tenth of a percent. Solution: Find the number of days

Month June July August Total

R= R=

Actual Number of Days 29 31 16 76

There are 76 days from June 1 to August 16.

I PT P365.22 76 ) (P 86,500)( 360

=0.0200(rounded)

The rate of interest is 2.0% Quick Check 4 A loan of P 37,000 made on February 4 results in interest of P 770.83. If the loan is due on May 15, find the rate to the nearest tenth of a percent.

OBJECTIVE 3 Find the time. The time (T) is found by dividing both sides of the simple interest equation I = PRT by PR. Note that time will be in years, or fraction of a year.

T (¿ years)=

Interest I ∨T = PR Principal x Rate

The preceding formula gives time in years, but we often need time in days or months. Find these as follows.

T ( ¿ days )=

I x 360 ( Using Ordinary Interest ) PR

T ( ¿ days )=

I x 365 ( Using Exact Interest ) PR

Example 5: Roberta Sanchez deposited P 18,600 in an account paying 3%, and she earned P217 in interest. Find the number of days that the deposit earned interest using Ordinary and Exact Interest method.

Solution: Ordinary Interest

I T ( ¿ days )= x 360 PR ¿

P 217 x 360 (P 18,600)( 0.03 ) =140 days

Exact Interest

T ( ¿ days )= ¿

I x 360 PR

P 217 x 365 (P 18,600)( 0.03 )

=142 days (rounded)

Quick Check 5 A loan for P 22,000 results in interest of P 1283.33 at 10.5%. Find the time to the nearest day.

Section 2 Exercises. Provide a short solution as shown by the solved problem in item 1. Find the principal in each of the following. Round to the nearest cent. Rate

Time (in days)

Interest

Principal 1.

10%